1
ADVANCED QUANTUM MECHANICS AND
INTRODUCTION TO GROUP THEORY(PHYS5000) LECTURE NOTES
Lecture notes based on a course given by Roman Koniuk.The course begins with a discussion on advanced quantum mechanics and then
moves to group theory, Hydrogen, and the Dirac equation
York University, 2012
Presented by: ROMAN KONIUK
LATEXNotes by: JEFF ASAF DROR
2012
YORK UNIVERSITY
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CONTENTS
I. Foundations of Quantum Mechanics 2A. Vector Spaces, Dual Spaces, and Scalar Products 2
1. Vector Spaces 22. Dual Spaces 33. Scalar Products 3
B. Tensor Product(Outer Product) 3C. Operators 4D. Matrix Representations 4E. Projection Operators 4F. Hermitian and Unitary Operators 5G. Continuum Basis 7H. Postulates of Quantum Mechanics 7
II. Compatibility, Incompatibility, and Uncertainty 8
III. Pure States and Mixtures: Density Matrix 8
IV. Group Theory for Quantum Mechanics 14A. Fundamentals 14B. Lie Groups 20C. Lie Algebra 22D. SU (3) 26
V. Accidental Degeneracies 27
VI. Dirac Equation 31A. Non-relativistic limit of the Dirac equation 41B. Covariance of the Dirac equation 47
VII. Hydrogen 50
I. FOUNDATIONS OF QUANTUM MECHANICS
A. Vector Spaces, Dual Spaces, and Scalar Products
1. Vector Spaces
In this class we will use the Bra and Ket spaces. Ket vector spaces obey the following:
|a〉+ |b〉 ∈ V (I.1)
|a〉+ |0〉 = |a〉 (I.2)
|a〉+ |−a〉 = |0〉 (I.3)
|a〉+ (|b〉+ |c〉) = (|a〉+ |b〉) + |c〉 (I.4)
|a〉+ |b〉 = |b〉+ |a〉 (I.5)
α |a〉 ∈ V (I.6)
α (β |a〉) = (αβ) |a〉 (I.7)
Vectors in this space are linearly independent. In other words
n∑i=1
ai |i〉 = 0 (I.8)
⇒ai = 0 ∀i (I.9)
Lecture 2 - Jan 6, 2012
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2. Dual Spaces
The adjoint of a vector |a〉 in the vector space defines a “bra” 〈a|. Such that if
|c〉 = α |a〉+ β |b〉 (I.10)
then
〈c| = α∗ 〈a|+ β∗ 〈b| (I.11)
3. Scalar Products
For any vectors |a〉 , |b〉 ∈ V we denote 〈a|b〉 is the scalar product “bra-ket”. The inner product is in general acomplex number. i.e.
〈a|b〉 ∈ C (I.12)
Inner products obey the relation
〈a|b〉 = (〈b|a〉)∗ (I.13)
The inner product of a vector with itself is a positive definite quantity:
〈a|a〉 ≥ 0 (I.14)
Further
〈a|a〉 = 0⇒ |a〉 = |0〉 (I.15)
The Schwartz inequality says that
|〈a|b〉|2 ≤ 〈a|a〉 〈b|b〉 (I.16)
The space of ket vectors and the dual space of bra vectors form a hilbert space. Hilbert space is a complete space.We define the norm of a vector as
|a| =√〈a|a〉 (I.17)
The set of vectors
|1〉 , |2〉 , ..., |n〉 (I.18)
Form an orthonormal basis.
B. Tensor Product(Outer Product)
Let∣∣a1⟩ represent the state of particle 1 and let
∣∣a2⟩ represent the state of particle 2. Then the two-particle systemis represented by the tensor product ∣∣a1⟩⊗ ∣∣a2⟩ =
∣∣a1⟩ ∣∣a2⟩ =∣∣a1a2⟩ (I.19)
Any operator Ω1 only operates on∣∣a1⟩ and any operator Ω2 only operates on
∣∣a2⟩. The commutator of operatorsthat act on different particles must be zero: [
Ω1,Ω2]
= 0 (I.20)
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C. Operators
Let
Ω |a〉 = |a′〉 (I.21)
If Ω is a linear operator then
Ωα |a〉 = αΩ |a〉 (I.22)
where α ∈ C and
Ω (α |a〉+ β |b〉) = αΩ |a〉+ βΩ |b〉 (I.23)
The product of two operators simply means to carry out the operators in sequence:
ΩΛ |a〉 = Ω (Λ |a〉) (I.24)
The inverse of an operator Ω is denoted by Ω−1:
Ω−1Ω = I = ΩΩ−1 (I.25)
D. Matrix Representations
Consider |a〉 and the operator Ω such that
Ω |a〉 = |b〉 (I.26)
the matrix form of Ω has the elements Ωij = 〈i|Ω |j〉. If we consider the expansions
|a〉 =∑i
ai |i〉 (I.27)
|b〉 =∑j
bj |j〉 (I.28)
If |a〉 is given then you can find ai since
〈i|a〉 = 〈i|∑n
|n〉 (I.29)
= ai (I.30)
bj =∑i
aiΩj,i (I.31)
The matrix elements of the identity operator are
I = 〈i| I |j〉 = δi,j (I.32)
E. Projection Operators
Consider the set of basis vectors |1〉 , |2, 〉 , ..., |n〉. The projection operator of any vector onto the ith basis vector isgiven by
Pi = |i〉 〈i| (I.33)
This is clear since
Pi |a〉 = 〈i|a〉 |i〉 (I.34)
= ai |i〉 (I.35)
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Now consider the following
|a〉 =∑i
ai |i〉 (I.36)
=∑i
〈i|a〉 |i〉 (I.37)
=∑i
|i〉 〈i|a〉 (I.38)
=∑i
Pi |a〉 (I.39)
Hence ∑i
Pi = I =∑i
|i〉 〈i| (I.40)
The projection operator squared is simply the projection operator since
P2 = |i〉 〈i| (|i〉 〈i|) (I.41)
= |i〉 〈i| (I.42)
= P (I.43)
The matrix representation of Pi is:
〈j |Pi| k〉 = 〈j |i〉 〈i| k〉 (I.44)
= δj,iδi,k (I.45)
(I.46)
This looks like 0 0 0 0 0
0. . . 0 0 0
0 0 1 0 0
0 0 0. . . 0
0 0 0 0 0
(I.47)
The identity is given by
I =∑i
Pi (I.48)
F. Hermitian and Unitary Operators
Hermitian operators obey
Ω = ∆† (I.49)
In other words operators for which the operator is equal to the hermitian conjugate.Unitary operators are operators for which
U† = U−1 (I.50)
In quantum mechanics observables are associated with Hermitian operators. For examples
x→ x (I.51)
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px →~i
∂
∂x(I.52)
The eigenvalues of Hermitian operators are all real (since they correspond to observables). The states correspondingto distinct eigenvalues are orthogonal.
Unitary operators are used to transform to another basis (called unitary since they preserve the scalar product).Proof:
〈b|a〉 = 〈Ub′|Ua′〉 (I.53)
=⟨b′∣∣U†U ∣∣ a′⟩ (I.54)
= 〈b′|a′〉 (I.55)
Hence the scalar product is conserved. Other things that are preserved by unitary transformations are trace, deter-minant, and algebraic equations involving matrices and vectors.Any operator in quantum mechanics doesn’t take you out of the Hilbert space.Let
A |a〉 = |a′〉 (I.56)
Then
U†A |a〉 = U† |a′〉 (I.57)
U†AUU† |a〉 = U† |a′〉 (I.58)
A′U† |a〉 = U† |a′〉 (I.59)
(I.60)
where A′ = UAU† is called a unitary-similarity transformation.Typically use a unitary-similarity transformation to diagonalize an operator. We most typically diagonalize the
Hamiltonian (then we can just read off the eigenvalues).
UΩU† = UD (The subscript D means diagonal) (I.61)
] The way we solve for U is by solving the characteristic equation (also called the secular equation).An important unitary operator is the time-evolution operator:
U(t) = e−iHt/~ = 1− iHt
~+
1
2
(−iHt)2
~22!+ ... (I.62)
where H is the Hamiltonian. We use this operator to evolve states.
U |α(0)〉 = |α(t)〉 (I.63)
If the Hamiltonian is diagonal in some basis then
U = e−iHt/~ (I.64)
1− iHt~
+1
2
(−iHt)2
~2(I.65)
This is easy to calculate since for a diagonal matrix
H =
λ1 0 0
0. . . 0
0 0 λm
(I.66)
Hn =
λn1 0 0
0. . . 0
0 0 λnm
(I.67)
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If we are acting on a basis vector α =
0...10...0
then
U |α〉 = 1− iλαt~
+1
2
(−iλ
2αt
2
~2
)+ ... (I.68)
G. Continuum Basis
So far we have dealt with the discrete basis. For a continuum basis, such as the position basis. A state of definiteposition is given by
|x′〉 (I.69)
x |x′〉 = x′ |x′〉 (I.70)
In the continuum basis we know that
I =
∫|x〉 〈x| (I.71)
〈x|x′〉 = δ (x− x′) (I.72)
The scalar product of a mix of discrete and continuous states is the wavefunction:
〈x|n〉 = ψn(x) (I.73)
|〈x|n〉|2 = P (x) (I.74)
H. Postulates of Quantum Mechanics
1. The state of a particle is represented by a “vector” is Hilbert space |α〉
2. Every observable corresponds to a Hermitian operator
3. Every measurement of the observable corresponding to the operator Ω results only in an eigenvalue
4. The average value of an observable is calculated in quantum mechanics by
〈α|Ω |α〉 (I.75)
and we call this the expectation value
5. The time evolution of a state is given by the Schrodinger equation:
i~d
dt|α(t)〉 = H |α(t)〉 (I.76)
Lecture 4 - Jan. 11, 2012
In measurement |α〉 may not be in an eigenstate of Ω but one can expand |α〉 =∑i |ωi〉
ci︷ ︸︸ ︷〈ωi|α〉. The coefficients 〈ωi|α〉
give the amplitude of finding the system in the eigenstate |ωi〉. The probability of being in this eigenstate is
P (ωi) = |〈ω|α〉|2 (I.77)
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Once a measurement is made the state is in that eigenstate (until the time evolution operator acts) “collapse of thewavefunction”. We of course have ∑
i
P (ωi) = 1 (I.78)
For continuous eigenspectra. For the position spectrum:∫all x
P (x) dx = 1 (I.79)
By dimensions we know that P (x) is a probability density.
II. COMPATIBILITY, INCOMPATIBILITY, AND UNCERTAINTY
1. Two variables are compatible if there corresponding operators commute. For example if
|Ω,Λ] = 0 (II.1)
then Ω and Λ are compatible. In this case we can have simultaneous eigenfunctions. In other words
Ω |α〉 = ωi |α〉 ; Λ |α〉 = λi |α〉 (II.2)
This implies exact knowledge of ωi and λi.
2. Two variables are incompatible of their corresponding operators don’t commute. In this case you cannot havesimultaneous eigenfunctions. Hence you cannot have exact knowledge of ω and λ simultaneously.
The uncertainty is defined as
∆A =⟨
(〈A− 〈A〉〉)2⟩1/2
(II.3)
= 〈(A− 〈A〉) (A− 〈A〉)〉1/2 (II.4)
=⟨A2 − 2A 〈A〉+ 〈A〉2
⟩1/2(II.5)
=(⟨A2⟩− 〈A〉2
)1/2(II.6)
This is a measure of how far on average A is away from the average. Noncommutivity is related to uncertainties by:
∆A∆B ≥ 1
2|〈[A,B]〉| (II.7)
As an example consider the uncertainty of x and p:
∆x∆p ≥ 1
2|〈[x, p]〉| (II.8)
≥ 1
2
∣∣∣∣⟨x~i ∂∂x − ~i
(1 +
∂
∂x
)⟩∣∣∣∣ (II.9)
≥ 1
2|〈i~〉| (II.10)
≥ ~2
(II.11)
III. PURE STATES AND MIXTURES: DENSITY MATRIX
Suppose you have an experiment. The beam is 50% spin up and 50% spin down. The state
1√2
(↑ + ↓) (III.1)
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is a pure state. It is a linear combination of states but we know the state exactly. Another possibility is literally 50%of the electrons and 50% are in the down state. How do we differentiate these two cases? We could use
〈Sz〉 =~2σz (III.2)
where
σ =
(0 11 0
),
(0 −ii 0
),
(1 00 −1
)(III.3)
In the case of a pure state we calculate this the expectation value by⟨1√2
(↑ + ↓)∣∣∣∣Sz ∣∣∣∣ 1√
2(↑ + ↓)
⟩=
~4
(0) (III.4)
= 0 (III.5)
In the second case we use
1
2〈↑|Sz |↑〉+
1
2〈↓|Sz |↓〉 = 0 (III.6)
Next we do the same for Sx. For the first case:⟨1√2
(↑ + ↓)∣∣∣∣Sx ∣∣∣∣ 1√
2(↑ + ↓)
⟩=
1
2
~2〈2〉 =
~2
(III.7)
For the second case
1
2〈↑|Sx |↑〉+
1
2〈↓|Sx |↓〉 = 0 (III.8)
Hence the two states are distinguishable since we can measure the spin in the x direction and see the difference.
