Chapter 6: SYMMETRY IN QUANTUM MECHANICS
Since the beginning of physics, symmetry considerations have provided us with an extremely
powerful and useful tool in o
ur effort in understanding nature. Gradually they become the
backbone of our theoretical formulation of physical laws.
—— T.D. Lee, Particle Physics and an Introduction to Field Theory
For example,
1. Theory of relativity: physical laws are invariant in any reference systems.
2. Quantum field theory: gauge invariance. In 1954, C.N. Yang and Robert Mills extended the
gauge theory of Abelian group for QED to non-Abelian group for QCD.
3. (Super)String theory: super-symmetry, or the one-to-one correspondence between boson and
fermion. This is an extension to symmetry groups in QFT towards combining QFT and general
relativity, i.e., a theory for everything —— Einstein’s dream.
4. The fundamental postulate of statistical mechanics: given an isolated system in equilibrium, it
is found with equal probability in each of its accessible microstates.
6.1 Continuous Symmetry
Time evolution invariance (time-independent H), space translation invariance, rotational
invariance, etc. Noether’s (first) theorem states that any differentiable (continuous) symmetry
in a physical system corresponds to a conservation law.
A continuous symmetry can be described by an infinitesimal translational operator
Ω̂(Δλ)Δλ→0 = 1 −i
Δλ
Ĝ

and the corresponding finite translational operator:
G
Ω(λ) = exp(−i λ), and Ω̂(λ) Ψ(0) = Ψ(λ) .
!
∂
G ≡ i
= H; U(t) = exp(−iHt / ) .
Time evolution:
∂t
Space translation: G ≡ −i∇r = p; T(r) = exp(−ip̂ ⋅ r / ) .
Rotation:
G ≡ −i
∂
 = J; D(n,ϕ) = exp[−i(Ĵ ⋅ n)ϕ / ] .
∂ϕ
(6.1)
(6.2)
(6.3)
(6.4)
(6.5)
Translational invariance means that Ω†H Ω = H ⇒ H Ω = ΩH ⇒
[G,H ] = 0 .
84
(6.6)
(1) Using Ehrenfest’s theorem, one obtain
d
G = [G,H ] = 0 .
dt
(2) In the Heisenberg picture, the equation of motion becomes
i
(6.7)
d
G = [G,H ] = 0 .
dt H
(3) G and H have the same set of eigenstates.
i
(6.8)
Assume n is an energy eigenstate (eigenket) with engenvalues of En , then
(
H Ωn
) = ΩH n
(
)
= En Ω n ,
(6.9)
so the translated state Ω(λ) n(0) = n(λ) is also an eigenstate of H with the same energy, i.e.,
states n and n(λ) are degenerate, and all the translated states are degenerate for an invariant
translation Ω .
One example is the rotation invariance, [D(R),H ] = 0 , where a rotation R is represented by
the rotation axis n and the rotation angle ϕ , and which implies that [J,H ] = 0 and [J2,H ] = 0 ,
Then we can form the common eigenkets of H, J2, and J z denoted by njm , and all the
rotated states
D(R) njm =
j
∑D
m ′=−j
(j )
m,m ′
(R) njm ′
(6.10)
have the same energy. The above expression suggests that (1) if the original state has only
J z = m component, then under rotation different m-values are mixed up. (2) An arbitrary
rotated state, which is a linear combination of (2j + 1) states of njm ′ , has the same energy,
then each of these (2j + 1) states are degenerate.
6.2 Time Reversion Invariance
We have discussed the continuous symmetry, which can be obtained by applying successively
infinitesimal symmetry operations. But not all symmetry operations are continuous, instead,
there are many important discrete symmetry operations, such as time reversion, space inversion,
crystal lattice translation and rotation, etc. These discrete symmetry operations also lead to
85
conservation laws, except for time-reversal invariance. However, its consequences for quantum
mechanics are indispensible.
Time-reversal means motion reversal. Let’s begin with classical physics. If a particle subject
to a force field F(r) = −∇V(r) , then reversing time (or reversing the motion) causes the particle
to go backward along the same trajectory as the forward motion.
Formally, it r(t) is a solution to
d 2r(t)
= F(r) = −∇V(r) ,
(6.11)
dt 2
then r(−t) is also a possible solution. Time-reversal invariance requires no dissipation force,
m
which depends on motion.
