90
Giuseppe E. Santoro Lectures on Quantum Mechanics ICTP Diploma Course Printed on: May 7, 2012
Giuseppe E. Santoro
Lectures on Quantum Mechanics
ICTP Diploma Course
Printed on: May 7, 2012
Preface
These are just short notes intended as a guide to the Diploma Students taking their course inAdvanced Quantum Mechanics.
Contents
1 The basics of Quantum Mechanics 6
1.1 A brief history 6
1.2 Gaussian wave-packets 7
1.3 The Schrodinger equation: free-particle 9
1.4 The Schrodinger equation 10
1.5 The time-independent Schrodinger equation 14
2 One-dimensional problems 17
2.1 Infinite square well 17
2.2 Scattering by a step 18
2.3 Finite square well 19
2.4 Delta-function potential 20
3 Harmonic oscillator 21
3.1 Commutators and Hermitean conjugates 21
3.2 Dimensionless variables 22
3.3 The algebraic solution 23
4 The hydrogen atom problem 25
4.1 Orbital angular momentum 25
4.2 Separation of variables 27
4.3 Dimensionless variables 28
4.4 The Coulomb problem bound states 29
5 Theory of angular momentum 32
5.1 Diagonalization of J2 and Jz 32
5.2 Construction of spherical harmonics 33
6 Symmetries 36
6.1 Space transformations 36
6.2 Transformations on scalar functions 36
4 Contents
6.3 Symmetry properties of Hamiltonians 38
6.4 Commuting observables 40
6.5 Selection rules 41
7 Measurement in Quantum Mechanics 42
7.1 Measurements in QM. 43
7.2 Expectation values of measurements. 44
8 Spin 46
8.1 Stern-Gerlach experiment: the spin operators and states 46
8.2 Experiments with Stern-Gerlach apparatus 48
8.3 A single spin-1/2 in a uniform field: Larmor precession 50
8.4 Heisenberg representation 50
9 Spin-orbit and addition of angular momenta 52
9.1 Relativistic effects 52
9.2 The origin of spin-orbit 52
9.3 Construction of J-states out of p-states 53
9.4 Generalization: addition of L and S 55
10 Density matrices 56
10.1 The density matrix for a pure state 56
10.2 The density matrix for a mixed state 57
10.3 Density matrices and statistical mechanics 57
10.4 Density matrices by tracing out the universe 58
11 Time-independent perturbation theory 60
11.1 Two-state problem 60
11.2 Non-degenerate case: general solution 61
11.3 Degenerate case: first order 66
11.4 Application: the fine-structure of Hydrogen 68
12 Time-dependent perturbation theory 71
12.1 General case 71
12.2 First-order solution 72
12.3 Electromagnetic radiation on an atom 75
Contents 5
13 Many particle systems 77
13.1 Many-particle basis states for fermions and bosons. 77
13.2 Two-electrons: the example of Helium. 81
Appendix A Second quantization: brief outline. 84
A.1 Why a quadratic Hamiltonian is easy. 88
1
The basics of Quantum Mechanics
1.1 A brief history
The history of quantum mechanics starts right at the beginning of the 1900. The years from 1900to 1930 might be really seen, as the title of a nice book by Gamow says, as the 30 years that shookphysics.
This story has to do with light (more generally, electromagnetic radiation) as well as electronsand atoms: while a full quantum treatment of light, however, necessarily requires a relativisticquantum theory (which we will not enter into), electrons and atoms can be very well described bya non-relativistic quantum theory, which will be the subject of our course. I urge you to read thefirst chapter of any of the many books in QM, for instance that of Gasiorowitz, for a full accountof this very interesting story. I will be here very sketchy.
The story itself, begins really with electromagnetic (e.m.) radiation, more precisely with thepuzzle of the black-body radiation, an idealized furnace (oven) where the spectrum of the radiationinside was experimentally well known but escaping any classical physics explanation. It was MaxPlanck, in 1900, who postulated (very boldly, at that time), that everything could be explained ifone assumed that the energy of an e.m. field of frequency ν was not arbitrarily and continuouslychangeable, but only multiples of the fundamental energy
E = hν = ~ω , (1.1)
h being the famous Planck’s constant (ω = 2πν, and ~ = h/2π) would be admitted. This fixed allthe difficulties that physicists had encountered in the black-body radiation problem. Constructingon this idea, soon after (in 1905) Einstein was able to explain in a very natural way the photoelectriceffect: electrons are kicked out from the metal by individual “collisions” with a single “quantum” ofradiation (a “photon”): if the energy E = hν of the photon is not larger than the work function W(by definition, the energy you have to provide to extract an electron from a metal), then no electronwill come out, no matter how “intense” you make the field (by shining very many photons on yourmetal). The final experimental confirmation of the “particle-like” nature of the e.m. radiation camewith Compton (in 1923), who performed a true scattering experiment of X-rays from electrons,where all the results could be rationalized by applying energy and momentum conservation to thecollision event; the momentum that appeared to be associated to the X-ray had to be:
p =E
c=hν
c=h
λ=
h
2π2πλ
= ~k , (1.2)
where k = 2π/λ is the wave-number.
In the atomic world, the first bricks of a quantum theory came with Bohr, in 1913 (at thattime a post-doc of Rutherford), who “solved” the mystery of the atomic spectra seen in emissionand adsorption. Rutherford had put forward a “planetary model” for the atoms, with the nucleusoccupying only a small central volume and the electrons orbiting around it. However, why thoseorbiting electrons would not “emit radiation” (being accelerated) as the electrons in a synchrotrondo, was a mastery. Bohr elaborated a model for “quantifying” the allowed orbits, and their energies
Gaussian wave-packets 7
En (which were, once again, discrete) in such a planetary system, with results which were insurprising agreement with spectral observations. In particular, the various radiation lines observedcould be accounted for as differences between the quantized energy values En that Bohr hadcalculated, as:
hνn,m = En − Em . (1.3)
In 1925 de Broglie realized that the orbits of Bohr’s atom appeared to possess a nice wave-likefeature: an orbit of energy En (with n = 1, 2, · · · ) had a radius rn and a circumference 2πrn whichappear to “contain” exactly n wavelengths λ = h/p (p being the momentum of the electron in theorbit):
2πrn = nλ . (1.4)
This mysterious wave-like nature of electrons was revealed later on (before 1930) in diffractionexperiments, by Davisson and Gerner, and, independently, by G. Thompson (son of the Thompsonwho discovered the electron!). Notice that electrons of energy E = 160 eV have momenta p ≈6.6× 10−19 (in c.g.s), and therefore the resulting λ ≈ 10−8 cm. de Broglie was, however, too slowin elaborating a full mathematical treatment of his electron waves. In 1926, Schrodinger came outwith a beautiful theory that put de Broglie insight on a firm ground, and explained in a verynatural way the results of Bohr.
1.2 Gaussian wave-packets
Question: how are we supposed to explain the motion of a material particle with a wave? A singleplane-wave eikx cannot certainly be representing anything localized, because it is periodic in space,with a period λ = 2π/k. Therefore, in order to look for something “localized”, which might beexchanged for a “point-particle”, at least from reasonably far away, we must superimpose manyplane waves and create a wave-packet. We do that, by taking a superposition of the form:
ψ(x) =∫ +∞
−∞
dk√2π
g(k)eikx . (1.5)
g(k) can in principle be any weight function representing how much of the wave eikx is included inthe superposition, for each k. In order to carry out the calculations explicitly, we will assume thatg(k) has the simple Gaussian form:
g(k) = Ce−α(k−k0)2 , (1.6)
where C is a normalization constant which we choose so that∫dk|g(k)|2 = 1. The reason why we
impose that the square of g(k) is normalized, and not g(k) itself, will be clear in a while.
Now, the simple Gaussian integral that you certainly have encountered many times is:∫dk√2π
e−ak22 =
1√a. (1.7)
Based on this, it is simple to show that:
C =(
2απ
)1/4
.
By defining σ2k = 1/(4α), it is also quite simple to show that σk is the width of the function |g(k)|2,
i.e., its second moment is:
〈(k − k0)2〉 =∫dk (k − k0)2|g(k)|2 = C2
∫dk (k − k0)2e
− (k−k0)2
2σ2k = σ2
k . (1.8)
Prove, as an exercise, the latter identity.
8 The basics of Quantum Mechanics
Still based on the Gaussian integral in Eq. (1.7), by completing the square, it is a simple matterto show the following generalization:∫
dk√2π
e−ak22 ekJ =
1√ae
J22a . (1.9)
There is no restriction on J being a real quantity, in this integral. So, if we substitute J = ix,where x is real, the equality still holds, and reads:∫
dk√2π
e−ak22 eikx =
1√ae−
x22a . (1.10)
This very useful identity, expressed in words by saying that the Fourier transform of a Gaussianin k-space is a Gaussian in x-space, with an inversely proportional width, will be used now tocalculated ψ(x). Indeed, a straightforward calculation leads to:
ψ(x) = Ceik0x∫ +∞
−∞
dk√2π
e−αk2eikx =
(1
2πα
)1/4
eik0xe−x24α , (1.11)
i.e., ψ(x) is a plane-wave eik0x with a Gaussian envelope. If we consider |ψ(x)|2, it will be a simpleGaussian with a width σ2
x = α. Notice that the widths of the distributions in k- and x-space arerelated by a simple inverse proportionality, i.e.:
σx · σk =√α · 1
2√α
=12. (1.12)
We will see in a while that this is a realization of a minimum Heisenberg-uncertainty-principlewave-packet (we will see that Heisenberg uncertainty principle requires ∆x∆p ≥ ~/2).
Let us now consider the time dependence of this wave-packet. For a single plane wave eikx thetime-dependence has the form:
ψk(x, t) = ei(kx−ωkt) ,
where ωk is the angular frequency of the wave. Such a form guarantees that the wave moves witha phase velocity
vk =ωkk,
since, indeed, ψk(x, t) = eik(x−vkt), i.e., the wave amplitude depends only on x−vkt. Now considerthe full time evolution of ψ(x), which naturally reads:
ψ(x, t) =∫ +∞
−∞
dk√2π
g(k)ei(kx−ωkt) =∫ +∞
−∞
dk√2π
g(k)ψk(x, t) . (1.13)
The crucial question is the choice of the k-dependence of ωk. For the case of an electromagneticwave in vacuum, we have ωk = kc, and the phase velocity of each plane-wave component is identicalto the speed of light: vk = c. As a consequence, it is very simple to prove that, in such a case,the wave would move rigidly, undistorted, with the speed of light, i.e., ψ(x, t) = ψ(x− ct, 0). Onerefers to such a case as non-dispersive. On the contrary, whenever each k-component moves with adifferent phase-velocity vk, the resulting wave has a dispersive behaviour: we will soon show thatthis implies a rigid motion with a group velocity vg = dωk/dk plus an inevitable spreading of theenvelope. Clearly, to possibly describe the motion of a non-relativistic particle, we have to chosethis second route. In order to carry on the calculation, we will need to Taylor expand ωk (which is ingeneral a non-linear function of k) around the peak wave-vector k0. To fully justify this expansion,
The Schrodinger equation: free-particle 9
let us assume that α is large, so that g(k) is quite narrow around its peak at k = k0. The Taylorexpansion of ωk around k = k0, up to second order, reads:
ωk = ωk0 + vg(k − k0) + β(k − k0)2 + · · · , (1.14)
where
vg =dωkdk
∣∣∣∣k0
, (1.15)
and β = (1/2)d2ωk/dk2|k0 . By substituting these expressions in ψ(x, t) we easily get:
ψ(x, t) = Cei(k0x−ωk0 t)
∫ +∞
−∞
dk√2π
e−(α+iβt)k2eikx =
C√2(α+ iβt)
ei(k0x−ωk0 t)e−(x−vgt)2
4(α+iβt) , (1.16)
where we have used that fact that the Fourier transform in Eq. (1.10) holds even when a is complex,as long as with a positive real part (so that the integrand does not blow up), which is the case here,since α > 0. Consider now |ψ(x, t)|2. It reads:
|ψ(x, t)|2 =1√2π
√α
α2 + β2t2e− α(x−vgt)2
2(α2+β2t2) . (1.17)
Notice that, since β 6= 0, there is a spreading of the wave-packet, i.e., its width inevitably increases:this is completely independent of the sign of β. Now, the velocity with which the center of thewave-packet moves should be related to the classical momentum in the usual non-relativistic way:
vg =dωkdk
∣∣∣∣k0
=p0
m=
~k0
m, (1.18)
where we have used de Broglie’s identification of the momentum with ~k0. It is then clear that theonly way this identity holds is that:
Ek = ~ωk =~2k2
2m=
p2
2m, (1.19)
which is the standard non-relativistic kinetic energy of a particle.
Exercise 1.1 Consider the problem of spreading of a Gaussian wave-packet for a free non-relativisticparticle. Calculate how much the width of the wave-packet changes in t = 1 second in the followingtwo cases: a) the particle is an electron (m = 9 × 10−31 kg) and the initial wave-packet has dimensions∆x = 10−5 m, or b) ∆x = 10−10 m; c) the particle has mass m = 10−3 kg and initial size ∆x = 10−3 m.It is convenient to calculate, in all cases, the fractional change of the width given by:
∆xt −∆x
∆x.
1.3 The Schrodinger equation: free-particle
Let us now verify that a wave-packet of the form given in Eq. (1.16) – or more generally oneobtained from Eq. (1.13), as long as ωk has the non-relativistic form in Eq. (1.19) – satisfies thefollowing Schrodinger equation
i~∂ψ(x, t)∂t
= −~2∇2
2mψ(x, t) , (1.20)
where ∇2 = ∂2/∂x2 +∂2/∂y2 +∂2/∂z2 is the Laplacian (in three-dimension, in one-dimension onlythe x-component survives). Verifying this statement is very simple. We just prove that it is true
10 The basics of Quantum Mechanics
for a single plane-wave ψk(x, t) = ei(k·x−ωkt), and then, by linearity of the Schrodinger equation, itwill be true also for any combination of plane-waves, as in Eq. (1.5). Verifying that ψk(x, t) solvesEq. (1.20) is very simple and is left as an exercise (done in class).
Moreover, if we define the momentum operator as:
p = −i~~∇ , (1.21)
it is simple to show that the plane-waves are eigenstates of the momentum operator (remember deBroglie):
p eik·x = (~k) eik·x . (1.22)
It is also simple to show that the ∇2 term appearing in Eq. (1.20) can be also written as:
−~2∇2
2m=
p2
2m, (1.23)
i.e., the same form as the kinetic energy of a free non-relativistic particle.
1.4 The Schrodinger equation
In general, for a particle which is subject to a potential V (x), the Schrodinger equation (SE) reads:
i~∂ψ(x, t)∂t
= Hψ(x, t) =(
p2
2m+ V (x)
)ψ(x, t) . (1.24)
The operator H appearing in this equation is the Hamiltonian of the system, which has the standardnon-relativistic form, except that it is an operator, acting of functions (wave-functions).
We must think of the problem in terms of a space of functions (the ψ’s) on which we act withlinear operators (p, x, H).
The SE is first-order in time. So, given any initial wave-function ψ(x, t = 0), we can in principlecalculate ψ(x, t) at any later time t > 0 in a perfectly deterministic way. There is a formal way ofexpressing this solution, in terms of the so-called evolution operator. Let us consider for simplicitythe case in which the Hamiltonian is time-independent. (If the Hamiltonian is time-dependent,as in the presence of external varying fields, the evolution operator can be still defined, but itsexpression is more complicated.) In general, the exponential of an operator O is defined as:
eOdef= 1 + O +
12!O2 + · · · =
∞∑n=0
1n!On , (1.25)
where 1 is the identity operator, i.e., multiplying by 1. One should be careful, because Some of theusual rules of ordinary exponential functions do not hold for exponential of operators, for instance:
eO1eO2 6= eO1+O2 , (1.26)
unless the two operators commute (i.e., O1O2 = O2O1). However, some other rules do apply heretoo. For instance:
d
dteOt = OeOt = eOtO . (1.27)
The proof of this equality (done in class) is very simple, starting from the Taylor expansion of theexponential.
The Schrodinger equation 11
Consider now the following operator:
U(t) = e−iHt
~ . (1.28)
It is known as evolution operator, for a reason that will be clear in a moment. First of all, oneimmediately has U(t = 0) = 1, and:
i~d
dte−
iHt~ = H e−
iHt~ . (1.29)
As a consequence, if we define the time-evolved state
ψ(x, t) = U(t)ψ(x, t = 0) = e−iHt
~ ψ(x, t = 0) , (1.30)
it is a very simple matter to prove (done in class) that it satisfies correctly the initial-time conditionat t = 0 and the SE.
We ask now how to define, given the fact that our basic objects are wave-functions, the prob-ability density P (x, t) for finding the particle at some position x at time t. Obvious requirementsfor P (x, t) are:
a) P (x, t) ≥ 0
b)d
dt
(∫dx P (x, t)
)= 0
c)∫dx P (x, t) = 1 . (1.31)
The first requirement is common to all probability densities (real and non-negative functions); thesecond tells us that the total probability is conserved in time. As a consequence of this conservation,we will be able to rescale P in such a way to impose the standard normalization of the totalprobability given in c) above. How is P (x, t) related to ψ(x, t)? For one thing, ψ(x, t) is inevitablya complex wave-function! Even if we start from a real ψ(x, t = 0) the SE-evolution will make ψ(x, t)complex. A possible guess would be to take P = |ψ(x, t)|, but we will show that any such guess willviolate the conservation requirement in b), except for the following one suggested by Max Born(in strict analogy with the energy density of the electromagnetic field, in terms of the field itself):
P (x, t) = |ψ(x, t)|2 = ψ∗(x, t)ψ(x, t) . (1.32)
To show this in a nice way, it is better to introduce a few formal tools, which will be anyway usefulin the following. Notice that this form immediately implies interference between waves, because|ψ1(x, t) + ψ2(x, t)|2 6= |ψ1(x, t)|2 + |ψ2(x, t)|2.
1.4.1 A few formal tools
First of all, we introduce in the space of wave-functions ψ a structure of a linear space, which israther obvious: if c1 and c2 are arbitrary constants, and ψ1, ψ2 wave-functions, then c1ψ1 + c2ψ2 isalso a possible wave-function. Moreover, we introduce in this space a scalar products, by analogywith the usual euclidean scalar product of ordinary vectors:
〈ψ1|ψ2〉def=∫dx ψ∗1(x)ψ2(x) = 〈ψ2|ψ1〉∗ . (1.33)
In this way, the norm ||ψ|| of a function is defined as:
〈ψ|ψ〉 =∫dx |ψ(x)|2 = ||ψ||2 . (1.34)
The mathematicians call this norm (there are many other possible scalar products and norms thatyou can introduce) the L2-norm. In this way, the linear space of all possible normalizable wave-functions (i.e., with ||ψ|| <∞), forms a space known as Hilbert space, which we will denote by H
12 The basics of Quantum Mechanics
(to distinguish it from the Hamiltonian H or H). Notice immediately that our beloved plane-wavesψk(x) = eik·x are immediately rather bizarre, being ||ψk|| = ∞. We will keep using them becausethey are useful as states on which we expand physical normalizable wave-functions (rememberthe wave-packet discussion, and the Fourier expansion); we will also soon learn how to regularizethem properly on a finite-volume, through a procedure very familiar in condensed-matter (periodicboundary conditions).
Of all the linear operators that one can imagine to act on the wave-functions, of particularimportance are the so-called Hermitean operators, which are defined as those operators such that,for any ψ1 and ψ2, satisfy:
〈ψ1|Oψ2〉 = 〈Oψ1|ψ2〉 , (1.35)
i.e., the operator O is Hermitean if it can move back-and-forth on the two sides of the scalarproduct. Many (indeed all!) of the operators encountered so far are Hermitean. To start with, theoperator x, which simply multiplies by x a ψ, is Hermitean because x is real, and therefore (trywriting both terms explicitly to fully convince yourself):
〈ψ1|xψ2〉 = 〈xψ1|ψ2〉 . (1.36)
Similarly, any real potential V (x) corresponds to an Hermitean operator. Perhaps surprisingly (atleast at a first sight, because of the −i), the momentum operator is Hermitean: the very simpleproof (done in class) uses integration by parts and the extra minus sign in it makes for the necessarycomplex conjugate of −i:
〈ψ1|pψ2〉 = 〈pψ1|ψ2〉 . (1.37)
Powers of an Hermitean operators are Hermitean (just move one operator at a time from the rightto the left of the scalar product). Therefore:
〈ψ1|p2ψ2〉 = 〈pψ1|pψ2〉 = 〈p2ψ1|ψ2〉 . (1.38)
Also, the sum of two Hermitean operators is Hermitean:
〈ψ1|(Oa + Ob)ψ2〉 = 〈(Oa + Ob)ψ1|ψ2〉 , (1.39)
as the linearity of the scalar product immediately shows. Finally, the Hamiltonian H of a particlein a potential is Hermitean because it is the sum of two Hermitean operators, p2 and V (x).
Exercise 1.2 Of the following operators acting on functions, tell which ones are linear operators, and, outof these, which ones are Hermitean operators: 1) O1ψ(x) = x3ψ(x), 2) O2ψ(x) = xdψ/dx, 3) O3ψ(x) =
λψ∗(x), 4) O4ψ(x) = eψ(x), 5) O5ψ(x) = dψ/dx+ a, 6) O6ψ(x) =R x−∞ dx′(x′ψ(x′)).
With these definitions and with the use of the SE and of its conjugate:
−i~∂ψ∗(x, t)∂t
=(Hψ(x, t)
)∗, (1.40)
it is very simple to show that:
i~d
dt
∫dx ψ∗(x, t)ψ(x, t) = 〈ψ|Hψ〉 − 〈Hψ|ψ〉 = 0 , (1.41)
where the last equality follows from the fact that H is Hermitean.
The Schrodinger equation 13
Before leaving this formal section, let us introduce the concept of expectation value of an operator.By analogy with classical probability, a natural choice for the average position at time t < x > (t)is:
< x > (t) =∫dx xP (x, t) =
∫dx ψ∗(x, t)xψ(x, t) = 〈ψ(t)|xψ(t)〉 .
Similarly, and more generally, the mean value (or expectation value) of an operator O on a stateψ(t) is defined as:
< O > (t) def= 〈ψ(t)|Oψ(t)〉 =∫dx ψ∗(x, t)Oψ(x, t) . (1.42)
Notice that O might involve derivatives, which should be applied only to the state to its immediateright, and therefore, generally speaking, expectations values of operators cannot be expressed asintegrals where the probability P (x, t) appears!
Exercise 1.3 Prove that the expectation value of an Hermitean operator is a real quantity (done in class).
Exercise 1.4 Write down the expectation value of the momentum operator, and observe explicitly that:1) it is real even though there is an imaginary unit i in its expression; 2) it cannot be expressed in termsof P (x, t).
A final word on the notation. The Dirac’s notation is strictly related to what we have writtenabove, except for adding an extra |:
〈ψ1|Oψ2〉 = 〈ψ1|O|ψ2〉 , (1.43)
where the right hand side is in the Dirac’s notation, with the operator sitting between a bra 〈ψ1|and the ket |ψ2〉.
Exercise 1.5 Define the uncertainty in x and p on a state ψ with the usual rules:
(∆x)2 = 〈ψ|x2|ψ〉 − 〈ψ|x|ψ〉2 , (∆p)2 = 〈ψ|p2|ψ〉 − 〈ψ|p|ψ〉2 .
Next define the following quantity:
D(α) =
Zdx |(x− 〈x〉)ψ(x) + iα(p− 〈p〉)ψ(x)|2 ≥ 0 ,
which is non-negative, by inspection, for every value of α. Expand the modulus square appearing in D(α)to show that:
D(α) = (∆x)2 + α2(∆p)2 − ~α ≥ 0 .
From the fact that this quadratic function of α is always non-negative, deduce something on the discrimi-nant of the parabola, and show that Heisenberg’s uncertainty principle follows:
∆x ∆p ≥ ~2.
14 The basics of Quantum Mechanics
1.4.2 The continuity equation and the current
To convince ourself that ψ(x, t) is the fundamental quantity of Quantum Mechanics, and notP (x, t), let us write down the partial differential equation that P obeys. By using the SE and itsconjugate it is immediate to show that:
∂
∂tP (x, t) =
∂
∂t(ψ∗(x, t)ψ(x, t)) =
1i~
(ψ∗(Hψ)− (Hψ)∗ψ
). (1.44)
Now, you can verify directly that the potential V (x) disappears completely from this expression,and only the kinetic term survives:
∂
∂tP (x, t) =
i~2m
(ψ∗(x, t)(∇2ψ(x, t))− (∇2ψ(x, t))∗ψ(x, t)
). (1.45)
It takes now a small calculus exercise, based on identities like
~∇ · (ψ∗~∇ψ) = ~∇ψ∗ · ~∇ψ + ψ∗∇2ψ ,
to show that the right hand side of the latter equation is nothing but the divergence of a currentJ(x, t):
∂
∂tP (x, t) = −~∇ · J(x, t)
J(x, t) = − i~2m
(ψ∗(x, t)(~∇ψ(x, t))− (~∇ψ(x, t))∗ψ(x, t)
)=
12m
(ψ∗(x, t)(pψ(x, t)) + (pψ(x, t))∗ψ(x, t)) . (1.46)
In this form, the equation for P expresses the differential form of the conservation of probability,and is nothing but the familiar continuity equation which you encounter in electrodynamics.
1.5 The time-independent Schrodinger equation
Suppose the system is isolated, so that its Hamiltonian is time independent, H. (If the system issubject to external fields, for instance an external electromagnetic field, then the Hamiltonian isexplicitly time-dependent, H(t), and the evolution operator is not the simple exponential we haveconsidered previously.). We already know that the formal solution of the SE, given an initial-timeψ(x, t = 0), can be written as:
ψ(x, t) = U(t)ψ(x, t = 0) = e−iHt
~ ψ(x, t = 0) . (1.47)
Our question, now, is if and when such a wave-function can factorized into a space and timefunction:
ψ(x, t) = f(t)φ(x) . (1.48)
It is now difficult to show (done in class) that the only possibility this can happen is that φ(x) =ψ(x, t = 0) is one of the eigenstates of H
Hφn(x) = Enφn(x) , (1.49)
and, as a consequence, the time-dependent solution factorizes with f(t) = e−iEnt/~, i.e.:
ψ(x, t) = e−iEnt/~φn(x) . (1.50)
Notice that, for these solutions, P (x) = |ψ(x, t)|2 = |φn(x)|2 is time-independent, and, therefore,the current must have zero divergence:
Jn(x) = 0 .
