J. Broida
UCSD Fall 2009
Phys 130B
QM II
Angular Momentum
1
Angular momentum in Quantum Mechanics
As is the case with most operators in quantum mechanics, we start from the classical definition and make the transition to quantum mechanical operators via the
standard substitution x → x and p → −i~∇. Be aware that I will not distinguish
a classical quantity such as x from the corresponding quantum mechanical operator
x. One frequently sees a new notation such as x̂ used to denote the operator, but
for the most part I will take it as clear from the context what is meant. I will also
generally use x and r interchangeably; sometimes I feel that one is preferable over
the other for clarity purposes.
Classically, angular momentum is defined by
L = r× p.
Since in QM we have
[xi , pj ] = i~δij
it follows that [Li , Lj ] 6= 0. To find out just what this commutation relation is, first
recall that components of the vector cross product can be written (see the handout
Supplementary Notes on Mathematics)
(a × b)i = εijk aj bk .
Here I am using a sloppy summation convention where repeated indices are summed
over even if they are both in the lower position, but this is standard when it comes
to angular momentum. The Levi-Civita permutation symbol has the extremely
useful property that
εijk εklm = δil δjm − δim δjl .
Also recall the elementary commutator identities
[ab, c] = a[b, c] + [a, c]b
and
[a, bc] = b[a, c] + [a, b]c .
Using these results together with [xi , xj ] = [pi , pj ] = 0, we can evaluate the commutator as follows:
[Li , Lj ] = [(r × p)i , (r × p)j ] = [εikl xk pl , εjrs xr ps ]
= εikl εjrs [xk pl , xr ps ] = εikl εjrs (xk [pl , xr ps ] + [xk , xr ps ]pl )
1
= εikl εjrs (xk [pl , xr ]ps + xr [xk , ps ]pl ) = εikl εjrs (−i~δlr xk ps + i~δks xr pl )
= −i~εikl εjls xk ps + i~εikl εjrk xr pl = +i~εikl εljs xk ps − i~εjrk εkil xr pl
= i~(δij δks − δis δjk )xk ps − i~(δji δrl − δjl δri )xr pl
= i~(δij xk pk − xj pi ) − i~(δij xl pl − xi pj )
= i~(xi pj − xj pi ) .
But it is easy to see that
εijk Lk = εijk (r × p)k = εijk εkrs xr ps = (δir δjs − δis δjr )xr ps
= xi pj − xj pi
and hence we have the fundamental angular momentum commutation relation
[Li , Lj ] = i~εijk Lk .
(1.1a)
Written out, this says that
[Lx , Ly ] = i~Lz
[Ly , Lz ] = i~Lx
[Lz , Lx ] = i~Ly .
Note that these are just cyclic permutations of the indices x → y → z → x.
Now the total angular momentum squared is L2 = L · L = Li Li , and therefore
[L2 , Lj ] = [Li Li , Lj ] = Li [Li , Lj ] + [Li , Lj ]Li
= i~εijk Li Lk + i~εijk Lk Li .
But
εijk Lk Li = εkji Li Lk = −εijk Li Lk
where the first step follows by relabeling i and k, and the second step follows by the
antisymmetry of the Levi-Civita symbol. This leaves us with the important relation
[L2 , Lj ] = 0 .
(1.1b)
Because of these commutation relations, we can simultaneously diagonalize L2
and any one (and only one) of the components of L, which by convention is taken
to be L3 = Lz . The construction of these eigenfunctions by solving the differential
equations is at least outined in almost every decent QM text. (The old book Introduction to Quantum Mechanics by Pauling and Wilson has an excellent detailed
description of the power series solution.) Here I will follow the algebraic approach
that is both simpler and lends itself to many more advanced applications. The
main reason for this is that many particles have an intrinsic angular momentum
(called spin) that is without a classical analogue, but nonetheless can be described
mathematically exactly the same way as the above “orbital” angular momentum.
2
In view of this generality, from now on we will denote a general (Hermitian)
angular momentum operator by J. All we know is that it obeys the commutation
relations
[Ji , Jj ] = i~εijk Jk
(1.2a)
and, as a consequence,
[J 2 , Ji ] = 0 .
(1.2b)
Remarkably, this is all we need to compute the most useful properties of angular
momentum.
To begin with, let us define the ladder (or raising and lowering) operators
J+ = Jx + iJy
J− = (J+ )† = Jx − iJy .
(1.3a)
Then we also have
Jx =
1
(J+ + J− )
2
and
Jy =
1
(J+ − J− ) .
2i
(1.3b)
Because of (1.2b), it is clear that
[J 2 , J± ] = 0 .
(1.4)
In addtion, we have
[Jz , J± ] = [Jz , Jx ] ± i[Jz , Jy ] = i~Jy ± ~Jx
so that
[Jz , J± ] = ±~J± .
(1.5a)
Furthermore,
2
2
[Jz , J±
] = J± [Jz , J± ] + [Jz , J± ]J± = ±2~J±
and it is easy to see inductively that
k
k
.
[Jz , J±
] = ±k~J±
(1.5b)
It will also be useful to note
J+ J− = (Jx + iJy )(Jx − iJy ) = Jx2 + Jy2 − i[Jx , Jy ]
= Jx2 + Jy2 + ~Jz
and hence (since Jx2 + Jy2 = J 2 − Jz2 )
J 2 = J+ J− + Jz2 − ~Jz .
(1.6a)
Similarly, it is easy to see that we also have
J 2 = J− J+ + Jz2 + ~Jz .
3
(1.6b)
Because J 2 and Jz commute they may be simultaneously diagonalized, and we
denote their (un-normalized) simultaneous eigenfunctions by Yαβ where
J 2 Yαβ = ~2 αYαβ
and
Jz Yαβ = ~βYαβ .
Since Ji is Hermitian we have the general result
2
hJi2 i = hψ|Ji2 ψi = hJi ψ|Ji ψi = kJi ψk ≥ 0
and hence hJ 2 i − hJz2 i = hJx2 i + hJy2 i ≥ 0. But Jz2 Yαβ = ~2 β 2 Yαβ and hence we must
have
β2 ≤ α .
(1.7)
Now we can investigate the effect of J± on these eigenfunctions. From (1.4) we
have
J 2 (J± Yαβ ) = J± (J 2 Yαβ ) = ~2 α(J± Yαβ )
so that J± doesn’t affect the eigenvalue of J 2 . On the other hand, from (1.5a) we
also have
Jz (J± Yαβ ) = (J± Jz ± ~J± )Yαβ = ~(β ± 1)J± Yαβ
and hence J± raises or lowers the eigenvalue ~β by one unit of ~. And in general,
from (1.5b) we see that
Jz ((J± )k Yαβ ) = (J± )k (Jz Yαβ ) ± k~(J± )k Yαβ = ~(β ± k)(J± )k Yαβ
so the k-fold application of J± raises or lowers the eigenvalue of Jz by k units of ~.
This shows that (J± )k Yαβ is a simultaneous eigenfunction of both J 2 and Jz with
corresponding eigenvalues ~2 α and ~(β ± k), and hence we can write
(J± )k Yαβ = Yαβ±k
(1.8)
where the normalization is again unspecified.
Thus, starting from a state Yαβ with a J 2 eigenvalue ~2 α and a Jz eigenvalue ~β,
we can repeatedly apply J+ to construct an ascending sequence of eigenstates with
Jz eigenvalues ~β, ~(β + 1), ~(β + 2), . . . , all of which have the same J 2 eigenvalue
~2 α. Similarly, we can apply J− to construct a descending sequence ~β, ~(β − 1),
~(β − 2), . . . , all of which also have the same J 2 eigenvalue ~2 α. However, because
of (1.7), both of these sequences must terminate.