The non-relativistic quantum mechanics allows one to formulate the time reversal in a way
close to classical mechanics, but the form of the result depends on the representation. Consider
the time-independent Schrödinger equation,
dψ(r,t)
= H ψ(r,t) ,
dt
Under time reversal given by t → −t , the equation becomes
i
(6.12)
dψ(r,−t)
= H ψ(r,−t) ,
(6.13)
dt
The solution ψ(r,−t) doesn’t satisfy the original Schrödinger equation (6.12). Take the complex
−i
conjugate of Eq. (6.13),
dψ*(r,−t)
= H *ψ*(r,−t) ,
(6.14)
dt
therefore when t → −t , ψ(r,t) → ψT (r,t) = ψ*(r,−t) and H → H T = H * . The off-diagonal
i
matrix element of H :
H 12 = ψ1(t) H ψ2(t) .
(6.15)
Then the time-reversed matrix element
H T12 = ψ2(−t) H ψ1(−t) .
(6.16)
The transformed Hamiltonian coincide with the original one if it is real, for example, the
ordinary one particle Hamiltonian in a real potential,
H =−
2 2
∇ +V(r) ,
2m
86
(6.17)
then the system is time-reversal invariant (TRI), and both ψ(r,t) and ψ*(r,−t) are solutions to
the same Schrödinger equation (6.12).
Time-reversal operator Θ defined as ψT ≡ Θ ψ , which is an anti-unitary operator. Thus,
we expect that pT = Θ p = e iφ −p . Let’s work out the explicit form of Θ . Since quantum
states are linear, i.e., ψ is a linear combination of other states, ψ = a α +b β , then
(
Θ a α +b β
)=a
*
αT +b* βT ,
(6.18)
Θ is antilinear. So we can write Θ =UK , where U is an unitary operator, while the complex
conjugate operator K has the following property:
(
)
K a α = a *K α .
(6.19)
Expressing state α in terms of a complete set of basis states,
α = ∑ an an α ,
(6.20)
n
we apply the time-reversal operator,
αT =UK α = ∑ an α *UK an = ∑ α an U an .
n
(6.21)
n
Similarly, we have β = ∑ an an β , then
n
βT = ∑ β an U an , and βT = ∑ an β an U † ,
n
(6.22)
n
Then taking the scalar product we find that
βT αT = ∑ ∑ α an am β am U †U an
m
(6.23)
n
= ∑ α an an β = α β ,
n
†
because U U = 1 , and am an = δmn .
Using the coordinate representation, the time-reversed state
ψT = Θ ψ = ∫ U r K r ψ d 3r = ∫ r ψ r d 3r .
(6.24)
The time-reversed wave function
*
ψT (r) = rT ψT = r ψT = ∫ r r′ ψ r′ d 3r′ = ψ r = r ψ = ψ*(r) .
Thus we reproduced our previous result of t → −t , ψ(r,t) → ψT (r,t) = ψ*(r,−t) .
87
(6.25)
The consequence of time reversal
A. The time-reversed states: Θ r = r , while Θ p = −p . (up to a phase)
The time-dependent wave function:
Ψ(t) = ∑ e
j =1
c j ψj → ΨT (t) = ∑ e
−iE j t/
j =1
−iE j t/ *
j
c ψ*j .
(6.26)
B. The time-reversed operator: Â → ÂT = Θ−1ÂΘ
Θ−1rΘ = r, Θ−1p̂Θ = −p̂ .
The corresponding observables are even and odd under time reversal, respectively.
p̂2
+V(r) ,
For a particle in a real potential, H =
2m
(6.27)
(6.28)
Θ−1HΘ = H, or HΘ = ΘH .
It is time-reversal invariant (TRI). TRI is violated in nature. However, this violation was only
observed as a small effect in specific process of decay of neutral K- and B-mesons.
The fundamental commutator ⎡⎢r̂i , p̂ j ⎤⎥ = iδij is preserved under time reversal:
⎣
⎦
Θ−1 ⎡⎢r̂i , p̂ j ⎤⎥ Θ = ⎡⎢r̂i ,−p̂ j ⎤⎥ = −iδij ,
⎣
⎦
⎣
⎦
−1
while Θ iδij Θ = −iδij . Similarly, in order to preserve ⎡⎢Jˆi , Jˆj ⎤⎥ = iεijkJ k , the angular
⎣
⎦
(6.29)
momentum operator must be odd under time reversal:
Θ−1ĴΘ = −Ĵ .