Lecture 5 - Jan 13, 2011Every state we have seen thus far has been a pure state. We say that the pure state is coherent while mixed statesare incoherent.The density matrix is used to handle both pure and mixed states. For a mixed state we say the probability(notamplitude) of being in state |αi〉 is equal to ωi and
∑i ωi = 1. Therefore the expectation value for a mixed state is
written as
〈A〉 =∑i
ωi 〈αi|A |αi〉 (III.9)
Important point: there are two averaging procedures in this equation. One is the usual quantum mechanical averageprocedure (〈αi|A |αi〉). e.g.
〈x〉 =
∫ψ∗(x)xψ(x)dx (III.10)
=
∫P (x)xdx (III.11)
The other is the classical averaging of multiplying the probability of being in a state by the value of being in thatstate. It’s important to realize that |αi〉 need not be orthogonal but they must be normalized. For example you mayhave the states 50% |↑〉 and 50% 1√
2(|↑〉+ |↓〉). For a mixed state 〈A〉 =
∑i ωi 〈αi|A |αi〉 can be written as follows
〈A〉 = Tr (ρA) ; where ρ =∑i
ωi |αi〉 〈αi| (III.12)
Recall that
Tr(B) =∑j
〈j|B |j〉 (III.13)
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Proof of the relation shown above:
Tr(ρA) =∑j
〈j| ρA |j〉 (III.14)
=∑j
〈j|∑i
ωi |αi〉 〈αi|A |j〉 (III.15)
=∑j
∑i
ωi 〈j|αi〉 〈αi|A |j〉 (III.16)
=∑i
ωi 〈αi|A∑j
|j〉 〈j|αi〉 (III.17)
=∑i
ωi 〈αi|A |αi〉 (III.18)
The operator ρ is called the density matrix operator which is defined as
ρ =∑i
ωi |αi〉 〈α| (III.19)
Remember that you can write any state |αi〉 as
|αi〉 =∑j
Cj |j〉 (III.20)
where here j is labeling the basis state and i is labeling state state in the mixture. We can now expand the densitymatrix as
ρ =∑i
ωi |αi〉 〈αi| (III.21)
=∑i
∑j
∑k
ωiCijC
i†k |j〉 〈k| (III.22)
=∑j,k
ρj,k |j〉 〈k| (III.23)
where ρj,k =∑i ωiC
ijC
i†
k . Traces are basis-independent quantities (since Tr(ABC) = Tr(BCA)). In particular Tr(ρA)is independent of basis. Next consider the trace of ρ:
Tr(ρ) =∑j
〈j|∑i
ωi |αi〉 〈αi|h〉 (III.24)
=∑j
∑i
ωi 〈j|αi〉 〈αi|j〉 (III.25)
=∑i
∑j
ωiCj∑k
C†k 〈k|j〉 (III.26)
=∑i
∑j
ωiCjC†j (III.27)
=∑i
ωi (III.28)
= 1 (III.29)
If we have a pure state the density matrix is just the projection operator
ρ =∑i
ωi |αi〉 〈αi| (III.30)
= |αi〉 〈αi| (III.31)
Hence in for a pure state
ρ2 = ρ (III.32)
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For a mixed state
ρ2 6= ρ (III.33)
Proof: We can go to basis which diagonalizes ρ and then we have
ρ =
ω1 0 00 ω2 0
0 0. . .
(III.34)
but for a mixed state all the ωi < 1 and hence the square of them is less than 1. For example find ρ for case the purestate |↑x〉 = 1√
2(|↑z〉+ |↓z〉). This can be done in several ways
ρ =∑i
ωi |αi〉 〈αi| (III.35)
=
∣∣∣∣ 1√2
(↑ + ↓)⟩⟨
1√2
(↑ + ↓)∣∣∣∣ (III.36)
=1
2(|↑〉 〈↑|+ |↑〉 〈↓|+ |↓〉 〈↑|+ |↓〉 〈↓|) (III.37)
(III.38)
ρj,k =∑i
ωiCijC
i†k (III.39)
= CjC†k (III.40)
but C1C†1 = C1C
†2 = C2C
†1 = C2C
†2 = 1
2 hence
ρj,k =1
2(III.41)
We use this result to find some expectation values:
〈Sz〉 = Tr(ρSz) (III.42)
1
2
~2
Tr
(1 11 1
)(1 00 −1
)(III.43)
=~4
Tr
(1 −11 −1
)(III.44)
= 0 (III.45)
and
〈Sx〉 = Tr(ρSx) (III.46)
1
2
~2
Tr
(1 11 1
)(0 11 0
)(III.47)
=~4
Tr
(1 11 1
)(III.48)
=~2
(III.49)
Assignment: to be handed in on Monday
1. Find ρ for case B (50% ↑ and 50% ↓)
2. Evaluate Tr(ρSz) and Tr ρSx
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1 2
3
R G
12
3
R G
Detectors Detectors
FIG. 1. The EPR experiment
Lecture 6 - January 20th, 2012Consider the EPR setup shown in figure 1. The experiment has two observations.
1. 100% of the time if the two detectors are pointing in the same directions then opposite colours flash. 100%anti-correlation of the colours if the pointers are pointing in the same direction
2. If we pay no attnetion to the direction of the pointers then there is no correlation (completely random)
Classical understanding of the experiment says if we consider observation 1 then the particles know prior to theexperiment what they will choose. This can be though of as the particles having genes. This is shown in table I.Consider the second row of table I. In total there are 9 possible pointer positions: 11, 12, 13, 23, ... (always true). For
TABLE I. The gene table of the EPR experiment
1 2 3 1 2 3
G G G R R R
G G R R R G
G R G R G R
R G G G R R
R G R G R G
G R R R G G
R R G G G R
R R R G G G
the second gene 11, 22, 33, 12, 21 gives us anti correlation (e.g. G⇒ R) while 13, 31, 32, 23 gives us perfect correlation(e.g. G → G). Looking at this gene we see that 5
9 of the time we get anti correlation. But there is nothing specialabout this gene. The only important feature of this gene is that one of the colours is different. In other words we’dget anti-correlation 5
9 of the time for genes 2, 3, 4, 5, 6, 7. The conclusion is that for 6 out of the 8 genes we get 59 of
the time. For genes 1 and 8(the remaining two) we have perfect anti-correlation. The final conclusion is:
P (anti-correlation) >5
9(III.50)
i.e. Bell’s theorem says that classically there is anti-correlation greater then 59 of the time. The experiment violates
Bell’s theorem!We now consider the quantum calculation. We consider the state of an entangled positron electron pair emitted by
a decaying π0 meson. This state is
χ =1√2
(|↑↓〉 − |↓↑〉) (III.51)
Comments:
1. This is a pure state.
2.
Jzχ = 0 (III.52)
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3.
J2χ = 0 (III.53)
4.
Sz1Sz2χ = −~2
4χ (III.54)
5. Define the correlator:
C(O1,O2) ≡ 〈O1O2〉√〈O2
1〉 〈O22〉
(III.55)
Note if O1 = O2 ⇒ C = 1.
6. It’s easy to see (or show) that we get perfect anti-correlation
C(Sz1Sz2 ) = −1 (III.56)
7. Further one can show that
C(Sx1Sx2 ) = C(Sy1S
y2 ) = −1 (III.57)
The state we are considering is an example of an entangled state. The state cannot be written as a simple product.This confirms observation 1 since the particles always have perfect anti-correlation. Next we need to show thatquantum mechanics can produce observation 2. Consider the explicit pointer directions shown in figure 2:
120 deg
FIG. 2. The EPR pointer directions
n1 = z (III.58)
n2 =
√3
2x− 1
2z (III.59)
n3 =
√3
2x− 1
2z (III.60)
We measure the correlation of the spins dotted into one of the pointer directions:
∑ij
〈χ|S1 · niS2 · nj |χ〉 = 〈χ|S1 ·
0∑i
ni
S2 ·
0∑j
nj
|χ〉 (III.61)
= 0 (III.62)
Lecture 7, January 25th, 2012
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IV. GROUP THEORY FOR QUANTUM MECHANICS
A. Fundamentals
The uses of studying group theory is that you can study the effects of symmetry. A symmetry is if we can performsome transformation and the system doesn’t change. For example if we have a triangle
Then there are 6 symmetry transformations. Three reflection symmetries, 2 rotations (120o and 240o), as well as theidentity transformation.
Symmetries lead to conservation laws
〈α|A |b〉 → 〈a′|A′ |b′〉 = 〈a|U−1A′U |b〉 (IV.1)
If
〈a|A |b〉 = 〈a′|A′ |b′〉 (IV.2)
then we have an invariance. U−1AU = A if [A,U ] = 0. For example consider the time translation operator,U(t) = e−iHt/~. Operators that commute with U will not change with time. For example
[H,U ] = 0⇒ Energy conservation (IV.3)
Conservation of energy is a result of time translation invariance. Next consider the spatial transformation operator
T (x) = e−ixp/~ (IV.4)
Commutation with this operator implies momentum conservation. Invariance with respect to rotations leads toangular momentum conservation. Invariance gauge transformation leads to charge conservation.
U(1)→ Charge conservation (IV.5)
SU(2)→Weak charge conservation (IV.6)
SU(3)→ Colour conservation (IV.7)
Lecture 8 - January 27th, 2012Suppose R is the rotation operator and R leaves the Hamiltonian invariant. For example
R|`m〉 (IV.8)
leaves all the m states invariant. For example all the states with a given ` invariant (2px, 2py, 2pz) are degenerate.The mathematics of symmetry is Group Theory. A group G is a set with a rule for assigning to every ordered pair
a third element satisfying the following:
1. Closure: If f, g ∈ G then h = fg ∈ G
2. Associativity: For f, g, h ∈ G, f(gh) = (fg)h
3. Identity: There is an identity element e such that ef = fe = f (where f is every element in the set)
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4. Inverse: Every element f ∈ G has an inverse which we call f−1 such that ff−1 = f−1f = e
A group really is a “multiplication table” satisfying all these properties. The table specifies g1g2 ∀g1, g2 ∈ G. If thegroup elements are discrete, then we can write the multiplication table.
e g1 g2 ...
e e g1 g2 ...
g1 g1 g2g1 ... ...
g2 g2...
. . .. . .
......
... . . .. . .
A representation of G is a mapping of the elements of G onto a set of linear operators with the following properties.
1. Let D be the representation
D(e) = I (IV.9)
where I is the identity operators in the space where the linear operator act.
2.
D(g1)D(g2) = D(g1g2) (IV.10)
In other words the group multiplication law is mapped onto the natural multiplication in the linear space onwhich the operators act.
When we write
D(g) |α〉 (IV.11)
We mean that D is the representation, g is the group element, and |α〉 is the state.Consider for example the group Z3.
Def 1. A group is finite if it has a finite number of elements. Otherwise it’s infinite.
Def 2. The number of elements in a finite group is called the order of the group.
Z3 is of order 3:
e a b
e e a b
a a b e
b b e a
Note that Z3 is commutative (g1g2 = g2g1).
Def 3. Commutative groups are called Abelian groups
A representation of Z3 is
D(e) = 1
D(a) = e2πi/3
D(b) = e4πi/3
The dimension of the representation is the dimension of the space on which the representation acts. This representationis two dimensional. This representation clearly represents the table since
D(a)D(b) = e2πi/3e4πi/3 = e2πi = 1 (IV.12)
The rest of the relations are easy to show. Another representation for the same group is
D(e) =
1 0 0
0 1 0
0 0 1
; D(a) =
0 0 1
1 0 0
0 0 1
; D(b) =
0 1 0
0 0 1
1 0 0
This is a 3 dimensional representation. This representation has a special name. This is called the regular represen-tation. The trick to construct the regular representation is as follows.
16
1. First we identify the orthonormal basis on which the presentation acts with group elements.
|e1〉 ≡ |e〉 , |e2〉 ≡ |a〉 , |e3〉 ≡ |b〉 (IV.13)
2. Now we can construct the representation with
[D(g)]i,j = 〈ei|D(g) |ej〉 (IV.14)
The dimension of any regular representation is equal the number of group elements.