A general mathematical theorem guarantees that the full set of eigenstates φn of H provides anorthonormal basis set for the Hilbert space: we will see several examples of this statement, with
The time-independent Schrodinger equation 15
n being a discrete set of labels, or even a continuous one. We will use a notation, for simplicity,where n is assumed to be discrete. The fact that the energy eigenstates form an orthonormal basismeans that:
〈φn|φm〉 = δn,m ,
and that every function ψ0(x) can be expanded in terms of the φn:
ψ0(x) =∑n
cnφn(x) .
Indeed, the expansion coefficients cn are simply given (by using the orthonormality of the basis)as:
cn = 〈φn|ψ0〉 .
Armed with the knowledge of the basis set φn, we can readily solve any time-dependentSchrodinger problem with assigned initial condition ψ0(x):
i~∂ψ(x, t)∂t
= Hψ(x, t)
ψ(x, t = 0) = ψ0(x) . (1.51)
Indeed, all we need to do is to expand the initial condition ψ0(x) on the eigenstates φn(x), ψ0(x) =∑n cnφn(x), and write down the state at time t as:
ψ(x, t) =∑n
cne−iEnt/~φn(x) . (1.52)
1.5.1 Free-particles and plane-wave normalization.
The simplest case it that of a free particle. Here the energy eigenvalues are simply Ek = ~2k2/(2m),associated to wave-functions φk(x) = eik·x. Notice that here k stands for the index n of the previoussection, and is in general a continuous multidimensional label. The spectrum of H, that is the setof all its possible eigenvalues, is here continuous, and extends in the region E ∈ [0,+∞).
A few observations: 1) while k univocally specifies φk, the same energy Ek = E is obtained by,in general, infinitely many k, all those with |k| = kE =
√2mE/~ (a sphere in 3d, a circle in 2d,
and 2 points in 1d). 2) All the φk are not normalizable in the infinite volume case. A trick oftenused in condensed matter is to consider the system as confined to a cube of side L, hence of volumeΩ = L3 in 3d (Ω = Ld, in general dimension d), with periodic boundary conditions (PBC), i.e., wepostulate that:
φ(x + Lei) = φ(x)
where ei is any of the three coordinate versors. This requirement (done in class) restricts thepossible allowed k-vectors to a discrete (although still infinite) set:
k =2πL
(n1, n2, n3) , (1.53)
with ni ∈ Z (i.e., positive and negative integers, or zero). The correctly normalized free-particleeigenstates with PBC in the cube will then be:
φk(x) =1√Ωeik·x , (1.54)
with the k given by Eq. 1.53. Notice that now the φk form a nice orthonormal basis for periodicfunctions in the cube of volume Ω, since:
〈φk|φk′〉Ω =1Ω
∫Ω
dx ei(k′−k)·x = δk,k′ .
16 The basics of Quantum Mechanics
The fact that φk forms a basis set, simply a restatement of the fact that any periodic functionψ(x + Lei) = ψ(x) can be expanded in Fourier series:
ψ(x) =∑k
φk(x)〈φk|ψ〉 =1Ω
∑k
eik·xψk , (1.55)
where the Fourier coefficient ψk is simply given by:
ψk =√
Ω〈φk|ψ〉 =∫
Ω
dx e−ik·xψ(x) . (1.56)
It is now quite simple to take the thermodynamic limit L → ∞. Consider, indeed, the wave-functions:
φk =1√
(2π)deik·x .
Their overlap, restricted to the volume Ω, tends towards the Dirac’s delta function:
〈φk|φk′〉Ω =Ω
(2π)d〈φk|φk′〉Ω =
Ω(2π)d
δk,k′Ω→∞−→ δ(k− k′) .
To prove this, you should start from the natural discretization of the integral∫dk, which is:(
2πL
)d∑k
Ω→∞−→∫dk ,
and show that, for any function f(k):
f(k) =(
2πL
)d∑k′
f(k′)Ω
(2π)dδk,k′
Ω→∞−→∫dk f(k′)δ(k− k′) .
As a consequence, you can take the limit of the Fourier sum in Eq. 1.55, obtaining:
ψ(x) =1Ω
∑k
eik·xψkΩ→∞−→
∫dk
(2π)deik·xψk . (1.57)
The Ω →∞ limit of the Fourier coefficient ψk in Eq. 1.56 gives:
ψk =∫dx e−ik·xψ(x) , (1.58)
which we recognize to be the Fourier transform ψk of the function ψ(x). Eq. 1.57 simply definesthe inverse Fourier transform. Notice that, when L→∞, k becomes a continuous vector.
2
One-dimensional problems
Let us begin with some general remarks on the time-independent SE. First of all, the spectrum ofthe Hamiltonian (i.e., the set of all its eigenvalues) is constrained to be not below the minimum ofthe potential (if one exists):
En ≥ Vmin .
The proof of this is simple (done in class). Quickly, one can show that if Hψn(x) = Enψn(x) foran eigenstate, then by taking the scalar product with ψn (assumed to be normalized) we get:
En =∫dx(−ψ∗n
~2∇2
2mψn + V (x)|ψn(x)|2
)=∫dx
(~2|~∇ψn|2
2m+ V (x)|ψn(x)|2
), (2.1)
from which it is immediate to conclude that En cannot be less then Vmin, because kinetic energyis in general positive, and the wave-function is never really localized on the minimum of V .
A second general consideration regards bound versus extended states. A bound state is associatedto a discrete eigenvalue of H, and to a normalizable wave-function, usually quite well localized inspace. On the contrary, an extended state, exemplified by the familiar plane-wave eik·x, is notnormalizable in the infinite volume, and associated to an eigenvalue belonging to a continuous partof the spectrum. We have already seen this for the free particle. We will see more examples of bothshortly.
A third general remark has to do with the continuity conditions of a solution of the SE at theboundary between regions were the potential has some discontinuity. Generally speaking, a solutionof the SE must be continuous, with a continuous first derivative (indeed, ψ should admit a secondderivative!). For simplicity, assume we are in one-dimension. Suppose that x0 is a point were thepotential has possibly a finite (jump) discontinuity ∆V . By integrating the SE over the interval[x0 − ε, x0 + ε], writing:
− ~2
2m
∫ x0+ε
x0−εdx ψ′′(x) = − ~2
2m[ψ′(x0 + ε)− ψ(x0 − ε)] =
∫ x0+ε
x0−εdx [E − V (x)]ψ(x) ,
and taking the limit ε→ 0 we can quickly prove that the first derivative ψ′(x) is continuous at x0,unless the potential V (x) has an infinite jump ∆V at x0, or some delta-function contribution.
2.1 Infinite square well
The first problem we have solved in class is that of an infinite square-well in the region [0, L]. Inother words, the potential is infinite for x < 0 and x > L, and V = 0 in [0, L]. Due to the fact thatψ(x) must be 0 for x < 0 and x > L and it must be continuous everywhere (no exception to thisrule), we concluded that the appropriate boundary conditions for ψ(x) are:
ψ(0) = ψ(L) = 0 .
We solved this problem by taking combinations of eikx and e−ikx (in the free-particle region [0, L])of the form:
18 One-dimensional problems
ψk(x) = Ak sin kx .
The boundary condition ψk(0) = 0 is automatically satisfied. As for the condition ψk(L) = 0, itimplies sin kL = 0, or:
k =nπ
Lwith n = 1, 2, · · · (2.2)
Negative n’s do not bring extra solutions, because sin−kx = − sin kx. The corresponding energyeigenvalue is
Ek =~2k2
2m=
~2π2n2
2mL2. (2.3)
Each solution is non degenerate, i.e., only one eigenstate is associated to each eigenvalue. Noticethat the spectrum is now purely discrete, as an effect of the confining potential. For L → ∞ werecover a dense continuous spectrum in [0,+∞).
We discussed some properties of the box wave-functions. First of all, we showed that 〈ψk|p|ψk〉 =0 (because you can always take the ψk to be real). Next, we calculated 〈ψk|p2|ψk〉 in terms of energyeigenvalues by using the fact that p2 = 2mH. We commented on the Heisenberg uncertaintyprinciple, showing that (∆x) · (∆p) grows with the quantum number n. We also commented onthe increasing number of zeroes of the eigenstates, and the corresponding kinetic energy increase,since you can always re-express the kinetic energy as an integral of |~∇ψ|2.
Exercise 2.1 Solve the same infinite square well problem by choosing the origin in the center of the region,which is therefore [−L/2.+ L/2], showing that now you have cos kx and sin kx solutions, alternating (theground state is even, the first excited state is odd, and so on). Plot the first eigenstates. Calculate theeigenstate normalization factors, and show the orthogonality of different eigenstates.
Obviously, the eigenfunctions of the Hamiltonian form a orthonormal basis in the chosen in-terval: they are the basis element of the Fourier series. For instance, with the symmetric choice[−L/2,+L/2] the Fourier series reads:
ψ(x) =∑
n=1,3,···an cos
(nπLx)
+∑
n=2,4,···bn sin
(nπLx). (2.4)
Exercise 2.2 Write the expressions for the Fourier coefficients an and bn, taking proper care of the correctnormalization of the basis functions cos kx and sin kx.
2.2 Scattering by a step
Next, we did a scattering problem: a potential step of V0 > 0 for x > 0. There is no loss of generalityin doing so: if V0 < 0 you can reset the zero of energy and view it as a step on the left of theorigin. Since the minimum of the potential is 0, we have E ≥ 0. We should distinguish two cases: i)E > V0 and ii) E ≤ V0. For the case i), E > V0 we have that in both x < 0 and x > 0 the problemis a free-particle problem, but with energy, respectively, E and E − V0 (because you can bring theconstant potential on the right-hand side of the SE). We therefore consider candidate solutions ofthe form:
ψk(x) =eikx + re−ikx for x ≤ 0teiqx for x ≥ 0 , (2.5)
with positive k and q, so that the ψk describe a wave incoming from the left of the origin, partiallyreflected (reflection amplitude r), and transmitted to the right of the origin (transmission amplitudet). The overall normalization factor is not important (this is why we put the coefficient of eikx to1) because the ψk is intrinsically not-normalizable, as any extended (unbound) state is. In order
Finite square well 19
to be a solution with energy E, we must have (verify) E = ~2k2/2m and E − V0 = ~2q2/2m. So,for any given E, k > 0 and q > 0 are univocally defined: ~k =
√2mE, and ~q =
√2m(E − V0).
The continuity of ψk (always true) and of ψ′k (because the potential has just a finite jump), leadto the following two equations:
t = 1 + r
tq = k(1− r) , (2.6)
whose solution gives:
t =2kk + q
and r =k − q
k + q. (2.7)
The other case, ii) E < V0, is solved in a similar way, but with the choice of an evanescent wavete−qx for x ≥ 0, with ~q =
√2m(V0 − E). The calculations are very similar to case i), and simply
require substituting q → iq, obtaining:
t =2k
k + iqand r =
k − iq
k + iq. (2.8)
We noted that r is a pure phase factor, and that, for V0 → +∞, we have q → +∞, r → −1 andt → 0, and the wave-function becomes 2i sin kx for x ≤ 0, and vanishes for x ≥ 0, i.e., it is stillcontinuous in 0, but its derivative becomes discontinuous in the origin (an infinite V0 does notpreserve continuity of ψ′.
We discussed the degeneracy of solutions, the physical meaning of the reflection and transmissionamplitudes r and t, the expression for the current (which must be constant everywhere).
2.3 Finite square well
Next we considered the case of a finite, symmetrical, potential well in [−a, a]: the potential is setto 0 in [−a, a], and is equal to V0 for |x| ≥ a. Once again we must have E ≥ 0, and we can havetwo cases: i) E < V0 (search of bound states), and ii) E > V0 (scattering states, with transmissionand reflection coefficients).
In case i), E < V0 we write the wave-function as:
ψk(x) =
aeqx for x ≤ −abeikx + ce−ikx for |x| ≤ ade−qx for x ≥ a
, (2.9)
with ~k =√
2mE, and ~q =√
2m(V0 − E). Next we discussed how to use parity, taking d = a andc = b for even solutions, and d = −a and c = −b for odd solutions. In both cases, once continuityof ψ and ψ′ are imposed at point x = −a, they are automatically guaranteed at the other pointx = +a. We arrived at the (graphical) solution for the bound-state eigenvalues, proving that thereis at least one bound state independently of a and V0, while the total number of bound statespresent (both even and odd) depends on the depth of the well.
Exercise 2.3 Solve the finite symmetric square well for the scattering states with E > V0, by setting up,as usual in scattering problems, ψk(x) = eiqx + re−iqx for x ≤ −a, ψk = aeikx + be−ikx for |x| ≤ a, and
ψk(x) = teiqx for x ≥ +a, with ~k =√
2mE, and ~q =p
2m(E − V0). Calculate the current everywhere,and deduce a relationship between t and r. Show that there are peculiar energy values (corresponding toresonances for the infinite square well) at which the transmission coefficient becomes 1 (resonant tunneling).
20 One-dimensional problems
Exercise 2.4 By exploiting the results of the symmetric finite potential well (in particular, the oddsolutions of it) solve the case of a square well with an infinite wall in the origin, i.e., such that the potentialis +∞ for x < 0, 0 for [0, a], and V0 for x > a.
2.4 Delta-function potential
Finally, we considered a rather singular potential, an attractive delta-function in the origin: V (x) =−V0δ(x) with V0 > 0. We discussed how the boundary condition for the derivate ψ′ should beimposed, and found the bound state of the problem.
Exercise 2.5 Discuss scattering solutions with E ≥ 0 for both the delta-function potential, in bothattractive and repulsive cases.
3
Harmonic oscillator
The problem we consider now is very important, because it arises in many contexts as an approx-imation to the real, more complicated, potential (close to a minimum). The Hamiltonian reads:
H =P2
2m+
12mω2X2 = Hx +Hy +Hz , (3.1)
where we have explicitly indicated the separation into three one-dimensional Hamiltonian eachreferring to a given Cartesian component (α = x, y, z):
Hα =P 2α
2m+
12mω2X2
α . (3.2)
Obviously, these three Hamiltonians commute among each other. We now illustrate how to solvethe 3D problem by the method of separation of variables. It is easy to prove (done in class) thatif we solve the one-dimensional problems:
Hαψ(α)n (Xα) = E(α)
n ψ(α)n (Xα) , (3.3)
then Ψn,m,l(X,Y, Z) = ψ(x)n (X)ψ(y)
m (Y )ψ(z)l (Z) is an eigenstates of H with energy En,m,l = E
(x)n +
E(y)m + E
(z)l . So, it is enough to solve a one-dimensional harmonic oscillator (indeed, one might
generalize a bit the 3D Hamiltonian and consider a different ωα for each component). Omitting allindices, we write the Hamiltonian as:
H =P 2
2m+
12mω2X2 . (3.4)
3.1 Commutators and Hermitean conjugates
The concept of commutator is central to QM. The order in which two operators are applied to awave-function matters. Even in the finite-dimensional case, when the operators are simple matrices,this is so. For instance, consider two matrices A, and B defined as follows:
A =(
1 00 −1
), B =
(0 11 0
). (3.5)
Then, it is easy to show that:
A ·B =(
0 1−1 0
), B ·A =
(0 −11 0
). (3.6)
We define the commutator of A and B as:
[A,B] def= A ·B −B ·A . (3.7)
Consider, for instance A = X and B = P . When acting on a generic wave-function ψ(X):
22 Harmonic oscillator
XPψ(X) = (−i~)Xd
dXψ(X) .
However:
P Xψ(X) = (−i~)d
dXXψ(X) = (−i~)
(ψ(X) +X
d
dXψ(X)
)= .
Therefore, on any ψ(X) we have that:
[X, P ] = i~ . (3.8)
Exercise 3.1 Prove that the following equalities hold:
[AB,C] = A[B,C] + [A,C]B ,
[A,BC] = [A,B]C +B[A,C] .
With these equalities, it is simple to calculated the following commutators:
Exercise 3.2 Show that [X, P 2] = 2i~P , [X, P 3] = 3i~P 2, and, in general, [X, Pn] = ni~Pn−1, or,
for functions f(P ) which can be expanded in Taylor series: [X, f(P )] = i~f ′(P ), where f ′ denotes the
derivative. Similarly, show that [Xn, P ] = ni~Xn−1, and [f(X), P ] = i~f ′(X).
An important mathematical concept related to operators is that of the Hermitean conjugate ofan operator A, denoted with A†. The definition is simple. For any two vectors ψ1 and ψ2 in theHilbert space, one has to have:
〈ψ1|Aψ2〉 = 〈A†ψ1|ψ2〉 . (3.9)
In words, you can bring an operator acting on the ket in such a way that it acts on the bra, ifyou take the Hermitean conjugate. Hermitean operators, evidently, are the Hermitean conjugateof themselves, i.e., A = A†.
Exercise 3.3 If A and B are Hermitean operators, show that:
(A+ iB)† = A† − iB† = A− iB .
3.2 Dimensionless variables
It is always a good practice, both in analytical work as well as in numerical implementations onthe computer, to work with dimensionless variables, so that parameters like the mass m, or thefrequency ω, or universal constants like ~, drop out of the problem. In the present case, for instance,it is a good idea to work with a length l such that the potential energy of the spring and the typicalconfinement energy for a particle in a box of the same length are equal:
~2
2ml2=
12mω2l2 .
This provides a typical “oscillator length”, defined by:
The algebraic solution 23
l =
√~mω
.
Define now a dimensionless length x = X/l and a dimensionless momentum p = P l/~. It is simpleto verify that their commutator is [x, p] = i, i.e., effectively as if ~ = 1. It is now simple to verifythat the Hamiltonian reads:
H = ~ωh with h =12(p2 + x2
), (3.10)
where we have defined a dimensionless Hamiltonian h = H/(~ω). It is now clear that this procedureis totally equivalent to setting m = 1, ω = 1, ~ = 1, as often stated in books or articles. You shouldalways be aware, however, of the exact meaning of this statement, and be ready to reintroduce theappropriate parameters and universal constants if asked to produce explicit numbers. Summarizing,we measure lengths in units of l, masses in units of m, energies in units of ~ω, momenta in unitsof ~/l, times in units of ~/(~ω) = 1/ω, etc.
3.3 The algebraic solution
Define now:a =
1√2
(x+ ip) . (3.11)
a is quite clearly not Hermitean (see exercise 3.3). Its Hermitean conjugate is:
a† =1√2
(x− ip) . (3.12)
In x-representation, both are quite simple operators:
a =1√2
(x+
d
dx
)and a† =
1√2
(x− d
dx
). (3.13)
The commutator of a and a† is immediately calculated:
[a, a†] = 1 .
It is simple to invert the definition of a and a† to re-express x and p in terms of a and a†,obtaining:
x =1√2
(a+ a†
)and p =
1√2i
(a− a†
). (3.14)
As done in class (do it as an exercise), one can then show that:
h =(a†a+
12
). (3.15)
Let us start finding a state |ψ0〉 which is annihilated by a, i.e. such that:
a|ψ0〉 = 0 .
Such a state has a simple expression in x-representation, ψ0(x) = 〈x|ψ0〉, since it obeys the equa-tion:
1√2
(x+
d
dx
)ψ0(x) = 0 ,
which can be rewritten as:d
dxψ0(x) = −xψ0(x) ,
and is solved by:ψ0(x) = Ce−x
2/2 ,
where C a normalization constant.
24 Harmonic oscillator
Exercise 3.4 Calculate the normalization constant C appearing in the state ψ0.
Evidently, |ψ0〉 is an eigenstate of h with eigenvalue 1/2:
h|ψ0〉 =12|ψ0〉 .
It is also easy to show that |ψ0〉 must be the ground state of h:
Exercise 3.5 Given any state |ψ〉, the expectation value of h satisfies:
〈ψ|„a†a+
1
2
«|ψ〉 =
1
2+ ||a|ψ〉||2 ≥ 1
2.
Construct now the state |ψ1〉 = a†|ψ0〉. Using the commutator [a, a†] = 1 it is simple to showthat |ψ1〉 is an eigenstate with energy 3/2. In real space, the representation of ψ1 is simple:
ψ1(x) = 〈x|ψ1〉 =1√2
(x− d
dx
)ψ0(x) .
Calculate and plot this wave-function.
More generally, we are now going to use an important property of the number operator
n = a†a . (3.16)
Exercise 3.6 Prove (for instance by recursion) that:
n“a†
”n|ψ0〉 = n
“a†
”n|ψ0〉 ,
or, in words, that the state with n operators a† applied to ψ0 is an eigenstate of the number operatorn = a†a, with eigenvalue n, i.e., n counts the number of times a† appears.
Based on this, and observing that h = n+1/2, we realize that the eigenstates of h are constructedas:
|ψn〉 = Cn(a†)n |ψ0〉 =⇒ h|ψn〉 =
(n+
12
)|ψn〉 ,
where Cn is a normalization constant. It is easy to show that a and a† allow us to move up anddown in the ladder of states |ψn〉.
Exercise 3.7 Prove that the states
|ψn〉 =(a†)n√n!
|ψ0〉 ,
are correctly normalized. Deduce also that: a†|ψn〉 =√n+ 1|ψn+1〉 and a|ψn〉 =
√n|ψn−1〉.
Notation: You often find |n〉 in place of |ψn〉 and |0〉 in place of |ψ0〉. The state |0〉 is called thevacuum of a.
4
The hydrogen atom problem
The hydrogen atom problem consists of a proton (more generally, a nucleus of charge +Ze), withan electron (charge −e) orbiting around it. The Hamiltonian is:
H =p2
1
2m1+
p22
2m2− Ze2
|r2 − r1|, (4.1)
where m1 indicates the mass of the nucleus, m2 that of the electron, and (r1, p1) and (r2,p2)denote coordinate and momenta for nucleus and electron, respectively. This is a typical two-bodyproblem, and, as in classical mechanics, one solves it by separating variables into center-of-massand relative motion. To this end, one introduces the center of mass R = (m1r1 +m2r2)/M and thetotal momentum P = p1 + p2. It is a simple matter to show that these are canonically conjugatevariables, i.e., their commutator is
[Rα, Pβ ] = i~δα,β .
Regarding the relative motion, an obvious coordinate is the relative coordinate r = r2−r1. However,the relative momentum is less trivial: the naive guess p = p2−p1 is wrong, since it is not canonicallyconjugate to r, and does not commute with R either, as it should. The way to go is to posep = β2p2 − β1p1 and determine β1 and β2 requiring canonical conjugation. The solution is:
p =m1
Mp2 −
m2
Mp1 . (4.2)
Therefore, the problem admits now separation of variables, since the total Hamiltonian is:
H =P2
2M+
p2
2µ− Ze2
|r|. (4.3)
The total wave-function can be written as Ψ(R, r) = Φ(R)ψ(r). The center-of-mass problem isjust a free-particle problem, with wave-function Φ(R) = eik·R and energy Ek = ~2k2/2M (there isa translation symmetry for the whole atom): we will neglect it from now on. The relative particlewave-function ψ(r) obeys the 3D problem:(
p2
2µ− Ze2
|r|
)ψ(r) = Eψ(r) . (4.4)
This would be still a quite intricate problem (we would need a computer if the potential was notspherically symmetric): luckily enough, we can still separate variables in spherical coordinates.This require introduction of a very important operator: the angular momentum.
4.1 Orbital angular momentum
The classical mechanics angular momentum is defined as:
L = r× p .
26 The hydrogen atom problem
In quantum mechanics, the corresponding operator is obtained by assuming that r and p areoperators. The three Cartesian components of L are immediate:
Lx = ypz − zpy
Ly = zpx − xpz
Lz = xpy − ypx . (4.5)
Commutators between the different components of L are easy to establish, using the usual rulesfor commutators, such as [AB,C] = A[B,C] + [A,C]B. The result (worked out in class) is:
[Lx, Ly] = i~Lz ,
and similar cyclic relations, which can be all summarized as:
[Lα, Lβ ] = i~εαβγLγ . (4.6)
where εαβγ is the totally antisymmetric tensor appearing also in the vector product of two ordinaryvectors: (a×b)γ = εαβγaαbβ (sum over repeated indices are always assumed). By changing variablesfrom Cartesian to spherical coordinates,
x = r sin θ cosφy = r sin θ sinφz = r cos θ , (4.7)
it is relatively simple to show that all components of L do not depend on r, but only on the anglesθ and φ:
Lx = i~(
+sinφ∂
∂θ+ cosφ cot θ
∂
∂φ
)Ly = i~
(− cosφ
∂
∂θ+ sinφ cot θ
∂
∂φ
)Lz = −i~ ∂
∂φ(4.8)
Exercise 4.1 Verify that the three components of angular momentum are given by Eq. 4.8. To this end,it might be useful to show that:0@ ∂
∂x∂∂y∂∂z
1A =
0@ sin θ cosφ cos θ cosφ − sinφsin θ
sin θ sinφ cos θ sinφ + cosφsin θ
cos θ − sin θ 0
1A 0@ ∂∂r
1r∂∂θ
1r∂∂φ
1A . (4.9)
Another important operator is:L2 = L2
x + L2y + L2
z , (4.10)
which can be expressed in spherical coordinates as:
L2 = −~2
(∂2
∂θ2+
1tan θ
∂
∂θ+
1sin2 θ
∂2
∂φ2
). (4.11)
Separation of variables 27
Exercise 4.2 Verify the expression in Eq. 4.11 for L2.
It is simple to show (done in class) that [L2, Lα] = 0. Therefore, we can find simultaneous eigen-functions of L2 and, say, Lz (we will prove this result, in general, later on). We will indeed show, inthe next lecture, that we can construct functions of θ and φ, known as spherical harmonics, whichare common eigenfunctions of L2 and Lz:
L2Yl,m(θ, φ) = ~2l(l + 1)Yl,m(θ, φ)LzYl,m(θ, φ) = ~mYl,m(θ, φ) . (4.12)
with l integers and m = −l, · · ·+ l.