Let the upper Jz eigenvalue be ~βu and the lower eigenvalue be −~βl . Thus, by
definition,
and
Jz Yαβl = −~βl Yαβl
(1.9a)
Jz Yαβu = ~βu Yαβu
with
J+ Yαβu = 0
and
J− Yαβl = 0
and where, by (1.7), we must have
βu2 ≤ α
and
4
βl2 ≤ α .
(1.9b)
By construction, there must be an integral number n of steps from −βl to βu , so
that
βl + βu = n .
(1.10)
(In other words, the eigenvalues of Jz range over the n intervals −βl , −βl + 1,
−βl + 2, . . . , −βl + (βl + βu ) = βu .)
Now, using (1.6b) we have
J 2 Yαβu = J− J+ Yαβu + (Jz2 + ~Jz )Yαβu .
Then by (1.9b) and the definition of Yαβ , this becomes
~2 αYαβu = ~2 βu (βu + 1)Yαβu
so that
α = βu (βu + 1) .
In a similar manner, using (1.6a) we have
J 2 Yαβl = J+ J− Yαβl + (Jz2 − ~Jz )Yαβl
or
~2 αYαβl = ~2 βl (βl + 1)Yαβl
so also
α = βl (βl + 1) .
Equating both of these equations for α and recalling (1.10) we conclude that
β u = βl =
n
:= j
2
where j is either integral or half-integral, depending on whether n is even or odd.
In either case, we finally arrive at
α = j(j + 1)
(1.11)
and the eigenvalues of Jz range from −~j to ~j in integral steps of ~.
We can now label the eigenvalues of Jz by ~m instead of ~β, where the integer or
half-integer m ranges from −j to j in integral steps. Thus our eigenvalue equations
may be written
J 2 Yjm = j(j + 1)~2 Yjm
Jz Yjm = m~Yjm .
(1.12)
We say that the states Yjm are angular momentum eigenstates with angular momentum j and z-component of angular momentum m. Note that (1.9b) is now
written
and
J− Yj−j = 0 .
(1.13)
J+ Yjj = 0
5
Since (J± )† = J∓ , using equations (1.6) we have
hJ± Yjm |J± Yjm i = hYjm |J∓ J± Yjm i = hYjm |(J 2 − Jz2 ∓ ~Jz )Yjm i
= ~2 [j(j + 1) − m2 ∓ m]hYjm |Yjm i
= ~2 [j(j + 1) − m(m ± 1)]hYjm |Yjm i .
We know that J± Yjm is proportional to Yjm±1 . So if we assume that the Yjm are
normalized, then this equation implies that
p
J± Yjm = ~ j(j + 1) − m(m ± 1) Yjm±1 .
(1.14)
If we start at the top state Yjj , then by repeatedly applying J− , we can construct all
of the states Yjm . Alternatively, we could equally well start from Yj−j and repeatedly
apply J+ to also construct the states.
Let us see if we can find a relation that defines the Yjm . Since Yjj is defined
by J+ Yjj = 0, we will only define our states up to an overall normalization factor.
Using (1.14), we have
p
p
J− Yjj = ~ j(j + 1) − j(j − 1) Yjj−1 = ~ 2j Yjj−1
or
1
Yjj−1 = ~−1 √ J− Yjj .
2j
Next we have
(J− )2 Yjj = ~2
or
p
p p
2j j(j + 1) − (j − 1)(j − 2) Yjj−2 = ~2 (2j)2(2j − 1) Yjj−2
1
Yjj−2 = ~−2 p
(J− )2 Yjj .
(2j)(2j − 1)2
And once more should do it:
p
p
(J− )3 Yjj = ~3 (2j)(2j − 1)2 j(j + 1) − (j − 2)(j − 3) Yjj−3
p
= ~3 (2j)(2j − 1)(2)(3)(2j − 2) Yjj−3
or
1
Yjj−3 = ~−3 p
(J− )3 Yjj .
2j(2j − 1)(2j − 2)3!
Noting that m = j − 3 so that 3! = (j − m)! and 2j − 3 = 2j − (j − m) = j + m, it
is easy to see we have shown that
s
(j + m)!
m
m−j
Yj = ~
(J− )j−m Yjj .
(1.15a)
(2j)!(j − m)!
6
And an exactly analogous argument starting with Yj−j and applying J+ repeatedly
shows that we could also write
s
(j − m)!
Yjm = ~−m−j
(J+ )j+m Yj−j .
(1.15b)
(2j)!(j + m)!
It is extremely important to realize that everything we have done up to this
point depended only on the commutation relation (1.2a), and hence applies to both
integer and half-integer angular momenta. While we will return in a later section to
discuss spin (including the half-integer case), for the rest of this section we restrict
ourselves to integer values of angular momentum, and hence we will be discussing
orbital angular momentum.
The next thing we need to do is to actually construct the angular momentum
wave functions Ylm (θ, φ). (Since we are now dealing with orbital angular momentum, we replace j by l.) To do this, we first need to write L in spherical coordinates.
One way to do this is to start from Li = (r×p)i = εijk xj pk where pk = −i~(∂/∂xk ),
and then use the chain rule to convert from Cartesian coordinates xi to spherical
coordinates (r, θ, φ). Using
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
so that
r = (x2 + y 2 + z 2 )1/2
θ = cos−1 z/r
φ = tan−1 y/x
we have, for example,
∂r ∂
∂θ ∂
∂φ ∂
∂
=
+
+
∂x
∂x ∂r ∂x ∂θ
∂x ∂φ
=
x ∂
∂
xz ∂
y
+ 3
− 2 cos2 φ
r ∂r r sin θ ∂θ x
∂φ
= sin θ cos φ
cos θ cos φ ∂
sin φ ∂
∂
+
−
∂r
r
∂θ r sin θ ∂φ
with similar expressions for ∂/∂y and ∂/∂z. Then using terms such as
∂
∂
Lx = ypz − zpy = −i~ y
−z
∂z
∂y
we eventually arrive at
∂
∂
− cot θ cos φ
Lx = −i~ − sin φ
∂θ
∂φ
∂
∂
− cot θ sin φ
Ly = −i~ cos φ
∂θ
∂φ
Lz = −i~
∂
.
∂φ
(1.16a)
(1.16b)
(1.16c)
7
However, another way is to start from the gradient in spherical coordinates (see
the section on vector calculus in the handout Supplementary Notes on Mathematics)
∇ = r̂
∂
∂
1 ∂
1
+ θ̂
+ φ̂
.
∂r
r ∂θ
r sin θ ∂φ
Then L = r × p = −i~ r × ∇ = −i~ r (r̂ × ∇) so that (since r̂, θ̂ and φ̂ are
orthonormal)
∂
∂
1 ∂
1
L = −i~r r̂ × r̂
+ r̂ × θ̂
+ r̂ × φ̂
∂r
r ∂θ
r sin θ ∂φ
1 ∂
∂
− θ̂
= −i~ φ̂
∂θ
sin θ ∂φ
If we write the unit vectors in terms of their Cartesian components (again, see the
handout on vector calculus)
θ̂ = (cos θ cos φ, cos θ sin φ, − sin θ)
φ̂ = (− sin φ, cos φ, 0)
then
∂
∂
∂
∂
∂
+ ŷ cos φ
+ ẑ
− cot θ cos φ
− cot θ sin φ
L = −i~ x̂ − sin φ
∂θ
∂φ
∂θ
∂φ
∂φ
which is the same as we had in (1.16).
Using these results, it is now easy to write the ladder operators L± = Lx ± iLy
in spherical coordinates:
∂
∂
.
(1.17)
± i cot θ
L± = ±~e±iφ
∂θ
∂φ
To find the eigenfunctions Ylm (θ, φ), we start from the definition L+ Yll = 0. This
yields the equation
∂Yll
∂Y l
+ i cot θ l = 0 .
∂θ
∂φ
We can solve this by the usual approach of separation of variables if we write
Yll (θ, φ) = T (θ)F (φ). Substituting this and dividing by T F we obtain
1 ∂T
1 ∂F
= −i
.