As an example we check D(a):
〈e1|D(a) |e1〉 = 〈e|a〉 (IV.15)
= 0 (IV.16)
〈e1|D(a) |e2〉 = 〈e|b〉 (IV.17)
= 0 (IV.18)
〈e1|D(a) |e3〉 = 1 (IV.19)
Do the rest in bundles:
〈e2|D(a) |e1,2,3〉 =
1 for e2
0 for e1 and e2(IV.20)
Assignments:Construct the multiplication table for Z2
1. Find the 1 dimensional representation
2. Find the 2D representation
3. Is 2 the regular representation
Lecture 9 - January 30th, 2012Recall the trick to find the regular representation. The reason it works is because the following
[D(g1g2)]i,j = [D(g1)D(g2)] (IV.21)
= 〈ei|D(g1)D(g2) |ej〉 (IV.22)
=∑k
〈ei|D(g1) |ek〉 〈ek|D(g2) |ej〉 (IV.23)
=∑k
Di,k(g1)Dk,j(g2) (IV.24)
We can transform the basis. A transformation on the states implies a transformation on the operators:
D(g)→ D′(g) = S−1D(g)S (IV.25)
The transformed operators (D′(g)) will have the same multiplication table so we say that D′ and D are equivalentbecause they only differ by a trivial change of basis.
Def 4. A representation is reducible if it has an invariant subspace. This means that the action D(g) on anyvector in the subspace is still in the subspace.
17
Every group has a trivial representation, D(g) = 1. e.g. for Z2,
D(e) = 1; D(a) = 1 (1D) (IV.26)
One can have a two dimensional representation:
D(e) =
(1 0
0 1
); D(a) =
(1 0
0 1
); (2D) (IV.27)
A third dimensional representation of Z2 is
D(e) =
1 0 0
0 1 0
0 0 1
; D(a) =
1 0 0
0 0 1
0 1 0
(IV.28)
Any action onto a vector
0
a
b
→ 0
a′
b′
stays inside the subspace. The same goes for the vector
c
0
0
→ c′
0
0
.
A representation is irreducible if it is not reducible. A representation is completely reducible if it is equivalentto a representation whose matrix elements have the following form:
D1(g) 0 ...
0 D2(g)...
... .... . .
(IV.29)
where Di(g) is irreducible. This is called block diagonal form. A representation in block diagonal form is said tobe the direct sum of the sub representations Dj(g):
D1 ⊕D2 ⊕ ... (IV.30)
e.g. Take our 3-D representation to Z3 and apply the similarity transformation
S =1
3
1 1 1
1 w2 w
1 w w2
(IV.31)
where w ≡ e2πi/3. With this process we find
D′(e) =
1 0 0
0 1 0
0 0 1
; D′(a) =
1 0 0
0 w 0
0 0 w2
; D′(b) =
1 0 0
0 w2 0
0 0 w
(IV.32)
If we act on any of the three vectors a
0
0
;
0
b
0
;
0
0
c
(IV.33)
The vectors will stay in the subspace hence this is an irreducible representation.In Quantum Mechanics the D(g) are unitary transformations. They map the Hilbert space to an equivalent one.
They reflect the symmetry of the problem if
[D(g), H] = 0 (IV.34)
This means we can always choose the energy eigenstates to transform like irreducible representations of the group.For example: Y00 doesn’t rotate so it transforms under the trivial representation. Y10 transform according to a 3dimensional representation (since |1,m〉 can only transform to |1,m′〉). Furthermore the Y2m transforms according to
18
a 5 dimensional representation.Consider the parity (reflection in a mirror) transformation, P
P2 = e (IV.35)
Hence
e Pe e PP P e
There is a trivial reprentation for this group, D(g) = 1 and a non-trivial representation, D(e) = 1 and D(P). If
[D(g), H] = 0 (IV.36)
e.g. 1D parity, x→ −x. Consider the harmonic oscillator potential. This does not means that all the eigenfunctionsare even. However it does mean that the even wavefunctions will transform trivially (they are already even) and theodd functions will transform according to the non-trivial representation.
Note if two groups have a different physical origin, e.g. Parity and rotations by π but have the same multiplicationtable we say the two groups are isomorphic.
Assignment (not to hand in): Construct the multiplication table for S3 (the group of permutations of 3 objects).For notational purposes call the elements, e, a, b, x, y, z
Lecture 10th - February 1st, 2012Consider the regular representation of Z2
D(e) =
(1 0
0 1
); D(a) =
(0 1
1 0
)(IV.37)
We can think of this D(a) as reflection in the y = x line (thinking of it as a 2D space).
• This suggests the similarity transformation that will reduce this reducible representation
• What about rotating by 45%? Try
S =1√2
(1 1
−1 1
)(IV.38)
D(a)→ D′(a) = S−1D(a)S =
(1 0
0 1
)(IV.39)
and
D(a) =
(1 0
0 −1
)(IV.40)
Consider the group S3. The elements are listed below. The notation is as follows the element (3, 1, 2) is element 1goes to position 2, element 2 goes to position 3 and element 3 goes to position 1.
e = (1, 2, 3)
a = (2, 3, 1)
b = (3, 1, 2)
x = (1, 3, 2)
y = (3, 2, 1)
z = (2, 1, 3)
19
Can think of these as the symmetry group of the equilateral triangle.
e→ Identity
a, b→ Rotation
x, y, z → Reflection
The multiplication table of this group is as follows
e a b x y z
e a b e z x y
b b e a y z x
x x y z e a b
y y z x b e a
z z x y a b e
We lost commutativity and hence this is a non-Abelian group. Notice that the top left corner of the table is closedwithin itself. Hence we say that e, a, b forms a subgroup. In fact this is the same group as Z3. Therefore thissubgroup is isomorphic with Z3. A two-dimensional representation of S3 is
D(e) =
(1 0
0 1
); D(a) =
(− 1
2 −√32√
32 − 1
2
); D(b) =
(− 1
2
√32
−√32 − 1
2
)
D(z) =
(−1 0
0 1
); D(x) =
(12
√32√
32 − 1
2
); D(y) =
(12 −
√32
−√32 − 1
2
)Since this is irreducible there is not similarity transformation that will diagonalize all the matrices in the representation.It is necessary that at least some of the irreps (irreducible representations) are matrices so that the non-commutativity(non-Abelian) can hold (numbers always commute!). Here’s a 3D rep (reducible representation) of S3
D(e) =
1 0 0
0 1 0
0 0 1
; D(a) =
0 0 1
1 0 0
0 1 0
; D(b) =
0 1 0
0 0 1
1 0 0
D(z) =
0 1 0
1 0 0
0 0 1
; D(x) =
1 0 0
0 0 1
0 1 0
; D(y) =
0 0 1
0 1 0
1 0 0
This particular represention (NOT the regular representation) is important because it is the defining representationfor the group (we call this the fundamental representation). It actually implements the permutation on the states.Note that a subscript on a representation given as
Dn (IV.41)
means the dimensionality of the representation. However a subscript in denoting the group:
Sm (IV.42)
means the number of objects the group is acting on. Consider the following where |1〉 denotes the states which arethe objects
D3(a) |1〉 =
3∑k=1
|k〉 [D3(a)]k,1 = |2〉 (IV.43)
D3(a) |2〉 = |3〉 =∑k
|k〉 [D3(a)]k,2 (IV.44)
20
D3(a) |3〉 = |1〉 =∑k
|k〉 [D3(a)]k,3 (IV.45)
This 3D representation decomposes into a direct sum of the irreducible representations
D3 = D1 ⊕D2 (IV.46)
The regular representation of Z2
(1 0
0 1
),
(0 1
1 0
)is reducible as
D2 = D1 ⊕D1 (IV.47)
Consider the following theorems
• All of the irreducible representations of a finite Abelian group are 1D. e.g. Z3, 3D → 1⊕ 1⊕ 1.
• If a Hermitian operator H, commutes with all the elements D(g) of a representation of a group G, then you canchoose the eigenstates of H to transform according to irreducible representations of G.
– If an irrep appears only once in the Hilbert space every state in the irrep is an eigenstate of with the sameeigenvalue.
Lecture 11th, February 2nd, 2012We have finished finite (point) groups. One can consider having shapes with more and more sides. Start with
a triangle then square... all the way to a circle. A circle is invariant under any rotation. This is an example of acontinuous group.
A rotation in 2-d can be represented by (cos θ sin θ
− sin θ cos θ
)(IV.48)
Because the parameter θ is continuous, the group is said to be continuous. We need to check that this is a group.This transformation clearly has an identity. Proof that it’s closed:(
cos θ sin θ
− sin θ cos θ
)(cosφ sinφ
− sinφ cosφ
)=
(cos θ cosφ− sin θ sinφ cos θ sinφ+ sin θ cosφ
− (cos θ sinφ+ sin θ cosφ) sin θ sinφ+ cos θ cosφ
)(IV.49)
=
(cos (θ + φ) sin (θ + φ)
− sin (θ + φ) cos(θ + φ)
)(IV.50)
This group clearly also has an inverse (if θ = −φ) you get the identity. This group is clearly Abelian (if we rotate oneway or another way the order doesn’t matter). This group is called SO(2). S denotes special, O denotes orthogonal,and 2 denotes 2D.
Consider multiplication by the complex numbers, eiθ. There is clearly an identity (θ = 0). The other properties are
eiθeiφ = ei(θ+φ)
eiθe−iθ = 1
This group rotates numbers in the complex plane. This group is called U(1) where U denotes unitary (U−1 = U†)and 1 denotes 1D. U(1) and SO(2) are clearly isomorphic (see figure 3). They have the same multiplication table,parameterized by 1 real number.
B. Lie Groups
Consider rotations in 3D, we could characterize these rotations with 3 parameters, e.g. θx, θy, θz where eachdefine a rotation about x, y, z respectively. Alternatively we can use the 3 Euler angles. The group members of
21
U(1) SO(2)
Im
Re Re
Im
FIG. 3. SO(2) and U(1) are isomorphic
rotations in 3 dimensions can be written as the set g(α) where α is a “vector” of 3 parameters. If the productg(α′′) = g(α)g(α′) then the group is a Lie group if α′′ = Γ (α, α′) where Γ is an analytic function. The groupelements depend smoothly on the set of parameters. By smooth we mean there is some notion of “closeness” on thegroup such that if two-elements of a group are “close together” in the space of the group the parameters that describethe elements are close. Another way to say this in a fancy-shmancy way is to say is that the group is a manifold. Amanifold looks like Rn locally. Thus in the neighborhood of the identity the group elements can be described by afunction of n real parameters. We now parameterize the group by
αa, a = 1, 2, ...n (IV.51)
g(α)
∣∣∣∣α=0
= e (IV.52)
Then if we find a representation we will parameterize the representation in a way such that
D(α)
∣∣∣∣α=0
= I (IV.53)
In some neighborhood of the identity we can Taylor expand (since it’s analytic) D(α)
D (δα) = I + iδαaXa (IV.54)
here we are using Einstein summation convention. The δαa are infinitesimal and the
Xa = −i ∂
∂αaD(a)
∣∣∣∣α=0
(IV.55)
The Xa are matrices for a = 1, 2...n (as many of them as the dimension of the group). These are called generatorsof the group.
On the first test, the first question will be to show that
Hψ = Eψ (IV.56)
Recall our previous discussion. The D(δα) form a group so we can multiply two elements
(I + iδαaXa) (I + δβbXb) (IV.57)
In particular we do this. We can write δαa = αak where k is large. Consider the following group element
limk→∞
(I + i
αakXa
)k= eiαaXa (IV.58)
Note the summation in the exponents. This is equal to the exponential by a definition of the exponential (not theTaylor expansion). Hence any finite representation can be written as an exponential.
22
C. Lie Algebra
Now in any particular direction the group multiplication is uncomplicated because we can parametrize our elementwith λ
U(λ) = eiλαaXa (IV.59)
U(λ1)U(λ2) = U (λ1 + λ2) = ei(λ1+λ2)αaXa (IV.60)
However it is complicated if we go into different directions.
eiαaXaeiβbXb 6= ei(αa+βa)Xa (IV.61)
but we have closure in the group and therefore
eiαaXaeiβbXb = eiγaXa (IV.62)
In other words the product is equal to some group element. We find that to next to lowest order, i.e. expanding bothsides of the equation above to next to lowest order
γc = αc + βc −1
2αaβbfabc (IV.63)
where
[Xa, Xb] = ifabcXc (IV.64)
where the fabc are called the structure constants of the group and [Xa, Xb] = ifabcXc is called the Lie Algebra forhe Lie Group. For unitary transformations the fabc are real.
The worry is that this is true to next to lowest order but we’ll need more to characterize group multiplication ateach order. The remarkable thing is that the fabc completely characterizes the group. The matrix generators satisfythe Jacobi identity
[Xa, [Xb, Xc]] = 0 (IV.65)
This is true for all cyclic permutations as well. The adjoint representation of an algebra is given by
[Ta]bc = −ifabc (IV.66)
i.e. the bc’th element of the a’th generator matrix is equal to −ifabc.The unitary group of order n, U(n) is the group associated with n× n unitary matrices. The matrices operate on
n− d complex vectors.