The reason why all this is useful is that p2 (i.e., the Laplacian) can be easily expressed in termsof L2 and derivatives with respect to r. Classically, consider the plane containing r and p, andconvince yourself that
L2 = r2p2⊥ = r2(p2 − p2
‖) = r2p2 − (r · p)2 .
Quantum mechanically, things are slightly more complicated because r and p do not always com-mute. You can still prove (done in class) that:
L2 = r2p2 − (r · p)2 + i~r · p . (4.13)
This equality can be shown either directly, by components, or by doing the following exercise:
Exercise 4.3 Using the fact that Li = εijkxjpk, with the convention of summing on repeated indices,construct L2 = LiLi, and use the fact that εijkεilm = (δjlδkm − δjmδkl) to prove the result for L2.
Moreover, if you calculate
∂
∂r=∂x
∂r
∂
∂x+∂y
∂r
∂
∂y+∂z
∂r
∂
∂z=
1rr · ~∇r ,
you can immediately conclude that:
r · p = −i~r ∂∂r
. (4.14)
Therefore, using Eqs. 4.13 and 4.14 we can easily rewrite the Laplacian as:
p2 = −~2
r2r∂
∂r
(r∂
∂r
)− ~2
r2r∂
∂r+
1r2L2
= −~2 ∂2
∂r2− 2~2
r
∂
∂r+
1r2L2 . (4.15)
4.2 Separation of variables
This expression for the Laplacian clearly separates the r-variable from the θ and φ variables, whichare contained only in L2. As a result of that, you can factorize the wave-function ψ(r) as
ψk,l,m(r) = Rk,l(r)Yl,m(θ, φ) ,
where k is an extra quantum number (still not specified) which labels the many possible solutionsof the equation for R, which reads:
28 The hydrogen atom problem
− ~2
2µ
(∂2
∂r2+
2r
∂
∂r
)Rk,l +
(~2l(l + 1)
2µr2− Ze2
r
)Rk,l = Ek,lRk,l . (4.16)
Notice that this is a one-dimensional differential equation, which, however, has not the standardform of a SE. The normalization condition is also quite different from a 1D normalization, since itreads: ∫ +∞
0
dr r2 |Rl,l(r)|2 = 1 .
It is therefore tempting to define a new function
uk,l(r) = rRk,l(r) ,
which, to start with, will have a standard normalization∫ +∞0
dr |uk,l(r)|2 = 1. Moreover, you canimmediately verify that the non-standard derivative terms for R become completely standard whenwritten in terms of derivatives of u:(
∂2
∂r2+
2r
∂
∂r
)Rk,l =
1r
∂2uk,l∂r2
.
When written in terms of uk,l the SE becomes therefore a standard 1D SE in the half-line [0,∞):
− ~2
2µ∂2uk,l∂r2
+(
~2l(l + 1)2µr2
− Ze2
r
)uk,l = Ek,luk,l , (4.17)
with a standard normalization condition:∫ +∞
0
dr |uk,l(r)|2 = 1 . (4.18)
4.3 Dimensionless variables
We now get rid of all the physical constants (~, e, µ, Z) by switching to dimensionless variables.The procedure is quite similar to that for the harmonic oscillator. Let us define a length a suchthat potential energy and kinetic energy of confinement are of the same order:
Ze2
a=
~2
µa2.
Solving this equation we get:
a = aB =~2
µe2Z, (4.19)
which we recognize to be the effective Bohr radius of the problem. The corresponding energy is:
EH = Ze2
aB=µe4Z2
~2. (4.20)
This quantity is, for Z = 1, the Hartree and is equal to EH ≈ Z2(27.2)eV . Measuring lengthsin units of aB , energies in units of EH and masses in units of µ defines a system of units calledatomic units, often abbreviated as “a.u.”. If ρ = r/aB is the dimensionless length and E = E/EHthe dimensionless energy, defining uk,l(ρ) =
√aBuk,l(r = aBρ), we get the dimensionless SE in the
form:
−12∂2uk,l∂ρ2
+(l(l + 1)
2ρ2− 1ρ
)uk,l = Ek,luk,l , (4.21)
with the normalization condition: ∫ +∞
0
dρ |uk,l(ρ)|2 = 1 . (4.22)
The Coulomb problem bound states 29
Notice that this looks like a standard 1D problem in the half-line, with an effective potential givenby:
V(l)eff (ρ) =
l(l + 1)2ρ2
− 1ρ, (4.23)
i.e., comprising an attractive Coulomb term plus a repulsive centrifugal term for l > 0, which triesto push electrons out of the nucleus (ρ = 0) the higher is the angular momentum (which is aconserved quantity). Notice also that exactly the same discussion would have been possible for anycentral potential V (r): the explicit form of V (r) would be substituted to the 1/r in the effectivepotential. Nevertheless, as we shall see, the 1/r form is rather special in being exactly solvable andhaving peculiar degeneracies of levels.
4.4 The Coulomb problem bound states
First of all, we denote simply by u the u and omit the tilde everywhere. We also denote by u′ andu′′ the first and second derivatives with respect to ρ. The first part of the analysis would apply toany reasonable potential, and looks for asymptotic behaviours of u for small and large ρ.
Small-ρ behaviour. Take a power law u(ρ) = A1ρα1 + · · · where the dots indicate less relevant
powers for ρ→ 0 (i.e., those with a larger exponent). Taking derivatives we have: u′ = A1α1ρα1−1+
· · · and u′′ = A1α1(α1 − 1)ρα1−2 + · · · . Substituting in the SE, keeping only the most relevantterms in each piece we have:
−12[A1α1(α1 − 1)ρα1−2 + · · ·
]+l(l + 1)
2[A1ρ
α1−2 + · · ·]−[A1ρ
α1−1 + · · ·]
= E [A1ρα1 + · · · ] .
The most relevant term is that coming out of u′′ and the l-centrifugal term, both going as ρα1−2.In order for this most relevant term to be satisfied, we need:
α1(α1 − 1) = l(l + 1) ,
which admits two solutions: i) α1 = l + 1 > 0 and ii) α1 = −l ≤ 0. The first case is certainlyadmissible: it predicts that u(ρ) ∼ A1ρ
l+1, i.e., the smaller the larger is l, and in particularu(0) = 0 (a kind of boundary condition) for all l. The possibility ii) turns out to be physicallyabsurd: it predict a larger and larger u close to the origin as l increases. And even for l = 0, itwould predict a constant u(0) (non-vanishing), which in turn implies a R(r) ∼ 1/r close to theorigin. But a wave-function behaving like 1/r close to the origin, would give rise to a delta functionterm when we act with the Laplacian on it, which would never be canceled by any other piece inthe SE. Therefore, we conclude that:
ukl(ρ) ∼ A1ρl+1 + · · · for ρ→ 0 . (4.24)
Large-ρ behaviour. Next consider the large ρ behaviour. There, the potential (including thecentrifugal repulsion) has decreased to zero, so that the SE becomes:
−12∂2uk,l∂ρ2
≈ Ek,luk,l ,
where Ek,l < 0 because we are considering bound states. A function which satisfies this approximateequation is the exponential:
uk,l ∼ e−λk,lρ ,
with λk,l =√−2Ek,l.
30 The hydrogen atom problem
Educated by these observations, let us consider the simple l = 0 case as a warm-up. Our guessfor the wave-function is:
ul=0 = Aρe−λρ .
By substitution in the SE (done in class) we see that this is a solution provided λ = 1, which,together with λ =
√−2E gives us:
El=0 = −12.
This is indeed the ground state (GS) energy for the hydrogen problem, in atomic units. The factthat it is indeed the GS, comes from the absence of nodes in the function.
Let us now set up the general solution. We write:
uk,l(ρ) = wk,l(ρ)e−λk,lρ ,
where the function wk,l(ρ) should start as ρl+1, and λk,l =√−2Ek,l. Calculating u′ and u′′ in
terms of w′ and w′′ we see that the Ek,l-term disappears, and the equation for wk,l comes out tobe:
−12∂2wk,l∂ρ2
+ λk,l∂wk,l∂ρ
+(l(l + 1)
2ρ2− 1ρ
)wk,l = 0 . (4.25)
Now, assume for wk,l the form:
wk,l(ρ) = ρl+1[c0 + c1ρ+ c2ρ
2 + · · ·]
=∞∑q=0
cqρq+l+1 , (4.26)
where the coefficients of the (possibly infinite) series cq have to be determined, and c0 6= 0. Calcu-lating w′ and w′′ and substituting in Eq. 4.25 we get:
−12
∞∑q=0
(q+ l+1)(q+ l)cqρq+l−1 +λk,l
∞∑q=0
(q+ l+1)cqρq+l+l(l + 1)
2
∞∑q=0
cqρq+l−1−
∞∑q=0
cqρq+l = 0 .
Notice that the l(l+1)-term cancels against a similar term coming from the first series. Therefore,we can rewrite:
−12
∞∑q=0
q(q + 2l + 1)cqρq+l−1 + λk,l
∞∑q′=1
(q′ + l)cq′−1ρq′+l−1 −
∞∑q′=1
cq′−1ρq′+l−1 = 0 . (4.27)
The q = 0 from the first term is actually multiplied by zero, and we can also relabel q′ → q.Therefore, we can finally rewrite:
−12
∞∑q=1
q(q + 2l + 1)cqρq+l−1 +∞∑q=1
[λk,l(q + l)− 1] cq−1ρq+l−1 = 0 . (4.28)
Term by term, this implies the following recursion relation for the coefficients cq:
cq = 2λk,l(q + l)− 1q(q + 2l + 1)
cq−1 , (4.29)
with c0 6= 0 determined only by the normalization condition.
By looking at this recursion formula, it would appear that we have solved our problem no matterwhat the value of the energy Ek,l (hence of the constant λk,l) is. Is it really so? This would implythat the bound states do not have a discrete spectrum. Where is the problem with this argument?The error is here: if the series goes on forever, then, for large q’s, the ratio:
cqcq−1
= 2λk,l(q + l)− 1q(q + 2l + 1)
−→≈ 2λk,lq
+ · · ·
But an infinite series with this behaviour for the coefficients is really representing a function wwhich diverges like w ∼ e2λk,lρ, i.e., u ∼ e+λk,lρ which is simply the other exponential solution,
The Coulomb problem bound states 31
which must be rejected because exploding at +∞. Indeed, by writing e2λk,lρ =∑q(2λk,l)
qρq/q!you can simply verify that the coefficients of this Taylor series do exactly that. Therefore, and hereis the key point for energy quantization, the only acceptable solutions have a power series thatterminates in a finite number, call it k ≥ 1, of terms, i.e., ck = 0. But this, in turn, implies:
λk,l(k + l)− 1 = 0 −→ λk,l =1
k + l, (4.30)
or, for the energy, restoring the tilde:
Ek,l = −12λ2k,l = − 1
2(k + l)2= − 1
2n2, (4.31)
where we have defined the principal quantum number n = k + l, while k is known as radialquantum number, because it determines the number of zeroes of the radial function. The groundstate corresponds to n = 1, with energy −1/2 (in a.u.), and must be associated to k = 1 and l = 0:it is known as 1s-state (“s” because of the l = 0 and “1” to remind us of the n = 1). The next levelhas n = 2 and energy −1/8 a.u.: it can be obtained either with (k = 2, l = 0) (2s-state) or with(k = 1, l = 1) (2p-states, there are 3 of them, because I still have m = −1, 0,+1 for an l = 1).
Exercise 4.4 Enumerate the quantum numbers (and degeneracy) associated to the n = 3 and n = 4states of the hydrogen problem, writing the corresponding energy.
Summarizing, for the radial functions we have:
Rk,l(ρ) = ρl(c0 + c1ρ+ · · · ck−1ρ
k−1)e−λk,lρ , (4.32)
where the coefficients of the polynomial in ρ are given by the recursive formula in Eq. 4.29, andλk,l = 1/(k + l). It is now clear what is the meaning of the quantum number k introduced at thebeginning of our discussion. Notice, however, that the hydrogen problem has extra degeneracies,with respect to a generic central potential, for which, for instance, the 2p and 2s states would notbe degenerate.
Exercise 4.5 By calculating explicitly, in each case, the coefficients ci, write down the radial wave-functions associated to all the energy levels up to n = 3, and draw them.
Exercise 4.6 Solve the free-particle problem in three dimensions by using spherical coordinates. (Obtaina recursion relation for the radial function, and check that this is related to Bessel functions.)
Exercise 4.7 Solve the spherical infinite potential well in three dimensions, i.e., the potential is infiniteoutside of a sphere of radius R, and zero inside. (Use again spherical coordinates, and try to find the radialfunctions that vanish at r = R.)
Exercise 4.8 Solve the three-dimensional harmonic oscillator with potential mω2(x2 + y2 + z2)/2 inspherical coordinates by adapting the procedure we have used for the hydrogen case. Compare the resultswith the solution obtained by separation of variables in Cartesian coordinates.
5
Theory of angular momentum
We devote this lecture to studying in some detail the theory of angular momentum. We willdenote by (Jx, Jy, Jz) the three components of our angular momentum, which we do not specify inadvance: it is not necessarily the single-particle orbital angular momentum introduced in discussingthe angular motion for the hydrogen atom. The same discussion applies to any three operatorsobeying (with ~ = 1):
[Jα, Jβ ] = iεαβγJγ .
5.1 Diagonalization of J2 and Jz
Our first aim is to construct eigenvalues and eigenvectors of J2 = J2x +J2
y +J2z and Jz, which com-
mute and can be therefore diagonalized together. Let us denote by |j,m〉 the common eigenvectors,and fj and m the eigenvalues of J2 and Jz:
J2|j,m〉 = fj |j,m〉Jz|j,m〉 = m|j,m〉 . (5.1)
For the time being, j and m are generic labels for the eigenvectors. We will soon show that j mustbe an integer of half-integer, that m runs from −j to j in integer steps, and that fj = j(j + 1).
5.1.1 Ladder operators
Given an eigenstate of Jz with eigenvalue m, we can easily construct eigenstates with m ± 1 andthe same fj . To do that, introduce the following ladder operators:
J± = Jx ± iJy . (5.2)
You can easily verify that [J2, J±] = 0 and [Jz, J±] = ±J±. Therefore:
Jz (J±|j,m〉) = ([Jz, J±] + J±Jz)|j,m〉 = (m± 1)|j,m〉 ,
while:J2 (J±|j,m〉) = J±J
2|j,m〉 = fj (J±|j,m〉) .
Therefore, apart from a possible normalization factor Cjm (see below):
J±|j,m〉 = Cjm,±|j,m± 1〉 .
5.1.2 Bounds on m
It is simple to prove that:
J2 = J2z +
12(J+J− + J−J+) . (5.3)
Construction of spherical harmonics 33
As it turns out, all the operators in the previous equation are positive operators (i.e., their ex-pectation value on any state is positive). Taking the expectation value over the state |j,m〉 wehave:
〈j,m|J2 − J2z |j,m〉 = fj −m2 =
12(||J−|j,m〉||2 + ||J+|j,m〉||2
)≥ 0 .
Therefore m cannot be increased (by J+) or decreased (by J−) forever, since m2 ≤ fj . There mustexist a maximum mmax and a minimum mmin above and below which we cannot continue to go,i.e. J+|j,mmax〉 = 0 and J−|j,mmin〉 = 0. But J−J+ = J2 − J2
z − Jz and J+J− = J2 − J2z + Jz, as
one can simply verify. Therefore:
0 = J−J+|j,mmax〉 = (J2 − J2z − Jz)|j,mmax〉 = (fj −m2
max −mmax)|j,mmax〉 = 0 ,
or, equivalently:fj = mmax(mmax + 1) .
Similarly:
0 = J+J−|j,mmin〉 = (J2 − J2z + Jz)|j,mmin〉 = (fj −m2
min +mmin)|j,mmin〉 = 0 ,
or, equivalently:fj = mmin(mmin − 1) .
The two expressions for fj imply that mmin = −mmax. Let us call j = mmax, so that mmin = −j.Then fj = j(j + 1) and m has to run in the interval:
−j ≤ m ≤ j .
We can now easily calculate the normalization constant Cjm,±. It is simple to show that:
Cjm,± =√j(j + 1)−m(m± 1) .
5.1.3 Values of j
We must reach |j,mmin = −j〉 by applying J− an integer number of times n to the state |j,mmax =j〉:
|j,−j〉 ∝ (J−)n|j, j〉 ,
which implies that −j = j − n, or n = 2j, or:
j =n
2. (5.4)
For n even we have that j is integer, while for n odd j must be half-integer. The total number ofstates obtained by applying J− repeatedly to |j, j〉 is evidently n+ 1 = 2j + 1, i.e., all values of mfrom +j down to −j in integer steps.
5.2 Construction of spherical harmonics
We now explicitly construct the spherical harmonics. First of all, we adopt the usual notation Lfor the orbital angular momentum, and use l instead of j, accordingly. Still with ~ = 1, we needto solve:
L2Yl,m(θ, φ) = l(l + 1)Yl,m(θ, φ)LzYl,m(θ, φ) = mYl,m(θ, φ) , (5.5)
where the spherical harmonics are Yl,m(θ, φ) = 〈θ, φ|l,m〉. First of all, we notice that Lz = −i∂/∂φso that the eigenfunctions of Lz must solve:
34 Theory of angular momentum
−i ∂∂φ
Yl,m(θ, φ) = mYl,m(θ, φ) .
The solution of this equation is evidently of the form:
Yl,m(θ, φ) = Pl,m(θ)eimφ .
Moreover, since the function Yl,m must be single-valued when φ → φ + 2π we must have that mis integer, and therefore l must also be an integer. Half-integer values of l are not allowed for theorbital angular momentum: we will see that they are characteristic of the spin.
Consider now the state with mmax = l, i.e. Yl,l. We must have:
L+Yl,l(θ, φ) = L+Pl,l(θ)eilφ = 0 .
But the differential expression for L+ = Lx + iLy is (see Eq. 4.8):
L+ = ieiφ[−i ∂∂θ
+cos θsin θ
∂
∂φ
].
Applying it to Yl,l we get a differential equation for Pl,l(θ):
∂Pl,l∂θ
= lcos θsin θ
Pl,l .
It is simple to verify that the solution of this differential equation is:
Pl,l(θ) = Cl sinl θ .
The normalization constant can be calculated to be:
Cl =(−1)l
2ll!
√(2l + 1)(2l)!
4π,
where the sign (−1)l is a convention. All other states Yl,m are obtained by applying L− to Yl,l acertain number of times, where:
L− = ie−iφ[i∂
∂θ+
cos θsin θ
∂
∂φ
].
5.2.1 Explicit l = 1 case: the p-states
The case of l = 0 (the so-called s-states) is trivial, since the only state is:
Y0,0 =1√4π
.
Let us study in more detail the case l = 1, the so-called p-states of an atom. Here:
Y1,1(θ, φ) = C1 sin θeiφ = −√
38π
sin θeiφ .
Applying L−, and recalling that L−Y1,1 =√
2Y1,0 we immediately get:
Y1,0 = −√
2C1 cos θ =
√34π
cos θ .
Finally, applying L− again we have:
Construction of spherical harmonics 35
Y1,−1 =1√2L−Y1,0 = −C1 sin θe−iφ =
√38π
sin θe−iφ = −Y ∗1,1 .
Notice that:
1√2(Y1,−1 − Y1,1) =
√34π
sin θ cosφ = |px〉
i√2(Y1,−1 + Y1,1) =
√34π
sin θ sinφ = |py〉
Y1,0 =
√34π
cos θ = |pz〉 . (5.6)
The meaning of px, py and pz, the Cartesian expression for the p-states, should be evident, if yourecall the expression for x, y and z in spherical coordinates. We discussed in class also the meaningof the polar plots typically used to depict them in textbooks.
Exercise 5.1 Derive and appropriately draw the five l = 2 spherical harmonics (d-states), staring fromY2,2 and repeatedly applying L−. Refer to a textbook to construct the Cartesian form of the d-states, andplot them with the usual polar plots.
Result: Y2,±2 =q
1532π
e±2iφ sin2 (θ), Y2,±1 = −q
158πe±iφ cos (θ) sin (θ), Y2,0 =
q5
16π(3 cos2 (θ)− 1).
6
Symmetries
We give here the theoretical framework of symmetries in QM, which we have already seen at playin a few of the problems encountered.
6.1 Space transformations
Let us start with well known transformations in ordinary space of three-dimensional vectors.
We start with translations, defined as:
Ta(r) = r′ = r + a .
Rotations are slightly more complicated. For instance, for a rotation by an angle θ around thez-axis we have:
TRθ,z(r) = r′ = Rθ,zr ,
where Rθ,z is the well known rotation matrix:
Rθ,z =
cos θ − sin θ 0sin θ cos θ 0
0 0 1
(6.1)
Rotations are represented by orthogonal matrices, Rt = R−1, whose determinant is +1.
Inversion is also simple to represent with a matrix I = −1, whose determinant is −1 in 3D. (In2D, inversion is really not distinct from a rotation by π.)
More generally, a roto-translation is defined by a rotation matrix R (proper, if detR = +1 orimproper, that is product of a proper rotation R times the inversion I, if detR = −1) and atranslation vector a:
TR,a(r) = r′ = Rr + a . (6.2)
All roto-translations form a group (prove this very simple fact) under the product rule:
TR2,a2 · TR1,a1 = TR2R1,R2a1+a2 .
6.2 Transformations on scalar functions
We now ask how these transformations can be seen as operators acting on wave-functions.
6.2.1 Translations
Starting with translation (for instance in a one-dimensional case) it is quite natural to say that atranslation by a of a function ψ(x) gives us (draw it to convince yourself):
Transformations on scalar functions 37
PTa(ψ(x)) = ψ(x− a) = ψ(T−1
a (x)) .
Therefore, upon acting on a function ψ(x), I obtain a new function ψ(x−a). PTais quite evidently
a linear operator on functions: which form has it? We now show that it is a quite familiar object!To show this, let us expand ψ(x− a) in powers of a around point x. We have:
ψ(x− a) = ψ(x) + (−a) ∂∂xψ(x) +
(−a)2
2!∂2
∂x2ψ(x) + · · · .
Now, it is simple to realize that this involves just the momentum operator, so that we can write:
PTa(ψ(x)) = ψ(x− a) = ψ(T−1
a (x))
=[1 + (−ia
~p) +
12!
(−ia~p)2 + · · ·
]ψ(x)
= e−iap/~ψ(x) . (6.3)
In other words, PTa = e−iap/~. More generally, for translations in 3D we have:
PTa = e−ia·p/~ ,
because translations along different coordinates (and corresponding components of momentum)commute. Notice the strong similarity with the time-evolution operator:
|ψ(t)〉 = e−iHt/~|ψ(0)〉 .
One expresses these equalities by saying that p/~ is the generator of translations in space, whileH/~ is the generator of translations in time. The meaning of “generator” is that the full operatoris obtained by taking the exponential of the generator, times a parameter (a or t) specifying theamount of translation.
6.2.2 Rotations
Slightly more complicated is the discussion of rotations. Consider the usual rotation Rθ,z. Onceagain, define the action on a wave-function by:
PRθ,z(ψ(r)) = ψ(R−1
θ,z(r)) .
It is simple to calculate R−1θ,z(r):
R−1θ,z(r) =
cos θ sin θ 0− sin θ cos θ 0
0 0 1
xyz
=
x cos θ + y sin θ−x sin θ + y cos θ
z
Therefore:
PRθ,z(ψ(r)) = ψ(x+ δx, y + δy, z) ,
where δx = x(cos θ−1)+y sin θ and δy = y(cos θ−1)−x sin θ. Assume now that θ is infinitesimallysmall, and expand (cos θ − 1) and sin θ to lowest order in θ. One finds that δx = θy + O(θ2) andδy = −θx+O(θ2). The Taylor expansion of ψ to first order in δx and δy, therefore, gives:
PRθ,z(ψ(r)) = ψ(x+ δx, y + δy, z) = ψ(r) + δx
∂ψ
∂x+ δy
∂ψ
∂y+ · · ·
= ψ(r) + θ
(y∂ψ
∂x− x
∂ψ
∂y
)+ · · ·
= ψ(r)− iθ
~Lzψ(r) +O(θ2) . (6.4)
We will soon prove that this is, once again, just the first term of the expansion of the exponential:
PRθ,z(ψ(r)) = e−i
θ~ Lzψ(r) . (6.5)
38 Symmetries
6.2.3 Generalities
Let us spell out some general properties of the operators PT . If T ∈ G, where G is a group of thetransformations T , then we can prove that PT , defined by the action:
PT (ψ(r)) = ψ(T−1(r)) , (6.6)
form a group of operators acting on scalar functions, and, moreover, such operators are unitary.The fact that the PT |T ∈ G form a group is simple, if we define the product of two PT by therule:
PT2 · PT1 = PT1·T2 .
(One can easily verify associativity, that the inverse is P−1T = PT−1 , etc.) To prove that the PT are
unitary, just write:
〈PTψ1|PTψ2〉 =∫dr ψ∗1(T−1(r))ψ2(T−1(r)) = 〈ψ1|ψ2〉 ,
where in the last equality we have changed variable to r′ = T−1(r), with a Jacobian-determinantequal to 1, since detT = ±1. Therefore:
〈ψ1|P †TPTψ2〉 = 〈ψ1|ψ2〉 ∀ψ1, ψ2 =⇒ P †
TPT = 1 .
Consider now a general rotation by an angle θ + ε around the z-axis, with θ finite but ε small.Since Rθ+ε,z = Rθ,z ·Rε,z = Rε,z ·Rθ,z, we deduce that:
PRθ+ε,z= PRε,z · PRθ,z
=(1− i
ε
~Lz +O(ε2)
)· PRθ,z
.