T cot θ ∂θ
F ∂φ
Following the standard argument, the left side of this is a function of θ only, and
the right side is a function of φ only. Since varying θ won’t affect the right side,
and varying φ won’t affect the left side, it must be that both sides are equal to a
constant, which I will call k. Now the φ equation becomes
dF
= ikdφ
F
8
which has the solution F (φ) = eikφ (up to normalization). But Yll is an eigenfunction of Lz = −i~(∂/∂φ) with eigenvalue l~, and hence so is F (φ) (since T (θ) just
cancels out). This means that
−i~
∂eikφ
= k~eikφ := l~eikφ
∂φ
and therefore we must have k = l, so that (up to normalization)
Yll = eilφ T (θ) .
With k = l, the θ equation becomes
dT
cos θ
d sin θ
= l cot θ dθ = l
dθ = l
.
T
sin θ
sin θ
This is also easily integrated to yield (again, up to normalization)
T (θ) = sinl θ .
Thus, we can write
Yll = cll (sin θ)l eilφ
where cll is a normalization constant, fixed by the requirement that
Z
Z π
2
2
Yll dΩ = 2π cll
(sin θ)2l sin θ dθ = 1 .
(1.18)
0
I will go through all the details involved in doing this integral. You are free to skip
down to the result if you wish (equation (1.21)), but this result is also used in other
physical applications.
First I want to prove the relation
Z
Z
n−1
1
n−1
n
x cos x +
sinn−2 x dx .
(1.19)
sin x dx = − sin
n
n
R
by parts (remember the formula u dv = uv −
RThis is done as an integration
v du) letting u = sinn−1 x and dv = sin x dx so that v = − cos x and du =
(n − 1) sinn−2 x cos x dx. Then (using cos2 x = 1 − sin2 x in the third line)
Z
Z
n
sin x dx = sinn−1 x sin x dx
n−1
= − sin
x cos x + (n − 1)
= − sinn−1 x cos x + (n − 1)
Z
Z
sinn−2 x cos2 x dx
sinn−2 x dx − (n − 1)
Z
sinn x dx .
Now move the last term on the right over to the left, divide by n, and the result is
(1.19).
9
We need to evaluate (1.19) for the case where n = 2l + 1. To get the final result
in the form we want, we will need the basically simple algebraic result
(2l + 1)!! =
(2l + 1)!
2l l!
l = 1, 2, 3, . . .
(1.20)
where the double factorial is defined by
n!! = n(n − 2)(n − 4)(n − 6) · · · .
There is nothing fancy about the proof of this fact. Noting that n = 2l + 1 is always
odd, we have
n!! = 1 · 3 · 5 · 7 · 9 · · · (n − 4) · (n − 2) · n
=
1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · · · (n − 4) · (n − 3) · (n − 2) · (n − 1) · n
2 · 4 · 6 · 8 · · · (n − 3) · (n − 1)
=
1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · · · (n − 4) · (n − 3) · (n − 2) · (n − 1) · n
n−1
(2 · 1)(2 · 2)(2 · 3)(2 · 4) · · · (2 · n−3
2 )(2 · 2 )
=
n!
2
n−1
2
( n−1
2 )!
.
Substituting n = 2l + 1 we arrive at (1.20).
Now we are ready to do the integral in (1.18). Since the limits of integration
are 0 and π, the first term on the right side of (1.19) always vanishes, and we can
ignore it. Then we have
Z π
Z π
2l
(sin x)2l+1 dx =
(sin x)2l−1 dx
2l + 1 0
0
Z π
2l
2l − 2
=
(sin x)2l−3 dx
2l + 1
2l − 1
0
Z π
2l
2l − 2
2l − 4
=
(sin x)2l−5 dx
2l + 1
2l − 1
2l − 3
0
2l − 2
2l − 4
2l
= ··· =
2l + 1
2l − 1
2l − 3
Z π
2l − (2l − 2)
sin x dx
× ···×
2l − (2l − 3)
0
=
2l l(l − 1)(l − 2) · · · (l − (l − 1))
2
(2l + 1)!!
=2
(2l l!)2
2l l!
=2
(2l + 1)!!
(2l + 1)!
10
(1.21)
Rπ
where we used 0 sin x dx = 2 and (1.20).
Using this result, (1.18) becomes
4π cll
and hence
cll = (−1)l
2
(2l l!)2
=1
(2l + 1)!
(2l + 1)!
4π
1/2
1
2l l!
(1.22)
where we included a conventional arbitrary phase factor (−1)l . Putting this all
together, we have the top orbital angular momentum state
1/2
1
l
l (2l + 1)!
(sin θ)l eilφ .
(1.23)
Yl (θ, φ) = (−1)
4π
2l l!
To construct the rest of the states Ylm (θ, φ), we repeatedly apply L− from equation (1.17) to finally obtain
1/2
1/2
1
(l + m)!
(2l + 1)!
Ylm (θ, φ) = (−1)l
4π
2l l! (2l)!(l − m)!
× eimφ (sin θ)−m
dl−m
(sin θ)2l .
d(cos θ)l−m
(1.24)
It’s just not worth going through this algebra also.
2
Spin
It is an experimental fact that many particles, and the electron in particular, have
an intrinsic angular momentum. This was originally deduced by Goudsmit and Uhlenbeck in their analysis of the famous sodium D line, which arises by the transition
from the 1s2 2s2 2p6 3p excited state to the ground state. What initially appears as a
strong single line is slightly split in the presence of a magnetic field into two closely
spaced lines (Zeeman effect). This (and other lines in the Na spectrum) indicates a
doubling of the number of states available to the valence electron.
To explain this “fine structure” of atomic spectra, Goudsmit and Uhlenbeck
proposed in 1925 that the electron possesses an intrinsic angular momentum in
addition to the orbital angular momentum due to its motion about the nucleus.
Since magnetic moments are the result of current loops, it was originally thought
that this was due to the electron spinning on its axis, and hence this intrinsic angular
momentum was called spin. However, a number of arguments can be put forth to
disprove that classical model, and the result is that we must assume that spin is a
purely quantum phenomena without a classical analogue.
As I show at the end of this section, the classical model says that the magnetic
moment µ of a particle of charge q and mass m moving in a circle is given by
q
µ=
L
2mc
11
where L is the angular momentum with magnitude L = mvr. Furthermore, the
energy of such a charged particle in a magnetic field B is −µ · B. Goudsmit and
Uhlenbeck showed that the ratio of magnetic moment to angular momentum of the
electron was in fact twice as large as it would be for an orbital angular momentum.
(This factor of 2 is explained by the relativistic Dirac theory.) And since we know
that a state with angular momentum l is (2l + 1)-fold degenerate, the splitting
implies that the electron has an angular momentum ~/2.
From now on, we will assume that spin is described by the usual angular momentum theory, and hence we postulate a spin operator S and corresponding eigenstates
|s ms i such that
[Si , Sj ] = iεijk Sk
2
(2.1a)
2
S |s ms i = s(s + 1)~ |s ms i
(2.1b)
Sz |s ms i = ms ~|s ms i
p
S± |s ms i = ~ s(s + 1) − ms (ms ± 1) |s ms ± 1i .
(2.1c)
(2.1d)
|χi = c+ |ẑ +i + c− |ẑ −i .
(2.2a)
(I am switching to the more abstract notation for the eigenstates because the states
themselves are a rather abstract concept.) We will sometimes drop the subscript s
on ms if there is no danger of confusing this with the eigenvalue of Lz , which we
will also sometimes write as ml . Be sure to realize that particles can have a spin s
that is any integer multiple of 1/2, and there are particles in nature that have spin
0 (e.g., the pion), spin 1/2 (e.g., the electron, neutron, proton), spin 1 (e.g., the
photon, but this is a little bit subtle), spin 3/2 (the ∆’s), spin 2 (the hypothesized
graviton) and so forth.