U†U = I (IV.67)
To parameterize a complex n× n matrix we require 2n2 numbers. The unitarity condition requires n2 constraints onthe matrix and therefore we are left with n2 free parameters. Therefore there are n2 generators.SU(n) is a subclass of U(n). The S stands for special which means that the determinant of U is +1. In other words
detU =∣∣eiθ∣∣ (IV.68)
Because U†U = I it implies that (I − iδαX†
)(I + iδαX) = I (IV.69)
⇒iδα(X† −X
)= 0 (IV.70)
X = X† (IV.71)
In other words the generators are Hermitian. Now we use a general property of matrices:
detA = exp Tr (lnA) (IV.72)
23
Consider
det (I + ε) = 1 + Trε (IV.73)
If we use that detU = 1 we have that
TrX = 0 (IV.74)
Thus the generators in SU(n) are traceless and Hermitian.
Lecture 13 - February 8, 2012U(n) is a unitary group. If we have SU(n) then |detU | = 1. U(n) has n2 generators while SU(n) has n2 − 1.U(n) and SU(n) have subgroups O(n) and SO(n). These are just the real versions of U(n) and SU(n). Thesetransformations act on real vectors and are said to correspond to the isometries in n-D (this means that they don’t
change the shape of objects). The n2 parameters of U(n) go down to n(n−1)2 for O(n) (O stands for the orthogonal
group). If
|detU | = 1 (IV.75)
for O(n) then we have detU = ±1 and this group is called SO(n). The unitarity condition for the orthogonal groupis just
OTO = OOT = 1 (IV.76)
Note that we often see this notation
U(n) = SU(n)⊗ U(1) (IV.77)
It means that U(n) is made up of SU(n) and U(1).Consider some particular groups.
U(1) : eiαQ (IV.78)
where Q is the charge operator. If we construct a Hamiltonian invariant under this U(1) then it implies that we havecharge conservation. We can generalize the above in the following way
U(1) : eiα(x)Q (IV.79)
The parameter controlling the phase change depends on x. In other words it depends on position in space. Hamil-tonians can be constructed that are invariant under U(1) = eiα(x)Q. This is called a local or gauge symmetry. Inorder to construct such a Hamiltonian we require a gauge field (which predicts the existence of the photon). Historydidn’t work this way for electromagnetism (the gauge symmetry was found after). Physicists later constructed gaugeHamiltonians and group was used to predict W+,W−, Z, and g.
Consider the SU(2) group. It acts on 2−D complex vectors. The defining representation is 2−D. In other wordstwo by two matrices. There are n2 − 1 generators and since n = 2 we have 3 generators. There are three directionsyou can go on the group manifold. Recall that the generators are traceless and hermitian. This can be encoded bywriting a general representation as (
a b∗
b −a
)(IV.80)
There are three parameters here: a,Re(b), Im(b). A non-unique but suitable representation are the set of generators
Ji =σi2
(IV.81)
where
σ1 =
(0 1
1 0
); σ2 =
(0 −ii 0
); σ3 =
(1 0
0 −1
)(IV.82)
and we know the structure constants since
[Ji, Jj ] = iεijkJk (IV.83)
where εijk is the Levi-Cevita symbol. Here we have εijk = fijk.
24
Def 5. A Casimir operator is an operator which commutes with all of the generators.
For SU(2) we need a Casimir such that
[ , J1] = [ , J2] = [ , J3] = 0 (IV.84)
One such operator is J2. The importance of a Casimir operator is that we can construct simultaneous eigenstates ofJ2 and Ji. It’s easy to show that (try at home) that
[J3, J±] = ±J± (IV.85)
where J± = J1 ± J2. We define eigenstates |j,m〉 such that
J2 |j,m〉 = j (j + 1) |j,m〉 (IV.86)
and
J3 |j,m〉 = m |j,m〉 (IV.87)
Recall from quantum mechanics that J+ and J− generate different states within a multiplet. In other words
J3J± |j,m〉 = (m± 1) J± |j,m〉 (IV.88)
Furthermore
J2J± |j,m〉 = j (j + 1) J± |j,m〉 (IV.89)
and
J± |j,m〉 =√
(j ∓m)(j ±m+ 1) |j,m± 1〉 (IV.90)
We use these results to form a two dimensional representation of SU(2) by choosing∣∣∣∣12 , 1
2
⟩=
(1
0
)(IV.91)
∣∣∣∣12 ,−1
2
⟩=
(0
1
)(IV.92)
We can now find the matrix elements of J3: ⟨1
2,
1
2
∣∣∣∣ J3 ∣∣∣∣12 , 1
2
⟩=
1
2(IV.93)⟨
1
2,−1
2
∣∣∣∣ J3 ∣∣∣∣12 , 1
2
⟩= 0 (IV.94)⟨
1
2,
1
2
∣∣∣∣ J3 ∣∣∣∣12 ,−1
2
⟩= 0 (IV.95)⟨
1
2,−1
2
∣∣∣∣ Je ∣∣∣∣12 ,−1
2
⟩= −1
2(IV.96)
Hence
J3 =1
2
(1 0
0 −1
)(IV.97)
Now we find the matrix elements of J+. Define ↑ and ↓ as usual. We use the following
J+ |↓〉 =
√(1
2+
1
2
)(1) = 1 (IV.98)
25
〈↑| J+ |↑〉 = 0 (IV.99)
〈↑| J+ |↓〉 = 1 (IV.100)
〈↓| J+ |↑〉 = 0 (IV.101)
〈↓| J+ |↓〉 = 0 (IV.102)
J+ =
(0 1
0 0
)(IV.103)
For J− = J†+ we get
J− =
(0 0
1 0
)(IV.104)
It’s now easy to find J1 and J2 using J± = J1 ± iJ2:
J1 =1
2
(0 1
1 0
); J2 =
1
2
(0 −ii 0
)(IV.105)
So we recover Ji = σi2 .
Lecture 14, February 10th 2012We can construct a 2-D representation of SU(2). The SU(2) generators are σi
2 and therefore the SU(2) group elementsare
U = eiθiσi/2 (IV.106)
(here we use summation notation), where θi are continuous parameters. Note that in U = eiθiσi/2 is implicit butit turns out that in SU(2) we can construct the explicit matrix. For convenience we write θi ≡ 2ωni where ni is anormal vector (determines how much of θi is in the direction ni. With this change in notation we write
U = eiωn·σ = I + iωn · σ − ω2 (n · σ)
2+ ... (IV.107)
To simplify this quantity we work out the following
(n · σ)2
= (n1σ1 + n2σ2 + n3σ3)2
(IV.108)
= n21 + n22 + n23 + n1n2σ1σ2 + n1n3σ1σ3 + n2n3σ2σ3 + n1n2σ2σ1 + n1n3σ3σ1 + n2n3σ3σ2 (IV.109)
= I (IV.110)
where we used σi, σj , i 6= j. We can now simplify U :
U = I + iωn · σ − ω2
2− iω3 n · σ
2+ ... (IV.111)
=
1− ω2
2+ω4
4!...
+ i
ω − ω3
3!+ω5
5!+ ...
n · σ (IV.112)
= cosω + i sinω (n · σ) (IV.113)
= cos
(θ
2
)+ in · σ sin
(θ
2
)(IV.114)
Now we work out the matrix
n · σ = n1
(0 1
1 0
)+ n2
(0 −ii 0
)+ n3
(1 0
0 −1
)(IV.115)
=
(n3 n1 − in2
n1 + in2 −n3
)≡=
(n3 n−n+ −n3
)(IV.116)
Thus we have explicitly: (cos(θ2
)+ n3i sin
(θ2
)in− sin
(θ2
)in+ sin
(θ2
)cos(θ2
)− in3 sin
(θ2
) ) (IV.117)
26
D. SU (3)
SU(3) is the group of Special unitary transformations acting on 3D complex “vectors”. For SU(n) there are n2− 1generators and therefore for SU(2) there are 8 generators:
X1, X2, ...X8 (IV.118)
One possible choice for these generators Xi = λi2 where the λi are the Gell-man matrices:
λ1 =
0 1 0
1 0 0
0 0 0
; λ2 =
0 −i 0
i 0 0
0 0 0
; λ3 =
1 0 0
0 −1 0
0 0 0
λ4 =
0 0 1
0 0 0
1 0 0
; λ5 =
0 0 −i0 0 0
i 0 0
; λ6 =
0 0 0
0 0 1
0 1 0
λ7 =
0 0 0
0 0 −i0 1 0
; λ8 =1√3
1 0 0
0 1 0
0 0 −2
These matrices obey
Tr (λiλj) = 2δij (IV.119)
They obey the Lie algebra relation
[Xi, Xj ] = fijkxk (IV.120)
However the fijk are not simple as in SU(2).Nature has an approximate SU(3) global symmetry, flavour: u
d
s
(IV.121)
but nature also has an exact SU(3) local (gauge) symmetry, colour: q1
q2
q3
(IV.122)
Lecture 15, February 13th, 2012Recall that for SU(2):
cosθ
2+ i (n · σ) sin
θ
2(IV.123)
The algebra is defined by [σi2,σj2
]= iεijk
σk2
(IV.124)
The defining representation is 2D. Consider SO(3):
[Li, Lj ] = iεijkLk (IV.125)
and the defining representation is 3D. SU(2) and SO(3) have the same Lie Algebra and they are locally isomorphic.For this to be true the group manifold must have the same dimension (3D). For SU(2)
θ = 0→ I
θ = 2π → −I
27
If you have some angle θ then θ and θ+ 2π gives minus the original transformation. This doesn’t occur in SO(3) (weknow that since we are familiar with orbital angular momentum).
In fancy shamncy words we say: The groups are locally isomorphic but the global topology of the two groups isdifferent. In group theory language we say
SO(3)→ SU(2)
Z2(IV.126)
V. ACCIDENTAL DEGENERACIES
Accidental degeneracies mean there are degeneracies in the spectrum in a QM system that are unexpected. Howeverthere is no such thing as a true accident so this is just a misnomer. We will consider Hydrogen. Consider theHamiltonian:
H =p2
2m− e2
r(V.1)
We see that this is rotationally invariant (nothing in this problem is picking out a direction). I.e.
[SO(3)(g), H] = 0 (V.2)
where SO(3)(g) is a rotation matrix. Therefore the eigenstates of H transform according to irreducible representationsof SO(3). Also note invariance is O(3) which includes Parity transformations (det = −1). Therefore eigenstates ofthe Hamiltonian will also be irreducible representations of Parity. The 1s forms a 1D irrep of SO(3), 2P is a 3D irrepof SO(3), 3d forms a 5D irrep of SO(3). This again is sloppy language the better way to say this is to say that thestates transform among each other under SO(3) if states transform among each other they are transformed by anirrep (i.e. block matrix). These are the expected degeneracies.
However in practice we see that the 2s is also degenerate which 2p and 3s, 3p are all degenerate with 3d, etc. Thesedegeneracies are called accidental. There is way more degeneracy then we expect. There must be more symmetrythen we naively think. Since symmetries produce conservation laws there must be some conservation law we are notseeing. Before we look at quantum mechanics can we see something in the classical problem? The classical problemis the Kepler problem where
E =p2
2m− GmM
r(V.3)
When we solve this problem we find the orbits are elliptical. The orbitals are shown in figure 4. We also notice that
Ellipses
FIG. 4. The elliptical orbit of the Kepler problem
the angular momentum is conserved and hence the orbit is planar. We summarize the interesting points below:
1. Orbits are ellipses
2. Angular momentum in conserved
3. Orbit is planar
28
4. Orbit closes (no precession)
The major axis points in the same direction are all times and has the same magnitude. We are getting hints of anotherconserved quantity. If the potential was not 1
r but 1r1+ε the orbit does not close! In fact in the real solar system the
orbit does not close. They almost do but they in fact don’t. The most famous example of this is Mercury. Wherethe parahelien precesses. The orbit of Mercury precesses because
1. There are other planets which also exert a force on Mercury (classical Newtonian effect)
2. Einstein’s theory of General Relativity modifies the 1r potential
Now back to quantum mechanics, we define a new quantity called
M =p× L
µ− e2
rr (V.4)
This quantity is a conserved quantity. To show that this is conserved we need to show that [M, H] = 0. Howeverthere is a problem, p and L don’t commute and M is not Hermitian. To make this Hermitian we do the following:
M =1
2µ(p× L− L× p)− e2
rr (V.5)
In assignment 9 we will show that [M, H] = 0. We will also show that L ·M = M · L = 0 as well as M2 =2Hµ
(L2 + ~2
)+ e4.