(6.30)
Then the angular momentum state l,m → l,−m under time reversal, which is consistent with
Yl m (θ,φ) →Yl m*(θ,φ) = (−1)mYl −m (θ,φ) .
(6.31)
where Yl m (θ,φ) is the wave function for state l,m : Yl m (θ,φ) = θ,φ l,m . Then
Θ l,m = (−1)m l,−m .
(6.32)
Time reversal for spin-half systems
A state with spin directed along a unit vector n with polar angle α and azimuthal angle β is
given in terms of the spin-up state along the z-axis + ≡ z;+ by
n;+ = e
n;− = e
−iSz β/ −iSy α/
e
+ ,
−iSz β/ −iSy (α+π)/
e
+ .
Under the time reversion, based on Eq. (6.32) we expect that
88
(6.33)
(6.34)
Θ n;+ = η n;− .
(6.35)
Comparing Eqs. (6.34) and (6.35),
Θ = ηe
−iπSy /
K.
(6.36)
Note that
e
−iπSy /
+ =+ −
and e
−iπSy /
− =− +
.
(6.37)
Question: how to prove the above equations?
Then for an arbitrary spin-half state χ = c+ + +c− − ,
Θ χ = ηc+* − − ηc−* +
(6.38)
Θ2 χ =| η |2 (−c+ ) + − | η |2 c− −
(
(6.39)
)
= − c+ + +c− − = − χ .
Hence for spin-half states we have the most usual property,
Θ2 = −1 .
More generally,
Θ2 = −1, for j = half-integer
(6.40)
(6.41)
2
Θ = +1, for j = integer
(6.42)
The time-reversal operator becomes
Θ = ηe
−iπJ y /
K.
(6.43)
For an arbitrary state with a well-defined angular momentum quantum number j (e.g., a rotated
eigenstate D(R) jm ), χ = ∑ jm
jm χ , we have
m
Θ χ = η∑e
−iπJ y /
jm χ jm
(6.44)
m
Θ2 χ = η *η ∑ e
−i 2πJ y /
jm
m
jm χ = ∑ e
−i 2πJ y /
jm
jm χ
(6.45)
m
Because (Question: how to derive this?)
e
−i 2πJ y /
jm = (−1)2j jm ,
(6.46)
we obtain the final result:
Θ2 χ = (−1)2j χ .
(6.47)
Question: But is the above result consistent with Eq. (6.32), Θ l,m = (−1)m l,−m ?
In general, one can take.
Θ j,m = (−1)m j,−m = i 2m j,−m .
89
(6.48)
6.3 Space Inversion and Parity
Another discrete symmetry if the Hamiltonian is important for the search and classification of
stationary states. The space inversion (parity transformation) operator Π changes the sign of
spatial coordinates so that the localization state of a particle r transforms as
Π r = −r .
(6.49)
Π p = −p ,
(6.50)
ψP (r) ≡ r Π ψ = −r ψ = ψ(−r) .
(6.51)
It follows that
Note: in one-dimension, ψ(−x) is the mirror image of ψ(x) about the origin.
From Eq. (6.49) one can derive Π2 r = Π −r = r , then
Π2 = I .
(6.52)
From this we obtain that
1) Π = Π−1 .
2) The eigenvalue of Π are ±1 .
3) Π is Hermitian and unitary.
†
4) Then Π−1 = Π = Π .
Parity (π) : eigenvalue of the operator Π . If a state has a definite parity, then
Π ψ = ψ P = πψ ψ ,
(6.53)
where πψ is either +1 or −1. ψ(−r) = −r ψ = r Π ψ = r πψ ψ = ± −r ψ = ±ψ(r) .
Space reversion of operators:
Π−1rΠ = −r ,
and
Π−1p̂Π = −p̂ ,
Which can proved by considering the infinitesimal space transformation
⎛
Π ⎜⎜I
⎜⎝
ΠT(dr) = T(−dr)Π ,
ip̂ ⋅dr ⎞⎟ ⎛⎜
ip̂ ⋅dr ⎞⎟
⎟⎟ = ⎜I +
⎟⎟ Π .
