Sourendu Gupta Quantum Mechanics 1 Ninth...

Outline Crystals Oscillator Landau levels 2D oscillator Keywords and References

Simple one-dimensional potentials

Sourendu Gupta

TIFR, Mumbai, India

Quantum Mechanics 1Ninth lecture

Sourendu Gupta Quantum Mechanics 1


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2 Energy bands in periodic potentials

3 The harmonic oscillator

4 A charged particle in a magnetic field

5 The isotropic two-dimensional harmonic oscillator

6 Keywords and References



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Bloch’s theorem for periodic potentials

For a periodic potential, V (x) = V (x + x0), the Hamiltoniancommutes with the translation, T (x0). If ψ(x) is one of thesimultaneous eigenfunctions, then T (x0)ψ(x) = λψ(x), whereλ = exp(iKx0), since T (x0) is unitary. Trivially, one gets Bloch’stheorem, which states that an eigenfunction in a period potentialhas the form ψ(x + nx0) = T n(x0)ψ(x) = exp(iKnx0)ψ(x). Thevalue of K depends on the boundary conditions one chooses.

The generalization to arbitrary dimensions is straightforward. Sincethe translation group is Abelian, translations in all directionscommute. As a result, all these translations can be diagonalizedtogether with a periodic Hamiltonian. So the eigenfunctions of theperiodic Hamiltonian must satisfy ψ(r + R) = exp(iK · R)ψ(r), forany lattice vector R.



Problem 8.1: the scattering matrix

A potential V (x) has a finite range, if V (x) = 0 for |x | > a. Astationary solution far from the potential is the superposition

ψ(y ; ρ) =

{

A1eiρy + B1e

−iρy (y < −1)

A2eiρy + B2e

−iρy . (y > 1)

The incoming waves are exp(iρy) for y < −1 and its complexconjugate on the right. The outgoing waves are the remainder.The scattering matrix relates the incoming waves to the outgoingwaves,

(

B1

A2

)

= S

(

A1

B2

)

.

Since the states are stationary, |ψ|2 is time independent. Showthat the probability current, J = (ψ∗pψ + ψp†ψ∗)/(2m) isconstant. Prove that, as a result, SS† = 1.



Problem 8.2: a transmission matrix

Define the transmission matrix, T , across the potential

(

A1

B1

)

= T

(

A2

B2

)

.

The parametrization of an unitary matrix (problem 4.1)

S =

(

z1 −wz∗2z2 wz∗1

)

, gives T =1

z2

(

1 −wz∗1z1 −w

)

,

where |z1|2 + |z2|2 = 1 and w is a pure phase, i.e., w = exp(iφ).Show that the transmission coefficient, T = |z2|2 and thereflection coefficient R = |z1|2. Unitarity of the S-matrix forcesT +R = 1. Resonances occur when T = 1, i.e., z1 = 0.



Problem 8.3: band formation

2a l

Consider a periodically repeating short ranged potential with rangeof 2a and distance l over which V (x) = 0 (hence x0 = l + 2a).The transmission matrix from one force-free region to another is

Q =

(

λ 00 λ∗

)

T =1

z2

(

λ −wz∗1λz1λ

∗ −wλ∗

)

, where λ = exp(iKl).

Show that the N-step transmission matrix, in the limit N → ∞ haszeroes on the diagonal (no transmission) except exactly atresonance. Show that energy bands arise as a result.



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The Hamiltonian for a harmonic oscillator

The harmonic oscillator is a system which obeys Hooke’s law: theforce is proportional to the displacement from equilibrium andpoints towards the equilibrium position. So the potential isV (x) = mω2x2/2, and the Hamiltonian is

H =1

2mp2 +

mω2

2x2.

This describes ellipses in phase space: this is the classical motionof harmonic oscillators.

In the quantum Hamiltonian p and x become operators. In termsof the dimensionless quantities

X =√mωx and P =

1√mω

p, one has H =ω

2

(

P2 + X 2)

.

Note that [X , P] = i .Sourendu Gupta Quantum Mechanics 1


Examples

1 Near the minimum of any smooth potential, its Taylorexpansion is: V (x) = V (x0) + V ′′(x0)(x − x0)

2/2 + · · · ,where x0 is position of the minimum. Small amplitude motionis always harmonic.

2 Low-lying vibrational modes of molecules show almostharmonic spectrum as a result of this general fact.Complicated many-body interactions in a nucleus can beexpanded in a similar Taylor series to give a central potentialwhich is approximately harmonic. The addition of simple extraterms in the Hamiltonian then explain the spectra of manycomplex nuclei.