Lecture 16, February 15th, 2012The vector M is always perpendicular to L. This is shown in figure 5. We have the commutation relations
M
L
FIG. 5. The direction of the vector M
[Li, Lj ] = iεijk~Lk (V.6)
[Li,Mj ] = i~εijkMk (V.7)
[Mi,Mj ] 2i~µHLkεijk (V.8)
Notice that [Li, Lj ] closes. It is just SO(3). Further note that [Li,Mj ] and [Li, Lj ] close. However [Mi,Mj ] doesn’tclose! Since H is there. However
[H,M] (V.9)
[H,L] (V.10)
So if we restrict ourselves to subspace that corresponds to a particular eigenvalue of H we can replace H by theeigenvalue E. If you replace H by a number then the relations all stay within M,L! Therefore in the subspace wehave closure. This implies that we have a Lie Algebra in the subspace. Thus we can write the commutator (it’sunderstood that we are staying in this subspace for which this is true)
[Mi,Mj ] = −2i~µELkεijk (V.11)
29
Therefore the algebra of all the Li and Mi close. For convenience we rescale M
M′ =(− µ
2E
)1/2M (V.12)
Remember that for bound state spectrum we have E < 0. Now r = (r1, r2, r3) and p = (p1, p2, p3) satisfy
[ri, pj ] = i~δij (V.13)
where i, j = 1, 2, 3. Lets extend i, j to 1, 2, 3, 4. In other words add r4, p4. We define
Lij = ripj − rjpi (V.14)
For example In this notation
L23 = r2p3 − r3p2 = Lx (V.15)
We call
M ′i = Li4 = rip4 − r4pi (V.16)
If we do this and still use [ri, pj ] = i~δij then you retrieve all the commutation relations of Li and Mi. Note thatr4 and p4 are NOT time and energy. r4 and p4 are artificial constructs. This is not 4D.
[Li, Lj ] = i~εijkLk (V.17)
is the Lie Algebra of SO(3). (L23, L31, L12) = (Lx, Ly, Lz) but (L14, L24, L34) = (M1,M2,M3). We know that SO(3)
is a subgroup and we also know that there are 6 generators. We also know that in SO(n) there are n(n−1)2 generators.
In fact we have the Lie Algebra of SO(4). We have rotations in 4D (but it’s not real 4D, just 3 real dimensions and1 artificial). We can construct the generators
Lnm = |n〉 〈m| − |m〉 〈n〉 (V.18)
Thus
L12 = |1〉 〈2| − |2〉 〈1| (V.19)
Warning: L12 is not a matrix element. This is the operator. Using the basis
|1〉 =
1
0
0
0
; |2〉 =
0
1
0
0
; ... (V.20)
we haveBy finding the matrix elements we have
L12 =
0 1 0 0
−1 0 0 0
0 0 0 0
0 0 0 0
(V.21)
We now find the commutator
[L23, L31] = (|2〉 〈3| − |3〉 〈2|) (|3〉 〈1| − |1〉 〈3|)− (|3〉 〈1| − |1〉 〈3|) (|2〉 〈3| − |3〉 〈2|) (V.22)
= |2〉 〈1| − |1〉 〈2| (V.23)
= −L12 (V.24)
= −Lz (V.25)
30
eiθL12 = 1 + θL12 + θ2L212
2!+ θ3
L312
3!(V.26)
= 1 +
0 θ 0 0
−θ 1 0 0
0 0 1 0
0 0 0 1
+θ2
2
−1 0 0 0
0 −1 0 0
0 0 0 0
0 0 0 0
+θ3
3!
0 −1 0 0
1 0 0 0
0 0 0 0
0 0 0 0
+ ... (V.27)
=
cos θ sin θ 0 0
− sin θ cos θ 0 0
0 0 1 0
0 0 0 1
(V.28)
In a fancy shancy way we can say
SO(4) =SU(2)⊗ SU(2)
Z2
We can define
I =1
2(L + M) , K =
1
2(L−M) (V.29)
Thus we get
[Ii, Ij ] = i~εijkIk (V.30)
[Ki,Kj ] = i~εijkKk (V.31)
and
[I,K] = [I, H] = [K, H] = 0 (V.32)
We have two casimir operators
I2 = i (i+ 1) ~2; K2 = k (k + 1) ~2 (V.33)
where i, k = 0, 12 , 1, .... There are two Casimir Invarients
C = I2 +K2 =1
2
(L2 +M
′2)
; C ′ = I2 −K2 = L ·M′ = 0 (V.34)
Hence
I2 = K2 (V.35)
which implies that i = k. Therefor the eigenvaleus of
C = I2 +K2 = 2I2 = 2k (k + 1) ~2 (V.36)
where k = 0, 12 , 1, .... We just said
C =1
2
(L2 +M
′2)
(V.37)
=1
2
(L2 − µ
2EM2)
(V.38)
=1
2
(L2 − µ
2E
(L2 + ~2
)+ e4
)(V.39)
=1
2
(−~2 − µ
4Ee4)
(V.40)
= −µe4
4E− ~2
2(V.41)
31
but C also has values of 2k (k + 1) ~2, k = 0, 12 , 1, .... Therefore
2k (k + 1) ~2 = −µe4
4E− ~2
2(V.42)
~2(
2k2 + 2k +1
2
)= −µe
4
4E(V.43)
~2 (2k + 1)2
= −µe4
2E(V.44)
E = −1
2µ
e4
~2 (2k + 1)2 (V.45)
where 2k + 1 = n where n is an integer. So we can finally write
E = −1
2
µ
~2e4
n2(V.46)
where the degeneracy is n2.
Lecture 17 - February 27th ,2012Test 1 consists of (the marks are in brackets)
1. Hydrogen eigenstates (6)
2. EPR (4)
3. Density matrix (6)
4. Groups (18)
• Know your definitions
The test is in TEL 0005, from 9:30-11:30.
VI. DIRAC EQUATION
This is the subject of relativistic quantum mechanics. We first consider the Lorentz group. In SO(3), the quantityr2 = x2 + y2 + z2 is invariant under rotations. In other words
Rr2 = r′2 = x
′2 + y′2 + z
′2 = r2 (VI.1)
In special relativity the invariant quantity is the length of the 4-vector (ct, x, y, z). In other words
c2t2 − x2 − y2 − z2 = c2t′2 − x
′2 − y′2 − z
′2 (VI.2)
This quantity measures the space-time distance between space -time events. Notation:
xµ = (t, r) = (t, x, y, z) =(t, ri
)(VI.3)
where from now on we use units such that c = 1. We also have
xµ = (t,−r) = (t,−x,−y,−z) =(t,−ri
)(VI.4)
We call xµ a contravariant quantity (i.e. a quantity which transforms as the position 4-vector) and we call xµ a
covariant quantity (quantity that transforms like the derivative of a 4-vector, ∂∂xµ ). Therefore we have
x2 = xµxµ (VI.5)
32
where here Einstein summation notation is used. We can also write
x2 = xµxµ = gµνxµxν (VI.6)
where gµν is called the metric tensor. In special relativity we have
gµν =
1 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 −1
(VI.7)
In general relativity gµν becomes a dynamical quantity (it satisfies (Einstein) field equations) and may have timedependence. A Lorentz transformation (LT) is a transformation that leaves the length of 4 − vectors invariant. Wewrite this as
X′µ = ΛµνX
ν (VI.8)
or
Xµ =(Λ−1
)µνX
′ν (VI.9)
we require that
X′2 = X2 (VI.10)
but we have
X′2 = X
′µgµνX′ν = ΛµαX
αgµνΛνβXβ (VI.11)
X2 = gαβXαXβ (VI.12)
Hence we demand that
gαβ = ΛµαgµνΛνβ (VI.13)
To rewrite this consider
gµνΛβν = gΛ (VI.14)
and
Λναgµν = ΛT g (VI.15)
Therefore our condition can be written in matrix notation as
g = ΛT gΛ (VI.16)
We will specialize to proper LT (LT without parity inversion). We will also specialize to orthochronous LT (LTwithout time reversal). These are the physical transformations. Hence we can be in the neighborhood of the identity:
Λ = 1 + ελ (VI.17)
We insert this into our condition
g =(1 + ελT
)g (1 + ελ) (VI.18)
g = g + ε(λT g + gλ
)+ ε2λTλ (VI.19)
To first order we have λT g + gλ = 0. Alternatively we can write
ΛT = −gλg (VI.20)
33
since g2 = 1. Consider the matrix:
λ =
λ11 λ12 λ13 λ14λ21 λ22 λ23 λ24λ31 λ32 λ33 λ34λ41 λ42 λ43 λ44
(VI.21)
Then using the equation above we have
λ =
λ11 λ21 λ31 λ41λ12 λ22 λ32 λ42λ13 λ23 λ33 λ43λ14 λ24 λ34 λ44
=
−λ11 λ12 λ13 λ14λ21 −λ22 −λ23 −λ24λ31 −λ32 −λ33 −λ34λ41 −λ42 −λ43 −λ44
(VI.22)
Clearly we have λii = −λii and hence each of these is zero. Furthermore λ12 = λ21, λ13 = λ31, λ14 = λ41. The restof the relations are λ23 = −λ32, λ42 = −λ24, .... Initially we required 16 parameters to specify this matrix. Howeverthese conditions show that there 16− 4− 3− 3 = 6 free parameters. The number of generators of the group are equalto the number of free parameters. Hence there are 6 generators of the LT group. To specify the generators we use theconventions that (1, 2, 3, 4) → (0, 1, 2, 3) and we called the generators ωµν where µ, ν = 0, 1, 2, 3. Note: DO NOTconfuse µ, ν with matrix indices, they are labeling a generator.
ω10 =
0 1 0 0
1 0 0 0
0 0 0 0
0 0 0 0
; ω20 =
0 0 1 0
0 0 0 0
1 0 0 0
0 0 0 0
; ω30 =
0 0 0 1
0 0 0 0
0 0 0 0
1 0 0 0
ω12 =
0 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 0
; ω23 =
0 0 0 0
0 0 0 0
0 0 0 −1
0 0 1 0
; ω13 =
0 0 0 0
0 0 0 −1
0 0 0 0
0 1 0 0
This can be written as
(ωµν)αβ = −1
2εµνλσε
λσαβ (VI.23)
where εµνλσ is the 4D Levi-Cevita Symbol. One can show that
ω01 = ω10, ω21 = −ω12 (VI.24)
The most general infinitesimal LT can be written
Λ = 1 + ξiωi0 +1
2θiεijkωjk (VI.25)
This group contains 6 parameters θi, i = 1, 2, 3 and ξi, i = 1, 2, 3. Therefore a finite LT can be written
Λ = exp
(ξiωi0 +
1
2θiεijkωjk
)(VI.26)
This is called the Lorentz group, denoted by SO (1, 3).
Lecture 18 - March 2nd, 2012Consider the LT for ξi = θi = 0 for all i except for θ3.
Λ(θ3) = exp
(1
2θ3ε3jkωjk
)(VI.27)
= exp
(1
2−ω21θ3 + ω12θ3
)(VI.28)
= exp
(1
2−2θ3ω21
)(VI.29)
= exp (ω12θ3) (VI.30)
34
we have
ω12 =
0 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 0
(VI.31)
(ω12)2
=
0 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 0
0 0 0 0
0 0 −1 0
0 1 0 0
0 0 0 0
=
0 0 0 0
0 −1 0 0
0 0 −1 0
0 0 0 0
(VI.32)
Since this is diagonal this is easy
(ω12)2
=
0 0 0 0
0 0 1 0
0 −1 0 0
0 0 0 0
= −ω12 (VI.33)
(ω12)4
=
0 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0
(VI.34)
and then the sequence repeats,i.e. ω512 = ω12 etc. We can now find Λ explicitly:
Λ (θ3) = I + θ12 +θ22ω212 +
θ3
3!ω312 + ... (VI.35)
=
0 0 0 0
0 1− θ2
2 −θ + θ3
3! + ... 0
0 θ + θ3
3! + .... 1− θ2
2 0
0 0 0 0
(VI.36)
=
1 0 0 0
0 cos θ − sin θ 0
0 sin θ cos θ 0
0 0 0 1
(VI.37)
We see that θ3 gives a rotation about the 3 axis. Since there is nothing special about θ3, θ1 and θ2 give us rotationsabout the x and y axes. If we set all the ξi = 0 then we have the group SO (3). Now lets set all parameters to zeroexcept ξ1.