Since the z component of electron spin can take only one of two values ±~/2, we
will frequently denote the corresponding orthonormal eigenstates simply by |ẑ ±i,
where |ẑ +i is called the spin up state, and |ẑ −i is called the spin down state.
An arbitrary spin state |χi is of the form
If we wish to think in terms of explicit matrix representations, then we will write
this in the form
(z)
(z)
χ = c + χ+ + c − χ− .
(2.2b)
Be sure to remember that the normalized states |s ms i belong to distinct eigenvalues
of a Hermitian operator, and hence they are in fact orthonormal.
To construct the matrix representation of spin operators, we need to first choose
a basis for the space of spin states. Since for the electron there are only two possible
states for the z component, we need to pick a basis for a two-dimensional space,
and the obvious choice is the standard basis
" #
" #
1
0
(z)
(z)
χ+ =
and
χ− =
.
(2.3)
0
1
12
With this choice of basis, we can now construct the 2 × 2 matrix representation of
the spin operator S.
Note that the existence of spin has now led us to describe the electron by a
multi-component state vector (in this case, two components), as opposed to the
scalar wave functions we used up to this point. These two-component states are
frequently called spinors.
(z)
The states χ± were specifically constructed to be eigenstates of Sz (recall that
we simultaneously diagonalized J 2 and Jz ), and hence the matrix representation
of Sz is diagonal with diagonal entries that are precisely the eigenvalues ±~/2. In
other words, we have
~
hẑ ± |Sz |ẑ ±i = ±
2
so that
"
#
0
~ 1
Sz =
.
(2.4)
2 0 −1
(We are being somewhat sloppy with notation and using the same symbol Sz to
denote both the operator and its matrix representation.) That the vectors defined
in (2.3) are indeed eigenvectors of Sz is easy to verify:
"
" #
"
" #
#" #
#" #
0
0
1
0
~ 1
~ 1
~ 1
~ 0
and
.
=
=−
2 0 −1
2 0
2 0 −1
2 1
0
1
To find the matrix representations of Sx and Sy , we use (2.1d) together with
S± = Sx ± iSy so that Sx = (S+ + S− )/2 and Sy = (S+ − S− )/2i. From
p
S± |ẑ ms i = ~ 3/4 − ms (ms ± 1) |ẑ ms ± 1i
we have
S+ |ẑ +i = 0
S− |ẑ +i = ~|ẑ −i
S+ |ẑ −i = ~|ẑ +i
S− |ẑ −i = 0 .
Therefore the only non-vanishing entry in S+ is hẑ + |S+ |ẑ −i, and the only nonvanishing entry in S− is hẑ − |S− |ẑ +i. Thus we have
"
#
"
#
0 1
0 0
S+ = ~
and
S− = ~
.
0 0
1 0
Using these, it is easy to see that
"
#
~ 0 1
Sx =
2 1 0
and
13
~
Sy =
2
"
0 −i
i
0
#
.
(2.5)
And from (2.4) and (2.5) it is easy to calculate S 2 = Sx2 + Sy2 + Sz2 to see that
"
#
3 2 1 0
2
S = ~
4
0 1
which agrees with (2.1b).
It is conventional to write S in terms of the Pauli spin matrices σ defined by
S=
where
σx =
"
0 1
1 0
#
σy =
"
~
σ
2
0
−i
i
0
#
σz =
"
1
0
0
−1
#
.
(2.6)
Memorize these. The Pauli matrices obey several relations that I leave to you to
verify (recall that the anticommutator is defined by [a, b]+ = ab + ba):
[σi , σj ] = 2iεijk σk
σi σj = iεijk σk
(2.7a)
for i 6= j
[σi , σj ]+ = 2Iδij
(2.7b)
(2.7c)
σi σj = Iδij + iεijk σk
(2.7d)
Given three-component vectors a and b, equation (2.7d) also leads to the extremely
useful result
(a · σ)(b · σ) = (a · b)I + i(a × b) · σ .
(2.8)
We will use this later when we discuss rotations.
From the standpoint of physics, what equations (2.2) say is that if we have
an electron (or any spin one-half particle) in an arbitrary spin state χ, then the
probablility is |c+ |2 that a measurement of the z component of spin will result in
2
+~/2, and the probability is |c− | that the measurement will yield −~/2. We can
say this because (2.2) expresses χ as a linear superposition of eigenvectors of Sz .
But we could equally well describe χ in terms of eigenvectors of Sx . To do so,
we simply diagonalize Sx and use its eigenvectors as a new set of basis vectors for
our two-dimensional spin space. Thus we must solve Sx v = λv for λ and then the
eigenvectors v. This is straightforward. The eigenvalue equation is (Sx − λI)v = 0,
so in order to have a non-trivial solution we must have
det(Sx − λI) =
−λ
~/2
~/2
−λ
= λ2 − ~2 /4 = 0
so that λ± = ±~/2 as we should expect. To find the eigenvector corresponding to
λ+ = +~/2 we solve
"
#" #
1
a
~ −1
=0
(Sx − λ+ I)v =
2
1 −1
b
14
(x)
so that a = b and the normalized eigenvector is (we now write v = χ± )
" #
1
1
(x)
.
χ+ = √
2 1
For λ− = −~/2 we have
~
(Sx − λ− I)v =
2
"
1
1
1
1
#" #
c
d
(2.9a)
=0
so that c = −d and the normalized eigenvector is now (where we arbitrarily choose
c = +1)
"
#
1
1
(x)
χ− = √
.
(2.9b)
2 −1
To understand just what this means, suppose we have an arbitrary spin state
2
2
(normalized so that |α| + |β| = 1)
" #
α
χ=
.
β
Be sure to understand that this is a vector in a two-dimensional space, and it exists
independently of any basis. In terms of the basis (2.3) we can write
" #
" #
1
0
(z)
(z)
χ=α
+β
= αχ+ + βχ−
0
1
2
so that the probability is |α| that we will measure the z component of spin to be
2
+~/2, and |β| that we will measure it to be −~/2.
Alternatively, we can express χ in terms of the basis (2.9):
" #
" #
"
# " √
√ #
α
1
a/ 2 + b/ 2
a 1
b
=√
+√
=
√
√
2 1
2 −1
β
a/ 2 − b/ 2
or
√
√
α = a/ 2 + b/ 2
√
√
β = a/ 2 − b/ 2 .
and
Solving for a and b in terms of α and β we obtain
1
a = √ (α + β)
2
so that
χ=
α+β
√
2
1
b = √ (α − β)
2
and
(x)
χ+ +
15
α−β
√
2
(x)
χ− .
(2.10)
2
Thus the probability of measuring the x component of spin to be +~/2 is |α + β| /2,
and the probability of measuring the value to be −~/2 is |α − β|2 /2.
(Remark: What we just did was nothing more than the usual change of basis in
a vector space. We started with a basis
" #
" #
1
0
e1 =
and
e2 =
0
1
(which we chose to be the eigenvectors of Sz ) and changed to a new basis
" #
"
#
1
1
1 1
and
ē2 = √
ē1 = √
2 1
2 −1
√
√
(which were the eigenvectors of Sx ). Since ē1 = (e1 +e2 )/ 2 and ē2 = (e1 −e2 )/ 2,
we see that this change of basis is described by the transition matrix defined by
ēi = ej pj i or
"
#
1
1
1
= P −1 .
P = √
2 1 −1
Then a vector
χ=
"
α
β
#
can be written in terms of either the basis {ei } as
" #
" #
1
0
χ=α
+β
= χ1 e 1 + χ2 e 2
0
1
or in terms of the basis {ēi } as χ = χ̄1 ē1 + χ̄2 ē2 where χ̄i = (p−1 )i j χj . This then
immediately yields
1
1
χ = √ (α + β)ē1 + √ (α − β)ē2
2
2
which is just (2.10).)