3 The motion of a charged particle in a magnetic field is a helix.The momentum components transverse to the field lines playthe roles of X and P .



Raising and lowering operators

We will factorize H using the non-Hermitean operators

a =1√2(X + i P), where [a, a†] = 1.

Consider the Hermitean operator N = a†a. One sees that[N, a] = −a. If |z〉 is an eigenvector of N with eigenvalue z , then[N, a] |z〉 = −a |z〉, and hence Na |z〉 = (z − 1)a |z〉. So |a〉 lowersthe eigenvalue of N by one unit.

For any |ψ〉, let |φ〉 = a |ψ〉. Since 〈φ|φ〉 ≥ 0 and 〈φ|φ〉 = 〈ψ|N|ψ〉,one finds that every eigenvalue of N must be greater than or equalto zero. If the eigenvalues of N are not integers, then there cannotbe a lower bound to the eigenvalues, since a will always lower theeigenvalue by one unit. On the other hand, if the eigenvalues areintegers, then for |0〉 such that N |0〉 = 0, one also has a |0〉 = 0.So, integer eigenvalues are allowed, and exist.



The eigenstates of the Hamiltonian

Since N = (X − i P)(X + i P)/2 = (X 2 + P2 − 1)/2, we find

H = ω

(

N +1

2

)

.

So the eigenvalues of H are E = ω(n + 1/2), for integer n ≥ 0.The ground state satisfies(

mωx +d

dx

)

ψ0(x) = 0, i .e., ψ0(x) =(mω

π

)1/4e−mωx2/2,

Since this is a first order differential equation, there is an uniquesolution. Thus, the lowest eigenvalue of N (and hence, of H) isunique. Then, by induction with a† we can show that none of thestates are degenerate.If |n − 1〉 = cna |n〉, then 〈n − 1|n − 1〉 = |cn|2n〈n|n〉. If |n − 1〉 isnormalized, then normalization of |n〉 requires cn = 1/

√n. Then

|n〉 = a† |n − 1〉 /√n = (a†)n |0〉 /√n!.



All wavefunctions at one go!

We construct a generating function for the wavefunctions–

G (z ; x) =∞∑

n=0

zn√n!ψn(x) =

∞∑

n=0

zn√n!

〈x | (a†)n√n!

|ψ0〉 = 〈x | eza† |ψ0〉 .

Now, from the Baker-Campbell-Hausdorff formula, we find thatexp(za†) = exp(zX/

√2) exp(−izP/

√2) exp(−z2/4). Hence

G (z ; x) = exp

(

−z2

4+ zx

√

mω

2

)

〈x | e−izP/√2 |ψ0〉 .

Then using the action of the exponential on the bra we find that

G (z ; x) =(mω

π

)1/4exp

(

−mωx2

2+√2mωzx − z2

2

)

.



The Hermite polynomials

Define the function f (x) = exp(−x2) and its n-th derivative,f (n)(x) = (−1)nHn(x) exp(−x2). The Hn(x) are called Hermitepolynomials (prove that they are polynomials). By directdifferentiation one can obtain the recurrence relation

Hn(x) =

[

2x − d

dx

]

Hn−1(x).

From the definition of the Hermite polynomials it is clear that

e−z2+2zx = ex2f (x − z)2 =

∞∑

n=0

zn

n!Hn(x).

From the recurrence relation we can write down

H0(x) = 1 H1(x) = 2x H2(x) = 4x2 − 2 · · ·Hn has exactly n zeroes. The zeroes of successive polynomials areinterleaved. Even numbered polynomials have even parity.



The harmonic oscillator wave functions

Comparing the recurrence relation for harmonic oscillator wavefunctions

G (z ; x) =(mω

π

)1/4exp

(

−mωx2

2+√2mωzx − z2

2

)

with that for Hermite polynomials,

e−z2+2zx = ex2f (x − z)2 =

∞∑

n=0

zn

n!Hn(x),

we find that

ψn(x) =(mω

π

)1/4 1√2nn!

Hn(X )e−X 2/2.