−
 ⎟⎠ ⎜⎝
 ⎟⎠
Then we obtain
90
(6.54)
(6.55)
(6.56)
(6.57)
Πp̂ + p̂Π = {Π, p̂} = 0, or equivalently Π−1p̂Π = −p̂ .
(6.58)
Angular momentum under space reversion:
Π−1LΠ = L, or [Π, L] = 0 ,
(6.59)
because the orbital angular momentum L̂ = r× p̂ .
Under rotations, vectors r , p and L behave in the same way; however, under space
reversion, parities of r and p are odd ( −1 ) while the parity of L is even (+1)! The former is
called polar vectors, while the latter is called axial vectors or pseudovector.
Consider the scalar operators S ⋅ r , S ⋅ L and r ⋅ p , under rotations they behave in the same
way like an ordinary scalar; however, under space reversion,
Π−1S ⋅ rΠ = −S ⋅ r, while Π−1S ⋅ LΠ = S ⋅ L and Π−1r ⋅ pΠ = r ⋅ p .
So the operator S ⋅ r is a pseudoscalar.
(6.60)
Space reversion in 3D can be written as a matrix,
⎛ −1 0
0
⎜⎜
⎜
R = ⎜⎜ 0 −1 0
⎜⎜⎝ 0
0 −1
Π
⎞⎟
⎟⎟
⎟⎟ ,
⎟⎟
⎟⎠
(6.61)
and it commutes with an arbitrary 3D real-space rotation: R ΠR = RR Π , therefore,
ΠD(R) = D(R)Π .
Since quantum rotation operator D(R) = e
−i
J⋅n
ϕ

(6.62)
, and an infinitesimal rotation D(R) = I −i J ⋅ n dϕ ,

we obtain that
(6.63)
Π−1JΠ = J, or [Π, J] = 0 .
Not all wave functions are eigenstate of Π , i.e., they might not have definite parities. An
eigenstate of orbital angular momentum is expected to have a definite parity (Q: why?) The
common eigenstate of L̂2 and L̂z is denoted as nlm , then the wave function
r nlm = Rn (r)Yl m (θ,φ) .
(6.64)
Under the space reversion transformation r → −r ,
r → r,
θ → π−θ
(cosθ → −cos θ)
φ → π+φ
Using the expression for spherical harmonics,
[e
91
imφ
m imφ
→ (−1) e
(6.65)
]
Yl m (θ,φ) = (−1)m
(2l + 1)(l −m)! m
Pl (cos θ)e imφ .
4π(l + m)!
(6.66)
For positive m,
l −|m|
(−1)m+l (l+ | m |)! −|m| ⎛⎜ d ⎞⎟
⎟⎟
Pl (cos θ) =
sin (θ)⎜⎜
2l l ! (l− | m |)!
⎝d(cos θ)⎟⎠
We can show that
Yl m → (−1)lYl m ,
|m|
sin 2l (θ) .
(6.67)
(6.68)
under space reversion as θ and φ change as in Eq. (6.65). Therefore, we find the parity for the
eigenstate of nlm
Π nlm = (−1)l nlm .
(6.69)
An alternative way to prove the above expression is to work with m = 0 and using the ladder
operator L± . Because Π and Lm± (m = 0, 1, 2, …, l) commute, Lm± l, 0 must all have the same
parity as that of l, 0 , which is
obviously equal to (−1)l .
Spontaneous symmetry breaking
Consider the double-well potential as
shown on the right. The two lowest
eigenstates are denoted as the symmetric
Fig. 12.1: Symmetric and antisymmetric states.
state S and the anti-symmetric state
A with eigenenergies of ES and E A . Question: which state has lower energy? Why?
We can form
R =
L =
1
2
1
2
(S
+ A ,
)
(6.70)
(S
−A .
)
(6.71)
Since S and A are not degenerate, R and L are not eigenstates of H , so they are nonstationary. If at t = 0 , ψ = R , then
92
ψ(t) =
1
(e
2
−iES t/
S +e
−iEAt/
)
A = e −iΩt ⎡⎢cos(ωt) R + sin(ωt) L
⎣
⎤,
⎥⎦
(6.72)
where Ω = (E A + ES ) / 2 and ω = (E A − ES ) / 2 , oscillating between R are L . The
system is NOT polarized in average, though it is polarized instantaneously.