Now we ask how to incorporate spin into the general solution to the Schrödinger
equation for the hydrogen atom. Let us ignore terms that couple spin with the
orbital angular momentum of the electron. (This “L–S coupling” is relatively small
compared to the electron binding energy, and can be ignored to first order. We will,
however, take this into account when we discuss perturbation theory.) Under these
conditions, the Hamiltonian is still separable, and we can write the total stationary
state wave function as a product of a spatial part ψnlml times a spin part χ(ms ).
Thus we can write the complete hydrogen atom wave function in the form
Ψnlml ms = ψnlml χ(ms ) .
Since the Hamiltonian is independent of spin, we have
HΨ = H[ψnlml χ(ms )] = χ(ms )Hψnlml = En [χ(ms )ψnlml ] = En Ψ
16
so that the energies are unchanged. However, because of the spin function, we have
doubled the number of states corresponding to a given energy.
A more mathematically correct way to write these complete states is as the
tensor (or direct) product
|Ψi = |ψnlml χ(ms )i := |ψnlml i ⊗ |χ(ms )i .
In this case, the Hamiltonian is properly written as H ⊗ I where H acts on the
vector space of spatial wave functions, and I is the identity operator on the vector
space of spin states. In other words,
(H ⊗ I)(|ψnlml i ⊗ |χ(ms )i) := H|ψnlml i ⊗ I|χ(ms )i .
This notation is particularly useful when treating two-particle states, as we will see
when we discuss the addition of angular momentum.
(Remark : You may recall from linear algebra that given two vector spaces V
and V ′ , we may define a bilinear map V × V ′ → V ⊗ V ′ that takes ordered pairs
(v, v ′ ) ∈ V × V ′ and gives a new vector denoted by v ⊗P
v ′ . Since this map
Pis bilinear
′
by definition,
if
we
have
the
linear
combinations
v
=
x
v
and
v
=
yj vj′ then
i i
P
′
′
′
v⊗v =
xi yj (vi ⊗ vj ). In particular, if V has basis {ei } and V has basis {e′j },
then {ei ⊗ e′j } is a basis for V ⊗ V ′ which is then of dimension (dim V )(dim V ′ )
and called the direct (or tensor) product of V and V ′ . Then, if we are given
two operators A ∈ L(V ) and B ∈ L(V ′ ), the direct product of A and B is the
operator A ⊗ B defined on V ⊗ V ′ by (A ⊗ B)(v ⊗ v ′ ) := A(v) ⊗ B(v ′ ).)
2.1
Supplementary Topic: Magnetic Moments
Consider a particle of charge q moving in a circular orbit. It forms an effective
current
q
qv
∆q
=
=
.
I=
∆t
2πr/v
2πr
By definition, the magnetic moment has magnitude
µ=
I
qv
qvr
× area =
· πr2 =
.
c
2πrc
2c
But the angular momentum of the particle is L = mvr so we conclude that the
magnetic moment due to orbital motion is
µl =
q
L.
2mc
(2.11)
The ratio of µ to L is called the gyromagnetic ratio.
While the above derivation of (2.11) was purely classical, we know that the
electron also possesses an intrinsic spin angular momentum. Let us hypothesize
that the electron magnetic moment associated with this spin is of the form
µs = g
−e
S.
2mc
17
The constant g is found by experiment to be very close to 2. (However, the relativistic Dirac equation predicts that g is exactly 2. Higher order corrections in
quantum electrodynamics predict a slightly different value, and the measurement
of g − 2 is one of the most accurate experimental result in all of physics.)
Now we want to show is that the energy of a magnetic moment in a uniform
magnetic field is given by −µ · B where µ for a loop of area A carrying current
I is defined to have magnitude IA and pointing perpendicular to the loop in the
direction of your thumb if the fingers of your right hand are along the direction of
the current. To see this, we simply calculate the work required to rotate a current
loop from its equilibrium position to the desired orientation.
Consider the figure shown below, where the current flows counterclockwise out
of the page at the bottom and into the page at the top.
B
µ
a/2
θ
θ
θ
FB
B
FB
a/2
B
Let the loop have length a on the sides and b across the top and bottom, so its area
is ab. The magnetic force on a current-carrying wire is
Z
FB = Idl × B
and hence the forces on the opposite “a sides” of the loop cancel, and the force on
the top and bottom “b sides” is FB = IbB. The equilibrium position of the loop is
horizontal, so the potential energy of the loop is theR work required to rotate it from
θ = 0 to some value θ. This work is given by W = F · dr where F is the force that
I must apply against the magnetic field to rotate the loop.
Since the loop is rotating, the force I must apply at the top of the loop is in the
direction of µ and perpendicular to the loop, and hence has magnitude FB cos θ.
Then the work I do is (the factor of 2 takes into account both the top and bottom
sides)
Z
Z
Z θ
W = F · dr = 2 FB cos θ(a/2)dθ = IabB
cos θ dθ = µB sin θ .
0
But note that µ · B = µB cos(90 + θ) = −µB sin θ, and therefore
W = −µ · B .
18
(2.12)
In this derivation, I never explicitly mentioned the torque on the loop due to B.
However, we see that
kNk = kr × FB k = 2(a/2)FB sin(90 + θ) = IabB sin(90 + θ)
= µB sin(90 + θ) = kµ × Bk
and therefore
Note that W =
R
N = µ× B.
(2.13)
kNk dθ. We also see that
dL
d
dp
= r×p =r×
=r×F
dt
dt
dt
where we used p = mv and ṙ × p = v × p = 0. Therefore, as you should already
know,
dL
= N.
(2.14)
dt
3
Mathematical Digression:
Rotations and Linear Transformations
Let’s take a look at how the spatial rotation operator is defined. Note that there
are two ways to view symmetry operations such as translations and rotations. The
first is to leave the coordinate system unchanged and instead move the physical
system. This is called an active transformation. Alternatively, we can leave the
physical system alone and change the coordinate system, for example by translation
or rotation. This is called a passive transformation. In the case of an active
transformation, we have the following situation:
x2
r̄
r
θ
φ
x1
Here the vector r is rotated by θ to give the vector r̄ where, of course, krk = kr̄k = r.
We define a linear transformation T by r̄ = T (r). (This is linear because it is easy
to see that rotating the sum of two vectors is the sum of the rotated vectors.) From
the diagram, the components of r̄ are given by
x̄1 = r cos(θ + φ) = r cos θ cos φ − r sin θ sin φ
= (cos θ)x1 − (sin θ)x2
19
x̄2 = r sin(θ + φ) = r sin θ cos φ + r cos θ sin φ
= (sin θ)x1 + (cos θ)x2
or
"
x̄1
x̄2
#
=
"
cos θ
− sin θ
sin θ
cos θ
#"
x1
x2
#
.
(3.1)
Since T is a linear transformation, it is completely specified by defining its values
on a basis because
X
X
xi T ei .
xi ei =
T (r) = T
i
i
2
But T ei is just another vector in R , and hence it can be expressed in terms of the
basis {ei } as
X
ej aji .
(3.2)
T ei =
j
Be sure to note which indices are summed over in this expression. The matrix (aji )
is called the matrix representation of the linear transformation T with respect
to the basis {ei }. You will sometimes see this matrix written as [T ]e .
It is very important to realize that a linear transformation T takes the ith basis
vector into the ith column of its matrix representation. This is easy to see if we
write out the components of (3.2). Simply note that with respect to the basis {ei },
we have
" #
" #
1
0
e1 =
and
e2 =
0
1
and therefore
T ei = e1 a1i + e2 a2i =
" #
1
0
a1i +
" #
0
1
a2i =
"
a1i
a2i
#
which is just the ith column of the matrix A = (aji ).