This expression allows us to write a differential equation for the operator PRθ,z. Indeed:
dPRθ,z
dθ= limε→0
PRθ+ε,z− PRθ,z
ε= − i
~LzPRθ,z
, (6.7)
and the solution of this differential equation is given, quite evidently, by:
PRθ,z= e−i
θ~ Lz . (6.8)
For a general rotation by an angle θ around an axis n one can show that:
PRθ,n= e−i
θ~ n·L . (6.9)
Notice, however, that n · L = nxLx + nyLy + nzLz is a sum of three non-commuting operators(unlike the linear momentum case) and therefore
e−iθ~ n·L 6= e−i
θ~nxLx · e−i θ
~nyLy · e−i θ~nzLz .
In order to express PRθ,nas a product of three independent rotations, one needs to resort to Euler
angles.
6.3 Symmetry properties of Hamiltonians
We say that a Hamiltonian H is invariant under a certain space transformation T if:
Hr′ = Hr , (6.10)
where Hr and Hr′ denote the Hamiltonian expressed using the variables r and r′ = Tr (includingtransformation of momenta, etc.). We will illustrate this definition in a moment, but we now stress
Symmetry properties of Hamiltonians 39
that all the transformation T which leave H invariant form a group GH , called invariance groupof the Hamiltonian. The fact that these transformation form a group is left as an exercise forthe reader.
Let us give a few examples, to illustrate this definition. First of all, it is simple to prove that thekinetic energy term p2/2m is always invariant with respect to any translation and rotation. Fortranslation, the proof is trivial, since from x′ = x+ a you immediately deduce that d/dx′ = d/dx,hence p′ = p (where the prime denotes the momenta in the variables r′). Consider now rotations.For a rotation around z we have:x′
y′
z′
=
cos θ − sin θ 0sin θ cos θ 0
0 0 1
xyz
, (6.11)
from which we immediately deduce that: ∂∂x′∂∂y′∂∂z′
=
cos θ − sin θ 0sin θ cos θ 0
0 0 1
∂∂x∂∂y∂∂z
. (6.12)
More generally:
Exercise 6.1 Given any rotation matrix R, and the rotated variables x′ = R ·x, show that ~∇r′ = R · ~∇r.
In words, we say that the momentum operator transforms under rotation, exactly as the vector rdoes:
r′ = R · r =⇒ p′ = R · p . (6.13)
Clearly, p2 is trivially invariant by rotation, as any scalar product is.
Consider now more realistic Hamiltonians. Our first example is a two-dimensional infinite po-tential well:
H1 =p2x + p2
y
2m+ V∞(x) + V∞(y)
where V∞(x) is zero in x ∈ [−a, a] and ∞ elsewhere. Our second example is a central potential in3D:
H2 =p2x + p2
y + p2z
2µ− e2
|r|.
As a third example, consider the hydrogen atom (proton plus electron):
H3 =P 2x + P 2
y + P 2z
2M+p2x + p2
y + p2z
2m− e2
|R− r|.
Obviously, H1 is not invariant under any translations, since, for instance for x′ = x + a, wewould have V∞(x′) = V∞(x + a) 6= V∞(x). Most of the rotations (around z axis, obviously) alsofail the test, except for R±π/2,z, which exchange the roles of x and y, and for Rπ,z which sendsx→ −x and y → −y. Therefore:
GH1 = 1, PRπ/2,z, PR−π/2,z
, PRπ,z .
Regarding H2, we know that all translations satisfy the kinetic term, but unfortunately 1/|r′| =1/|r+a| 6= 1/|r|. All rotations, on the contrary, including inversion (hence also improper rotations)
40 Symmetries
satisfy both the kinetic term and the 1/|r| potential. Therefore (O(3) is the standard mathematicalsymbol for the group of 3D rotations):
GH2 = All proper and improper rotations R = O(3) .
Finally, H3 is closely connected to H2 (recall the center-of-mass transformation) but now transla-tions of both coordinates by the same amount a does not change the potential:
e2
|R′ − r′|=
e2
|(R + a)− (r + a)|=
e2
|R− r|.
Therefore:GH3 = All proper and improper roto-translations TR,a .
We now prove the following important theorem:
Theorem 6.1 Let GH be the invariance group of the Hamiltonian H, and consider the operatorsPT |T ∈ GH. One can prove that: i) this is a group of unitary operators isomorphic to GH , andii) all such operators commute with the Hamiltonian, PT H = HPT , or:
[PT , H] = 0 .
The proof is very simple. The fact that PT are unitary operators has already been shown, and thefact that it is a group is very simple (exercise left to the reader). The only thing we need to proveis ii), i.e, commutation with H. Define Hrψ(r) = φ(r). Then:
PTHrψ(r) = PTφ(r) = φ(T−1r) = HT−1rψ(T−1r) .
On the other hand, since T ∈ GH we have that Hr = HT−1r:
HrPTψ(r) = Hrψ(T−1r) = HT−1rψ(T−1r) .
Therefore, since the equality of the two sides holds for any wave-function ψ(r), we conclude thatPT H = HPT .
6.4 Commuting observables
We now explain the importance of having a set of Hermitean operators that commute with theHamiltonian. An important theorem (proof not given) states that a Hermitean operator can bealways diagonalized, i.e., we can find a set of states |φa〉 which forms an orthonormal basis ofthe Hilbert space and such that A|φa〉 = a|φa〉. The collection of the eigenvalues a, which canbe proven to be real, forms the spectrum of the operator, and can be discrete, continuum, or both.Next, we say that two observables A and B (i.e., Hermitean operators, associated to physicalquantities that can be measured) are compatible if [A,B] = 0, and incompatible if [A,B] 6= 0. Wewill fully appreciate the meaning of this when we will talk about the process of measurement inQM. For the time being, take it as a pure definition. An important theorem (proof sketched inclass), tells us that:
Theorem 6.2 If A and B are compatible observables (i.e., if [A,B] = 0), then they can diag-onalized simultaneously, i.e. one can find a common set of states |φa,b〉 which are simultaneouseigenvectors of A and B:
A|φa,b〉 = a|φa,b〉 and B|φa,b〉 = b|φa,b〉 .
Selection rules 41
In the second example analyzed previously (hydrogen case), we saw that rotations were allcommuting with H2. But we know that rotations are represented by PRθ,n
= e−iθ~ n·L. From the
fact that these operators commute with H2, taking a derivative of the commutator with respect toθ we have:
0 =∂
∂θ
(H2e
−i θ~ n·L − e−i
θ~ n·LH2
).
Taking the derivative of the exponential, and setting then θ = 0, we immediately deduce that:
[H2, n · L] = 0 ,
i.e., H2 commutes with any component of the angular momentum operator (since you can take nin any direction). Out of the three components of the angular momentum, we know that we cansimultaneously diagonalize only L2 and, say, Lz. In the end, we have proved that the following threeoperators, H2,L2, Lz are all commuting among each other. By a generalization of the previoustheorem, one can show that it is possible to find common eigenvectors to all of them, and this isjust what we have done in solving the hydrogen atom problem: rotational symmetry was essential,therefore, in diagonalizing the problem analytically.
This example is paradigmatic: symmetries imply operators which commute with the Hamilto-nian, which, in turn, can be used to diagonalize the problem in a simpler way.
6.5 Selection rules
There are several cases where one needs to know if certain matrix elements of operators vanishor not. Symmetry properties of the relevant states and operator are important to answer suchquestions in a simple way. As a paradigmatic example, consider 〈ψ2|p|ψ1〉 and the parity operatorPI (I denotes the inversion r → −r). We know that PIψ(r) = ψ(−r), and that eigenstates of PIare functions which are either even (PIψ(r) = ψ(−r) = ψ(r), eigenvalue +1), or odd (PIψ(r) =ψ(−r) = −ψ(r), eigenvalue -1). If the Hamiltonian is invariant under inversion, it commutes withparity: [PI , H] = 0. This, in turn, implies that we can always classify the eigenstates of H as evenor odd under parity: assume that ψ1 and ψ2 are two such states, PI |ψi〉 = (−1)Pi |ψi〉, where Pi = 0if parity is even, Pi = 1 if parity is odd. It is also simple to show that
P †I pPI = −p ,
or, in words, that the momentum is odd under parity (the same is true for r, while L is even underparity: prove all these statements). Now, inserting 1 = PIP
†I we have:
〈ψ2|p|ψ1〉 = 〈ψ2|PIP †I pPIP
†I |ψ1〉 = −(−1)P1+P2〈ψ2|p|ψ1〉 ,
which clearly shows that the matrix element vanishes if the two states have the same parity.
As an example, consider hydrogen atomic states. We know that states with l even are even underparity, while states with l odd are odd under parity. Therefore, we immediately conclude that, forinstance:
〈ns|p|1s〉 = 0 .
Similarly, we can immediately conclude that 〈nd|p|1s〉 = 0, while, in general, 〈np|p|1s〉 is notforced to vanish (indeed, it doesn’t).
Parity is by no means the only operator useful in this game. Indeed, consider:
〈nf |p|1s〉 .
According to parity, this matrix element might be non-vanishing, since the two states have oppositeparity (f states have l = 3, and are therefore odd). However, angular momentum would teach usthat this matrix element vanishes. The reason which this is so has to do with a very importanttheorem, Wigner-Eckart theorem, which we will not state or prove, which states that an operatorlike p can only connect states whose angular momentum differs by at most 1.
7
Measurement in Quantum Mechanics
Observables and Hermitean operators. One of the basic postulates of QM is that to everyphysical observable A we can associate an Hermitean operator A: example seen, include the energy,the momentum, the position, the angular momentum.
Completeness of eigenstates. The eigenstates |φan〉 of an Hermitean operator A with eigen-value an (which are real, as you should try to prove) form a complete orthonormal basis of theHilbert space. Allowing for a possible degeneracy gn ≥ 1 of the eigenvalue an we write:
A|φ(i)an〉 = an|φ(i)
an〉 for i = 1, · · · , gn .
Projectors. To simplify the notation, assume for a while that the complete orthonormal basisis denoted by φn. We can define a family of operators Pn such that:
Pm|φn〉 = δn,m|φm〉 ,
whose matrix elements are therefore very simple:
〈φn′ |Pm|φn〉 = δn,mδn,n′ .
Such operators are useful because, if a general state |ψ〉 =∑n φn|φn〉 (with φn = 〈φn|φ〉) is acted
upon with Pm, we end up with obtaining only the φm-component of it, i.e.:
Pm|ψ〉 = ψm|φm〉 .
So, Pm projects out all components except for the mth one. A formal but very common expressionfor Pm, in terms of Dirac’s ket and bra is:
Pm = |φm〉〈φm| ,
which simply means “act on any state |ψ〉 to the right by taking the scalar product of |ψ〉 with thebra 〈φm|”, which we can immediately verify to give the correct result. Notice that:
P 2m = Pm ,
and that we can represent the identity operator 1 as:∑m
Pm =∑m
|φm〉〈φm| = 1 .
Position and momentum eigenstates. Examples seen until now include the eigenstates ofposition |x〉 (for which sums are changed into integrals, and Kronecker delta δn,m is changed intoa Dirac’s delta 〈x′|x〉 = δ(x− x′)), with the associated projectors Px = |x〉〈x|, such that:
|ψ〉 =∫dx|x〉〈x|ψ〉 =
∫dx|x〉ψ(x) ,
Measurements in QM. 43
where we recognize the usual wave-function ψ(x) = 〈x|ψ〉. An alternative choice is to use momen-tum eigenstates |k〉, whose wave-function reads:
φk(x) = 〈x|k〉 =eik·x√(2π)3
,
where we used the correct normalization on the continuum, 〈k′|k〉 = δ(k−k′). Using completenesswe can now write:
|ψ〉 =∫dk|k〉〈k|ψ〉 ,
where we see the momentum-space wave-function 〈k|ψ〉 appearing. Explicitly, inserting a com-pleteness on the x, we can show that:
〈k|ψ〉 =∫dx〈k|x〉〈x|ψ〉 =
∫dx
e−ik·x√(2π)3
ψ(x) =1√
(2π)3ψ(k) .
where we recognize the Fourier transform ψ(k) =∫dxe−ik·xψ(x) of the ψ(x). Substituting, we get:
|ψ〉 =∫
dk√(2π)3
|k〉ψ(k) .
Notice also that, by taking the scalar product of the last expression with 〈x|, we immediatelyobtain the inverse Fourier transform relation:
ψ(x) = 〈x|ψ〉 =∫
dk(2π)3
eik·xψ(k) .
Expanding a general state in eigenstates of A. Let us turn again to the general case of anarbitrary observable A. Constructing the projectors on the corresponding eigenstates, and usingcompleteness,
1 =∑n
gn∑i=1
|φ(i)an〉〈φ(i)
an| ,
we can write, for any state |ψ〉:
|ψ〉 =∑n
gn∑i=1
|φ(i)an〉〈φ(i)
an|ψ〉 . (7.1)
Notice that, since |ψ〉 is assumed to be normalized, we have that:∑n
gn∑i=1
∣∣∣〈φ(i)an|ψ〉∣∣∣2 = 1 . (7.2)
7.1 Measurements in QM.
Measuring A on a state |ψ〉. Here comes one of the crucial postulates of the theory of mea-surement in QM. If we perform a measurement of the observable A on a system in state |ψ〉, then,the result of each individual measurement can only be, according to QM, one of the eigenvaluesan of the Hermitean operator A associated to the observable which is measured. Performing severalrepeated measurements on the system prepared in the identical state |ψ〉 I can obtain, as a result,different eigenvalues an, which gives to QM a probabilistic aspect. According to the postulates ofQM, the probability Pan that I obtain an as a result of the measurement is given by:
Pan =gn∑i=1
∣∣∣〈φ(i)an|ψ〉∣∣∣2 , (7.3)
which is nicely normalized to unity, according to Eq. 7.2,∑n Pan = 1.
44 Measurement in Quantum Mechanics
Collapse of |ψ〉 after measuring A. Another postulate of QM is that, immediately afterhaving measured A, obtaining an eigenvalue an of the associated A, the wave-function is projectedinto its an-component (appropriately normalized, again):
|ψ〉 Measured an−→ |ψ′〉 = Ngn∑i=1
|φ(i)an〉〈φ(i)
an|ψ〉 . (7.4)
This is known as collapse of the wave-function after measurement. It contains the probabilisticaspect of QM, since the final state |ψ′〉 depends on the eigenvalue an measured, which in turn isobtained with a probability Pan .
Measuring A again on |ψ〉. If we measure again A after having measured it, obtaining an(with probability Pan), we in effect are performing a measurement of A on the collapsed state |ψ′〉,and therefore we get again an, this time with certainty.
Measuring two commuting observables A and B on |ψ〉, one after the other. Whentwo observables A and B commute, one can find a common basis of eigenvectors that diagonalizesboth simultaneously (we proved this). If I measure A on |ψ〉, getting an with probability Pan andcollapsing the state to |ψ′〉, a subsequent measurement of B on |ψ′〉 will give me only one of theeigenvalues of B in the gn-dimensional subspace associated to an, collapsing the state to some |ψ′′〉which is still made of an components only. A subsequent measurement of A after B will thereforeconfirm the result an with certainty, i.e., the measurement of B has not disrupted that of A: wesay that two commuting observables are compatible. On the contrary, if the two observables do notcommute, measuring B disrupts completely the an-subspace, and a subsequent measurement of Agives no longer necessarily the same an.
7.2 Expectation values of measurements.
If we repeatedly measure A on the same state |ψ〉, we can construct the mean value of our mea-surements, each always given by some eigenvalue an obtained however with a probability Pan
. Aswe know from elementary statistics, the mean is given by:
〈A〉 =∑n
anPan.
Inserting the expression in Eq. (7.3) for Pan , after simple algebra (done in class), we proved thatsuch a mean value coincides with the expectation value 〈ψ|Aψ〉:
〈A〉 = 〈ψ|A|ψ〉 . (7.5)
We stress, however, that in each individual measurement we obtain an, and not 〈A〉.
We can also construct the mean-square:
〈A2〉 def=∑n
a2nPan ,
and show that〈A2〉 = 〈ψ|A2|ψ〉 . (7.6)
Exercise 7.1 Verify Eq. (7.6).
Given the mean and the mean-square, we can construct the root-mean-square-deviation:
∆A def=√〈A2〉 − 〈A〉2 . (7.7)
Expectation values of measurements. 45
Exercise 7.2 After having proved that, in general 〈A2〉 ≥ 〈A〉2, so that ∆A is well defined, prove that,
given two operators A and B, then
∆A ·∆B ≥ 1
2|〈ψ|[A, B]|ψ〉| ,
which is a generalization of Heisenberg’s uncertainty principle to arbitrary operators. [Hint: to provethis, use Schwartz inequality of scalar products, |〈α|β〉|2 ≤ 〈α|α〉〈β|β〉.]
8
Spin
The spin is the new operator we are going to present. Introducing the spin is strongly demandedby a crucial experiment done in the early days of quantum mechanics by Stern and Gerlach (inthe following, SG).
8.1 Stern-Gerlach experiment: the spin operators and states
The experiment consists in sending a collimated beam of Ag atoms (coming out of a furnace)along a direction (call it y) inside a specially designed magnet having a magnetic field along z,but also having ∂Bz/∂z 6= 0, and then registering the spots observed on a screen perpendicularto the beam. Much to their surprise, two spots — call them N1 and N2 — where observed in allcircumstances, with a 50% abundance of atoms ending up in each spot. The same result if foundif the apparatus (i.e., the magnet) is rotated in any direction.
To fully appreciate why this is surprising, let us discuss a bit the theoretical background. Eachatom, as a whole, obeys to a very good approximation Newton’s equations, due to its large mass.Classically, a magnetic field induces a potential term V (R) = −~µ ·B(R), and hence a force whosecomponent j (j = x, y, z) is:
Fj(R) = ~µ · ∂B∂Rj
,
where ~µ is the magnetic moment of the atom and R indicates its (center-of-mass) position. Thepresence of the gradient of the magnetic field explains why the design of the poles of the magnet wasspecially conceived by Stern and Gerlach so as to produce a non-uniform field with ∂Bz/∂z 6= 0: thiswas achieved through a sharp corner protruding from the Nord-pole (sketched at the blackboard).But, what is the magnetic moment of the atom? Think, for simplicity, of a beam of hydrogenatoms (Ag atoms are not very different in that respect, since they have an electronic configuration[Kr]4d105s1 with a single unpaired electron in the 5s shell). A contribution to the magnetic momentcertainly comes from the electron which orbits around the nucleus. As in classical electrodynamicsone can show that:
~µ = − e
2mcL = − e~
2mcL~
= −µBL~.
Notice that, when L = ~, this gives a magnetic moment − e~2mc = −µB , equal to the Bohr magneton.
If this was the only source of magnetic moment, then the experiment cannot be explained. For onething, L = 0 for an electron in the s-shell. And even if this was not the case, L would have 2l + 1possible values, due to m = −l, · · · , l, while only two spots are seen in the experiment!
All we have said above can be quite firmly justified quantum mechanically, by resorting to aBorn-Oppenheimer approach to separate the motion of the nucleus from that of the electrons. Togive an idea of the origin of the L · B term, consider the Hamiltonian of the single electron of ahydrogen atom in a magnetic field. So far, we would simply write:
H =1
2m
(p +
e
cA)2
− e2
r,
Stern-Gerlach experiment: the spin operators and states 47
where A is the vector potential. For a uniform field Bz directed along z, we can take A =(Bz/2)(−y, x, 0), which has zero divergence, ∇ · A = 0. Expanding the square in the kineticenergy we can rewrite H (done in class) as:
H =p2
2m− e2
r+ µBLzBz +
e2
8mc2B2z(x
2 + y2) ,
from which we see that there is an extra potential term µBLzBz as predicted from classical physics(as well as a quadratic Bz term, which one can easily estimate to be of the order of (µBBz)2/Ry,a very small quantity, since µBBz ≈ 5.8× 10−5eV for a field Bz of 1 Tesla).
So, there is no way of having two spots only in a SG experiment without invoking some othersource of intrinsic magnetic moment, not due to the orbital motion of the electrons. Indeed, wesaw from the general theory of angular momentum, that in principle half-integer values of j arepossible in principle. So, let us postulate that there exist an intrinsic angular momentum S, calledspin, which has S = 1/2 and hence two values for Sz, Sz = ±1/2. The spin enters the Hamiltonianin the presence of magnetic field in the following way:
H =1
2m
(p +
e
cA)2
− e2
r+ g0µBS ·B ,
i.e., in a very similar way to what L does, but with g0 ≈ 2. A full derivation of such expressionrequires taking the non-relativistic limit of the Dirac’s equation, and we will not do it. What isimportant for us is to explore the experimental implications of such form. Before continuing, weneed to discuss what happens to the states of the system. The spin has its own Hilbert space,generated, for instance, by the two eigenstates of Sz, which we denote by | ± 1
2 〉z or simply |±〉 ifthere is no ambiguity. The most general spin state can therefore be written as
|ψ〉spin = α+|+〉+ α−|−〉 ,
with α± complex constants. Normalization of the state requires |α+|2+|α−|2 = 1. The total Hilbertspace of the system must be therefore supplemented by such a spin part, in a way that is known astensor product of Hilbert spaces: H = Horb ⊗Hspin. In practice that simply means the, if |ψn〉orbdenotes a complete basis for the orbital Hilbert space (the one we have used so far), then the mostgeneral state of the system can be written as:
|Ψ〉 =∑n
|ψn〉orb (cn,+|+〉+ cn,−|−〉) .
Notice that this does not imply that |Ψ〉 = |ψ〉orb|ψ〉spin, which is only a special (although, inabsence of spin-orbit coupling, see next chapter, quite frequent) case occurring when the coefficientsαn,± have the factorized form cn,± = cn α±.
At this point, we can proceed in two ways. We can go on guided by theory, and directly writedown the matrix elements of Sx, Sy, Sz in the basis of eigenstates of Sz, using the expressions forS+ and S−. Or, alternatively, resort to the experiment.
The first route uses that S+|−〉 = |+〉 and S−|+〉 = |−〉 (recall the expressions for J± andspecialize them to j = 1/2), and arrives at writing:
〈σ|Sx|σ′〉 =12
(0 11 0
)=
12σx , (8.1)
〈σ|Sy|σ′〉 =12
(0 −ii 0
)=
12σy , (8.2)
〈σ|Sz|σ′〉 =12
(1 00 −1
)=
12σz . (8.3)
48 Spin
Exercise 8.1 Deduce the matrix elements of the spin operators given above by using the fact that Sz|±〉 =± 1
2|±〉 and S±|∓〉 = |±〉.
Here the notation used is that row and column indices are labeled by σ and σ′ which take the values± with the ordering convention that a state |ψ〉spin = α+|+〉+α−|−〉 is denoted as a column vector(α+
α−
). The three very important matrices σx,y,z previously defined are known as Pauli matrices.
The second route requires the following series of conceptual experiments. Prepare a beam ofatoms in the eigenstate | + 1
2 〉x by rotating the SG-apparatus with the magnet axis in the x-direction and making a hole in the point N1 of the screen, so that only atoms directed towardsN1 pass the screen. Next, let the beam of “surviving atoms” go through a second SG-apparatus,this time with the magnet axis along z. According to the prescription of QM, this will measure theprobability of having Sz = ± 1
2 in the state |+ 12 〉x, i.e., |〈±|+ 1
2 〉x|2. The (surprising) experimental
result would be that there is a 50% abundance in the two spots, implying that |〈+| + 12 〉x|
2 = 12
and |〈−|+ 12 〉x|
2 = 12 . These expressions immediately imply that:
|+ 12 〉x = 1√
2
(|+〉+ eiδx |−〉
)| − 1
2 〉x = 1√2
(|+〉 − eiδx |−〉
) , (8.4)
where δx is an arbitrary phase factor that we cannot, so far, determine (an overall phase factor infront is, on the contrary, irrelevant, and has been not included). By a similar conceptual experimentwe can write the following expressions for the states | ± 1
2 〉y:
|+ 12 〉y = 1√
2
(|+〉+ eiδy |−〉
)| − 1
2 〉y = 1√2
(|+〉 − eiδy |−〉
) , (8.5)
where another unknown phase factor δy has been included. Now, let us “close the circle”, andperform a measurement of Sy after having prepared |±1/2〉x. The results of the experiments wouldtell us that |y〈± 1
2 |+12 〉x|
2 = |y〈± 12 | −
12 〉x|
2 = 12 , which, together with the explicit expressions for
the state in Eqs. (8.4,8.5), lead after simple algebra to the conclusion that:
cos (δy − δx) = 0 =⇒ δy − δx = ±π2.
Exercise 8.2 Verify that the experimental finding that |y〈± 12| + 1
2〉x|2 = |y〈± 1
2| − 1
2〉x|2 = 1
2, together
with the explicit expressions for the states in Eqs. (8.4,8.5) imply that cos (δy − δx) = 0.
A possible phase convention is that δx = 0 (i.e., | ± 12 〉x states have real coefficients) but that
necessarily implies that δy = ±π/2, i.e., complex numbers are mandatory in the |± 12 〉y states! The
correct choice for a right-handed system turns out to be δy = π/2.
One can easily verify (done in class) that the form of the matrix elements of Sx,y,z coming outof this form of the states is exactly the one given above in terms of the Pauli matrices. One canalso easily verify that [Sx, Sy] = iSz, as expected. Similarly S2 = S2
x + S2y + S2
z = (3/4)1.
8.2 Experiments with Stern-Gerlach apparatus
One can easily prepare, experimentally, a spin-state pointing along any direction in space. If ndenotes a direction, associated to two angles (θ, φ), all one needs to do is to rotate the SG apparatus
Experiments with Stern-Gerlach apparatus 49
so that the magnetic field axis is along n and then collect only the atoms passing through N1 (whilethe other spot is blocked by the screen). The spin along n is associated to the operator
n · S = sin (θ) cos (φ)Sx + sin (θ) sin (φ)Sy + cos (θ)Sz =12
(cos (θ) sin (θ)e−iφ
sin (θ)e+iφ − cos (θ)
),
where the last equality applies, strictly speaking, to the matrix elements of such an operator onthe standard basis and is deduced by using the form of the Pauli matrices. By diagonalizing the2 × 2 matrix for n · S = sin (θ) it is possible to show that the eigenvalues of n · S are ± 1
2 withassociated eigenstates given by:
|+ 12 〉n = cos
(θ2
)|+〉+ sin
(θ2
)eiφ|−〉
| − 12 〉n = sin
(θ2
)|+〉 − cos
(θ2
)eiφ|−〉 . (8.6)
Exercise 8.3 Verify the expression (8.6) for the eigenstates of n · S and convince yourself that the usualform of the eigenstates of Sx and Sy are obtained as particular cases.