Λ (ξ1) = exp (ξ1ω10) (VI.38)
Recall that
ω10 =
0 1 0 0
1 0 0 0
0 0 0 0
0 0 0 0
Hence we have
ω210 =
1 0 0 0
0 1 0 0
0 0 0 0
0 0 0 0
35
and hence
ω310 = ω10
Hence we have
Λ (ξ) = I + ξω10 +ξ2
2(ω10)
2+ξ3
3!(ω10)
3+ ... (VI.39)
=
1 + ξ2
2 + ... ξ + ξ3
3 + ... 0 0
ξ + ξ3
3 + ... 1 + ξ2
2 + ... 0 0
0 0 1 0
0 0 0 1
(VI.40)
=
cosh ξ sinh ξ 0 0
sinh ξ cosh ξ 0 0
0 0 1 0
0 0 0 1
(VI.41)
Hence under the action of the Lorentz Transformation, Λ (ξ1) we see thatt′
x′
y′
z′
=
cosh ξ sinh ξ 0 0
sinh ξ cosh ξ 0 0
0 0 1 0
0 0 0 1
t
x
y
z
(VI.42)
Explicitely we have
t′ = cosh ξt+ sinh ξx
x′ = sinh ξt+ cosh ξx
y′ = y
z′ = z
If we define ξ = tanh−1 v which is called the rapidity. This gives
v = tanh ξ
v2 = tanh2 ξ
= 1− sech2ξ
= 1− 1
cosh2 ξ
cosh2 ξ =1
1− v2
We define γ = cosh ξ. We can write
tanh ξ =sinh ξ
cosh ξ=
v/c√1− v2/c2
and we get
t′ = γ (1 + βx)
x′ = γ (1 + βct)
We see that we have a subgroup of boosts. So all together the Lorentz group has 3 rotations and 3 boosts. If wechoose a more familiar parameterization
U(θ) ≈ 1 +i
2θµνJ
µν (VI.43)
36
where we relate these Jµν ’s to our angular momentum operators by
Ji =1
2εijkJjk (VI.44)
and we define
Ki = J0i (VI.45)
This gives
[Ji, Jj ] = iεijkJk (VI.46)
[Ji,Kj ] = iεijkKk (VI.47)
[Ki,Kj ] = −iεijkJk (VI.48)
This forms the algebra of SO(3, 1)
Lecture 19 - March 5th, 2012We have finished with the Lorentz group. Now we move on to the Dirac equation. The Schrodinger equation is givenby (
p2
2m+ V
)ψ = Eψ (VI.49)
(K + V )ψ = Eψ (VI.50)
where K is the kinetic energy operator, V is the potential energy operator and E is the energy operator. This
equation is Galiliean covariant but obviously not Lorentz covariant. This is because E = p2
2m +V is not the relativisticEnergy-momentum (dispersion) relation. The relativistic energy-momentum relation is what we get from constructinga Lorentz-invariant quantity. This quantity is p2, where pµ is the Lorentz 4-vector equal to pµ =
(p0,p
)= (E,p).
p2 = pµpµ = gµνpµpν (VI.51)
p2 = E2 − p2 = m2 ⇒ E2 = p2 +m2 (VI.52)
Thus
E =(m2 + p2
)1/2(VI.53)
If we demand the correct dispersion relation then we would have
i~∂
∂tψ =
(m2 − ~2∇2
)1/2ψ (VI.54)
= m
(1− ∇
2
2m− ∇
4
8m4+ ...
)(VI.55)
Even though this equation works many problems are encountered using this method and this route was not historicallytaken. Alternatively we can not taken the square root:
E2 = m2 + p2 (VI.56)
where E2 =(i~ ∂∂t
)2= −~ ∂2
∂t2 . Recall that p2 = −~ ∂2
∂x2 . We now introduce useful notation that says
= gµνµν (VI.57)
with this we have ( +m2
)ψ = 0 (VI.58)
This is called the Klien Gordan equation. Solutions to this equation take the form
φ±n = Nei(k·r−Ent) (VI.59)
where En =√m2 + k2n. However there two problems with this solution
37
• Some states turn out to have a negative norm.
• This generates states with negative energy
The origin of the negative energy can traced back to the fact that we have the square of the energy. Is there anequation that is first order in time AND first order in space so that one can have manifest Lorentz symmetry? Thisis a fancy way of saying symmetry between time and space. With this in mind Dirac wrote
i~∂
∂tψ = Hψ = (α · p + βm)ψ (VI.60)
where α and β are at this point two unknowns. However we do know that
E2 = p2 +m2 (VI.61)
(we demand this to be true). We now implement this on our equation(i~∂
∂t
)2
= (α · p + βm)2ψ (VI.62)
=(p2 +m2
)ψ (VI.63)
This requires
(α · p + βm)2
= p2 +m2 (VI.64)
(αipi + βm)2
= p2 +m2 (VI.65)
αipiαjpj + αiβpim+ βαipim+ β2m2 = p2 +m2 (VI.66)
(VI.67)
This tells us that α and β can’t be just numbers because in order to get the two terms first order in p to be zero werequire that
αiβ + βαi ≡ αi, β = 0 (VI.68)
Continuing we can write the left hand side as
α2i p
2i +
1
2αj , αi pipj + αj , βmpj + β2m2 (VI.69)
Now demanding that this equal p2 +m2 we must have
α2i = 1; αi, αj = 2δij ; αi, β = 0; β2 = 1
We have the following Lemmas:
1. β and αi are traceless:
αiβ + βαi = 0 (VI.70)
αiβ = −βαi (VI.71)
βαiβ = −β2αi (VI.72)
βαiβ = −α2i (VI.73)
Tr (βα, β) = −Tr (αi) (VI.74)
Tr (αi) = −Tr (α) (VI.75)
Hence the αi are traceless (it’s similar to prove that β is traceless) but also we have α2i = 1 and β2 = 1.
2. Now we can diagonalize the αi and β. If we have a diagonal matrix that the square is equal to the identity wehave
x
x
x
x
=
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
(VI.76)
This requires that we have eigenvalues of ±1.
38
3. The matrices are n × n where n is even (since the eigenvalues must add to 0 and they are ±1)
The dimensionality of the matrices is even. Is it possible for n to equal 2? There are n2 independent Hermitian matrices(n2 parameters). However the identity is not traceless so it can’t be a candidate for αi or β. Thus subtracting theidentity gives n2 − 1 matrices. Thus we have n2 − 1 = 3 possible independent matrices. However we need at least 4matrices to have one for all of αi, β. n = 4 gives n2 − 1 = 15 matrices which is more then enough.
Conclusion: We can find 4 × 4 independent Hermitian matrices that satisfy all our requirements. Since n = 4gives many matrices we can choose which we want to use. We make the choice:
β =
(1 0
0 −1
); αi =
(0 σiσi 0
)(VI.77)
where each of these entries is a 2 by 2 entry.
Lecture 20, March 7th, 2012As it stands this equation doesn’t look manifestly covariant (though it is). We define
γµ =(β, βαi
)(VI.78)
where γ0 = β and γi = βαi. Thus
γ0 =
(1 0
0 −1
); γi =
(0 σi−σi 0
)(VI.79)
This gives the anticommutation relation:
γµ, γν = gµν (VI.80)
This gives (iγµ
d
dxν−m
)ψ = 0 (VI.81)
Note that ψ has the form
ψ1
ψ2
ψ3
ψ4
, but it is not a four-vector! It is a four-spinor. We will learn later how ψ transforms
under rotations and boosts (i.e. LT).Recall that the Klien-Gordan equation had problems. One problem was there were states with negative norm andthe other problem was that it had states of negative energy. We shouldn’t have negative states because if there werethen electrons would fall to these negative energy states emitting energy which we don’t observe.
Consider the following
i∂
∂tψb = Hψb ⇒ ψ†i
∂
∂tψb = ψ†aHψb (VI.82)
and
− i ∂∂tψ†a = (Hψa)
† ⇒ −i ∂∂tψ†aψb = (Hψa)
†ψb (VI.83)
subtracting these equations gives
i∂
∂t
(ψ†aψb
)− iψ†aαi
[−→∇i +
←−∇i]ψb (VI.84)
where the arrows denote direction over which they act. This gives
i∂
∂t
(ψ†aψb
)+ i∇i
(ψ†aαiψb
)= 0 (VI.85)
39
i∂
∂xµ[ψ†aγ
0γµψb]
= 0 (VI.86)
If we define ψ†γ0 ≡ ψ we have
i∂
∂xµ[ψaγ
µψb]
= 0 (VI.87)
Comparing with the known equation ∂∂xµ J
µ = 0, this is just the continuity equation! Recall in electromagnetism wehave
Jµ = (ρ,J) (VI.88)
In electromagnetism we have
∂
∂tρ = −∇ · J (VI.89)
Integrating over some volume
∂
∂t
∫ρdV =−
∫V
∇ · JdV (VI.90)
∂
∂tQ = −
∫S
J · dA (VI.91)
Back to Quantum mechanics. If charge was conserved before what is conserved now? What is this object
ψγµψ ? (VI.92)
This is the probability current. Since this satisfies the continuity equation probability is conserved. Therefore normis conserved. The Dirac equation can be written in covariant form as(
iγµ∂
∂xµ−m
)ψ = 0 (VI.93)
This is the free Dirac equation (no potential), written in covariant form. To make it non-free we need to add a potential.However the idea of a potential is non relativistic. The way we add interactions is by adding a four-potential:[
γµ(i∂
∂xµ− eAµ
)−m
]ψ = 0 (VI.94)
where Aµ = (φ,A). Note that the Dirac equation is mainly good for electromagnetism. We try free-Dirac equation
solutions of the form ψ±p (r, t) = Npep·r∓Ept
(χ
η
). χ and η are 2×1 column vectors. ± refer to positive and negative
energy solutions Ep > 0. We substitute the positive energy solution into the free-Dirac equation:(iγµ
∂
∂xµ−m
)Npe
p·r−Ept
(χ
η
)(VI.95)
Lecture 21 -Missed a lecture
Lecture 22 - March 12th, 2012We have the Dirac equation (
iγµ∂
∂xµ−m
)ψ = 0 (VI.96)
40
We try the solution ψ+(r, t) = Ne−ip·x
(χ
η
)(not ψ† but ψ+, where p · x = pµxµ = p0x0 − p · x = Et − p · x. We
found a solution terms of arbitrary χ but we require
η =σ · pE +m
χ (VI.97)
For normalization we require ∫ψ†ψd3r =
∫ (χ†η†
)( χ
η
)N2 (VI.98)
= N2(χ†η2
)( χ
η
)L3 (VI.99)
= N2L3(χ†χ+ η†η
)(VI.100)
= N2L3
(χ†χ+ χ†
(σ · p)2
(E +m)2χ
)(VI.101)
but (σ · p)2
= p2 (see notes from previous lecture). Thus we have (σ·p)2
(E+m)2= p2
(E+m)2. Inserting in this result gives
∫d3rψ†ψ = N2L3
(χ†χ+ χχ†
p2
(E +m)2χ
)(VI.102)
Since χ is just a number we have ∫d3rψ†ψ = N2L3χ†χ
((E +m)
2+ p2
(E +m)2
)(VI.103)
= N2L3χ†χ
(2E2 + 2Em
(E +m)2
)(VI.104)
= N2L3 2E
E +mχ†χ (VI.105)
We will choose χ†χ = 1. Thus we require
N =
(E +m
2EL3
)1/2
(VI.106)
It is conventional to choose
u (p, s) =√E +m
(1
σ·pE+m
)χs (VI.107)
where χ1/2 =
(1
0
)and χ−1/2 =
(0
1
). Note that more explicitly we have
u (p, s) =√E +m
χs 0 0 0
0 χs 0 0
0 0 χs
E+mpzχs
E+m (px − ipy)
0 0 χs
E+m (px + ipy) − χs
E+mpz
(VI.108)
Putting this together we have
ψ+p,s =
1√2EL3
u(p, s)e−p·x (VI.109)
41
similarly we have
ψ−−p,−s(x) =1√
2EL3v(p, s)eip·x (VI.110)
where v(p, s) =√E +m
(σ·pE+m
1
)(−iσ2χs). Notice that the zero momentum solutions take the form of
1
0
0
0
,
0
1
0
0
,
0
0
0
1
, and
0
0
−1
0
.
Consider the following operation
ψ → ψc = Cβψ∗ (VI.111)
where C = −iα2 =
(0 −σ2−iσ2 0
)and we have C2 = −1. Other properties of C are
C = iγ2γ0
C = −C−1
C = −C†
C = CT
CαC−1 = −α∗
CγµC−1 = γµT
Consider the Dirac equation
i∂
∂tψ =
(αi(−i∇i − eAi
)+ eA0 + βm
)ψ (VI.112)
with Aµ as the vector potential and in non-manifestly covariant form.
− i ∂∂tψ =
(α∗i(i∇i − eAi
)+ eA0 + βm
)ψ∗ (VI.113)
− i ∂∂tψ∗ =
(αi(i∇i − eAi
)+ eA0 − βm
)ψc (VI.114)
i∂
∂tψ∗ =
(αi(−i∇i + eAi
)− eA0 + βm
)ψc (VI.115)
This is the original Dirac equation but e→ −e. ψc satisfies the Dirac equation but with opposite charge. We alsofind that Cβu∗ (p, s) = v (p, s). Therefore we have Cβψ−∗−p,−s = ψ+
p,s(x).