As an example, let V have a basis {v1 , v2 , v3 }, and let T : V → V be the linear
transformation defined by
T v1 = 3v1
+ v3
T v2 = v1 − 2v2 − v3
T v3 =
v2 + v3
Then the representation of T (relative to this basis) is


3
1 0
[T ]v =  0 −2 1  .
1 −1 1
Now let V be an n-dimensional vector space, and let W be a subspace of V .
Let T be an operator on V , and suppose W has the property that T w ∈ W for
20
every w ∈ W . Then we say that W is T-invariant (or simply invariant when the
operator is understood).
What can we say about the matrix representation of T under these circumstances? Well, let W have the basis {w1 , . . . , wr }, and extend this to a basis
{w1 , . . . , wr , v1 , . . . , vn−r } for V . Since T w ∈ W for any w ∈ W , we must have
T wi =
r
X
wj aji
j=1
for some set of scalars {aji }. But for any vi there is no such restriction since all we
know is that T vi ∈ V . So we have
T vi =
r
X
wj bji +
j=1
n−r
X
vk cki
k=1
for scalars {bji } and {cki }. Then since T takes the ith basis vector to the ith column
of the matrix representation [T ], we must have
"
#
A B
[T ] =
0 C
where A is an r × r matrix, B is an r × (n − r) matrix, and C is an (n − r) × (n − r)
matrix. Such a matrix is said to be a block matrix, and [T ] is in block triangular
form.
Now let W be an invariant subspace of V . If it so happens that the subspace
of V spanned by the rest of the vectors {v1 , . . . , vn−r } is also invariant, then the
matrix representation of T will be block diagonal (because all of the bji in the above
expansion will be zero). As we shall see, this is in fact what happens when we add
two angular momenta J1 and J2 and look at the representation of the rotation
operator with respect to the total angular momentum states (where J = J1 + J2 ).
By choosing our states to be eigenstates of J 2 and Jz rather than J1z and J2z , the
rotation operator becomes block diagonal rather than a big mess. This is because
rotations don’t change j, so for fixed j, the (2j + 1)-dimensional space spanned by
{|j − ji, |j j − 1i, . . . , |j ji} is an invariant subspace under rotations. This change
of basis is exactly what Clebsch-Gordan coefficients accomplish.
Let us go back to our specific example of rotations. If we define r̄ = T (r), then
on the the one hand we have
X
x̄j ej
r̄ =
j
while on the other hand, we can write
XX
X
ej aji xi .
xi T (ei ) =
r̄ = T (r) =
i
i
21
j
Since the ej are a basis, they are linearly independent, and we can equate these last
two equations to conclude that
X
aji xi .
(3.3)
x̄j =
i
Note which indices are summed over in this equation.
Comparing (3.3) with (3.1), we see that the matrix in (3.1) is the matrix representation of the linear transformation T defined by
(x̄1 , x̄2 ) = T (x1 , x2 ) = ((cos θ)x1 − (sin θ)x2 , (sin θ)x1 + (cos θ)x2 ) .
Then the first column of [T ] is
T (e1 ) = T (1, 0) = (cos θ, sin θ)
and the second column is
T (e2 ) = T (0, 1) = (− sin θ, cos θ)
so that
[T ] =
"
cos θ
− sin θ
sin θ
cos θ
#
as in (3.1).
Using (3.3) we can make another extremely important observation.
Since
P the
P
length of a vector is unchanged under rotations, we must have j x̄j x̄j = i xi xi .
But from (3.3) we see that
X
X
x̄j x̄j =
aji ajk xi xk
j
and hence it follows that
X
j
i,k
aji ajk =
X
aTij ajk = δik .
(3.4a)
j
In matrix notation, this is
AT A = I .
Since we are in a finite-dimensional space, this implies that AAT = I also. In
other words, AT = A−1 . Such a matrix is said to be orthogonal. In terms of
components, this second condition is
X
X
aij akj = δik .
(3.4b)
aij aTjk =
j
j
It is also quite useful to realize that an orthogonal matrix is one whose rows (or
columms) form an orthonormal set of vectors. If we let Ai denote the ith row of A,
then from (3.4b) we have
X
aij akj = δik .
Ai · Ak =
j
22
Similarly, if we let Ai denote the ith column of A, then (3.4a) may be written
X
aji ajk = δik .
Ai · Ak =
j
Conversely, it is easy to see that any matrix whose rows (or columns) form an
orthonormal set must be an orthogonal matrix.
the above arguments using kxk2 =
P In∗ the case of complex matrices, repeating
†
†
†
∗T
i xi xi , it is not hard to show that A A = AA = I where A = A . In this case,
the matrix A is said to be unitary. Thus a complex matrix is unitary if and only if
its rows (or columns) form an orthonormal set under the standard Hermitian inner
product.
To describe a passive transformation, consider a linear transformation P acting
on the basis vectors. In other words, we perform a change of basis defined by
X
ēi = P (ei ) =
ej pji .
j
In this situation, the linear transformation P is called the transition matrix.
Suppose we have a linear operator A defined on our space. Then with respect to
a basis {ei } this operator has the matrix representation (aij ) defined by (dropping
the parenthesis for simplicity)
X
ej aji .
Aei =
j
And with respect to another basis {ēi } it has the representation (āij ) defined by
X
ēj āji .
Aēi =
j
But ēi = P ei , so the left side of this equation may be written
X
X
X
X
pji Aej =
pji ek akj =
ek akj pji
ej pji =
Aēi = AP ei = A
j
j
while the right side is
X
ēj āji =
X
j,k
(P ej )āji =
j
j
X
j,k
ek pkj āji .
j,k
Equating these two expressions and using the linear independence of the ek we have
X
X
akj pji
pkj āji =
j
j
which in matrix notation is just P Ā = AP . Since each basis may be written in
terms of the other, the matrix P must be nonsingular so that P −1 exists, and hence
we have shown that
Ā = P −1 AP .
(3.5)
23
This extremely important equation relates the matrix representation of an operator A with respect to a basis {ei } to its representation Ā with respect to a basis
{ēi } defined by ēi = P ei . This is called a similarity transformation. In fact, this
is exactly what you do when you diagonalize a matrix. Starting with a matrix A
relative to a given basis, you first find its eigenvalues, and then use these to find the
corresponding eigenvectors. Letting P be the matrix whose columns are precisely
these eigenvectors, we then have P −1 AP = D where D is a diagonal matrix with
diagonal entries that are just the eigenvalues of A.
4
4.1
Angular Momentum and Rotations
Angular Momentum as the Generator of Rotations
Before we begin with the physics, let me first prove an extremely useful mathematical result. For any complex number (or matrix) x, let fn (x) = (1 + x/n)n , and
consider the limit
x n
L = lim fn (x) = lim 1 +
.
n→∞
n→∞
n
Since the logarithm is a continuous function, we can interchange limits and the log,
so we have
x n
x n
ln L = ln lim 1 +
= lim ln 1 +
n→∞
n→∞
n
n
x
ln(1 + x/n)
= lim n ln 1 +
= lim
.
n→∞
n→∞
n
1/n
As n → ∞, both the numerator and denominator go to zero, so we apply
l’Hôpital’s rule and take the derivative of both numerator and denominator with
respect to n:
x
(−x/n2 )/(1 + x/n)
= lim
= x.
2
n→∞
n→∞
−1/n
1 + x/n
ln L = lim
Exponentiating both sides, we have proved
x n
lim 1 +
= ex .
n→∞
n
(4.1)
As I mentioned above, this result also applies if x is an n × n complex matrix
A. This is a consequence of the fact (which I state without proof) that for any such
matrix, the series
∞
X
An
n!
n=0
converges, and we take this as the definition of eA . (For a proof, see one of my
linear algebra books or almost any book on elementary real analysis or advanced
calculus.)
24
Now let us return to physics. First consider the translation of a wave function
ψ(x) by an amount a.