Think now of sending the beam of atoms which have been prepared in the state |+ 12 〉n through a
second SG-apparatus along z. The probability of measuring ± 12 for Sz are evidently given by:
P+ 12 (z) = cos2
(θ
2
), P− 1
2 (z) = sin2
(θ
2
),
which are in principle subject to experimental test. Notice that this is the first time we find a resultwhich is not 50% abundance in each spot!
Exercise 8.4 Calculate the probability P± 12 (x) of measuring ± 1
2for Sx on the state + 1
2〉n.
Consider now preparing |+ 12 〉n as before, and measuring each component of S (in turn, x, y, and
z) with a second SG apparatus appropriately rotated, repeatedly, over and over again, accumu-lating statistics for the calculation of probabilities, as usual. Each individual measurement of anycomponent of S will always give either + 1
2 or − 12 . But the probabilities of each outcome is different,
and you can therefore ask what is the mean value of the spin in the ordinary statistical manner. Asdiscussed previously (see lecture on Measurement) this requires calculating the expectation valueof S on |+ 1
2 〉n. A simple calculation shows that:
n〈+12|Sx|+
12〉n =
12
sin θ cosφ , (8.7)
n〈+12|Sy|+
12〉n =
12
sin θ sinφ , (8.8)
n〈+12|Sz|+
12〉n =
12
cos θ , (8.9)
or, more compactly:
n〈+12|S|+ 1
2〉n =
12n . (8.10)
Exercise 8.5 Verify the previous expressions for the mean values of S.
Summarizing, while individual spin measurements are quite surprising for our classical intuition,the mean values behave in a classically well understandable way.
50 Spin
8.3 A single spin-1/2 in a uniform field: Larmor precession
Consider now the usual hydrogen atom in a magnetic field, and write the Hamiltonian as
H = H0 + g0µBBSz ,
where H0 collects all the terms which do not depend on the spin (as we will see later on, we arenot including the spin-orbit coupling in H). Each eigenvalue En of H0 is now associated to twopossible eigenstates, |ψn〉|+〉, and |ψn〉|−〉, which are perfectly degenerate in absence of B. In thepresence of B, the two states are split in the following way:
En,±(B) = En ± µBB , (8.11)
where we have taken g0 ≈ 2. The splitting between the two levels is ∆E = 2µBB. Think now ofpreparing the system, in absence of magnetic field, in the state, at t = 0:
|ψ(t = 0)〉 = |ψn〉|+12〉n = |ψn〉
(cos(θ02
)|+〉+ sin
(θ02
)eiφ0 |−〉
).
At t = 0 we now turn on the magnetic field and let the state evolve. The state at time t is|ψ(t)〉 = e−iHt/~|ψ(t = 0)〉 which one can easily calculate to be:
|ψ(t)〉 = e−iHt/~|ψ(t = 0)〉 = e−iEnt/~|ψn〉e−iµBBt/~(
cos(θ02
)|+〉+ sin
(θ02
)eiφt |−〉
),
withφt = φ0 +
2µBBt~
= φ0 +∆E~t = φ0 + ωct .
This shows that, in the presence of B along z, the spin precesses around the z-axis with a frequency
ωc =∆E~
=eB
mc. (8.12)
Such a frequency, known as Larmor frequency (or cyclotron frequency), is entirely classical (no~ appears) and corresponds to the frequency with which a classical charged particle rotates in auniform magnetic field.
If we measure S at time t we get:
〈ψ(t)|S|ψ(t)〉 =12n(t) , (8.13)
where n(t) denotes the direction (θ0, φt).
8.4 Heisenberg representation
Given an arbitrary operator A, the expectation value on the state |ψ(t)〉 is
〈ψ(t)|A|ψ(t)〉 = 〈ψ(t = 0)|eiHt/~Ae−iHt/~|ψ(t = 0)〉 . (8.14)
The expression on the right-hand side defines the Heisenberg representation of the operator A:
AH(t) = eiHt/~Ae−iHt/~ , (8.15)
in terms of which we have:
〈ψ(t)|A|ψ(t)〉 = 〈ψ(t = 0)|AH(t)|ψ(t = 0)〉 . (8.16)
One can write a differential equation satisfied by AH(t) by simply taking the derivative with respectto time. It is simple to show that:
i~d
dtAH(t) = eiHt/~[A,H]e−iHt/~ = [AH(t),H] , (8.17)
an equation known as Heisenberg equation of motion for the operator AH(t).
Heisenberg representation 51
As a simple application of this approach, consider the previous problem of Larmor precessionand calculate the Heisenberg equations of motion for S. It is immediate to show that:
i~d
dtSz,H(t) = eiHt/~[Sz,H]e−iHt/~ = 0 , =⇒ Sz,H(t) = Sz ,
5 because Sz commutes with H. This illustrates the concept of constant of motion, which appliesto each operator that commutes with the Hamiltonian. A slightly more complicated calculation,which uses the commutation rules [Sx, Sy] = iSz, shows that:
d
dtSx,H(t) = −ωcSy,H(t) (8.18)
d
dtSy,H(t) = +ωcSx,H(t) . (8.19)
Similar equations are obeyed by the expectation values of the spin, and clearly describe the uniformrotation around the z-axis of 〈SH(t)〉.
9
Spin-orbit and addition of angularmomenta
Until now, an hydrogen atom in a magnetic field is described by:
H =1
2m
(p +
e
cA)2
− e2
r+ g0µBB · S ,
where µB = e~/2mc, and g0 ≈ 2. The spin appears only due to the Zeeman term (last term in H),and is totally decoupled from the orbital variables (r, p, L). As a consequence of such decoupling,wave-functions can be written, at this level, as products of the orbital wave-function times thespin wave-function, |ψorb〉 ⊗ |ψspin〉, where, e.g., ψorb(r) = Rnl(r)Ylm(θ, φ) (in absence of B), and|ψspin〉 = α+|+〉+ α−|−〉.
9.1 Relativistic effects
However, there are important relativistic effects that should be added to the previous H. A firstrelativistic contribution comes from kinetic energy. By expanding the relativistic dispersion E(p) =√p2c2 +m2c4 we have:
E(p) =√p2c2 +m2c4 = mc2 +
p2
2m− 1
2mc2
(p2
2m
)2
+ · · · .
Notice that the last term, considering that p2/2m ≈ Ry, is smaller with respect to the ordinarynon-relativistic kinetic energy by a factor of order Ry/mc2. Such a dimensionless factor is veryimportant:
2Rymc2
=Hartreemc2
=(e2
~c
)2
= α2 ,
where α = e2/(~c) ≈ 1/137 is known as fine-structure constant. There are other corrections to thesame order in α. These can be derived by taking the non-relativistic limit of the (fully relativistic)Dirac’s equation (including the B · S term).
9.2 The origin of spin-orbit
We concentrate here only on a term that couples S and L. The rigorous derivation is complicated,but the order-of-magnitude and form of the term can be understood easily. Suppose there is noB-field. The electron travels in the E-field due to the nucleus, E = (e/r3)r. However, in the frameat rest on the electron, the nucleus moves with velocity −v, and this generates a current = −evwhich in turn generates a B-field:
Beff = −vc×E .
This B-field seen by the electron in its rest-frame affects its spin with a term
Construction of J-states out of p-states 53
µBBeff · S =µBc
(E× v) · S =µBmc
(E× p) · S .
Notice that we have not included the factor g0 here! (This is just an argument, not a rigorousderivation.) In conclusion, we can anticipate a term
Hso =µBmc
(E× p) · S =µBmc
e
r3(r× p) · S = λso(r) L · S ,
where we have extracted ~ from the definition of L and defined:
λso(r) =e~
2mce~mc
1r3
= 2µ2B
r3.
Let us estimate the order of magnitude of this term by using r ∼ aB = ~2/(me2). We get:
λso ∼ 2µ2B
a3B
=me4
2~2
e4
c2~2= (Ry) α2 .
Once again, we see that the order of magnitude is the same as the correction to the kinetic energy:order α2. Numerically, one might expect this small effect to be not very important. However, effectsof this order are spectroscopically measurable and determine, for instance, the fine structure of thep-levels of hydrogen. Moreover, it is simple to realizes that the spin-orbit coupling becomes moreand more effective as the nuclear charge Ze is increased (i.e., for heavier atoms). Indeed, theelectric field due to a nucleus of charge Ze is E = (Ze/r2), and the factor 1/r3 appearing in λso(r)is now important at r ∼ aB/Z (the effective Bohr radius is reduced by Z). Therefore, we anticipateλ
(Z)so ≈ Z4(Ry)α2.
Summarizing, even in absence of an external magnetic field we should write a Hamiltonian forthe hydrogen atom where the spin operator explicitly appears through the spin-orbit term Hso:
H =p2
2m− e2
r+ λso(r)L · S +Hother relativistic .
9.3 Construction of J-states out of p-states
The full treatment of a Hamiltonian including relativistic effects like the previous one requiresusing perturbation theory (which will be covered later on in the course). We only concentrate hereon the symmetry aspects of the problem and learn how to construct the appropriate combinationsof orbital and spin states which make the effect of the L ·S term simple. Consider, to fix the ideas,the 2p-states (l = 1) of the unperturbed hydrogen problem. We classify them (including spin) inthe following way:
|n = 2, l = 1,m; s =12, sz = ±1
2〉 = |n = 2, l = 1,m〉 ⊗ |s =
12, sz = ±1
2〉 .
From now on we will omit the ⊗ sign, and simply indicate the spin states with |±〉 or with |sz〉.Similarly, we indicate the orbital part of the state with |1,m = 0,±1〉. There are therefore a totalof 3× 2 = 6 states, which we list below:
|1,+1〉|+〉|1, 0〉|+〉 , |1,+1〉|−〉|1, 0〉|−〉 , |1,−1〉|+〉|1,−1〉|−〉
(9.1)
Observe thatL · S = LzSz +
12
(L+S− + L−S+) ,
which implies that, when applied to the states we have listed, L · S mixes them (through theoff-diagonal term L+S− + L−S+).
54 Spin-orbit and addition of angular momenta
The idea, now, is that L ·S can also be viewed as the cross term appearing in (L + S)2. Indeed,let us introduce the operator J = L + S, which represents the total angular momentum, sum ofthe orbital plus spin parts. Notice that the two terms of J act on different Hilbert spaces and,more formally, we should write J as L ⊗ 1spin + 1orb ⊗ S, but we will not insist on this notation.First of all J is a legitimate “angular momentum”, as you can easily verify (done in class) that[Jx, Jy] = iJz (and so on for the cyclic combinations). Next, consider
J2 = (L + S)2 = L2 + S2 + 2L · S .
As a consequence:
L · S =12(J2 − L2 − S2
).
Therefore, if we were able to construct common eigenstates of J2, L2 and S2, the effect of L · Son such states would be very simple to calculate. The product states |1,m〉|sz〉 are obviouslyeigenstates of L2 (with eigenvalue l(l + 1) = 2), of S2 (with eigenvalue s(s + 1) = 3
4 ), and ofJz = Lz + Sz (with eigenvalue jz = m+ sz).
Consider now the top-most state with highest eigenvalue of jz = +32 , associated to the product
state |1,+1〉|sz = + 12 〉. Question: Is this an eigenstate of J2, and, if so, with what eigenvalue?
It is not difficult to show that the answer is positive: |1,+1〉|sz = + 12 〉 is an eigenstate of J2 with
a value of j = 32 . This can be seen in at least two ways. First, by direct calculation, by writing
J2 = L2 + S2 + 2LzSz + (L+S− + L−S+). Second, by the following argument: j cannot be lessthan 3/2 because otherwise jz = 3/2 would not be possible; j cannot be larger than 3/2 becauseotherwise there should be some value of jz = j > 3/2, which is evidently impossible (the highestjz, by direct inspection, is 3/2). In conclusion we have the first eigenstate of J2, Jz, L2, and S2 inthe simple form:
|j =32, jz = +
32; l = 1, s =
12〉 = |1,+1〉|sz = +
12〉 . (9.2)
To calculate the other states belonging to the j = 32 multiplet (2j + 1 = 4 state in total), we just
need to apply J− repeatedly. A simple calculation gives:
|j =32, jz = +
12; l = 1, s =
12〉 =
1√3(L− + S−)|j =
32, jz = +
32; l = 1, s =
12〉
=
√23|1, 0〉|+ 1
2〉+
√13|1,+1〉| − 1
2〉 . (9.3)
Similarly, we get:
|j =32, jz = −1
2; l = 1, s =
12〉 =
√13|1,−1〉|+ 1
2〉+
√23|1, 0〉| − 1
2〉 , (9.4)
|j =32, jz = −3
2; l = 1, s =
12〉 = |1,−1〉| − 1
2〉 . (9.5)
Notice that the two states |j = 32 , jz = ± 1
2 ; l = 1, s = 12 〉 are not simple product states, but sum of
two different product states!
In this way we have constructed 4 states, out of the 6 product states available. Where are themissing 2 states? Obviously, by combining |1, 0〉| + 1
2 〉 and |1,+1〉| − 12 〉 we can form two states,
while Eq. (9.3) represents only a single combination. There is a further combination which we canform, still with the same jz = 1
2 , which we write as:
|j =?, jz = +12; l = 1, s =
12〉 = α|1, 0〉|+ 1
2〉+ β|1,+1〉| − 1
2〉 ,
with |α|2 + |β|2 = 1. This new combination must be chosen to be orthogonal to the previous one,which requires
Generalization: addition of L and S 55√23α+
√13β = 0 .
Normalizing, we get:
|j =?, jz = +12; l = 1, s =
12〉 =
√13|1, 0〉|+ 1
2〉 −
√23|1,+1〉| − 1
2〉 .
The only remaining point is the value of j, which we have indicated by j =?. It is not hard to showthat j = 1
2 in this case.
Exercise 9.1 Apply J+ to the state |j = 12, jz = + 1
2; l = 1, s = 1
2〉 showing that the result vanishes.
The final state missing is the second state of the multiplet j = 12 , with jz = − 1
2 , which is
|j =12, jz = −1
2; l = 1, s =
12〉 =
√23|1,−1〉|+ 1
2〉 −
√13|1, 0〉|+ 1
2〉 .
The coefficients appearing in the previous expressions are known as Clebsch-Gordan coefficients.
9.4 Generalization: addition of L and S
Exercise 9.2 Construct J-states out of L = 2 and S = 1/2, as you would need for the fine structure ofd-states.
Exercise 9.3 For more then one electron, you find cases where you need to add the two spins S1 and S2
of each electron. Determine the eigenstates (triplet and singlet, so called) and eigenvalues of S2 and Szwhere S = S1 + S2 is the total spin of the two electrons in terms of the basis elements | ± 1
2〉1| ± 1
2〉2.
10
Density matrices
Let us ask ourselves the following question: what is the spin state |ψspin〉 of the silver atoms comingout of the furnace before the Stern-Gerlach (SG) measuring apparatus? Unfortunately, none of thestates
|ψspin〉 = α+|+〉z + α−|−〉zdescribes the experiment. Indeed, I can always write one such state as a spin pointing into anarbitrary direction n:
|+〉n = cosθ
2|+〉z + sin
θ
2eiφ|−〉z .
This, in turn, would imply that by rotating the SG apparatus in the direction n I should be ableto measure 100% of atoms in one spot, which is, experimentally, not the case: no matter how theSG apparatus is rotated, you always get 50% of atoms in each of the two spots.
How to describe such a state entering the SG apparatus?
10.1 The density matrix for a pure state
Let us go back to the expectation value of an operator A and reinterpret it, in terms of the projectorPψ = |ψ〉〈ψ|, after having inserted to identities (in terms of a general basis)::
〈A〉 def= 〈ψ|A|ψ〉 =∑α,α′
〈ψ|φα〉〈φα|A|φα′〉〈φα′ |ψ〉
=∑α,α′
〈φα|A|φα′〉〈φα′ |ψ〉〈ψ|φα〉 =∑α,α′
〈φα|A|φα′〉〈φα′ |Pψ|φα〉
= Tr[A][Pψ] . (10.1)
In words, the expectation value of any operator A on |ψ〉 can be expressed as the trace of theproduct of the matrix representing A with that representing the projector Pψ. It is therefore usefulto say that the state is univocally assigned once we know the operator ρ = Pψ, whose matrixelements on a given basis |φα〉 = |α〉 are:
〈α|ρ|α′〉 = 〈α′|ψ〉〈ψ|α〉 = ψα′ψ∗α .
Examples: In real space |α〉 = |x〉, and:
ρ(x′,x) = ψ(x′)ψ∗(x) .
For a one-dimensional harmonic oscillator, for instance, ψ(x) = Ce−x2/(2l2), and therefore:
ρ(x′, x) = |C|2e−x2+x′2
2l2 .
For a plane-wave state: ψ(x) = 1√Veik·x so that:
The density matrix for a mixed state 57
ρ(x′,x) =1Veik·(x
′−x) .
For a spin-1/2 state |ψspin〉 = α+|+〉z + α−|−〉z, we have:
ρ = |ψspin〉〈ψspin| = |α+|2|+〉〈+| + |α−|2|−〉〈−| + α+α∗−|+〉〈−| + α−α
∗+|−〉〈+| ,
so that the matrix elements of ρ are:
[ρ] =[|α+|2 α+α
∗−
α−α∗+ |α−|2
]=[
cos2 θ2 cos θ2 sin θ
2e−iφ
cos θ2 sin θ2eiφ sin2 θ
2
], (10.2)
where the last equality assumes that the spin is directed along n, parameterized by θ and φ.
Notice the following two properties of ρ = Pψ: i) Trρ = 1, and ii) ρ2 = P 2ψ = Pψ = ρ. The
first property is a consequence of the normalization of ψ, while ii) follows from the properties ofprojectors.
10.2 The density matrix for a mixed state
Suppose that the system we are trying to describe is found in any of the pure states |ψi〉, withi = 1, · · · , Nρ, with a certain probability wi (clearly,
∑i wi = 1). What is the natural probabilistic
way of calculating the mean value of any operator A? Simply sum all averages with pure stateswith their weight wi, obtaining:
〈A〉 =Nρ∑i=1
〈ψi|A|ψi〉 =Nρ∑i=1
wi∑α,α′
〈φα|A|φα′〉〈φα′ |Pψi|φα〉 = Tr[A][
Nρ∑i=1
wiPψi] .
It is therefore clear that, the density matrix describing such a state (called mixed state, as opposedto a pure state described by a single ψ) is:
ρ =Nρ∑i=1
wi|ψi〉〈ψi| , (10.3)
in terms of which averages are expressed, once again, as:
〈A〉 = Tr[A][ρ] . (10.4)
Simple properties of ρ are that, once again, Trρ = 1. However, it is simple to show (done in class)that:
ρ2 =∑i
w2i |ψi〉〈ψi| 6= ρ ,
unless the state is pure, i.e., only one of the wi is equal to 1, all the others being zero. In particular:
Tr ρ2 =∑i
w2i ≤ 1 .
10.3 Density matrices and statistical mechanics
Suppose you have a quantum system which is enclosed into a thermostat which holds it at temper-ature T . This is the so-called canonical ensemble in statistical mechanics (no exchange of particles
58 Density matrices
is possible, but heat can flow in and out of the system to maintain the constant temperature). Thequantum system is not described by a pure state, but rather by a density matrix which is
ρT =∑n
pn|ψn〉〈ψn| , (10.5)
where pn = e−βEn/Z, β = (kBT )−1, Z =∑n e
−βEn is the partition function, and |ψn〉 denotethe energy eigenstates of the system Hamiltonian H (which form a basis of the Hilbert space ofthe system). It is very simple to show that you can also rewrite:
ρT =1Ze−βH . (10.6)
As a simple application, consider a one-dimensional quantum oscillator of Hamiltonian
H =~ω2
(p2 + x2) ,
(p and x being, as usual, dimensionless momentum and coordinate) in thermal equilibrium witha reservoir at temperature T . Expressing H in terms of the annihilation and creation operators aand a†, as H = (~ω)
(a†a+ 1
2
), one can easily calculate many averages, as you will learn by doing
the following exercise.
Exercise 10.1 Thermal averages for an harmonic oscillator. (1) Calculate the thermal average 〈a†a〉 =Tr[ρT a
†a]. [Hint: Calculate first the partition function Z.] (2) Next, calculate the average potential energy~ωx2/2 and the average kinetic energy ~ωp2/2 and plot these quantities as a function of kBT/(~ω). (3)Finally, calculate the specific heat CV = ∂U/∂T , U being the total internal energy, and plot it as a functionof kBT/(~ω). (4) At what temperatures strong deviations from the expected classical result (state whatit is that you expect, classically) are seen, due to quantum effects?
10.4 Density matrices by tracing out the universe
Suppose that you have a system A (Hilbert space HA with a basis set |φα〉A) surrounded by therest of the universe B (Hilbert space HB with basis set |Φβ〉B). The total wave-function |Ψ〉 ofthe combined system will leave in the tensor space HA ⊗HB , which by definition has a basis set|φα〉A|Φβ〉B (we omit the ⊗ sign between the states of A and B, as notation is unambiguous).The total wave-function, therefore, is generally a combination of those basis states
|Ψ〉 =∑α,β
cα,β |φα〉A|Φβ〉B , (10.7)
with appropriate coefficients cα,β . Notice that, in general, the state |Ψ〉 cannot be written as aseparable state, i.e. a single product of a state of A times a state of B (as the basis states are):
|Ψ〉 6= |φ〉A|Φ〉B .
This occurs only for some special choices of the coefficients cα,β , for instance cα,β = aαbβ . Wheneverthe state |Ψ〉 cannot be written as a separable state one says that the state is entangled. A simpleexample of such a case is, for two spin-1/2 particles, a singlet state.
Suppose I want to calculate the expectation value of an operator A which involves only systemA. Then the result is (simple proof, which uses orthogonality of |Φβ〉B state):
〈Ψ|A|Ψ〉 =∑α,α′
∑β
c∗α′,βcα,β
〈φα′ |A|φα〉 , (10.8)
Density matrices by tracing out the universe 59
which you immediately recognize it can be expressed as:
〈Ψ|A|Ψ〉 = Tr[A][ρ] ,
where [A]α′,α = 〈φα′ |A|φα〉, and:[ρ]α,α′ =
∑β
c∗α′,βcα,β . (10.9)
One can easily prove that the matrix ρ so defined is indeed a density matrix. To prove it, observethat it is an Hermitean matrix, and hence can be diagonalized by an appropriate change of basis.Moreover, its trace is obviously 1:
Tr[ρ] =∑αβ
c∗α,βcα,β =∑αβ
|cα,β |2 = 1 .
Let |ψn〉 be the orthonormal basis (in the space HA) which diagonalizes [ρ]. Therefore, we canwrite:
ρ =∑n
wn|ψn〉〈ψn| ,
where, evidently, the eigenvalues wn are real, and∑n wn = 1, since the trace has to be 1. To
conclude that ρ is indeed a density matrix you only have to prove that its eigenvalues are non-negative: wn ≥ 0. Indeed, consider the operator A = Pφn
= |ψn〉〈ψn| (the projector on state φn).Then, on one hand you can write:
〈Ψ|A|Ψ〉 = Tr[A][ρ] = wn .
But, on the other hand, from the original expression of |Ψ〉 you can easily show that:
wn = 〈Ψ|A|Ψ〉 =∑β
∣∣∣∣∣∑α
〈ψn|φα〉cα,β
∣∣∣∣∣2
≥ 0 ,
which clearly shows that wn ≥ 0.
11
Time-independent perturbation theory
Until now, we have dealt only with exactly solvable models. The list of such models is, however,rather short. It is therefore very important to learn approximate techniques to tackle more compli-cated problems which cannot be solved exactly. We begin with the treatment of time-independentperturbation theory. That is, we assume we know how to solve:
H0|φ(0)n 〉 = E(0)
n |φ(0)n 〉 , (11.1)
where the unperturbed states |φ(0)n 〉 could be degenerate or non-degenerate, and aim at solving the
perturbed problem(H0 + λV ) |φn(λ)〉 = En(λ)|φn(λ)〉 . (11.2)
where V is a time-independent Hermitean operator, known as the perturbation. The couplingcoefficient λ is included only for convenience, so that for λ = 0 one recovers the unperturbedproblem, and also as a way of accounting for different orders in the perturbation V .
11.1 Two-state problem
To get a feeling for the general theory, let us start considering a two-state problem, i.e., a two-dimensional Hilbert space with two states |φ(0)
1,2〉 with unperturbed energies E(01,2. The system is
degenerate if E(0)1 = E
(0)2 , and non-degenerate otherwise.
Solving this problem exactly amounts to diagonalizing the 2× 2 matrix
[H0 + λV ] =
(E
(0)1 + λV11 λV12
λV21 E(0)2 + λV22
)(11.3)
which represents the Hamiltonian H0 + λV in the unperturbed basis. We find the two perturbedeigenvalues E(λ)
1,2 by imposing det(H0 + λV − E1) = 0, which gives, after very simple algebra:
E± =E
(0)1 + λV11 + E
(0)2 + λV22
2±
√√√√[ (E(0)2 + λV22)− (E(0)
1 + λV11)2
]2
+ λ2|V12|2 . (11.4)
11.1.1 Non-degenerate case
Consider first the non-degenerate in which E(0)2 > E
(0)1 . Rewrite it as:
E± =(E(0)
2 + λV22) + (E(0)1 + λV11)
2±
(E(0)2 + λV22)− (E(0)
1 + λV11)2
√1 +
4λ2|V12|2
[(E(0)2 + λV22)− (E(0)
1 + λV11)]2.