This behavior is a dynamical manifestation
of quantum tunneling. Now if the barrier
becomes infinitely high (V0  E A,ES ), as
shown on the right. Then S and A are
Fig. 12.2: Symmetric and antisymmetric
states when the potential barrier is infinity.
degenerate ( ω becomes very small, and the
oscillation period becomes extremely long),
which means that R and L also eigenstates of H . Then both R and L are stationary
states – once the system stays in either R or L , it stays there forever, and it is polarized.
A symmetric potential might lead to asymmetric (no well-defined parity) ground-state due to
degeneracy – the spontaneous symmetry breaking. It is crucial in nature, responsible for many
fundamental mechanisms, such as the formation of magnets, superconductors, and Higgs mechanisms.
One example is the optical isomerism. Two forms of the same chemical compound,
isomers, were found to rotate polarized light in two different directions − one to the left, the
other to the right. Isomers are essentially identical chemical compounds. They have the same
number and type of atoms and the same structure, almost. The difference in the two isomers of a
compound is that one is the mirror image of the other, so that it is not superposable on its mirror
image. They are called chiral objects, and its chirality is designated as left-handed or righthanded. Achrial objects are superposible to their mirror images, such as a right circular cone.
Fig. 6.3: The ammonia molecules.
Fig. 6.4: Two generic amino acids.
93
One example is the ammonia molecule, as seen above. They correspond to the R and L
states, and the common eigenstates of parity and Hamiltonian ( S and A ) are the superpositions of them. The oscillation frequency f = ω / 2π = 24, 000 MHz, with wave length λ ≈ 1 cm.
Other chiral organic molecules, such as amino acids and sugar, have very longer oscillation
time, on the order of 104 – 106 years. In lab we always produce equal mixtures of R-type and Ltype such organic molecules; but interestingly, living organisms were able to synthesize and use
only one isomer and never the other.
Violation of Parity Conservation
In 1947 Cecil F. Powell photographed cloud chamber tracks of charged particles and
identified the pi meson − the particle postulated twelve years earlier by the physicist Hideki
Yukawa, as the intermediary for the nuclear force. Two years later, he identified the τ-meson,
which disintegrated into three pions, and the θ-meson, which disintegrated into two pions.
Both particles have exactly the same mass and lifetime (within experimental error). However,
since the pion has parity of -1, two pions would combine to produce a net parity of (-1)(-1) = +1,
and three pions would combine to have
total parity of (-1)(-1)(-1) = -1. Hence, if
conservation of parity holds, the θmeson should have parity of +1, and the
τ-meson of -1. This is the θ-τ puzzle.
The conservation of parity was
confirmed experimentally for the strong
and electromagnetic interactions, and
scientists naturally believed that it is also
conserved for weak interactions those
particles are involved. In 1956, T.D. Lee
and C.N. Yang pointed out that there
was no experimental evidence of parity
conservation for weak interactions. They
94
argued that θ- and τ-mesons would be identical if parity is violated in weak interactions. A few
months later, C.S. Wu et al. confirmed their idea.
They looked at the change of pseudoscalar S ⋅ p under mirror reflection, which shall just
change the sign if parity is conserved. Therefore, for the beta-decay process of
60
Co →
60
Ni +e − + υe , if
60
Co is polarized, then number of up and down electrons emitted
must be equal. Instead, they found perfect correlation with asymmetric beta-decay signals to the
degree of nuclear polarization. Thus parity conservation in weak interaction is violated.
6.4 Berry’s phase, or geometric phase
References: 1. M. V. Berry, “Quantum Phase Factors Accompanying Adiabatic Changes,” in
Proc. Roy. Soc. London, Series A 392, 45 (1984). 2. B. R. Holstein, “The Adiabatic Theorem and
Barry’s Phase,” Am. J. Phys. 57, 1079 (1989).
The arbitrary phase of a quantum state, δ (phase factor e iδ ), is obviously cannot be
determined experimentally. Mathematically we set it to be some values for consistency. But
Berry’s phase is uniquely defined by a contour integration.