ψ ′ (x)
ψ(x)
x
x+a
This results in a new function ψ ′ (x + a) = ψ(x), or
ψ ′ (x) = ψ(x − a) .
For infinitesimal a, we expand this in a Taylor series to first order in a:
ψ(x − a) = ψ(x) − a
dψ
.
dx
Using p = −i~(d/dx), we can write this in the form
ia
ψ(x − a) = 1 − p ψ(x) .
~
For a finite translation, we consider a sequence of n infinitesimal translations
a/n and let n → ∞. Applying (4.1) we have
n
iap/~
ψ(x) = e−iap/~ ψ(x) .
(4.2)
ψ ′ (x) = ψ(x − a) = lim 1 −
n→∞
n
Thus we see that translated states ψ ′ (x) = ψ(x − a) are the result of applying the
operator
Ta = e−iap/~
to the wave function ψ(x). In the case of a translation in three dimensions, we have
the general translation operator
Ta = e−ia·p/~ .
(4.3)
We say that p generates translations.
Note that the composition of a translation by a and a translation by b yields a
translation by a + b as it should:
Ta Tb = e−ia·p/~ e−ib·p/~ = e−i(a+b)·p/~ = Ta+b .
Also, noting that T−a Ta = Ta T−a = T0 = 1, we see that Ta−1 = T−a . Thus we
have shown that the composition of translations is a translation, that translating
by 0 is the identity transformation, and that every translation has an inverse. Any
25
collection of objects together with a composition law obeying closure, the existence
of an identity, and the existence of an inverse to every object in the collection is
said to form a group, and this group in particular is called the translation group.
The composition of group elements is called group multiplication.
In the case of the translation group, is also important to realize that Ta Tb =
Tb Ta . Any group with the property that multiplication is commutative is called
an abelian group. Another example of an abelian group is the set of all rotations
about a fixed axis in R3 . (But the set of all rotations in three dimensions is most
definitely not abelian, as we are about to see.)
Now that we have shown that the linear momentum operator p generates translations, let us show that the angular momentum operator is the generator of rotations.
In particular, we will show explicitly that the orbital angular momentum operator
L generates spatial rotations in R3 . In the case of the spin operator S which acts
on the very abstract space of spin states, we will define rotations by analogy.
Let me make a slight change of notation and denote the rotated position vector
by a prime instead of a bar. If we rotate a vector x in R3 , then we obtain a new
vector x′ = R(θ)x where R(θ) is the matrix that represents the rotation. In two
dimensions this is
" ′# "
#" #
x
cos θ − sin θ
x
=
.
y′
sin θ
cos θ
y
If we have a scalar wavefunction ψ(x), then under rotation we obtain a new wavefunction ψR (x), where ψ(x) = ψR (R(θ)x) = ψR (x′ ). (See the figure below.)
ψR (x)
x′
ψ(x)
θ
x
Alternatively, we can write
ψR (x) = ψ(R−1 (θ)x) .
Since R is an orthogonal transformation (it preserves the length of x) we know
that R−1 (θ) = RT (θ) (this really should be written as R(θ)−1 = R(θ)T , but it
rarely is), and in the case where θ ≪ 1 we then have
"
#" # "
#
1 θ
x
x + θy
−1
R (θ)x =
=
.
−θ 1
y
−θx + y
Expanding ψ(R−1 (θ)x) with these values for x and y we have (letting ∂i = ∂/∂xi
for simplicity)
ψR (x) = ψ(x + θy, y − θx) = ψ(x) − θ[x∂y − y∂x ]ψ(x)
26
or, using pi = −i~∂i this is
i
ψR (x) = ψ(x) − θ[xpy − ypx ]ψ(x) =
~
i
1 − θLz ψ(x) .
~
Thus we see that angular momentum is indeed the generator of rotations.
For finite θ we exponentiate this to write ψR (x) = e−(i/~)θLz ψ(x), and in the
case of an arbitrary angle θ in R3 this becomes
ψR (x) = e−(i/~)θ·L ψ(x) .
(4.4)
In an abstract notation we write this as
|ψR i = U (R)|ψi
where U (R) = e−(i/~)θ·L . For simplicity and clarity, we have written U (R) rather
than the more complete U (R(θ)), which we continue to do unless the more complete
notation is needed.
What we just did was for orbital angular momentum. In the case of spin there
is no classical counterpart, so we define the spin angular momentum operator S to
obey the usual commutation relations, and then the spin states will transform under
the rotation operator e−(i/~)θ·S . (We will come back to justifying this at the end
of this section.) It is common to use the symbol J to stand for any type of angular
momentum operator, for example L, S or L + S, and this is what we shall do from
now on. In the case where J = L + S, J is called the total angular momentum
operator.
(The example above applied to a scalar wavefunction ψ, which represents a
spinless particle. Particles with spin are described by vector wavefunctions ψ, and in
this case the spin operator S serves to mix up the components of ψ under rotations.)
The angular momentum operators J 2 = J · J and Jz commute, and hence they
have simultaneous eigenstates, denoted by |jmi, with the property that (with ~ = 1)
J 2 |jmi = j(j + 1)|jmi
and
Jz |jmi = m|jmi
where m takes the 2j + 1 values −j ≤ m ≤ j. Since the rotation operator is given
by U (R) = e−iθ·J we see that [U (R), J 2 ] = 0. Then
J 2 U (R)|jmi = U (R)J 2 |jmi = j(j + 1)U (R)|jmi
so that the magnitude of the angular momentum can’t change under rotations.
However, [U (R), Jz ] 6= 0 so the rotated state will no longer be an eigenstate of Jz
with the same eigenvalue m.
Note that acting to the right we have the matrix element
hj ′ m′ |J 2 U (R)|jmi = hj ′ m′ |U (R)J 2 |jmi = j(j + 1)hj ′ m′ |U (R)|jmi
while acting to the left gives
hj ′ m′ |J 2 U (R)|jmi = j ′ (j ′ + 1)hj ′ m′ |U (R)|jmi
27
and therefore
hj ′ m′ |U (R)|jmi = 0
unless j = j ′ .
(4.5)
We also make note of the fact that acting with J 2 and Jz in both directions yields
hj ′ m′ |J 2 |jmi = j ′ (j ′ + 1)hj ′ m′ |jmi = j(j + 1)hj ′ m′ |jmi
and
hjm′ |Jz |jmi = m′ hjm′ |jmi = mhjm′ |jmi
so that (as you should have already known)
hj ′ m′ |jmi = δj ′ j δm′ m .
(4.6)
In other words, the states |jmi form a complete orthonormal set, and the state
U (R)|jmi must be of the form
X
X
(j)
U (R)|jmi =
|jm′ ihjm′ |U (R)|jmi =
|jm′ iDm′ m (θ)
(4.7)
m′
where
m′
(j)
Dm′ m (θ) := hjm′ |U (R)|jmi = hjm′ |e−iθ·J |jmi .
(4.8)
(Notice the order of susbscripts in the sum in equation (4.7). This is the same as
the usual definition of the matrix
P representation [T ]e = (aij ) of a linear operator
T : V → V defined by T ei = j ej aji .)
Since for each j there are 2j + 1 values of m, we have constructed a (2j +
1) × (2j + 1) matrix D (j) (θ) for each value of j. This matrix is referred to as the
jth irreducible representation of the rotation group. The word “irreducible”
means that there is no subset of the space of states {|jji, |j, m − 1i, . . . , |j, −ji}
that transforms into itself under all rotations U (R(θ)). Put in another way, a
representation is irreducible if the vector space on which it acts has no invariant
subspaces.
Now, it is a general result of the theory of group representations that any representation of a finite group or compact Lie group is equivalent to a unitary representation, and any reducible unitary representation is completely reducible. Therefore,
any representation of a finite group or compact Lie group is either already irreducible or else it is completely reducible (i.e., the space on which the operators act
can be put into block diagonal form where each block corresponds to an invariant
subspace). However, at this point we don’t want to get into the general theory
of representations, so let us prove directly that the representations D (j) (θ) of the
rotation group are irreducible.