ψ−c−p,−s describes an antiparticle with positive energy with momentum and spin. Dirac’s interpretation (which waseventually overtaken by QFT) is that electrons satisfy the Pauli exclusion principle so he says that the world has allthe negative energy states filled. Dirac says that ψ−c correspond to holes in the negative energy sea. Pair creationcan be thought of as knocking out an electron in a negative electron state into the positive energy states.
Lecture 23, March 16th, 2012
A. Non-relativistic limit of the Dirac equation
One reason to do this is to make contact with the Schrodinger equation and the Pauli equation. The Pauli equationis just the Schrodinger equation with two component spinors.
This will give us intuition on what a 4-spinor is. Recall that we have
ψ+ =1√
2EL3u (p, s) e−ip·x (VI.116)
42
where u(p, s) =√E +m
(1
σDpE+mχ
s
). Consider the v → 0 limit. In this limit we have p→ 0 and p = (m,0). Hence
ψ+(s =1
2) =
1√2mL3
u (0, s) e−imt =1√
2mL3
√2m
(1
0
)(1
0
)e−imt (VI.117)
=1√L3
1
0
0
0
e−imt (VI.118)
while we have
ψ+
(s = −1
2
)=
1√L3
0
1
0
0
e−imt (VI.119)
Similarly the negative energy solutions are
ψ−(s =
1
2
)=
1√2mL3
√2m
(0
1
)(0 −1
1 0
)(1
0
)eimt =
1√L3
0
0
0
1
eimt (VI.120)
ψ−(s = −1
2
)=
1√L3
0
0
−1
0
eimt (VI.121)
This is the spinors at rest. To gain more insight on the problem we need to boost these spinors. We don’t yet knowhow to do this since we only know the transformations in the fundamental representation. However in 1950 Folbyand Woothuysen came up with a technique to the non-relativistic reduction properly. They showed that you couldexpand the spinors in a p
m expansion.We call χ the large components of this expansion and η the small components (this is clear by the results above, for
electrons the η components, σ·pE+m are 0). What Folby and Woothuysen found was to find a way to write the spinors
in terms of χ’s in terms of Pauli spinors.This trick is difficult. We start with a practice problem. Consider you have the following Hamiltonian H =
σxBx + σzBz (a H that just involves pauli matrices). The Hamiltonian is a 2 × 2 matrix. This Hamiltonian woulddescribe Pauli spinors interacting with the magnetic field. Also consider the unitary transformation
U = eiσyθ/2 = cosθ
2+ iσy sin
θ
2(VI.122)
H ′ = UHU−1 = eiσyθ/2He−iσyθ/2 (VI.123)
=
(cos
θ
2+ iσy sin
θ
2
)H
(cos
θ
2− σy sin
θ
2
)(VI.124)
= cos2θ
2(σxBx + σzBz) + i sin
θ
2cos
θ
2([σy, σx]Bx + [σy, σz]Bz) + sin2 θ
2(σyσxσyBx + σyσzσyBz) (VI.125)
= σx (cos θBx − sin θBz) + σz (cos θBz + sin θBx) (VI.126)
If tan θ = BxBz
then the coefficient of σx is zero. Hence we have diagonalized H. We have decoupled the upper andlower components.
43
We now do a more difficult problem (however still not the full problem). Consider the free Dirac equation (novector potential).
H = α · p + βm (VI.127)
Here we want to decouple the large and small components of the Dirac equation (due to low velocity limit). Recallthat
αi =
(0 σiσi 0
)Hence in block diagonal form we have
H =
(0 σ · p
σ · p 0
)+
(1 0
0 −1
)m (VI.128)
It turns out that eis where is = βα · pθ(p) does the trick. This can be written as
eis = cos (|p| θ) + βα · p|p|
sin (|p| θ) (VI.129)
If we transform H then we have
H ′ = UHU−1 (VI.130)
= β (m cos 2 (|p| θ) + |p| sin (|p| θ)) (VI.131)
if we choose tan (|p| θ) = |p|m then we have
H ′ = β√m2 + p2 (VI.132)
=
E
E
−E−E
(VI.133)
Lecture 24 - March 19th, 2012We are not responsible for the Foldy-Wouthuysen online but we are responsible for the toy example, Dirac freeexample, and the March 19th (today) example.
We now continue with FW. The general case is
H = α · (p− eA) + βm+ eφ (VI.134)
where A0 = φ and Ai = α. So far we were able to completely decouple large and small components of the Dirac
equation but only in the free case (i.e. Aµ = 0). In the general case this is impossible to all orders in |p|m (this tells us
how relativistic the system is). However Foldy-Wouthuysen tells us how to decouple to any desired finite order in |p|m .Recall we are interested in decoupling so that we have an equation which is Schrodinger-like for the Pauli 2-spinorχ. We could continue with a systematic expansion through FW (which is done in the posted notes). Instead ofdeveloping a general systematic approach, we will use hindsight to efficiently get to the next level of desired accuracy.
First lets define two quantities:
T = E −m (VI.135)
(roughly speaking this is the kinetic energy). Further we define
V µ = eAµ (VI.136)
(just to simplify notation). We can now write the Dirac equation in first order with a Hamiltonian by
T
(χ
η
)=(−m+ V 0 + α · (p−V) + βm
)( χ
η
)= H
(χ
η
)(VI.137)
44
Consider
U = U† = Aβ +λ
mα · p (VI.138)
where A =√
1− λ2p2
m2 with λ as an undetermined parameter.
UU† =
(Aβ +
λ
mα · p
)(Aβ +
λ
mα · p
)= A2β2λ
A
m(α · pβ + βα · p) +
λ2
m2(α · p)
2
= A2β2 +λApim
(︷ ︸︸ ︷αiβ + βαi
)+λ2
m2pipj
1
2
δij︷ ︸︸ ︷αiαj + αjαi
(VI.139)
= A2 +λ2p2
m2(VI.140)
= 1 (VI.141)
Hence U is unitary. We have
T
(χ
η
)= H
(χ
η
)(VI.142)
UT
(χ
η
)UHU−1U
(χ
η
)(VI.143)
T
(χ′
η′
)= H ′
(χ′
η′
)(VI.144)
Note that we used T ′ = U(E −m)U−1 = T We now find H ′ = UHU−1
U (−m)U−1 = −m (VI.145)
We go on now
UV 0U−1 = AV 0A+βλ
m
(AV 0α · p−α · pV 0A
)+λ2
m2α · pV 0α · p (VI.146)
The third term is
Uα · (p−V)U−1 = −Aα · (p−V)A+βλ
m(Aα · (p−V)α · p) (VI.147)
and the last term is
UmpU−1 = mβA2 + 2λAα · p− β λ2p2
2m2(VI.148)
Remember our goal is to limit off-diagonal terms. To lowest order the off-diagonal terms all come from the α ·(p−V)term.
H ′off−diag = −α · (p−V) + 2λα · p (VI.149)
We now tune λ to λ = 12 . Our new off diagonal part of the Hamiltonian is
H ′off−diag = α ·V (VI.150)
The coupled equations become
Tχ′ = H ′11χ+ α ·Vη′ (VI.151)
45
Tη′ = σ ·Vχ′ − 2mη′ (VI.152)
The key step is to say that Tη′ is must smaller then 2mη′ (since we are in the nonrelativistic limit). This allows usto write
η′ =σ · p2m
χ′ (VI.153)
We have decoupled these equations to some order. The error we are making is higher-order in |p|m . Now the equationsare decoupled and we have one equation:
Tχ′ = H ′11χ′ +
σ ·Vσσ ·V2m
χ′ (VI.154)
where
H ′11 = V 0 − p2
8m2V 0 − V 0 p2
8m2+
σ · pV 0σ · p4m2
+1
2m(σ · (p−Vσ · p) + σ · pσ · (p−V))− p4
8m3− p2
2m(VI.155)
One can simplify this equation and you find that (and we put eφ = V 0 and eA = V)
H ′11 +V 2
2m=
(p− eA)2
2m+ eφ− e
2mσ ·B− p4
8m3+ e
(∇2φ
)8m2
+e
8m2σ · ([∇φ]× p) (VI.156)
Lecture 25 - March 21st, 2012Recall we are doing the non-relativistic reduction of the Dirac equation. However we are not doing a full reductionas we are keeping important relativistic effects. We are doing it systematically so that we keep all relativistic effects
to some order in |p|m . Foldy-Woothuysen shows how you to do this in general. We will just do this to lowest order.
However in practice for hydogen our results will be highly accurate, namely up to α4.Recall we were taking the Dirac equation and applying a unitary transformation
U = Aβ +λ
mα · p , (VI.157)
where A =√
1− λ2p2
m2 . This transformation let us to this coupled system
Tχ′ = H ′11χ′ + σ ·Vη′ (VI.158)
Tη′ = σ ·Vχ′ − 2mη′ (VI.159)
If we have T 2m then
Tχ′ = H ′11χ′ +
σ ·Vσ ·V2m
χ′ , (VI.160)
where
H ′11 = V 0 − p2
8m2V 0 − V 0 p2
8m2+
σ · pV 0σ · p4m2
+1
2m(σ · (p−V(σ · p)) + σ · pσ · (p−V))− p4
8m3− p2
2m(VI.161)
We simplify this term by term
σ · (p−V)σ · p + σ · p (σ · (p−V)) = 2p2 − σ · pσ ·V − σ ·Vσ · p (VI.162)
= 2p2 − p ·V −V · p− iσ · (p×V)− iσ · (V × p) (VI.163)
= (p−V)2
+ p2 − V 2 − σ · [∇×V] (VI.164)
where we have used A×B = εijkAjBk and square brackets,[ ] means that the operators inside act only inside. Ourpotential could have been any vector potential. We note choose the electromagnetic force:
V = eA (VI.165)
46
This gives (with B = ∇×A)
σ · (p−V)σ · p + σ · p (σ · (p−V)) = (p−V)2
+ p2 − V 2 − eσ ·B (VI.166)
This shows you that the electric interacts with the magnetic field with the expression eσ · B! Thus this actuallypredicts that the electron has a magnetic moment. Continuouing with the reduction we have
p2V 0 + V 0p2 =[p0V 0
]+ 2
[pB0
]· p + 2V 0p2 (VI.167)
and
σ · pV 0σ · p = σ ·[pV 0
]σ · p + V 0p2 (VI.168)
= σ ·[pV 0
]σ · p + V 0p2 (VI.169)
=[pV 0
]p + iσ ·
([pV 0
]× p
)+ V 0p2 (VI.170)
so we see that
− 1
8m2
(p2V 0 + V 0p2
)+
σ · pV 0σ · p4m2
= −[p2V 0
]8m
+iσ([
pV 0]× p
)4m2
(VI.171)
If we now input in
p = −i∇V = eA
V 0 = eφ
then we have
H ′11 +V 2
2m=
(p− eA)2
2m+ eφ− e
2mσ ·B− p4
8m3+ e
[∇2φ
]8m2
+e
8m2σ · ([∇φ]× p) (VI.172)
if we assume that φ is spherically symmetric then we have φ = φ(r) and thus ∇φ = rrdφdr then we finally have
H11 =
(1)︷ ︸︸ ︷(p− eA)
2
2m+
(2)︷︸︸︷eφ −
(1)︷︸︸︷p4
8m3−
(4)︷ ︸︸ ︷e
2mσ ·B +
(5)︷ ︸︸ ︷e
[∇2φ
]8m3
+
(6)︷ ︸︸ ︷3
4m2r
dφ
drσ · L (VI.173)
The first term can be rearranged: (we assume that A2 is small)
(p− eA)2
2m=
p2
2m− e
2m(p ·A + A · p) (VI.174)
=p2
2m+
ie
2m(∇ ·A + A · ∇) (VI.175)
=p2
2m+ie
mA · ∇ (VI.176)
=p2
2m− ie
2m(r×B) · ∇ (VI.177)
=p2
2m+
ie
2mB · (r×∇) (VI.178)
=p2
2m− e
2mB · L (VI.179)
p2
2m is the kinetic energy and e2mB · L is the Zeeman term. Thus we have obtains a field-orbit coupling term without
explicitly putting it in again! Now consider the second term. This is the non-relativistic potential. This is just theColoumb potential for hydrogen
eφ = −e2
r(VI.180)
47
The third term
p4
8m3(VI.181)
is just the next order relativistic correction to the kinetic energy. Now consider the fourth term:
− e
2mσ ·B = − e
mS ·B (VI.182)
= − e
mS ·B (VI.183)
This is a prediction that the gyromagnetic ratio of the electron is equal to 2(g = 2)! The full Zeeman effect can bewritten − e
2mB · (L + 2S). Now consider the fifth term:
e
[∇2φ2
]8m2
= − e2
8m2∇ ·E (VI.184)
= − e2
8mδ (R) (VI.185)
(with φ = − e2
r ). This term is called “Zetterbeuwegung”. The sixth term is
e
4m2r
dφ
drS · L (VI.186)
This is spin-orbit coupling which is because the electron sees (φ, 0, 0, 0)LT−−→ (φ′,A′) under a Lorentz transformation.