Non-degenerate case: general solution 61
Then, expanding the square-root up to second-order in λ, i.e.,√
1 + αλ2 ≈ 1 + αλ2/2 + · · · , andneglecting terms of order λ3, we have:
E2(λ) = E+ ≈ E(0)2 + λV22 + λ2 |V12|2
E(0)2 − E
(0)1
E1(λ) = E− ≈ E(0)1 + λV11 − λ2 |V12|2
E(0)2 − E
(0)1
. (11.5)
These results anticipate those we will derive from the general theory, up to second order.
11.1.2 Degenerate case
In the case E(0)2 = E
(0)1 = E(0) the previous expressions are not well defined because of a zero in
the denominator of the second-order term. Luckily, we can solve the problem exactly, in this case,the solution being:
E±(λ) = E(0) + λ
V11 + V22
2±
√(V22 − V11
2
)2
+ |V12|2
. (11.6)
So, in this case what we have to do is to diagonalize the perturbation V in the basis of thedegenerate unperturbed eigenstates of H0. We will see that this is general.
11.2 Non-degenerate case: general solution
To solve the general non-degenerate case, write:
En(λ) = E(0)n + λE(1)
n + λ2E(2)n + · · ·
|φn(λ)〉 = |φ(0)n 〉+ λ|φ(1)
n 〉+ λ2|φ(2)n 〉+ · · · , (11.7)
where E(i)n and |φ(i)
n 〉 denote, respectively, the i-th order correction to the energy eigenvalue andeigenstate, and impose that the perturbed Schrodinger problem is solved when terms of a givenorder in λ are considered. To zeroth-order in λ, the problems is solved by Eq. 11.1.
11.2.1 First-order correction
Collecting all the first-order terms in λ we get:
H0|φ(1)n 〉+ V |φ(0)
n 〉 = E(1)n |φ(0)
n 〉+ E(0)n |φ(1)
n 〉 ,
or, equivalently:(H0 − E(0)
n )|φ(1)n 〉+ (V − E(1)
n )|φ(0)n 〉 = 0 . (11.8)
Multiplying to the left by 〈φ(0)n |, we get:
E(1)n = 〈φ(0)
n |V |φ(0)n 〉 =⇒ En(λ) = E(0)
n + λ〈φ(0)n |V |φ(0)
n 〉+O(λ2) . (11.9)
As for the correction to the eigenstate, multiply to the left by a general 〈φ(0)m |, to get:
(E(0)m − E(0)
n )〈φ(0)m |φ(1)
n 〉+ 〈φ(0)m |V |φ(0)
n 〉 − E(1)n δm,n = 0 . (11.10)
If m = n, this equation gives exactly (11.9). When m 6= n, we can deduce:
〈φ(0)m |φ(1)
n 〉 =〈φ(0)m |V |φ(0)
n 〉E
(0)n − E
(0)m
=⇒ |φn(λ)〉 = |φ(0)n 〉+
∑m6=n
|φ(0)m 〉 〈φ
(0)m |λV |φ(0)
n 〉E
(0)n − E
(0)m
+O(λ2) , (11.11)
that is the nth state gets a small component of all other states with a coefficient given by thematrix element of the perturbation divided by the energy gap.
62 Time-independent perturbation theory
11.2.2 Second-order correction
Collecting all the second-order terms in λ we get:
H0|φ(2)n 〉+ V |φ(1)
n 〉 = E(2)n |φ(0)
n 〉+ E(1)n |φ(1)
n 〉+ E(0)n |φ(2)
n 〉 ,
or, rearranging the terms:
(H0 − E(0)n )|φ(2)
n 〉+ (V − E(1)n )|φ(1)
n 〉 = E(2)n |φ(0)
n 〉 .
Multiplying to the left by 〈φ(0)n |, we get:
E(2)n = 〈φ(0)
n |V |φ(1)n 〉 ,
and, upon inserting the expression (11.11) for |φ(1)n 〉,
E(2)n = 〈φ(0)
n |V |φ(1)n 〉 =
∑m6=n
〈φ(0)n |V |φ(0)
m 〉〈φ(0)m |V |φ(0)
n 〉E
(0)n − E
(0)m
. (11.12)
Therefore, up to second-order we have:
En(λ) = E(0)n + 〈φ(0)
n |λV |φ(0)n 〉+
∑m6=n
|〈φ(0)m |λV |φ(0)
n 〉|2
E(0)n − E
(0)m
+O(λ3) . (11.13)
This is a very important expression, often used in calculations. The second-order correction to theeigenstate, on the contrary, is very seldom used, and we will not derive it here.
11.2.3 Harmonic oscillator plus a linear potential
Consider, as a warm-up exercise, a one-dimensional harmonic oscillator under the action of auniform electric field E (imagine the particle has a charge q). The extra potential felt by theparticle is V (x) = −qEx, and the Hamiltonian reads:
H = H0 + V =p2
2m+
12mω2x2 − qEx .
Luckily, this problem can be solved exactly in a very simple way, by completing the square:
12mω2x2 − qEx =
12mω2
(x− qE
mω2
)2
− q2E2
2mω2.
The exact eigenvalues are given by:
En = ~ω(n+
12
)− q2E2
2mω2.
If we denote by φ(0)n (x) the eigenstates of the unperturbed harmonic oscillator (recall that they
are Hermite polynomials times the Gaussian e−x2/(2l2), where l =
√~/(mω) is the oscillator
characteristic length-scale), the exact eigenstates in the presence of the field are given by:
φn(x) = φ(0)n
(x− qE
mω2
),
i.e., they are the same functions, but shifted by a quantity − qEmω2 . To do the perturbative calcula-
tion, go to dimensionless quantities, and express
Non-degenerate case: general solution 63
V = −qEl xl
= −qEl√2(a+ a+) .
Next, observe that the expectation value of the perturbation is vanishing to first order: 〈φ(0)n |V |φ(0)
n 〉because both a and a+ change n to n−1 and n+1, respectively, and the bra then gives zero overlap.For the second-order calculation you need:
〈φ(0)m |V |φ(0)
n 〉 = −qEl√2
(δm,n+1
√n+ 1 + δm,n−1
√n).
The second-order expression for the energies gives:
En = E(0)n +
∑m6=n
|〈φ(0)m |V |φ(0)
n 〉|2
E(0)n − E
(0)m
+ · · ·
= ~ω(n+
12
)− q2E2l2[(n+ 1)− n]
2~ω+ · · ·
= ~ω(n+
12
)− q2E2
2mω2,
which coincides with the exact result: evidently, all the higher order corrections cancel exactly inthis case, at least for the eigenvalues. Indeed, verify that this is not the case for the eigenstates:
Exercise 11.1 Calculate the correction up to first order for the ground state of the harmonic oscillatorin the electric field. Compare the approximate expression with the exact one.
11.2.4 An-harmonic oscillator
Consider now a perturbation to the usual one-dimensional harmonic oscillator that is of the form:
V = −λ~ω(xl
)3
. (11.14)
Cubic terms come on quite general ground when you expand a generic potential close to a minimum(example, the inter-atomic potential of a diatomic molecule). The peculiarity of a potential thatstops at this level, however, is that it is necessarily unbounded in one direction (x→ +∞ if λ > 0),and therefore, strictly speaking, it always provides a dramatic perturbation to the spectrum, nomatter how small the coefficient λ is. The perturbation series is, in this case, not a regular powerseries (it is only an asymptotic series, but don’t worry about this): no bound state survives in thepresence of V , and the particle is always able to tunnel away (to x→ +∞) from close to the origin.Nevertheless, if λ is small enough, the time it takes for a particle to tunnel away is large enoughthat, for all practical purposes, the perturbative calculation makes sense.
Working with the a, a+, and n = a+a, we have:
V = −λ~ω2√
2(a+ a+)3 = −λ~ω
2√
2
(a3 + (a+)3 + 3na+ + 3(n+ 1)a
),
where we have carefully expanded the cube (a + a+)3 paying due attention to the fact that theterms do not commute, and used that an = (n + 1)a and a+n = (n − 1)a+. Once again, we haveno effect to first-order in λ, because V changes the n in the ket and the bra gives zero overlap. Forthe second-order calculation you need:
64 Time-independent perturbation theory
〈φ(0)n+3|V |φ(0)
n 〉 = −λ~ω2√
2
√(n+ 3)(n+ 2)(n+ 1)
〈φ(0)n−3|V |φ(0)
n 〉 = −λ~ω2√
2
√n(n− 1)(n− 2)
〈φ(0)n+1|V |φ(0)
n 〉 = −λ~ω2√
23√
(n+ 1)3
〈φ(0)n−1|V |φ(0)
n 〉 = −λ~ω2√
23√n3 .
The energy denominators are very simple: E(0)n − E
(0)m = ~ω(n −m). Putting all the ingredients
together we get:
En = E(0)n +
∑m6=n
|〈φ(0)m |V |φ(0)
n 〉|2
E(0)n − E
(0)m
+ · · ·
= ~ω(n+
12
)− 15λ2~ω
4
(n+
12
)2
− 7λ2~ω16
+ · · · .
Notice that the energy separation between two neighboring levels is now:
En − En−1 = ~ω(
1− 152λ2n+ · · ·
),
i.e., it decreases with increasing n from the equispaced harmonic levels, and shows a clear signalof the fact that the perturbative series is dangerous, here, for high values of n: it changes sign forlarge n, somewhat worrying for a quantity which is positive by definition!
Exercise 11.2 Calculate the correction up to first order for the ground state of the harmonic oscillatorin presence of the x3 correction. Show that the average value of x on such a perturbed state is now notvanishing, but equal to 〈φ0|x|φ0〉 = 3λl/2.
11.2.5 Van-der-Waals attraction between two H-atoms
Consider two hydrogen atoms far away from each other, with the two nuclei at a distance R. Whatwill be the potential V (R) between them? Each of the two atoms being neutral, there cannot beany 1/R term in V . The goal of the present exercise is to show that the first contribution to V (R)comes from dipole-dipole terms, and leads to a V (R) having an attractive 1/R6 tail, known as vander Walls attraction, i.e.,
V (R) = − C
R6.
The Hamiltonian of the two atoms, with protons sitting at positions R1 and R2, is:
H =(
P21
2M+
p21
2m− e2
r1
)︸ ︷︷ ︸
H1
+(
P22
2M+
p22
2m− e2
r2
)︸ ︷︷ ︸
H2
+ VCoulomb
VCoulomb =e2
|R2 −R1|+
e2
|(R2 + r2)− (R1 + r1)|
− e2
|(R2 + r2)−R1|− e2
|R2 − (R1 + r1)|.
We will consider here H1 + H2, the Hamiltonian of the two isolated atoms, as our unperturbedHamiltonian H0 = H1 +H2. The last four terms, accounting for the Coulomb repulsion between
Non-degenerate case: general solution 65
the two protons and the two electrons, and the attraction of each electron from the proton ofthe other atom, will be treated as a perturbation. Such a perturbation is small if the separationR = R2 −R1 between the two protons is much larger than the typical scale, the Bohr radius aB ,over which each electronic coordinate r1 and r2 varies. It is convenient to expand the last fourCoulomb term assuming |r1|, |r2|, |r2 − r1| R. The typical expression appearing is:
1|R + d|
=1√
R2 + d2 + 2R · d=
1R
[1 +
d2 + 2R · dR2
]−1/2
.
Using the Taylor expansion (1+x)−1/2 ≈ 1− (1/2)x+(3/8)x2 + · · · , and neglecting terms of orderd3 and higher, we get:
1|R + d|
≈ 1R
[1− 1
2d2 + 2R · d
R2+
38
(2R · d)2
R4+ · · · .
]Applying this expansion to VCoulomb we get:
VCoulomb ≈ Vdip−dip =e2
R3
(r1 · r2 − 3(R · r1)(R · r2)
)=
e2
R3(x1x2 + y1y2 − 2z1z2) ,
where we have assumed that the direction between the two atoms to be in the z-direction, R =R/R = z.
The ground state of H0 is given by φ(0)0 (r1, r2) = ψ100(r1)ψ100(r2), with unperturbed energy
E(0)0 = 2E1, where E1 = −e2/(2aB) = −Ry is the ground state energy of a single hydrogen atom.
This is the first time we encounter a problem with more than one electron. The wave-functionwritten above is symmetric under exchange of the two electrons, and we will see later on that afundamental principle of physics demands that the wave-function of a system of identical fermions(as the electrons are) should be anti-symmetric under exchange of the particles. We will also seethat the way out is given by the fact that there is actually a spin part to the wave-function, whichis anti-symmetric in the present case, and makes the whole wave-function anti-symmetric:
φ(0)0 = ψ100(r1)ψ100(r2)
1√2(| ↑↓〉 − | ↓↑〉) .
Such a spin-part is, however, inessential in the present case, because the Hamiltonian never actsupon it, and the corresponding spin-bra makes the spin-ket to disappear. So, let us neglect thispotential complication for the time being.
As is often the case, the first-order contribution of Vdip−dip to the ground-state energy vanisheshere, because of parity reasons. For instance, the x1x2 term would give a contribution:
〈ψ100ψ100|x1x2|ψ100ψ100〉 = 〈ψ100|x1|ψ100〉〈ψ100|x2|ψ100〉 = 0 ,
because 〈ψ100|x1|ψ100〉 = 〈ψ100|x2|ψ100〉 = 0 by parity (ψ100(r) is even, which x is odd underparity). A similar fate occurs to y1y2 and z1z2.
The second order contribution requires, in principle, all the excited states (treat the two electronsas distinguishable, for simplicity)
φ(0)n′l′m′,n′′l′′m′′(r1, r2) = ψn′l′m′(r1)ψn′′l′′m′′(r2)
of unperturbed energy En′+En′′ , where En = E1/n2 is the unperturbed energy of a single hydrogen
atom:
E0 = E(0)0 − e4
R6
6=(100,100)∑(n′l′m′,n′′l′′m′′)
|〈ψn′l′m′ψn′′l′′m′′ |(x1x2 + y1y2 − 2z1z2)|ψ100ψ100〉|2
En′ + En′′ − 2E1.
66 Time-independent perturbation theory
This immediately leads us to conclude that V (R) = −C/R6, as anticipated. The matrix elementswhich now appear are, for instance:
〈ψn′l′m′ψn′′l′′m′′ |z1z2|ψ100ψ100〉 = 〈ψn′l′m′ |z1|ψ100〉〈ψn′′l′′m′′ |z2|ψ100〉 ,
and similar ones for x1x2 and y1y2. Parity immediately tells us that l′ and l′′ should be both odd, inorder for the matrix elements not to vanish. This is an example of a selection rule (i.e., a vanishingof a matrix element) due to a symmetry (parity, in the present case). There is, however, anothersymmetry in the present case, i.e., rotational symmetry, which provides a further selection rule.Indeed, remember that ψ100(r) = R10(r)Y00(θ, φ) is rotationally symmetric, because Y00 = 1√
4π
does not depend on (θ, φ). Also, recall that z = r cos θ = r√
4π/3 Y10, and that 〈Yl′m′ |Ylm〉 =δl,l′δm,m′ . Therefore:
〈ψn′l′m′ |z|ψ100〉 =∫ ∞
0
drr2Rn′l′(r)R10(r)r
√4π3
∫ π
0
dθ
∫ 2π
0
dφ Y ∗l′m′Y10Y00
=∫ ∞
0
drr2Rn′l′(r)R10(r)r√3〈Y ∗l′m′ |Y10〉
= δl′,1δm′,0
∫ ∞
0
drr2Rn′l′(r)R10(r)r√3.
A similar argument applies to the x and y matrix elements, which can be expressed in terms ofY1,+1 and Y1,−1. Therefore, rotational symmetry implies a further, stronger, selection rule tellingus that:
l′ = 1 and l′′ = 1 . (11.15)
So, only p-states need to be included in the second-order calculation. The calculation might proceednow with the evaluation of all the necessary integrals. It is, however, rather long and boring.An approximate estimate for this second-order contribution is obtained by noticing that |En′ | =|E1|/(n′)2 |E1|, so that we can approximately neglect En′ + En′′ in the energy denominator.The advantage of this is that one can then perform the sum over (n′l′m′, n′′l′′m′′) exactly, becauseit leads to appearance of the identity:∑
(n′l′m′,n′′l′′m′′)
|ψn′l′m′ψn′′l′′m′′〉〈ψn′l′m′ψn′′l′′m′′ | = 1 .
Therefore:
E(2)0 ≈ − e4
R6
〈ψ100ψ100|(x1x2 + y1y2 − 2z1z2)2|ψ100ψ100〉−2E1
≈ −6e2aBR6
(〈ψ100|x2
1|ψ100〉)2
, (11.16)
where the factor 6 comes from 1+1+4, due to the terms x21x
22, y
21y
22 and 4z2
1z22 , all giving the same
contribution by rotational symmetry, and we used the fact that −2E1 = e2/aB . The necessaryintegral can be easily evaluated:
〈ψ100|x21|ψ100〉 = a2
B ,
so that, finally:
V (R) ≈ E(2)0 ≈ −6e2a5
B
R6. (11.17)
11.3 Degenerate case: first order
Until now we have considered only the case in which the level we are considering is non-degenerate,i.e., there is a unique eigenstate |φ(0)
n 〉 with energy E(0)n . Consider now the case in which there are
Degenerate case: first order 67
gn > 1 unperturbed states |φ(0)n,α〉, α = 1 · · · gn with energy E(0)
n . Clearly, this brings a potentialproblem in all formulas that have energy denominators, because different states with the sameenergy would give a divergent contribution.
On the other hand, the situation in which gn > 1 is quite common: think, for instance, of thehydrogen atom, where the first excited state at energy E2 = E1/22 is four-fold degenerate (g2 = 4,without accounting for the spin), the second excited state at E3 = E1/32 has g3 = 9, and so on.(You can show that the the level En has gn = n2.)
Clearly, you can mix the levels |φ(0)n,α〉, α = 1 · · · gn with a unitary gn × gn matrix Uα,α′ ,
obtaining states:
|φ(0)n,α〉′ =
gn∑α′=1
Uα,α′ |φ(0)n,α′〉 ,
which still have the same unperturbed energy E(0)n , and are as good as the original ones. Which,
among the infinite many possible choices obtained with arbitrary unitary rotations U , is the “best”possible choice? The answer is that the optimal choice is given by those states |φ(0)
n,α〉, α = 1 · · · gnwhich diagonalize the gn × gn matrix representing the perturbation V in the degenerate subspace,i.e., such that:
〈φ(0)n,α|V |φ
(0)n,α′〉 = E(1)
n,αδα,α′ .
To prove this, imagine that we pretend that all quantities are continuous as a function of λ forsmall λ, writing, as usual:
En,α = E(0)n + λE(1)
n,α + · · ·
|φn,α〉 = |φ(0)n,α〉+ λ|φ(1)
n,α〉+ · · · , (11.18)
and substituting in the Schrodinger equation (H0 + λV )|φn,α′〉 = En,α′ |φn,α′〉. The zero-th orderterms are automatically satisfied. The first-order terms read (just add the index α to Eq. (11.8):
(H0 − E(0)n )|φ(1)
n,α′〉+ (V − E(1)n,α′)|φ
(0)n,α′〉 = 0 . (11.19)
Multiply on the left by 〈φ(0)n,α|, and again observe that the first term (H0 − E
(0)n ) vanishes. The
second term, on the other hand, using the fact that the |φ(0)n,α〉, α = 1 · · · gn are assumed to be
orthonormal, gives:
〈φ(0)n,α|V |φ
(0)n,α′〉 = E
(1)n,α′〈φ
(0)n,α|φ
(0)n,α′〉 = E(1)
n,αδα,α′ ,
i.e., the selected states should diagonalize the corresponding matrix [V ] in order for the expansionto be continuous for small λ. In words, when you turn on an infinitesimal λ, the perturbation V“prefers” immediately the states that diagonalize it, out of the infinite possible combinations.
11.3.1 Linear Stark effect
Let us start applying this ideas to the problem of an hydrogen atom in the presence of an electricfield in the z-direction. The Hamiltonian is:
H =(
p2
2m− e2
r
)︸ ︷︷ ︸
H0
+ eEz . (11.20)
Consider first the ground state |φ(0)1 〉 = |ψn=1,l=0,m=0〉. Clearly, there is no first-order contribution
to the ground state energy, because, by parity, 〈ψ100|z|ψ100〉 = 0.
68 Time-independent perturbation theory
Exercise 11.3 Calculate (approximately) the second-order shift (down) of the ground state energy inpresence of the electric field E , by applying the second-order formula. The calculations are similar to thoseoccurring in the van der Walls problem for two distant hydrogen atoms.
Consider now the 4 levels corresponding to the first excited states, which are the 2s (n = 2, l =0,m = 0), and the three 2p (n = 2, l = 1,m = 0/ + 1/ − 1). It is simple to show that the matrixrepresenting the perturbation V = eEz in this 4-dimensional space is:
[V ] = eE
〈ψ200|z|ψ200〉 〈ψ200|z|ψ210〉 〈ψ200|z|ψ21+1〉 〈ψ200|z|ψ21−1〉〈ψ210|z|ψ200〉 〈ψ210|z|ψ210〉 〈ψ210|z|ψ21+1〉 〈ψ210|z|ψ21−1〉〈ψ21+1|z|ψ200〉 〈ψ21+1|z|ψ210〉 〈ψ21+1|z|ψ21+1〉 〈ψ21+1|z|ψ21−1〉〈ψ21−1|z|ψ200〉 〈ψ21−1|z|ψ210〉 〈ψ21−1|z|ψ21+1〉 〈ψ21−1|z|ψ21−1〉
= eE
0 ∆ 0 0∆ 0 0 00 0 0 00 0 0 0
.
Exercise 11.4 Calculate explicitly the quantity ∆ appearing in the matrix [V ] above, using the fact that
ψ200(r) = 1√8πa3
B
“1− r
2aB
”e−r/(2aB) and ψ210(r) = 1√
32πa3B
“zaB
”e−r/(2aB), to show that ∆ = 3aB .
So, the levels ψ21±1 are not affected by the perturbation, to first order, while the 2s state and the2p state with m = 0 are mixed by an off-diagonal term. By diagonalizing the small 2× 2 block ofthe matrix [V ] we realize that the correct states to consider are:
|ψ2,±〉 =1√2
(|ψ200〉 ± |ψ210〉) , (11.21)
which are split linearly in the field E :
E2,± = E(0)2 ± eE∆ . (11.22)
This linear splitting is also known as linear Stark effect. The corresponding optical transition linetowards the ground state is consequently split, in presence of an electric field.
11.4 Application: the fine-structure of Hydrogen
There are several relativistic corrections to the hydrogen atom Hamiltonian H0 which we haveencountered so far:
H =(
p2
2m− e2
r
)︸ ︷︷ ︸
H0
+e2
2m2c2r3L · S︸ ︷︷ ︸
Hspin−orbit
− 12mc2
(p2
2m
)2
︸ ︷︷ ︸Hkinetic−energy
+π~2e2
2m2c2δ(r)︸ ︷︷ ︸
HDarwin
+ · · · . (11.23)
The origin of the spin-orbit term has already been discussed. The kinetic energy correction comesfrom the expansion of the relativistic energy dispersion:√
p2c2 +m2c4 = mc2√
1 + p2/m2c2 ≈ mc2 +p2
2m− p4
8m3c2+ · · · ,
where the first constant term, mc2, is never included in condensed matter, p2/2m is the usualnon-relativistic expression, and the third term is the promised relativistic correction. The final
Application: the fine-structure of Hydrogen 69
term, called Darwin correction, is very difficult to justify: it comes from the limit of the Dirac’srelativistic equation, precisely from terms of the form ∇2(1/r), that is all I can tell you.
Let us estimate the order of magnitude of these three terms. As for the spin-orbit term we have:
〈Hspin−orbit〉〈H0〉
≈ e2~2/(m2c2a3B)
e2/aB=
~2
m2c2a2B
= α2 ,
where α = e2/(~c) ≈ 1/137 is the so-called fine-structure constant. Similarly:
〈Hkinetic−energy〉〈H0〉
≈ 〈(p2/2m)〉2mc2
≈ e2
aBmc2= α2 .
Finally, since |ψ(0)|2 ≈ 1/a3B :
〈HDarwin〉〈H0〉
≈ e2~2|ψ(0)|2/(m2c2)e2/aB
≈ α2 .
So, all three terms are of the same order of magnitude, i.e. α4mc2 = 2α2Ry.
Let us consider the n = 2 multiplied of the 2s-2p states. These are 4 orbitals, i.e., 4 × 2 = 8states, including spin. The Darwin term shifts only the 2s state, because ψ(0) = 0 for the 2p states.We can calculate the effect of HDarwin to first-order:
E(1,Darwin)2s = 〈ψ200|
π~2e2
2m2c2δ(r)|ψ200〉 =
π~2e2
2m2c2|ψ200(0)|2 .
Using |ψn00(0)|2 = 1/(πn3a3B), we get:
E(1,Darwin)2s =
116mc2α4 .
Let us continue with the 2s states. The effect of the spin-orbit term is obviously zero, becauseLψ200 = 0 for the 2s (l = 0) state. As for the kinetic-energy correction we have, in general:
E(1,kinetic−energy)nlm = − 1
2mc2〈ψnlm|
(p2
2m
)2
|ψnlm〉 = − 12mc2
〈ψnlm|(H0 +
e2
r
)2
|ψnlm〉 .
Expanding the square: (H0 +
e2
r
)2
= H20 +
e4
r2+H0
e2
r+e2
rH0 ,
and noticing that H0|ψnlm〉 = En|ψnlm〉 (even when H0 acts to the left, on the bra), we can easilycalculate:
E(1,kinetic−energy)nlm = − 1
2mc2
(E2n + 2En〈ψnlm|
e2
r|ψnlm〉+ 〈ψnlm|
e4
r2|ψnlm〉
).