Classical-quantum correspondence

Since X = (a+ a†)/√2 and P = −i(a− a†)/

√2, and the operators

a and a† have vanishing diagonal elements in the eigenbasis of N,it is clear that 〈n|X |n〉 = 〈n|P |n〉 = 0. Squaring each of theseoperators, we find that 〈n|X 2|n〉 = 〈n|P2|n〉 = n + 1/2. Clearly,then one has 〈V 〉 = 〈T 〉 = E/2. These relations are the same asfor a classical harmonic oscillator.The commutators [H,X ] = −iP and [H,P] = iX follow from thecommutators of N with a and a†. Then

d〈X 〉dt

=d

dt〈ψ| eiHtX e−iHt |ψ〉 = i〈[H,X ]〉

= ω〈P〉d〈P〉dt

= −ω〈X 〉.

These equations are the same as the classical Hamilton’s equations.



Schrodinger and Heisenberg pictures

We have chosen the Schrodinger representation of quantum states:the operators corresponding to time-independent classical variablesremain time independent, and the states evolve by the action ofthe unitary time-evolution operator.

Since 〈ψ′(t)|O|ψ(t)〉 = 〈ψ′(0)|U†(t)OU(t)|ψ(0)〉, physics remainsunchanged if we use the Heisenberg picture. In this states are timeindependent and operators evolve with time through the adjointaction of the unitary evolution operator.

If O evolves, then its eigenstates evolve. Since these form a basis,the basis evolves, whereas the state remains fixed. So thedifference between these pictures is the same as the active andpassive views of transformations in a vector space. Show that thetime evolution of wavefunctions is independent of the picture.



The thermal density matrix

For a single harmonic oscillator placed inside a heat bath, one findsthe partition function

Z (β) = Tr exp(−Hβ) = e−ωβ/2∞∑

n=0

e−ωβn =exp(−ωβ/2)1− exp(−ωβ) ,

where β = 1/kT . Since ρ(T ) = exp(−Hβ)/Z , the expectationvalue of the energy is

〈H〉 = 1

ZTrH exp(−βH) = − 1

Z

dZ

dβ= −d logZ

dβ.

Using the expression for Z above, we get

〈H〉 = 1

2ω + ωexp(ωβ)− 1.

The Planck spectrum begins to emerge.Sourendu Gupta Quantum Mechanics 1


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The classical theory

A particle of charge e and mass m moves in a magnetic field.

H =1

2m(P − eA)2, B = ∇× A.

The classical equations of motion are

r = p/m, p = ep× B/m.

Taking the B in the z-direction, we obtain, pz(t) = pz(0), and

d

dt

(

pxpy

)

= iωσ2

(

pxpy

)

,

where the cyclotron frequency is ω = eB/m. Scaling the time by afactor ω, it becomes clear that these equations of motion can beobtained from a fictitious Hamiltonian H ′ = (p2x + p2y )/2. Writing

p± = (px ± ipy )/√2, the solutions are p±(t) = p±(0) exp(∓iωt).

Since the phase space is 6d and the motion is integrable, there arethree conserved quantities: E , pz and |p±|.



Quantization

Introduce P = p− eA. Then [Pj , Pk ] = ieǫjklBl . With B in the

z-direction, [Pj , Pz ] = 0 for all j . Since H = P2/2m, this implies

that [H, Pz ] = 0.

The remainder of the Hamiltonian is like a harmonic oscillator,

H ′ = (P2x + P2

y )/2m, where [Px , Py ] = ieB .

By rescaling, P ′x ,y = Px ,y/

√eB, the Hamiltonain becomes the

same as that of a harmonic oscillator with ω = eB/m. Theseeigenvalues are called Landau levels. Eigenvalues of the fullHamiltonian are

E (N, kz) =1

2mk2z + ω

(

N +1

2

)

.



Gauge invariance

Classically, gauge invariance means that for any function φ(r),A′ = A+∇φ gives the same B = ∇× A′ = ∇× A. In thequantum theory, the algebra of P depends only on B , and hence isgauge invariant. However, a = (Px + i Py )/

√2 is translated by a

scalara′ = a− e√

2(φx + iφy ),

where the subscripts denote derivatives. This has no effect on thealgebra of operators, but may affect states.

Problem 8.4: Gauge invariance

Find the eigenstates with any choice of gauge, then investigatewhat happens under gauge transformations. If a |0〉 = 0, anda′ |0′〉 = 0, then what can one say about 〈0′|0〉? Using thegenerating function for the other states, what can one say aboutthe effect of gauge invariance on states?