Lecture 26 - March 23rd, 2012Missed this lecture.
Lecture 27 - March 25th, 2012
B. Covariance of the Dirac equation
We are trying to prove that
γµ (pµ − eAµ)−mψ = 0 (VI.187)
⇒γµ(p′µ − eA′µ
)−m
ψ′ = 0 (VI.188)
Contravarient vectors transform as
p′µ = Λµνpν (VI.189)
and the controvarient vectors transform as
p′µ =(Λ−1
)µνpν (VI.190)
We have
p2 = p′µp′µ = pµpµ (VI.191)
We insert in
ψ′ = Sψ (VI.192)
S is the spinor representation of the Lorentz group. After inserting Λ and S into preed equation we get
S−1 (Λ) γµS(Λ) = Λµνγν (VI.193)
48
For Lie groups, to find the generators we only have to satisfy this equation to infinitesimal order and then its true toall orders
Λ = 1 + εiωi0 +1
2θiεijkωjk (VI.194)
we have 6 parameters (εi, θi) and 6 generators ωi0, ωjk (note these are NOT indices but labels!). Whatever S(Λ) is ininfinitesimal form it will look like
S(Λ) = 1 + εiBi +1
2θiεijkRjk (VI.195)
where εi and εijk are the same parameters as for Λ. At this point we don’t know what the generators of the spinorrepresentation are, Bi, Rjk. We insert infinitesimal forms into S−1γµS = Λµνγ
ν . Last time we finished class at (heleft this as an exercise)
− [Bi, γµ] = (ωi0)
µν γ
ν (VI.196)
− [Rjk, γµ] = (ωjk)
µν γ
ν (VI.197)
The new exercise is to verify that
Bi =1
2γ0γi ← Boosts (VI.198)
Rjk =1
2γjγk ← Rotations (VI.199)
This appears non-covariant. This can be written in covariant form
σµν =i
2[γµ, γν ] (VI.200)
where σ0i = 2iBi and σij = 2iRij (these are not the Pauli matrices! We are just using σ as some matrix). Notehowever that this is not a convenient form of the generators. Bi are the generators of boosts for spinors and Rjk arethe generators for rotations for spinors. Let’s introduce some more notation.
Σj ≡ σj =
(σj 0
0 σj
)(VI.201)
and
γ5 ≡
(0 1
1 0
)(VI.202)
We now have (using the α matrix we had before)
Bi =1
2αi (VI.203)
Rjk =i
2εjk`γ
5α` (VI.204)
A boost in Dirac-space is
S(Λξ) = 1 +ξ ·α
2+
1
2
(ξ ·α
2
)2
+ ... (VI.205)
= cosh
(ξ
2
)+ ξ ·α sinh
(ξ
2
)(VI.206)
(VI.207)
where ξ is the rapidity, tanh(ξ) = vc , E
m = cosh(ξ), cosh( ξ2 ) =√
E+m2m , and sinh( ξ2 ) =
√E+m2m
(p
E+m
). For a spinor
at rest we ahve
ψ0,s(x) =e−imt√L3
(χs
0
)=
e−imt√2mL3
u (0, s) (VI.208)
49
remember that
α ≡
(0 σ
σ 0
)(VI.209)
and therefore
coshξ
2+ ξ ·α sinh
ξ
2=
(cosh ξ
2 ξ ·α sinh ξ2
ξ ·α sinh ξ2 cosh ξ
2
)(VI.210)
Lets apply S(Λξ) to
(Xs
0
).
S(Λξ)
(Xs
0
)=
cosh
(ξ
2
)+ ξ ·α sinh
(ξ
2
)(Xs
0
)(VI.211)
=
(cosh
(ξ2
)Xs
sinh ξ2σ · ξχ
s
)(VI.212)
=
√E +m
2m
(1
ξ · σ(
pE+m
) )χs (VI.213)
=
√E +m
2m
(1
σ·pE+m
)χs (VI.214)
we have retrieved that solutions for arbitrary momentum. Now consider rotations in Dirac space.
S(Λθ) = e−(i/2)γ5θ·α (VI.215)
= cosθ
2− iγ5θ ·α sin
θ
2(VI.216)
Consider the rotation below
S(Λθ)u(p, s) (VI.217)
Recall from our study of SU(2) we know how Pauli spinors transform under rotations.
χs′
= S(θ)χs =
(cos
θ
2− iσ · θ sin
θ
2
)χs (VI.218)
Let’s pick to rotate about the z axis. Back to our rotation in Dirac space we have
S(Λθ)u(p, s) =√E +m
(1
S(θ)σ·pS†(θ)E+m
)χs
′(VI.219)
(check this at home) where if we are specializing to rotations in the z axis we have
S(θ) = cosθ
2+ iσz sin
θ
2(VI.220)
To simplify S(Λθ)u(p, s) we need to simplify the following
S(θ)σ · pS†(θ) =
(cos
θ
2− iσz sin
θ
2
)σ · p
(cos
θ
2+ σz sin
θ
2
)(VI.221)
= cos2θ
2σ · p− i sin
θ
2cos
θ
2(σzσ · p− σ · pσz) + sin2 θ
2σzσ · pσz (VI.222)
= σ · p′ (VI.223)
where p′ = (pz cos θ − py sin θ, py cos θ + px sin θ, pz). Note that we skipped about a half page of steps in the last step.This is also left as an exercise.
50
Lecture 28 - March 28th, 2012For the test:
• Some qualitative questions
• In accidental degeneracy prove some commutation relation or identities
– Nothing too long, but take a good look at the assignment with M
• Short-ish proofs from the Dirac equation, similar to as done in the notes
• Formula sheet will be posted
VII. HYDROGEN
One can obtain exact solutions to the hydrogen Dirac problem. The Dirac equation takes the form[γµ(i∂
∂xµ− eAµ
)− µ
]ψ = 0 (VII.1)
where Aµ = (eφ,0) with eAµ = −e2r . We have chosen the rest frame of the proton. It no longer looks covariant.
That’s okay as long as you stay in that frame. In this case the Dirac equation becomes (exercise) (where we haveused β = γ0 and β2 = I).
Eψ =
[−iαi
∂
∂xi+ βm− e2
r
]ψ (VII.2)
One should ask where is the time dependence? It is in γ0i ∂∂t . V = V (r), thus just like the Schrodinger equation
Ψ(r, t) = ψ(r)e−iEt/~ (VII.3)
The time dependence in the wave function is just a phase rotation (ψ†ψ doesn’t change with time). Derivative termjust gives the energy term on the left, where E is the constant (an eigenvalue). Now we have
L = r× p = −i~ (r×∇)
S =1
2γ5α
[Li, H
]= [−iεijkrj∂k,−iα`∂`] = εijkαj∂k 6= 0 (VII.4)
This is another exercise. Note that one should memorize Li = −iεijkrj∂k, useful expression. That means that L isnot conserved! (L is not a good quantum number).[
Si, H]
=1
2
[γ5cai,−iαj∂j
]= −εijkαj∂k 6= 0 (VII.5)
(exercise) This is bad news! We lost all our good quantum numbers. However the good news is that the commutator[Li + Si, H
]= 0 (VII.6)
Hence the total angular momentum is conserved. J is a good quantum number.
S2 = SiSieq1
4α ·α =
3
4(VII.7)
( The magnitude of |S| is constant. This tells us we can label states by good quantum numbers, J2, Jz, S2, L2. All
these operators commute with H and therefore we can have simultaneous eigenfunctions of J2, Jz, S2, L2. Now we
write the 4-spinor as
ψ(r) =
(F (r)
G(r)
)(VII.8)
51
We introduce the following new object yjm(r).
yjm(r) =
⟨`,m− 1
2
∣∣∣∣⊗⟨1
2,
1
2
∣∣∣∣⊗ |j,m〉Y`,m− 12↑ +
⟨`,m+
1
2
∣∣∣∣⊗⟨1
2,−1
2
∣∣∣∣⊗ |j,m〉Y`,m+ 12(r) ↓ (VII.9)
Equivalently one often uses the notation:
yjm(r) =
⟨`,m− 1
2,
1
2,
1
2
∣∣∣∣j,m⟩Y`,m− 12↑ +
⟨`,m+
1
2,
1
2,−1
2
∣∣∣∣j,m⟩Y`,m+ 12(r) ↓ (VII.10)
These are just the Clebsh-Gordan coefficients. It is useful to define an operator,
K = (σ · L + 1) (VII.11)
since
K±`,m = ±ky±j,m (VII.12)
where
k =
j + 1
2 if j = `− 12
−(j + 1
2
)if j = `+ 1
2
(VII.13)
We can write the solution as
ψkj,m(r) =
(fkj (r)ykj,m(r)
igki (r)y−kj,m(r)
)(VII.14)
Note that this needs to be verified. At this point we have solved the angular problem. This is equivalent to ψ(r) =R(r)Y`,m(θ, φ) but not the radial problem. Notice the angular part is solved with the familiar spherical harmonics.
Lecture 29 - March 30th, 2012To check this solutions we need the following simplifications to our expressions (prove these)
− iσ · ∇ = −iα · r ∂∂r
+ iσ · rσ · Lr
(VII.15)
σ · Lykj,m(r) = − (k + 1) ykj,m(r) (VII.16)
σ · rykj,m(r) = y−kj,m(r) (VII.17)
−iσ · ∇ykj,m(r) = i∂y−kj,m(r)
∂r+ i
(k + 1)
ry−kj,m(r) (VII.18)
Separation of variables is successful and we find find two coupled linear 1st order homogenous equations
Efk(r) =
(m− e2
r
)fk(r)− gk(r)
dr− 1− k
rgk(r) (VII.19)
Egk(r) =
(−m− e2
r
)gk(r) +
dfk(r)
dr+
1 + k
rfk(r) (VII.20)
To solve these solutions we rescale our solutions (just as in the S.E. solutions)
fR =f
r, gR =
g
r
ρ =√m2 − E2r
ε =
√m− Em+ E
52
In natural units we call e2 = α. This gives two equations:
dfRdρ
+k
ρfR −
(1
ε+α
ρ
)gR = 0 (VII.21)
dgRdρ− k
ρgR −
(ε− e2
ρ
)fR = 0 (VII.22)
We solve with a power series:
fR = ρν∞∑n=0
Anρne−ρ (VII.23)
gρ = ρν∞∑n=0
Bnρne−ρ (VII.24)
Substitute forms in. There is a non trivial solution if ν =√k2 −m2. These differential equations have a continuous
set of solutions. However to enforce normalizability we must force the solutions to terminate the infiite series so thatsolutions are well behaved. This discrete subset are our physical solutions. We can force termination by demanding
2ν + 2N − α
ε
(1− ε2
)= 0 (VII.25)
where N is an integer. As always this is where the quantization of E comes from (comes from the quantization of ε).Amazingly one can solve this problem in terms of confluent hypergeometric function.
En,j = m
1− α2
n2 + 2(n−
(j + 1
2
))(√(j + 1
2
)2 − α2 −(j + 1
2
))1/2
(VII.26)
where we have now translated from k → j. We can expand terms of α (since in Hydrogen pm is of order α, p
m α).
En,j = m− mα2
2n2− mα4
2n4
(n
n+ 12
− 3
4
)(VII.27)
Note that we have lost a lot of our degeneracies. What about real hydrogen? there also have a proton with a spin,
. This produces a magnetic moment,
µ =ge
2mS . (VII.28)
This creates a spin -spin interaction which splits the 1S1/2 state into two states. The triplet and single states. Thisis the famous 21cm line. The energy difference is very simple and its given by
E = m√
1− α2 . (VII.29)
This is exact in the Dirac equation. The 1s1/2 state from the Dirac equation is
ψ 12 ,s=+ 1
2= A0ρ
γ−1e−ρ
1
0
− 1−γα cos θ
− 1−γα sin θeiφ
(VII.30)
53
where γ =(1− α2
)1/2. Notice that even the ground state has angular dependence! We also have
ψ 12 ,s=−
12
= A0ργ−1e−ρ
0
1
− 1−γα sin e−iφ
− 1−γα cos θ
(VII.31)
We can write this in more convenient notation:(cos θ
sin θeiφ
)=
z
x+ iy(VII.32)
and
σ · r =
(z x− iy
x+ iy −z
)(VII.33)
Hence
σ · r
(1
0
)=
(cos θ
sin eiφ
)(VII.34)
This gives
ψ 12 ,s=
12
= A0ργ−1e−ρ
(1
σ · r(1−γα
)χ1/2
)(VII.35)