For a time-dependent Schrödinger equation,
i
The general solution is
∂
ψ (t) = H ψn (t) = En ψn (t)
∂t n
ψn (t) = e iδe
−iEnt/
ψn (0)
Then under adiabatic variation, i.e., Hamiltonian varies extremely slowly, we can write
∂
i
ψ (t) = H(t) ψn (t) = En (t) ψn (t)
∂t n
The general solution is
iγ (t ) −iα (t )
ψn (t) = e n e n φn (t)
(6.73)
(6.74)
(6.75)
(6.76)
Here the dynamical phase is
αn (t) =
and the geometric (Berry’s) phase is
1 t
E(t ′)dt ′
 ∫0
(6.77)
∂
(6.78)
φ (t ′) dt ′
0
∂t n
The variation in time is dictated by a parameter or a set of parameter, for example, position r,
t
γn (t) = ∫ i φn (t ′)
95
then using the chain rule,
∂φ ∂φ ∂r
=
, one obtains
∂t
∂r ∂t
rf
γn (t) = ∫ i φn (r) ∇r φn (r) ⋅dr
ri
(6.79)
In the case of the vector r traces a closed contour, i.e., ri = r(0) = r(T ) = rf , the contour
integration determines uniquely the geometric phase
γn (C ) = i ∫
 φn (r) ∇r φn (r) ⋅d r
C
(6.80)
Now the geometric phase γn (C ) depends only on the contour integral, independent of arbitrary
quantum phase. That is when φn (r,t) → e iδ(r) φn (r,t) , Eq. (6.80) is invariant.
1
Berry’s phase for spin-1/2 system: γ±(C ) =  Ω , where Ω is the solid angle subtended by
2
the path through the vector r(t) travels. This is a significant result due to ultra-relativity.
Observation of Berry's Phase in a Solid-State Qubit,
Science 318, 1889 (2007)
See Phys. Rev. Lett. 61, 2030 (1988)
96
6.5 Wigner-Eckart Theorem
Early this semester we studied the quantum radiation theory, i.e., the interaction of electromagnetic
field with atoms. It is often necessary to evaluate matrix elements of vector and tensor operators
with respect to angular-momentum eigenstates.
For example, within the electric dipole approximation, we need to evaluate
f p̂ i = imω fi f r i .
(6.81)
However, when the first-order electric dipole transition vanishes (forbidden), the second-order
magnetic dipole and electric quadrupole approximations are responsible for the much weaker
optical transitions. The corresponding tensor operators are




1
1
M̂ + = [(k ⋅ r)(ξ ⋅ p̂) + (ξ ⋅ r)(k ⋅ p̂)], M̂ − = [(k ⋅ r)(ξ ⋅ p̂)−(ξ ⋅ r)(k ⋅ p̂)].
2
2
These are Cartesian tensor operators, defined as
(6.82)
(6.83)
Tijk =U iVjWk 
where U, V, W are vector operators. A rank k Cartesian operator T (k ) has 3k components;
under rotation R (a 3×3 matrix) it transforms as the following,
Tijk → ∑ ∑ ∑ Ri,i ′Rj, j ′Rk,k ′ Ti ′, j ′,k ′
i′
j′
(6.84)
k′
6.5.1 Irreducible spherical tensors
But the Cartesian tensor ( k > 1 ) is reducible, i.e., it can be decomposed to a number of
irreducible spherical tensors, which are denoted as Tq(k ) . For example, the rank 2 tensor,
U iVj =
U V −U jVi ⎛⎜U iVj +U jVi U ⋅ V ⎞⎟ ,
U⋅V
δij + i j
+ ⎜⎜
−
δ ⎟⎟
⎜⎝
3
2
2
3 ij ⎟⎟⎠
(6.85)
is decomposed into a scalar, vector, and a rank 2 spherical tensor, corresponding to 3×3 = 1
+ 3 + 5.
A rank k spherical tensor operator Tq(k ) has 2k + 1 ( q runs from −k to k ) components;
under rotation, it transform as
D †(R)Tq(k )D(R) =
k
∑D
q ′=−k
(k )* (k )
q,q ′ q ′
T
.
(6.86)
Considering an infinitesimal rotation, one can verify that the above condition is equivalent to
[J z ,Tq(k ) ] = qTq(k ); [J ±,Tq(k ) ] = (k  q)(k ± q + 1) Tq(k ) .