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The energy eigenvalues and eigenvectors

The isotropic harmonic oscillator in D dimensions has theHamiltonian

H =ω

2(P

2+ X

2),

where X = x√mω and P = p/

√mω. Introducing the ladder

operators aj = (Xj + i Pj)/√2, as before, there are D number

operators Nj = a†j aj . One can write

H = ω

D

2+

D∑

j=1

Nj

.

The energy eigenstates can be specified in the form |n1, · · · , nD〉where ni are the eigenvalues of Ni . Writing N = n1 + · · ·+ nD , theenergies are E = ω(N + D/2). Since the ni do not individuallyenter the expression for the energy, there could be a high degree ofdegeneracy. So the Hamiltonian must have symmetries.



Extended symmetry for D=2

For D = 2, n2 is fixed once N and n1 are given. However, for eachN, n1 can take any value from 0 to N. Hence the level is(N + 1)-fold degenerate.

One can increase n2 by 1 and decrease n1 by 1 without changingthe energy. This can be done by the operator a†1a2. In fact, the

Hermitean operators s1 = a†1a2 + a

†2a1 and s2 = ia

†1a2 − ia

†2a1

acting on a level |N, n1〉 produce linear combinations of |N, n1 − 1〉and |N, n1 + 1〉.

Problem 8.5: the symmetry algebra

Evaluate [H, s1], [H, s2], and [s1, s2]. Complete the symmetryalgebra by forming commutators of all the new operators formed.Continue the process until no new operators can be generated.



Matrix representations of algebras

Given an algebra, such as A2 = 1, one can construct manydifferent sizes of matrices which represent A. For example, A = −I

where I is the n × n identity matrix represents the algebra.Different representations of the symmetry algebra of the 2Dharmonic oscillator can be obtained by acting on states of differentN. For example, acting on the N = 0 space, we find therepresentations by integers: H = 1 and s1 = s2 = 0.The N = 2 space of states has |1〉 = |1, 0〉 and |2〉 = |0, 1〉. Thisgives the matrix representations

H = ω

(

1 00 1

)

, s1 = ω

(

0 11 0

)

, s2 = ω

(

0 −i

i 0

)

.

For any N, the representation of H is the (N + 1)× (N + 1)identity matrix. And the other operators?



The symmetry group SU(2)

Any linear combination of the degenerate eigenstates of anisotropic 2-dimensional harmonic oscillator is generated by the

unitary matrix U = exp(

i∑

j θjsj

)

. Note that detU = 1 (because

the trace of its logarithm is zero). Since this mixes states with thesame energy, all these U must commute with the Hamiltonian.In particular, this is true of the two-dimensional subspace withN = 1. 2× 2 unitary matrices with unit determinant form a group.This is called the group SU(2). Since all these matrices commutewith H, the symmetry group of H is SU(2). The higherdimensional matrices which commute with H are not all possiblelarger unitary matrices, but a subgroup which is isomorphic toSU(2). These matrices of different sizes are called differentrepresentations of SU(2). The Hermitean operators s1, s2 and s3are called the generators of SU(2), or elements of the algebra su(2).



A problem

Consider the isotropic harmonic oscillator in three dimensions. Inanalogy with the construction we have presented here, find thecomplete group of symmetries of this problem: it is called SU(3).

1 Construct the complete algebra of operators from Hermiteancombinations of the bilinears of the shift operators whichleave the energy unchanged.

2 Find the commutators of these operators, and construct thecompletion of this algebra. How many operators are there inthe algebra?

3 Find a complete set of commuting operators among these.4 In the degenerate space of eigenstates corresponding to the

energy eigenvalue E = 5ω/2, construct the representations ofthe elements of the algebra.

5 Construct the representation of the algebra in the space ofenergy eigenstates with eigenvalue E = 7ω/2.



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Keywords and References

Keywords

Bloch’s theorem, lattice vector, finite range potential, scatteringmatrix, transmission matrix, resonances, harmonic oscillator,Rayleigh coefficient, ground state, generating function, Hermitepolynomials, recurrence relation, Schrodinger representation,Heisenberg picture, cyclotron frequency, Landau levels, the groupSU(2), representations of SU(2), generators of SU(2).

References

Quantum Mechanics (Non-relativistic theory), by L. D. Landauand E. M. Lifschitz, chapters 3, 15.Quantum Mechanics (Vol 1), C. Cohen-Tannoudji, B. Diu and F.Laloe, chapter 5.Solid State Physics, by N. W. Ashcroft and N. D. Mermin.Classical groups for Physicists, by B. G. Wybourne.


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