So, all we need is 〈1/r〉 and 〈1/r2〉. Let us give the general results for these averages:
〈ψnlm|1r|ψnlm〉 =
1n2aB
(11.24)
〈ψnlm|1r2|ψnlm〉 =
1n3a2
B(l + 1/2)(11.25)
〈ψnlm|1r3|ψnlm〉 =
1n3a3
Bl(l + 1/2)(l + 1). (11.26)
(The last term will be useful in calculating the spin-orbit term.) Putting all the terms together,we get, after simple algebra:
70 Time-independent perturbation theory
E(1,kinetic−energy)nlm = − 1
mc2
(e2
aB
)2 16− 3(2l + 1)128(2l + 1)
. (11.27)
For the 2s state, therefore:
E(1),kinetic−energy2s = − 13
128mc2α4 ,
and, summing up all three terms we get:
E(1)2s = E
(1)
2j= 12 l=0s= 1
2= − 5
128mc2α4 .
where we have labeled the state with its total j = 0 + 12 = 1
2 .
Consider now the 2p states. The Darwin term, as mentioned, has no effect. The kinetic-energycorrection term gives:
E(1,kinetic−energy)2p = − 7
384mc2α4 . (11.28)
The spin-orbit term is the most delicate one. As previously discussed, the spin-orbit term (and,indeed, the full H) commutes with L2, S2, J2 and Jz, where J = L + S. Indeed, recall that:
L · S =J2 − L2 − S2
2.
Therefore, rather then using the 3× 2 = 6 product states |ψ21m〉|s = 12 ,±
12 〉, the most convenient
basis to use is that of the eigenstates of L2, S2, J2 and Jz: |l = 1, s = 12 , j, jz〉. As discussed
previously, the possible values of J are j = l + 12 = 3
2 (4 states), usually denoted by 4p 32, and
j = l − 12 = 1
2 (2 states), denoted by 2p 12. Therefore, for general l:
E(1,spin−orbit)njls =
e2
2m2c21
n3a3Bl(l + 1/2)(l + 1)
j(j + 1)− l(l + 1)− s(s+ 1)2
.
In particular, for the 2p states, where we have l = 1 and s = 12 , we get:
E(1,spin−orbit)
2jl=1s= 12
=e2
m2c21
96a3B
(j(j + 1)− 11
4
).
Summing together the kinetic-energy correction and the spin-orbit term we get, explicitly:
E(1)
2j= 12 l=1s= 1
2=(− 1
48− 7
384
)mc2α4 = − 5
128mc2α4
E(1)
2j= 32 l=1s= 1
2=(
+196− 7
384
)mc2α4 = − 1
128mc2α4 .
Notice that the two j = 12 doublets, i.e., the 2s 1
2(derived from l = 0 and s = 1
2 ) and the 2p 12
(derived from l = 1 and s = 12 ) are perfectly degenerate after accounting for the relativistic
corrections. This degeneracy survives all the relativistic corrections, to all orders in perturbationtheory, and is really due to the fact that the eigenvalues of the Dirac’s equation depend on j, ratherthan l or s. The degeneracy between 2s 1
2and 2p 1
2is only removed (the so-called Lamb shift) by
quantum electrodynamics corrections. Comment on the spectroscopic fingerprints of the resultingfine-structure spectrum, which is experimentally measurable with high accuracy.
12
Time-dependent perturbation theory
Until now, we have dealt with time-independent perturbations V . There are many cases, however,where the perturbation one considers represents an external time-dependent field (for instancean electromagnetic field). Therefore, we have to learn how to solve a time-dependent Schrodingerproblem
i~d
dt|ψ(t)〉 = [H0 + λV (t)] |ψ(t)〉. (12.1)
developing a systematic expansion in λ.
12.1 General case
Let |φ(0)n 〉 represent the unperturbed eigenstates associated to H0, with energies E0
n, whichform a complete orthonormal basis for expanding a generic state. The state |ψ(t)〉 can then bewritten as:
|ψ(t)〉 =∑n
cn(t)e−iE0nt/~|φ(0)
n 〉 . (12.2)
where we have explicitly included the time-dependence associated to H0, in such a way that forλ = 0 the coefficients cn would not depend on time, and we imagine them to be “slowly varying”for small λ. (Technically, this is equivalent to the so-called interaction picture representation.)By substituting this expansion in the time-dependent Schrodinger problem (12.1), and taking thescalar product with 〈φn| we get:
i~dcn(t)dt
= λ∑m
〈φn|V (t)|φm〉 ei(E0n−E
0m)t/~ cm(t) = λ
∑m
Vnm(t) eiωnmt cm(t) , (12.3)
where, as a shorthand, we have defined
ωnm =E0n − E0
m
~, (12.4)
to be the frequency associated to the difference between the unperturbed energies relative to levelsn and m, and
Vnm(t) = 〈φn|V (t)|φm〉 , (12.5)
to be the corresponding matrix element of the perturbation V (t). As it is, this is a complicatedset (infinite-dimensional, in general) of time-dependent, first-order differential equations for thecn(t). Assume, as usual, a series expansion in powers of λ:
cn(t) = c(0)n (t) + λc(1)n (t) + λ2c(2)n (t) + · · · =∞∑q=0
λqc(q)n (t) , (12.6)
and substitute this series in Eq. (12.3). Noticing that the right-hand side of it has an extra factorλ, we get:
72 Time-dependent perturbation theory
i~dc
(0)n (t)dt
= 0
i~dc
(1)n (t)dt
=∑m
Vnm(t) eiωnmt c(0)m
· · · = · · ·
i~dc
(q)n (t)dt
=∑m
Vnm(t) eiωnmt c(q−1)m (t) . (12.7)
In words, one can solve the problem iteratively.
12.2 First-order solution
Suppose that the system is found in the initial state with n = i for t ≤ t0 (very often, the groundstate, with n = i = 0), and that we switch-on the perturbation abruptly at t = t0. The zeroth-orderterm is clearly time-independent: c(0)n (t) = c
(0)n = δn,i. Up to first order, we get:
i~dc
(1)n (t)dt
= Vni(t) eiωnit =⇒ c(1)n (t) =1i~
∫ t
t0
dt′ Vni(t′) eiωnit′. (12.8)
So, the transition probability for the system to go from state i at t0 to a state n 6= i at time t is:
Pi→n 6=i(t) = |c(0)n + λc(1)n (t) + · · · |2 ≈ P(1)i→n 6=i(t)
P(1)i→n 6=i(t) = |λc(1)n (t)|2 =
1~2
∣∣∣∣λ ∫ t
t0
dt′ Vni(t′) eiωnit′∣∣∣∣2 . (12.9)
Let us consider the case in which H0 is the usual one-dimensional harmonic oscillator. At timet0 = −∞ the harmonic oscillator is in its ground state |φ(0)
0 〉; then we switch on an electric field Ewhich is subsequently switched-off as t→ +∞:
λV (t) = qExe−t2/τ2
.
We ask what is the probability that the system is left in the n-th excited state, |φ(0)n 〉, at t = +∞.
We have here:λVni(t) = qEe−t
2/τ2 l√2〈φ(0)n |a+ a+|φ(0)
0 〉 .
Calculating the matrix element and the time-integral (Fourier transform of a Gaussian,∫dt e−t
2/τ2eiωt =
(πτ2)1/2e−ω2τ2/4) we get:
P(1)0→n(t) = δn,1
πq2E2τ2
2m~ωe−ω
2τ2/2 . (12.10)
There is a non-zero probability of exciting the system to the first-excited state (the only one whichthe perturbation couples the ground state with); such a probability of excitation goes to zero bothin the extreme adiabatic limit τ →∞, and in the fast switching-on-and-off limit τ → 0.
As a second application, imagine that, starting from any H0, we turn on a perturbation λVslowly in time from t0 = −∞ to time t = 0 in the following way:
λV (t) = λV et/τ ,
so that the Hamiltonian at time t = 0 is H(t = 0) = H0 + λV . Every initial state |φ(0)n 〉 is mixed
with other states |φ(0)m 〉, for an amount which is exactly given by cm(t = 0). To first order in λ the
resulting state at time t = 0 is:
First-order solution 73
|φn(t = 0)〉 = |φ(0)n 〉+
∑m6=n
λc(1)m (0)|φ(0)m 〉 ,
where
c(1)m (0) =1i~
∫ 0
−∞dt′ Vmn e
t′/τeiωmnt′=
1i~Vmn
1iωmn + τ−1
.
In the limit in which the perturbation is turned on infinitely slowly (τ →∞) we see that we recoverexactly the same first-order perturbation result obtained previously:
|φn(λ)〉 = |φ(0)n 〉+
∑m6=n
|φ(0)m 〉 〈φ
(0)m |λV |φ(0)
n 〉E
(0)n − E
(0)m
+ · · · . (12.11)
12.2.1 Periodic perturbation
A particularly important case is represented by a perturbation which is periodic in time, with afrequency ω:
Vnm(t) = Vnm cos (ωt) = Vnmeiωt + e−iωt
2.
The time integrals can now be done explicitly:
P(1)i→n 6=i(t) =
|Vni|2
4~2
∣∣∣∣1− ei(ωni+ω)t
ωni + ω+
1− ei(ωni−ω)t
ωni − ω
∣∣∣∣2 , (12.12)
where we have dropped any reference to λ, by setting λVni → Vni. Now suppose that ωni > 0(as indeed is the case if the systems starts from the ground state i = 0). Then, quite clearly,the second term is resonant for ω ≈ ωni, and dominates over the first term. Indeed, if we have afrequency ω = ωni + ∆ω, the first term is proportional to 1/(2ωni + ∆ω), the second to 1/(−∆ω)and dominates over the first if |∆ω| 2|ωni|. (For ωni < 0 the resonance occurs in the first term,but the conclusions will be identical.) Keeping only the resonant term, we have:
P(1,res)i→n 6=i (t) ≈
|Vni|2
4~2
∣∣∣∣ei(ωni−ω)t/2 − e−i(ωni−ω)t/2
ωni − ω
∣∣∣∣2 =|Vni|2
4~2
sin2 (ωni − ω)t/2[(ωni − ω)/2]2
. (12.13)
Notice that the last expression is a function of ω which is strongly peaked at ω = ωni, with a widthin frequency of ±|∆ω| = ±2π/t. As mentioned above, the validity of the resonance approximationrequires
2|ωni| |∆ω| = 2πt
=⇒ t 2π|ωni|
≈ 2πω,
i.e., the perturbation should act for a time t which is long enough that the system experiencesmany cycles, 2π/ω, of the perturbation, in order to be able to do a “Fourier analysis” of it. Next,at resonance, i.e., for ω = ωni, the transition probability behaves as:
P(1,res)i→n (t) ≈ |Vni|2t2
4~2 1 =⇒ t ~
|Vni|, (12.14)
where the inequality P 1,res)i→n (t) 1 is justified because the t2 behaviour would lead to a divergence
for large t, which is clearly an artifact of our first-order approximation. Therefore, both inequalitiesimply:
2π|ωni|
≈ 2πω t ~
|Vni|=⇒ ~|ωni| |Vni| , (12.15)
that is, in words, that the perturbation matrix elements |Vni| should be much smaller than thebare excitation energy ~|ωni|, i.e.: ∣∣∣∣∣ 〈φ(0)
n |λV |φ(0)i 〉
E(0)n − E
(0)i
∣∣∣∣∣ 1 ,
which is indeed the requirement that the first-order correction to the initial eigenstate is “small”(see Eq. 12.11).
74 Time-dependent perturbation theory
12.2.2 Fermi Golden Rule
Let us consider again the expression for the transition probability in the resonance approximation.The resonance approximation is equivalent, for ωni > 0, to assuming that the time-dependence ofthe perturbation is simply V e−iωt/2, neglecting the other term V e+iωt/2. It is therefore customaryto assume from the beginning a time-dependence
H(t) = H0 + V e−iωt ,
without any factor 1/2. Repeating the steps done in the previous section, we would find andidentical result (without any need to drop the non-resonant term), without the factor 4 in thedenominator (coming from 1/22):
P(1,res)i→n 6=i (t) ≈
|Vni|2
~2πt
sin2 (ωni − ω)t/2πt[(ωni − ω)/2]2
, (12.16)
where we have multiplied and divided by a factor πt, which will be useful in a second. Define nowx = (ωni − ω)/2. The sine-factor reads:
sin2 tx
πtx2,
which is a function whose value in x = 0 is t/π, with a width |x| = π/t, and whose integral overall x is 1: ∫ +∞
−∞dx
sin2 tx
πtx2= 1 .
Therefore, as t grows towards +∞, the peak of the function is sharper and higher, but the integralremains 1, i.e., the function approaches a delta function:
sin2 tx
πtx2
t→+∞−→ δ(x) .
Now, let us define the transition rate
Ri→n =P
(1,res)i→n 6=i (t)
t,
to be the transition probability per unit time. Then, for long times we have:
Ri→n ≈|Vni|2
~2πδ((ωni − ω)/2) =
2π~|Vni|2δ(E(0)
n − E(0)i − ~ω) , (12.17)
where we have used the property that δ(αx) = (1/α)δ(x) to put the factors ~ and 2π in the properplace. Notice that, strictly speaking, the expression involving the delta-function, expressing in astrict way the requirement of “energy conservation” between the two energy levels involved in thetransition when the energy ~ω of the perturbing field is taken into account, is justified only if wehave a continuum of states with a given density of states: for any discrete spectrum, it is necessaryto leave a certain width to the energy conservation requirement (i.e., leave the sine-factor). Forinstance, when we will deal with an electromagnetic field impinging on an hydrogen atom, thesituation in which ~ω > 13.6eV, which would lead to exciting the electron to the continuum ofE > 0 states, is quite appropriately described by the delta-function expression: this is the so-calledphotoelectric effect, in which the impinging electromagnetic field takes off the electron from theatom.
In a situation in which you have a continuum of energy states E(0)n towards which we are exciting
the system, it is meaningful to integrate over all final states E(0)n having an energy in the interval
Electromagnetic radiation on an atom 75
[Ef −∆E/2, Ef + ∆E/2], in the limit in which ∆E → 0. Since there are ρ(Ef )×∆E such states,by the very definition of the density of states ρ(Ef ), we can calculate:
Ri→f =Ef +∆E/2∑
E(0)n =Ef−∆E/2
Ri→n →∫ Ef +∆E/2
Ef−∆E/2
dEf ρ(Ef ) Ri→n
≈ 2π~|Vfi|2ρ(E(0)
i + ~ω) , (12.18)
where we have set Ef = E(0)i + ~ω, as dictated by the energy-conserving delta function, and
assumed that the matrix element Vni does not depend much on the wave-function |φ(0)n 〉 of the
final state, but only on the final energy, and can be therefore written in terms of a representativestate |φ(0)
f 〉 at the final energy.
12.3 Electromagnetic radiation on an atom
Suppose that you have an hydrogen atom in a region of space in which there is a monochromaticelectromagnetic field of frequency ω. We will describe the em-field classically, and in the Coulombgauge (~∇ ·A = 0, and φ = 0), as a plane wave described by a vector potential:
A(r, t) =12
[A0e
i(k·r−ωt) + c.c.],
where the fact that ~∇ ·A = 0 implies k ·A0 = 0, i.e., the vector A0 is in a plane perpendicularto the direction of propagation k, which we will assume to be in the y-direction, k = ky. TakingA0 = A0z, with A0 real, we get the following expression for the electric and magnetic field in thewave:
A = = A0z cos (ky − ωt)
E = −1c
∂
∂tA = −E z sin (ky − ωt)
B = ∇×A = (kA0)x sin (ky − ωt) , (12.19)
where E = A0ω/c. The Hamiltonian for the hydrogen atom (neglecting all relativistic corrections)reads:
H =1
2m
(p +
e
cA)2
− e2
r+
2µB~
B · S
= H0 +e
mcA · p +
2µB~
B · S +e2
2mc2A2 . (12.20)
It is easy to estimate the order of magnitude of these three perturbing terms. We have, for instance:
〈 2µB
~ B · S〉〈 emcA · p〉
=e~mckA0
emcA0
~aB
= kaB = 2πaBλ 1 ,
where we used k = 2π/λ, and assumed an em-field with a wavelength λ ≈ 10−6 m, i.e., in thevisible range. In a similar way, one can show that the A2 term is negligible with respect to theA · p term. We are therefore lead to considering the following perturbation problem:
H(t) = H0 +e
mc
A0
2[eikye−iωt + e−ikyeiωt
]z · p , (12.21)
where we easily recognize the structure we have studied so far, with the e−iωt-term being resonantwhen we consider the excitation process (absorption), in which E(0)
n > E(0)i , and the eiωt-term being
76 Time-dependent perturbation theory
resonant when the de-excitation process (emission), E(0)n < E
(0)i , is considered. Both processes lead
to an identical result, in this approximation. Considering the excitation process, E(0)n > E
(0)i , and
keeping only the resonant term, we have:
Ri→n =|Vni|2
4~2
sin2 (ωni − ω)t/2[(ωni − ω)/2]2t
Vni =eA0
mc〈φ(0)n |eikypz|φ(0)
i 〉 . (12.22)
There is now an important approximation (the so-called dipole approximation) we can use incalculating the last matrix element. Indeed, for visible light eiky ≈ 1, because kaB 1, andtherefore:
Vni ≈eA0
mc〈φ(0)n |pz|φ(0)
i 〉 .
Next, by using the fact that the commutator
[z,H0] = [z,p2z
2m] =
i~mpz ,
and that H0 when applied to its unperturbed eigenstates gives the corresponding eigenvalues, weget:
Vni ≈ −eEi
E(0)n − E
(0)i
~ω〈φ(0)n |z|φ(0)
i 〉 . (12.23)
where we have used E = A0ω/c. Notice that at resonance:
E(0)n − E
(0)i = ~ω ,
and therefore:|V resni | ≈ eE
∣∣∣〈φ(0)n |z|φ(0)
i 〉∣∣∣ . (12.24)
In summary, the relevant matrix element for the transition is given by the same dipole matrixelement considered in the Stark effect. In particular, the same selection rules apply:
l′ = l ± 1m′ = m (12.25)
where the m′ = m is due to our choice of A along the z direction: had we chosen A = A0x wewould get m′ = m± 1. This selection rule is known as dipole selection rule and applies to opticaltransition in which a photon is absorbed or emitted by an atom. For an hydrogen atom in the 1sground state, l = m = 0, and therefore only p-states can be reached.
A few comments are in order. First, we have treated the em-field as classical; a proper quantumtreatment would require quantization of A in terms of photon creator and annihilation operators;in that case, one would get that the emission process has a slightly different rate (∝ n + 1) fromthe absorption one (∝ n), because of spontaneous emission (in absence of any field). Second, as afunction of ω, the absorption rate from the ground state (1s) would show several sharp peaks incorrespondence with the 1s → 2p, 1s → 3p, etc. transitions. Third, one can calculate the photo-ionization process when the excitation occurs towards E > 0 final states and the electron is takenaway from the nucleus, i.e. for ~ω > 13.6 eV, by using the Fermi golden-rule formula: this shows-upas a continuum contribution in the absorption spectrum.
13
Many particle systems
Dealing with a system of N interacting quantum particles, either fermions or bosons, is a com-plicated business. The required extension of the quantum mechanical formalism, i.e., learning thebasic properties of the many-particle states which are physically allowed, will be treated next, withan application to a variational approach to the Helium atom. A sketch of the second-quantizationformalism is included in the appendix, for those of you who are interested in getting deeper intothe subject.
13.1 Many-particle basis states for fermions and bosons.
Suppose we have a system of N identical particles. Consider an appropriate basis set of single-particle states |α〉. Generally speaking, each state is a so-called spin-orbital, composed of anorbital part |α〉o and a spin part |σα〉s
|α〉 = |α〉o|σα〉s (13.1)
which one could express in terms of real-space wave-functions by taking the scalar product withthe position-spin eigenstates |r〉 = |r〉o|σ〉s, 1
ψα(r) def= 〈r|α〉 = φα(r)χσα(σ) . (13.2)
Quite clearly, φα(r) = 〈r|α〉o is simply the orbital part of the wave-function, and χσα(σ) = 〈σ|σα〉 =
δσ,σα is the spin part. 2 The single particle spin-orbitals are often assumed, as we will henceforth,to form an orthonormal basis of the single particle Hilbert space. 3
Examples of such single particle spin-orbitals might be the plane waves, α = (k, σα) and φk(r) =eik·r, the hydrogenoid atomic orbitals, α = (nlm, σα) and φnlm(r) = R(r)Ylm(Ω), the Bloch or theWannier states of a crystalline solid, etc. whichever is more appropriate to describe the systemunder consideration.
A state for N identical particles may be written down as product of N single particle states
|α1, α2, · · · , αN ) = |α1〉 · |α2〉 · · · |αN 〉 , (13.3)
or, in terms of wave-functions,Product states
(r1, r2, · · · , rN |α1, α2, · · · , αN ) = ψα1(r1)ψα2(r2) · · · ψαN(rN ) .
However, such product wave-functions are not allowed for systems of identical particles.
1We will often use the symbol r to mean the combination of r, the position of a particle, and σ, its spin projectionalong the z-axis, i.e., whenever you see r, just read r = r, σ.
2We will often omit the subscripts “o” or “s” for the orbital and spin parts of a state whenever there is noambiguity.
3The requirement of orthonormality might be relaxed, if necessary, at the price of introducing the so-calledoverlap matrix.
78 Many particle systems
Consider a system consisting of N = 2 identical particles. Its Hamiltonian will read somethinglike:
H =p2
1
2m+ V (r1) +
p22
2m+ V (r2) + Vint(r1, r2) = H1 +H2 + Vint ,
where V (ri) and Vint(r1, r2) are the most generic external and interaction potentials, which coulddepend on the spins of the particles. The fact that the two particles are identical implies that themasses are the same, m1 = m2 = m, that the form of V (r) is identical for both particles, and thatVint(r1, r2) = Vint(r2, r1). If we denote by H(1, 2) the Hamiltonian we just wrote, then evidentlyH(2, 1) = H(1, 2). Formally, we can introduce the operator P12 that exchanges the two particles,an operator whose definition on a generic wave-function Ψ(r1, r2) is given by:
P12Ψ(r1, r2)def= Ψ(r2, r1) .
Quite clearly, the square of P12 is just the identity: P 212 = 1. Moreover, using H(2, 1) = H(1, 2),
you can easily convince yourself that:
Exercise 13.1 By applying the definition of P12 on the action of H on a generic wave-function:
P12 H(1, 2) Ψ(r1, r2) = H(2, 1) Ψ(r2, r1) ,
show that P12 commutes with H, i.e.,
P12 H(1, 2) = H(1, 2) P12 .
Now, the possible eigenvalues of P12 are just ±1, since their square should be 1, and therefore, thecorresponding eigenstates are, quite clearly, even or odd under exchange of the two particles:
P12Ψeven(r1, r2)def= Ψeven(r2, r1) = Ψeven(r1, r2)
P12Ψodd(r1, r2)def= Ψodd(r2, r1) = −Ψodd(r1, r2) .
Moreover, since [H,P12] = 0 we can diagonalize simultaneously H and P12, i.e., it is possible toclassify all eigenstates of H as even or odd under exchange of the two particles. Suppose nowthat the two particles do not interact, i.e., Vint = 0. Then, if ψa(r) and ψb(r) are eigenstates ofp2/(2m) + V (r), the product state ψa(r1)ψb(r2) is formally a good candidate eigenstate of thenon-interacting Hamiltonian H1 +H2, with eigenvalue Ea +Eb, but certainly not an eigenstate ofP12. How can we construct good eigenstates of P12, out of ψa and ψb? Consider the “symmetrizedstates”:
Ψa,b(r1, r2) =1√2
[ψa(r1)ψb(r2)± ψb(r1)ψa(r2] .
Evidently, they both correspond to the same “energy” Ea + Eb, and they are good eigenstatesof P12. Which one is the correct physical eigenstate, depends on the statistics of the particlesunder consideration. Physics restricts the possible eigenstates to be totally antisymmetric underpermutations of the particle labels, for fermions, or totally symmetric for bosons (see below).Therefore, the - sign applies to two fermions, and the + sign is good for two bosons.
Let us extend this concept to a generic number N of identical particles. The permutation sym-metry that physics dictates is easy to enforce by applying, essentially, a projection operator to theproduct states. Consider a permutation P : (1, · · · , N) → (P1, · · · , PN ), and define a permutationoperator P to act on product states in the following way:
P |α1, α2, · · · , αN ) def= |αP1 , αP2 , · · · , αPN) . (13.4)
Define now (−1)P to be the parity of the permutation P . This will appear shortly in constructingthe antisymmetric states for fermions. In order to have a notation common to both fermion and
Many-particle basis states for fermions and bosons. 79
boson cases, we introduce the symbol ξ = −1 for fermions and ξ = +1 for bosons, and form thecorrectly symmetrized states 4
Symmetrized
states|α1, α2, · · · , αN 〉def=
1√N !
∑P
ξPP |α1, α2, · · · , αN ) =1√N !
∑P
ξP |αP1 , αP2 , · · · , αPN) , (13.5)
where the sum is performed over the N ! possible permutations of the labels. Indeed, it is verysimple to verity that
Exercise 13.2 The symmetrized states |α1, α2, · · · , αN 〉 satisfy the relationship
P|α1, α2, · · · , αN 〉 = |αP1 , αP2 , · · · , αPN 〉 = ξP |α1, α2, · · · , αN 〉 . (13.6)
This is the precise formulation of what one means in saying that a state is totally antisymmetric (forfermions, ξ = −1) or totally symmetric (for bosons, ξ = 1).
Perhaps a more familiar expression is obtained by writing down the corresponding real-spacewave-functions: 5
ψα1,α2,··· ,αN(r1, r2, · · · , rN ) def= (r1, r2, · · · , rN |α1, α2, · · · , αN 〉
=1√N !
∑P
ξPψαP1(r1)ψαP2
(r2) · · ·ψαPN(rN ) . (13.7)
For the case of fermions, this is the so-called Slater determinantSlater
determinantψα1,α2,··· ,αN(r1, r2, · · · , rN ) =
1√N !
∑P
(−1)PψαP1(r1)ψαP2
(r2) · · ·ψαPN(rN )
=1√N !
det
ψα1(r1) ψα2(r1) · · · ψαN
(r1)ψα1(r2) ψα2(r2) · · · ψαN
(r2)... · · ·
...ψα1(rN ) ψα2(rN ) · · · ψαN
(rN )
. (13.8)
The corresponding expression for Bosons is called a permanent, but it is much less nicer then thedeterminant, computationally.