97
(6.87)
Eq. (6.87) suggests a one-to-one correspondence between a spherical tensor Tq(k ) and
spherical harmonics Yl m (θ,φ) , when k = l and q = m . Here (θ,φ) defines a unit vector
⎛x y z ⎞
n = ⎜⎜⎜ , , ⎟⎟⎟ with r = x 2 + y 2 + z 2 . We can write down Yl m (θ,φ) as
⎝ r r r ⎟⎠
Y11(θ,φ) = −
3
sin(θ)e iφ
8π
=−
3 x + iy
,
8π r
Y10(θ,φ) =
3
sin(θ)e iφ
4π
=
3 z
,
4π r
Y1−1(θ,φ) =
3
sin(θ)e −iφ =
8π
(6.88)
3 x −iy
.
8π r
Similarly, we can define a rank 1 spherical tensor for an arbitrary vector V as the following:
V + iVy
V −iVy
(1)
T1(1) = − x
≡V+1, T0(1) =Vz ≡V0, T−1
= x
≡V−1 ,
2
2
(6.89)
where V±1 and V0 are alternative components of the vector V .
Then a rank 2 Cartesian tensor operator T (2) for two vectors U and V , Tij =U iVj , is
decomposed into three spherical tensor operators with rank = 0, 1, and 2, respectively,
T0(0) = −
(U×V)q
U ⋅ V U +1V−1 +U −1V+1 −U 0V0
=
, Tq(1) =
,
3
3
i 2
(2)
(2)
T±2
=U ±1V±1, T±1
=
U ±1V0 +U 0V±1
2
, T0(2) =
U +1V−1 + 2U 0V0 +U −1V+1
6
(6.90)
.
Systematically, we construct irreducible tensor Tq(k ) by multiplying two irreducible tensor
(k )
(k )
operators Xq 1 and Z q 1 :
1
1
Tq(k ) = ∑ ∑ k1k2;q1q 2 k1k2;kq Xq 1 Zq
(k )
q1
1
q2
(k2 )
,
(6.91)
2
Here k1k2;q1q 2 k1k2;kq is the Clebsch-Gordan coefficients, | k1 −k2 |≤ k ≤ k1 + k2 , q = q1 + q 2 ,
runs from –k to k. Obviously, Eq. (6.91) corresponds to
l1l2;lm = ∑ ∑ l1l2;m1m2 l1l2;lm l1m1 ⊗ l2m2 .
m1
(6.92)
m2
Therefore, one can construct tensor operators of higher or lower ranks from two tensor operators.
6.5.2 Wigner-Eckart theorem
This is one of the most important theorems in quantum mechanics, which states that the matrix
elements of a tensor operator with respect to angular-momentum eigenstates satisfy
98
j ′m ′ Tq(k ) jm = jk;mq jk; j ′m ′ j ′ T (k ) j ,
(6.93)
where the reduced matrix elements j ′ T (k ) j depend only on k , j and j ′ . Eq. (6.93) tells us
that, for a given set of k , j and j ′ , if the matrix elements j ′m ′ Tq(k ) jm for one specific
combination of m ′ , m and q is known, then the rest of the matrix elements are determined by
the CG coefficients jk;mq j ′m ′ . It allows considerable simplifications in calculating the
matrix elements and leads to many interesting selection rules for non-zero matrix elements.
Proof of the Wigner-Eckart theorem. From Eq. (6.87) we obtain
(k )
j ′m ′ [J ±,Tq(k ) ] jm =  (k  q)(k ± q + 1) j ′m ′ Tq±1
jm .
(6.94)
On the other hand, the left-hand side is
j ′m ′ [J ±,Tq(k ) ] jm =  ( j ′ ± m ′)( j ′  m ′ + 1) j ′, m ′  1 Tq(k ) jm
−  (j  m)(j ± m + 1) j ′m ′ Tq(k ) j,m ± 1 .
(6.95)
Combining Eqs. (6.94) and (6.95), the recursion relations are derived
( j ′ ± m ′)( j ′  m ′ + 1) j ′, m ′  1 Tq(k ) jm =
(k )
(j  m)(j ± m + 1) j ′m ′ Tq(k ) j,m ± 1 + (k  q)(k ± q + 1) j ′m ′ Tq±1
jm .