For Fermions, it is very simple to show that the same label cannot appear twice in the state,an expression of the Pauli principle. Indeed, suppose α1 = α2, for instance. Then, by applying Pauli principle
the permutation P : (1, 2, 3, · · · , N) → (2, 1, 3, · · · , N) which transposes 1 and 2, we have, on onehand, no effect whatsoever on the labels
P|α1, α1, α3, · · · , αN 〉 = |αP1 , αP2 , αP3 , · · · , αPN〉 = |α1, α1, α3, · · · , αN 〉 ,
but, on the other hand, the action of P must result in a minus sign because the state is totallyantisymmetric
P|α1, α1, α3, · · · , αN 〉 = (−)P |α1, α1, α3, · · · , αN 〉 = −|α1, α1, α3, · · · , αN 〉 ,
and a state equal to minus itself is evidently zero. Perhaps more directly, we can see the result fromthe expression of the corresponding Slater determinant, which has the first two columns which areidentical, and therefore vanishes, by a known properties of determinants. So, if we define
nα = number of times the label α appears in (α1, · · · , αN ) , (13.9)
then Pauli principle requires nα ≤ 1. For Bosons, on the contrary, there is no limit to nα.
4We will consistently use the symbol | · · · ), with round parenthesis, for the product states, and the ket notation| · · · 〉 for symmetrized states.
5Observe that we take the scalar product with the product state |r1, · · · , rN ).
80 Many particle systems
A few examples should clarify the notation. Consider two particles occupying, in the plane-wavebasis, α1 = k1, ↑, and α2 = k2, ↓ or α2 = k2, ↑ (we will do both spin calculations in one shot). Thecorrectly symmetrized states are:
(r1, r2|k1 ↑,k2 ↓ (↑)〉 =1√2
[φk1(r1)φk2(r2)χ↑(σ1)χ↓(↑)(σ2)∓ φk2(r1)φk1(r2)χ↓(↑)(σ1)χ↑(σ2)
],
where the upper sign refers to fermions, and φk(r) = eik·r. Notice that, in general, such a state isnot an eigenvector of the total spin. Exceptions are: i) when both spins are ↑, in which case youobtain a maximally polarized triplet state:
(r1, r2|k1 ↑,k2 ↑〉 =1√2
[φk1(r1)φk2(r2)∓ φk2(r1)φk1(r2)]χ↑(σ1)χ↑(σ2) , (13.10)
or when the two orbital labels coincide. In general, good eigenstates of the total spin are ob-tained only by linear combination of more than one symmetrized state. For the example above, forinstance, it is simple to show that
1√2
[(r1, r2|k1 ↑,k2 ↓〉 ± (r1, r2|k1 ↓,k2 ↑〉] (13.11)
is a triplet (upper sign) or a singlet (lower sign). The triplet (S = 1) and singlet (S = 0) spin statesare the result of the addition of two s = 1/2 spin states for the two particles, with a procedurethat is virtually identical (but much simpler) to that used to construct j = 3/2 and j = 1/2 statesout of l = 1 and s = 1/2. This is just another elementary example of the general addition of twoangular momenta.
As a further example, consider three identical particles occupying three states ψa, ψb and ψc.The resulting physically allowed state is:
Ψa,b,c(r1, r2, r3) =1√6[ + ψa(r1)ψb(r2)ψc(r3) + ψb(r1)ψc(r2)ψa(r3) + ψc(r1)ψa(r2)ψb(r3)
± ψb(r1)ψa(r2)ψc(r3)± ψc(r1)ψb(r2)ψa(r3)± ψa(r1)ψc(r2)ψb(r3)] ,
where the upper (+) sign applies to three bosons, the lower (-) sign to three fermions.
The normalization, or more generally the scalar product, of symmetrized states involves a bitNormalization of permutation algebra.
Exercise 13.3 Show that the scalar product between two correctly-symmetrized states |α1, α2, · · · , αN 〉and |α′1, α′2, · · · , α′N 〉 is non-zero only if the set of labels (α′1, α
′2, · · · , α′N ) is a permutation of (α1, α2, · · · , αN ).
Verify then that:
〈α′1, α′2, · · · , α′N |α1, α2, · · · , αN 〉 = ξPYα
nα!
where P is a permutation that makes the labels to coincide, and nα is the number of times a certain labelα appears.
As a consequence, fermionic symmetrized states (Slater determinants) are normalized to 1, sincenα ≤ 1, while bosonic ones not necessarily. One can easily normalize bosonic states by defining
|nα〉 =1√∏α nα!
|α1, · · · , αN 〉 (13.12)
where we simply label the normalized states by the occupation number nα of each label, since theorder in which the labels appear does not matter for Bosons.
We conclude by stressing the fact that the symmetrized states constructed are simply a basis setfor the many-particle Hilbert space, in terms of which one can expand any N -particle state |Ψ〉
General |Ψ〉
Two-electrons: the example of Helium. 81
|Ψ〉 =∑
α1,α2,··· ,αN
Cα1,α2,··· ,αN|α1, α2, · · · , αN 〉 , (13.13)
with appropriate coefficients Cα1,α2,··· ,αN. The whole difficulty of interacting many-body systems is
that the relevant low-lying excited states of the systems often involve a large (in principle infinite)number of symmetrized states in the expansion.
13.2 Two-electrons: the example of Helium.
Suppose we would like to tackle the He atom problem, which has N = 2 electrons. More generally,we can address a two-electron ion like Li+, Be2+, etc. with a generic nuclear charge Z ≥ 2. TheHamiltonian for an atom with N electrons is evidently:
H =N∑i=1
(p2i
2m− Ze2
ri
)+
12
∑i 6=j
e2
|ri − rj |, (13.14)
where we have assumed the N interacting electrons to be in the Coulomb field of a nucleus of chargeZe, which we imagine fixed at R = 0. We have neglected spin-orbit effects and other relativisticcorrections. We plan to treat here the N = 2 case. Evidently, a good starting basis set of single-particle functions is that of hydrogenoid orbitals which are solutions of the one-body Hamiltonianh = p2/2m−Ze2/r. We know that such functions are labeled by orbital quantum numbers (nlm),and that (
p2
2m− Ze2
r
)φnlm(r) = εnφnlm(r) . (13.15)
The eigenvalues εn are those of the Coulomb potential, and therefore independent of the angularmomentum quantum numbers lm:
εn = −Z2
2e2
aB
1n2
, (13.16)
where aB = ~2/(me2) ≈ 0.529rA is the Bohr radius, and e2/aB = EHartree ≈ 27.2 eV. Thewave-functions φnlm(r) are well known. For instance, the 1s orbital (n = 1, l = 0,m = 0) is:
φ1s(r) =1√π
(Z
aB
)3/2
e−Zr/aB .
It is very convenient here two adopt atomic units, where lengths are measured in units of the Bohrlength aB , and energies in units of the Hartree, EH = e2/aB .
If we start ignoring the Coulomb potential, we would put our N = 2 electrons in the lowest twoorbitals available, i.e., 1s ↑ and 1s ↓, forming the Slater determinant:
Ψ1s↑,1s↓(r1, r2) = φ1s(r1)φ1s(r2)1√2
[χ↑(σ1)χ↓(σ2)− χ↓(σ1)χ↑(σ2)] ,
i.e., the product of a symmetric orbital times a singlet (antisymmetric) spin state. The one-bodypart of the Hamiltonian contributes an energy
E(one−body) = 〈1s ↑, 1s ↓ |h|1s ↑, 1s ↓〉 = 2ε1s = −Z2 (a.u.) . (13.17)
For He, where Z = 2, this is −4 (a.u.) a severe underestimate of the total energy, which is knownto be E(exact,He) ≈ −2.904 (a.u.). This can be immediately cured by including the average of theCoulomb potential part, which is simply the direct Coulomb integral K1s,1s of the 1s orbital, often Coulomb
integral
82 Many particle systems
indicated by U1s in the strongly correlated community,
〈1s ↑, 1s ↓ |V |1s ↑, 1s ↓〉 =∫dr1dr2 |φ1s(r2)|2
e2
|r1 − r2||φ1s(r1)|2
def= K1s,1s = U1s . (13.18)
The calculation of Coulomb integrals can be performed analytically by exploiting rotational invari-ance. We simply state here the result:
U1s =58Z (a.u.) . (13.19)
Summing up, we obtain the average energy of our Slater determinant as
E1s↑,1s↓ =(−Z2 +
58Z
)(a.u.) = −2.75 (a.u.) , (13.20)
where the last number applies to He (Z = 2).
To improve over this result, let us use the variational method, which states that the true groundstate energy is the minimum of the average energy over all allowed states:
Eg.s. = Min|Ψ〉〈Ψ|H|Ψ〉〈Ψ|Ψ〉
. (13.21)
The validity of this statement can be easily proven by expanding |Ψ〉 in the basis of eigenfunctionsof H, and calculating the resulting average energy. We will write a guess for the wave-functionmade by choosing the same orbital wave-function for the two opposite spin states
ψαo↑(r) = φαo(r)χ↑(σ) =⇒ ψαo↓(r) = φαo(r)χ↓(σ) . (13.22)
What is the best possible choice for such an orbital φαo(r)? The proper answer to such a questionrequires going through the Hartree (or, more generally, Hartree-Fock) mean-field approach. Wewill simply assume here this Hartree approach as very reasonable, without deriving the optimalitycondition. So, going back to our He exercises, we will put two electrons, one with ↑ spin, the otherwith ↓ spin, in the lowest orbital solution of the Hartree equation(
−~2∇2
2m− Ze2
r+ Vdir(r)
)φ1s(r) = ε1sφ1s(r) , (13.23)
where now the 1s label denotes the orbital quantum numbers of the lowest energy solution of arotationally invariant potential Vdir(r). The form of the direct (Hartree) potential Vdir(r) is:
Vdir(r) =∫dr′
e2
|r− r′||φ1s(r′)|2 , (13.24)
because the electron in φ1s feels the repulsion of the other electron in the same orbital. We im-mediately understand that the Hartree potential partially screens at large distances the nuclearcharge +Ze. Indeed, for r →∞ we can easily see that
Vdir(r) ∼e2
r
∫dr′ |φ1s(r′)|2 =
e2
r, (13.25)
so that the total effective charge seen by the electron at large distances is just (Z − 1)e, and notZe. This is intuitively obvious. The fact that we chose our original φ1s orbital as solution of thehydrogenoid problem with nuclear charge Ze gives to that orbital a wrong tail at large distances.This is, primarily, what a self-consistent mean-field calculation needs to adjust to get a betterform of φ1s. Indeed, there is a very simple variational scheme that we can follow instead of solving
Two-electrons: the example of Helium. 83
Eq. (13.23). Simply consider the hydrogenoid orbital φ(Zeff)1s which is solution of the Coulomb
potential with an effective nuclear charge Zeffe (with Zeff a real number),(p2
2m− Zeffe
2
r
)φ
(Zeff)1s (r) = ε1sφ
(Zeff)1s (r) , (13.26)
and form the Slater determinant |1s ↑, 1s ↓;Zeff〉 occupying twice φ(Zeff)1s , with Zeff used as a single
variational parameter. The calculation of the average energy E1s↑,1s↓(Zeff) is quite straightforward,6 and gives:
E1s↑,1s↓(Zeff) = 〈1s ↑, 1s ↓;Zeff |H|1s ↑, 1s ↓;Zeff〉 = Z2eff − 2ZZeff +
58Zeff . (13.27)
Minimizing with respect to Zeff this quadratic expression we find Zopteff = Z − 5/16 = 1.6875 and
E1s↑,1s↓(Zopteff ) = −(Z − 5/16)2 (a.u.) ≈ −2.848 (a.u.) ,
where the numbers are appropriate to Helium. This value of the energy in not too far form theexact Hartree-Fock solution, which would give
EHF ≈ −2.862 (a.u.) (for Helium).
In turn, the full Hartree-Fock solution differs very little from the exact non-relativistic energyof Helium, calculated by accurate quantum chemistry methods (Configuration Interaction). Aquantity that measures how far the Hartree-Fock mean-field energy differs from the exact answeris the so-called correlation energy, defined simply as:
Ecorrdef= Eexact − EHF , (13.28)
which, for Helium amounts to Ecorr ≈ −0.042 (a.u.), a bit more than 1% of the total energy.
6You simply need to calculate the average of p2/2m and of 1/r over φ(Zeff )1s .
Appendix A
Second quantization: brief outline.
It is quite evident that the only important information contained in the symmetrized states|α1, α2, · · · , αN 〉 is how many times every label α is present, which we will indicate by nα; therest is automatic permutation algebra. The introduction of creation operators a†α is a device bywhich we take care of the labels present in a state and of the permutation algebra in a direct way.
Given any single-particle orthonormal basis set |α〉 we define operators a†α which satisfy theCreation
operators
following rule: 1
a†α|0〉def= |α〉
a†α|α1, α2, · · · , αN 〉def= |α, α1, α2, · · · , αN 〉
. (A.1)
The first is also the definition of a new state |0〉, called vacuum or state with no particles, whichThe vacuum should not be confused with the zero of a Hilbert space: in fact, we postulate 〈0|0〉 = 1. It is formally
required in order to be able to obtain the original single-particle states |α〉 by applying an operatorthat creates a particle with the label α to something: that “something” has no particles, andobviously no labels whatsoever. The second equation defines the action of the creation operator a†αon a generic correctly-symmetrized state. Notice immediately that, as defined, a†α does two thingsin one shot: 1) it creates a new label α in the state, 2) it performs the appropriate permutationalgebra in such a way that the resulting state is a correctly-symmetrized state. Iterating the creationrule starting from the vacuum |0〉, it is immediate to show that
a†α1a†α2
· · · a†αN|0〉 = |α1, α2, · · · , αN 〉 . (A.2)
We can now ask ourselves what commutation properties must the operators a†α satisfy in order toenforce the correct permutation properties of the resulting states. This is very simple. Since
|α2, α1, · · · , αN 〉 = ξ|α1, α2, · · · , αN 〉
for every possible state and for every possible choice of α1 and α2, it must follow that
a†α2a†α1
= ξa†α1a†α2
, (A.3)
i.e., creation operators anti-commute for Fermions, commute for Bosons. Explicitly:
a†α1, a†α2
= 0 for Fermions (A.4)[a†α1
, a†α2
]= 0 for Bosons , (A.5)
with A,B = AB +BA (the anti-commutator) and [A,B] = AB −BA (the commutator).
The rules for a†α clearly fix completely the rules of action of its adjoint aα = (a†α)†, since itDestruction
operators1It might seem strange that one defines directly the adjoint of an operator, instead of defining the operator aα
itself. The reason is that the action of a†α is simpler to write.
Second quantization: brief outline. 85
must satisfy the obvious relationship
〈Ψ2|aαΨ1〉 = 〈a†αΨ2|Ψ1〉 ∀Ψ1,Ψ2 , (A.6)
where Ψ1,Ψ2 are correctly-symmetrized many-particle basis states. First of all, by taking theadjoint of the Eqs. (A.4), it follows that
aα1 , aα2 = 0 for Fermions (A.7)[aα1 , aα2 ] = 0 for Bosons . (A.8)
There are a few simple properties of aα that one can show by using the rules given so far. Forinstance,
aα|0〉 ∀α , (A.9)
since 〈Ψ2|aα|0〉 = 〈a†αΨ2|0〉 = 0, ∀Ψ2, because of the mismatch in the number of particles. 2 Moregenerally, it is simple to prove that that an attempt at destroying label α, by application of aα,gives zero if α is not present in the state labels,
aα|α1, α2, · · · , αN 〉 = 0 if α /∈ (α1, α2, · · · , αN ) . (A.10)
Consider now the case α ∈ (α1, · · · , αN ). Let us start with Fermions. Suppose α is just in firstposition, α = α1. Then, it is simple to show that
aα1 |α1, α2, · · · , αN 〉 = |α1, α2, · · · , αN 〉 = |α2, · · · , αN 〉 , (A.11)
where by α1 we simply mean a state with the label α1 missing (clearly, it is a N −1 particle state).To convince yourself that this is the correct result, simply take the scalar product of both sidesof Eq. (A.11) with a generic N − 1 particle state |α′2, · · · , α′N 〉, and use the adjoint rule. The lefthand side gives:
〈α′2, · · · , α′N |aα1 |α1, α2, · · · , αN 〉 = 〈α1, α′2, · · · , α′N |α1, α2, · · · , αN 〉 ,
which is equal to (−1)P when P is a permutation bringing (α2, · · · , αN ) into (α′2, · · · , α′N ), and 0otherwise. The right hand side gives
〈α′2, · · · , α′N |α2, · · · , αN 〉 ,
and coincides evidently with the result just stated for the left hand side. If α is not in first position,say α = αi, then you need to first apply a permutation to the state that brings αi in first position,and proceed with the result just derived. The permutation brings and extra factor (−1)i−1, sinceyou need i− 1 transpositions. As a result:
aαi |α1, α2, · · · , αN 〉 = (−1)i−1|α1, α2, · · · , αi, · · · , αN 〉 . (A.12)
The bosonic case needs a bit of extra care for the fact that the label α might be present more thanonce.
Exercise A.1 Show that for the bosonic case the action of the destruction operator is
aαi |α1, α2, · · · , αN 〉 = nαi |α1, α2, · · · , αi, · · · , αN 〉 . (A.13)
Armed with these results we can finally prove the most difficult commutation relations, thoseinvolving a aα with a a†α′ . Consider first the case α 6= α′ and do the following.
2A state which has vanishing scalar product with any state of a Hilbert space, must be the zero.
86 Second quantization: brief outline.
Exercise A.2 Evaluating the action of aαa†α′ and of a†α′aα on a generic state |α1, · · · , αN 〉, show that
aα, a†α′ = 0 for Fermions (A.14)haα, a
†α′
i= 0 for Bosons . (A.15)
Next, let us consider the case α = α′. For fermions, if α /∈ (α1, · · · , αN ) then
aαa†α|α1, · · · , αN 〉 = aα|α, α1, · · · , αN 〉 = |α1, · · · , αN 〉 ,
whilea†αaα|α1, · · · , αN 〉 = 0 .
If, on the other hand, α ∈ (α1, · · · , αN ), say α = αi, then
aαia†αi|α1, · · · , αN 〉 = 0 ,
because Pauli principle forbids occupying twice the same label, while
a†αiaαi |α1, · · · , αN 〉 = a†αi
(−1)i−1|α1, · · · , αi, · · · , αN 〉 = |α1, · · · , αN 〉 ,
because the (−1)i−1 is reabsorbed in bringing the created particle to the original position. Sum-marizing, we see that for Fermions, in all possible cases
aαa†α + a†αaα = 1 (A.16)
Exercise A.3 Repeat the algebra for the Bosonic case to show that
aαa†α − a†αaα = 1 (A.17)
We can finally summarize all the commutation relations derived, often referred to as the canonicalcommutation relations.
Canonical
commutatorsFermions =⇒
a†α1, a†α2
= 0aα1 , aα2 = 0aα1 , a
†α2 = δα1,α2
Bosons =⇒
[a†α1
, a†α2
]= 0
[aα1 , aα2 ] = 0[aα1 , a
†α2
]= δα1,α2
. (A.18)
Before leaving the section, it is worth pointing out the special role played by the operator
nαdef= a†αaα (A.19)
often called the number operator, because it simply counts how many times a label α is presentNumber
operator in a state.
Exercise A.4 Verify thatnα|α1, · · · , αN 〉 = nα|α1, · · · , αN 〉; , (A.20)
where nα is the number of times the label α is present in (α1, · · · , αN ).
Clearly, one can write an operator N that counts the total number of particles in a state by
Ndef=∑α
nα =∑α
a†αaα . (A.21)
Second quantization: brief outline. 87
A.0.1 Changing the basis set.
Suppose we want to switch from |α〉 to some other basis set |i〉, still orthonormal. Clearly there isa unitary transformation between the two single-particle basis sets:
|i〉 =∑α
|α〉〈α|i〉 =∑α
|α〉Uα,i , (A.22)
where Uα,i = 〈α|i〉 is the unitary matrix of the transformation. The question is: How is a†i de-termined in terms of the original a†α? The answer is easy. Since, by definition, |i〉 = a†i |0〉 and|α〉 = a†α|0〉, it immediately follows that
a†i |0〉 =∑α
a†α|0〉 Uα,i . (A.23)
By linearity, one can easily show that this equation has to hold not only when applied to thevacuum, but also as an operator identity, i.e.,
a†i =∑α a
†α Uα,i
ai =∑α aα U
∗α,i
, (A.24)
the second equation being simply the adjoint of the first. The previous argument is a convenientmnemonic rule for re-deriving, when needed, the correct relations.
A.0.2 The field operators.
The construction of the field operators can be seen as a special case of Eqs. (A.24), when we takeas new basis the coordinate and spin eigenstates |i〉 = |r, σ〉. By definition, the field operator Ψ†
σ(r)is the creation operator of the state |r, σ〉, i.e.,
Ψ†σ(r)|0〉 = |r, σ〉 . (A.25)
Then, the analog of Eq. (A.22) reads:
|r, σ〉 =∑α
|α〉〈α|r, σ〉 =∑α
|α〉δσ,σαφ∗α(r) , (A.26)
where we have identified the real-space wave-function of orbital α as φα(r) = 〈r|α〉o, and used thefact that 〈σα|σ〉 = δσ,σα . Assume that the quantum numbers α are given by α = (αo, σ), where αodenote the orbital quantum numbers. The analog of Eqs. (A.24) reads, then,
Ψ†σ(r) =
∑αo
φ∗αo,σ(r) a†αo,σ (A.27)
Ψσ(r) =∑αo
φαo,σ(r) aαo,σ .
These relationships can be easily inverted to give:
a†αo,σ =∫dr φαo,σ(r) Ψ†
σ(r) (A.28)
aαo,σ =∫dr φ∗αo,σ(r) Ψσ(r) .
88 Second quantization: brief outline.
A.0.3 Operators in second quantization.
We would like to be able to calculate matrix elements of a Hamiltonian like, for instance, that ofN interacting electrons in some external potential v(r),
H =N∑i=1
(p2i
2m+ v(ri)
)+
12
∑i 6=j
e2
|ri − rj |. (A.29)
In order to do so, we need to express the operators appearing in H in terms of the creation anddestruction operators a†α and aα of the selected basis, i.e., as operators in the so-called Fock space.Observe that there are two possible types of operators of interest to us:1) one-body operators, like the total kinetic energy
∑i p
2i /2m or the external potential
∑i v(ri),
One-body
operators
which act on one-particle at a time, and their effect is then summed over all particles in a totallysymmetric way, generally
U1−bodyN =
N∑i=1
U(i) ; (A.30)
2) two-body operators, like the Coulomb interaction between electrons (1/2)∑i 6=j e
2/|ri−rj |, whichinvolve two-particle at a time, and are summed over all pairs of particles in a totally symmetricway,Two-body
operatorsV 2−bodyN =
12
N∑i 6=j
V (i, j) . (A.31)
The Fock (second quantized) versions of these operators are very simple to state. For a one-bodyoperator:
U1−bodyN =⇒ UFock =
∑α,α′
〈α′|U |α〉a†α′aα , (A.32)
where 〈α′|U |α〉 is simply the single-particle matrix element of the individual operator U(i), forinstance, in the examples above,
〈α′|p2/2m|α〉 = δσα,σα′
∫dr φ∗α′(r)
(−~2∇2
2m
)φα(r) (A.33)
〈α′|v(r)|α〉 = δσα,σα′
∫dr φ∗α′(r)v(r)φα(r) .
For a two-body operator:
V 2−bodyN =⇒ VFock =
12
∑α1,α2,α′1,α
′2
(α′2α′1|V |α1α2) a
†α′2a†α′1
aα1aα2 , (A.34)
where the matrix element needed is, for a general spin-independent interaction potential V (r1, r2),
(α′2α′1|V |α1α2) = δσα1 ,σα′1
δσα2 ,σα′2
∫dr1dr2 φ
∗α′2
(r2)φ∗α′1(r1)V (r1, r2)φα1(r1)φα2(r2) . (A.35)
We observe that the order of the operators is extremely important (for fermions).
The proofs are not very difficult but a bit long and tedious. We will briefly sketch that for theone-body case.
A.1 Why a quadratic Hamiltonian is easy.
Before we consider the Hartree-Fock problem, let us pause for a moment and consider the reasonwhy one-body problems are considered simple in a many-body framework. If the Hamiltonian is
Why a quadratic Hamiltonian is easy. 89
simply a sum of one-body terms H =∑Ni=1 h(i), for instance h(i) = p2
i /2m+ v(ri), we know thatin second quantization it reads
H =∑α,α′
hα′,αa†α′aα , (A.36)
where the matrix elements are
hα′,α = 〈α′|h|α〉 = δσα,σα′
∫dr φ∗α′(r)
(−~2∇2
2m+ v(r)
)φα(r) . (A.37)
So, H is purely quadratic in the operators. The crucial point is now that any quadratic problem canbe diagonalized completely, by switching to a new basis |i〉 made of solutions of the one-particleSchrodinger equation 3
h|i〉 = εi|i〉 =⇒(−~2∇2
2m+ v(r)
)φi(r) = εiφi(r) . (A.38)
Working with this diagonalizing basis, and the corresponding a†i , the Hamiltonian simply reads:
H =∑i
εia†iai =
∑i
εi ni , (A.39)
where we assume having ordered the energies as ε1 ≤ ε2 ≤ · · · . With H written in this way, wecan immediately write down all possible many-body exact eigenstates as single Slater determinants(for Fermions) and the corresponding eigenvalues as sums of εi’s,
|Ψi1,··· ,iN 〉 = a†i1 · · · a†iN|0〉
Ei1,··· ,iN = εi1 + · · ·+ εiN . (A.40)
So, the full solution of the many-body problem comes automatically from the solution of thecorresponding one-body problem, and the exact many-particle states are simply single Slater de-terminants.
3Being simple does not mean that solving such a one-body problem is trivial. Indeed it can be technically quiteinvolved. Band theory is devoted exactly to this problem.
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