(6.96)
This is exactly the same as Eq. 4.42, if we make the following substitutions
j ′ → j, m ′ → m, j → j1, m → m1, k → j 2, q → m2 .
(6.97)
Therefore j ′m ′ Tq(k ) jm is proportional to the CG coefficient jk;mq jk; j ′m ′ by a constant λ ,
which depends only on k , j and j ′ . We can set λ = j ′ T (k ) j , the reduced matrix elements.
This proves the Wigner-Eckart theorem.
6.5.3 Applications of the Wigner-Eckart theorem
Example 1. Prove that j ′m ′ Tq(k ) jm = 0 , unless j ′ − j ≤ k ≤ j ′ + j and q = m ′ −m .
Example 2. Selection rules for optical transitions within the electric dipole approximation.
Here the transition matrix elements are j ′m ′ p jm , and the operator p is a rank 1 tensor,
(1)
which can be written as a spherical tensor Tq=±1,0
. Thus we obtain the selection rules,
⎧
⎪ ±1
.
Δm ≡ m ′ −m = ±1, 0; Δj ≡ j ′ − j = ⎪
⎨
⎪
⎪
⎩ 0
99
(6.98)
The only exception is that the j ′ = 0 → j = 0 transition is forbidden (Why?)
Example 3. The projection theorem when j ′ = j :
jm J ⋅ V jm
jm ′ Vq jm =
(6.99)
jm ′ J q jm ,
 2 j(j + 1)
where V is an arbitrary vector operator, and J the angular momentum. Using the spherical tensor
notation, q = ±1, 0 , k = 1 . Using the Wigner-Eckart theorem, we obtain
jm ′ Vq jm
jm ′ J q jm
=
j V j
j J j
.
(6.100)
We need to show that jm J ⋅ V jm is proportional to j V j . Eq. (6.90) ⇒
J ⋅ V = J 0V0 −J +1V−1 −J −1V+1 ,
where J 0 = J z , and J ±1 = 
(6.101)
2
2
(J x ± iJ y ) = 
J , based on Eq. (6.89). Since operators J z
2
2 ±
and J ± do not change j and jm ′ Vq jm ∝ j V j , jm J ⋅ V jm = c jm j V j .
Furthermore, c jm is independent of m because J ⋅ V is a scalar operator. Therefore,
jm J ⋅ V jm = c j j V j ,
(6.102)
For an arbitrary vector operator V . If we set V = J , then jm J2 jm = c j j J j , and Eq.
(6.100) becomes
jm ′ Vq jm
jm ′ J q jm
=
jm J ⋅ V jm
jm J2 jm
=
jm J ⋅ V jm
 2 j(j + 1)
.
Example 4. A magnetic moment operator

µ = g1J1 + g 2 J2 ,
(6.103)
(6.104)
where J1 + J2 = J , the total angular momentum. We can evaluate jj µ0 jj using the
projection theorem,
jj µ0 jj =

where J ⋅ µ = g1J ⋅ J1 + g 2 J ⋅ J2 , and

jj J ⋅ µ jj
 2 j(j + 1)
J ⋅ J1 = (J2 + J12 − J22 ) / 2,
jj J 0 jj ,
J ⋅ J2 = (J2 − J12 + J22 ) / 2 .
100
(6.105)
(6.106)
Here the jj state is the addition of eigenstates of J12 and J22 , i.e., jj = j1 j 2; j,m = j , and
jj J 0 jj = j , we finally obtain
jj µ0 jj =

⎡ j(j + 1)(g + g ) + j (j + 1)(g − g ) + j (j + 1)(g − g )⎤ . (6.107)
1
2
1 1
1
2
2 2
2
1 ⎦⎥
2(j + 1) ⎣⎢
Lande g-factor:
!
µ = gJ ,
(6.108)
jj µ0 jj = gj!.
(6.109)
where J = J1 + J2 . We obtain
Combining Eqs. (6.107) and (6.109), we find that
g=
j(j + 1) + j1(j1 + 1)− j 2(j 2 + 1)
2j(j + 1)
g1 +
101
j(j + 1) + j 2(j 2 + 1)− j1(j1 + 1)
2j(j + 1)
g2 .
(6.110)