Recall that the raising and lowering operators J± are defined by
p
J± |jmi = (Jx ± iJy )|jmi = j(j + 1) − m(m ± 1) |j, m ± 1i .
In particular, the operators J± don’t change the value of j when acting on the states
|jmi. (This is just the content of equation (1.4).)
28
Theorem 4.1. The representations D (j) (θ) of the rotation group are irreducible.
In other words, there is no subset of the space of states |jmi (for fixed j) that
transforms among itself under all rotations.
Proof. Fix j and let V be the space spanned by the 2j + 1 vectors |jmi := |mi.
We claim that V is irreducible with respect to rotations U (R). This means that
given any |ui ∈ V , the set of all vectors of the form U (R)|ui (i.e., for all rotations
U (R)) spans V . (Otherwise, if there exists |vi such that {U (R)|vi} didn’t span V ,
then V would be reducible since the collection of all such U (R)|vi would define an
invariant subspace.)
To show V is irreducible, let Ve = Span{U (R)|ui} where |ui ∈ V is arbitrary but
fixed. For infinitesimal θ we have U (R(θ)) = e−iθ·J = 1 − iθ · J and in particular
U (R(εb
x)) = 1 − iεJx and U (R(εb
y)) = 1 − iεJy . Then
1
1
J± |ui = (Jx ± iJy )|ui =
[1 − U (R(εb
x))] ± i
[1 − U (R(εb
y))]
|ui
iε
iε
=
1
± [1 − U (R(εb
y))] − i + iU (R(εb
x)) |ui ∈ Ve
ε
by definition of Ve and vector spaces. Since J± acting on |ui is a linear combination
of rotations acting on |ui and this is in Ve , we see that (J± )2 acting on |ui is again
some other linear combination of rotations acting on |ui and hence is also in Ve . So
in general, we see that (J± )n |ui is again in Ve .
By definition of V , we may write (since j is fixed)
X
X
|ui =
|jmihjm|ui =
|mihm|ui
m
m
= |mihm|ui + |m + 1ihm + 1|ui + · · · + |jihj|ui
6 0 (and not all of
where m is simply the smallest value of m for which hm|ui =
the terms up to hj|ui are necessarily nonzero). Acting on this with J+ we obtain
(leaving off the constant factors and noting that J+ |ji = 0)
J+ |ui ∼ |m + 1ihm|ui + |m + 2ihm + 1|ui + · · · + |jihj − 1|ui ∈ Ve .
Since hm|ui =
6 0 by assumption, it follows that |m + 1i ∈ Ve .
We can continue to act on |ui with J+ a total of j − m times at which point we
will have shown that |m + j − mi = |ji := |jji ∈ Ve . Now we can apply J− 2j + 1
times to |jji to conclude that the 2j + 1 vectors |jmi all belong to Ve , and thus
Ve = V . (This is because we have really just applied the combination of rotations
(J− )2j+1 (J+ )j−m to |ui, and each step along the way is just some vector in Ve .)
As the last subject of this section, let us go back and show that e−(i/~)θ·S really
represents a rotation in spin space. Let R denote the unitary rotation operator, and
29
let A be an arbitrary observable. Rotating our system, a state |ψi is transformed
into a rotated state |ψR i = R|ψi. In the original system, a measurement of A
yields the result hψ|A|ψi. Under rotations, we measure a rotated observable AR in
the rotated states. Since the physical results of a measurement can’t change just
because of our coordinate description, we must have
hψR |AR |ψR i = hψ|A|ψi .
But
hψR |AR |ψR i = hψ|R† AR R|ψi
and hence we have R† AR R = A or
AR = RAR† = RAR−1 .
(4.9)
(Compare this to (3.5). Why do you think there is a difference?)
Now, what about spin one-half particles? To say that a measurement of the spin
b can only yield one of two possible results means that the
in a particular direction m
b has only the two eigenvalues ±~/2. Let us denote the corresponding
operator S · m
b ±i. In other words, we have
states by |m
~
b ±i .
b m
b ±i = ± |m
(S · m)|
2
(This is just the generalization of (2.1c) for the case of spin one-half. It also applies
b m
b ms i = ms |m
b ms i.)
to arbitrary spin if we write (S · m)|
b by θ to obtain another unit vector n
b , and
Let us rotate the unit vector m
consider the operator
b (i/~)θ·S .
e−(i/~)θ·S (S · m)e
b ±i we have
Acting on the state e−(i/~)θ·S |m
b (i/~)θ·S ]e−(i/~)θ·S |m
b ±i = e−(i/~)θ·S (S · m)|
b m
b ±i
[e−(i/~)θ·S (S · m)e
~
b ±i .
= ± e−(i/~)θ·S |m
2
Therefore, if we define the rotated state
b ±i
|b
n ±i := e−(i/~)θ·S |m
(4.10a)
b := e−(i/~)θ·S (S · m)e
b (i/~)θ·S
S·n
(4.10b)
we see from (4.9) that we also have the rotated operator
with the property that
~
b )|b
n ±i
(S · n
n ±i = ± |b
2
(4.10c)
as it should. This shows that e−(i/~)θ·S does indeed act as the rotation operator on
the abstract spin states.
30
b ±i of the spin
What we have shown then, is that starting from an eigenstate |m
b we rotate by an angle θ to obtain the rotated state |b
operator S · m,
n ±i that is an
b with the same eigenvalues.
eigenstate of S · n
For spin one-half we have S = σ/2 (with ~ = 1), and using (2.8), you can show
that the rotation operator becomes
e−iσ·θ/2 =
∞
X
(−iσ · θ/2)n
n!
m=0
= I cos
θ
θ
− iσ · θb sin .
2
2
(4.11)
Example 4.1. Let us derive equations (2.9) using our rotation formalism. We start
b = ẑ and n
b = x̂, so that θ = (π/2)ŷ. Then the rotation operator is
by letting m
π
π
− iσy sin
4
4
"
"
#
cos π/4 − sin π/4
1 1
=
= √
2 1
sin π/4
cos π/4
R(θ) = e−iπσy /4 = I cos
−1
1
#
.
Acting on the state
we have
1
R(θ)|ẑ +i = √
2
"
b +i = |ẑ +i
|m
1 −1
1
1
#" #
1
0
1
=√
2
" #
1
1
= |x̂ +i
which is just (2.9a). Similarly, it is also easy to see that
"
"
#" #
#
1 −1
1
0
−1
1
= |x̂ −i
= √
R(θ)|ẑ −i = √
2 1
2 −1
1
1
which is the same as (2.9b) up to an arbitrary phase.
So far we have shown that the rotation operator R(θ) = R((π/2)ŷ) indeed takes
the eigenstates of Sz into the eigenstates of Sx . Let us also show that it takes
Sz = σz /2 into Sx = σx /2. This is a straightforward computation:
"
#"
#
"
#
1
0 1
1 1
1 1 −1
−iπσy /4
iπσy /4
√
e
σz e
= √
2 1
2 −1 1
1
0 −1
"
# "
#
0 1
1 0 2
=
=
= σx
2 2 0
1 0
as it should.
31
4.2
Spin Dynamics in a Magnetic Field
As we mentioned at the beginning of Section 2, a particle of charge q and mass m
moving in a circular orbit has a magnetic moment given classically by
µl =
q
L.
2mc
This relation is also true in quantum mechanics for orbital angular momentum, but
for spin angular momentum we must write
µs = g
q
S
2mc
(4.12)
where the constant g is called a g-factor. For an electron, g is found by experiment
to be very close to 2. (The Dirac equation predicts that g = 2 exactly, and higher
order correction in QED show a slight deviation from this value.) And for a proton,
g = 5.59. Furthermore, electrically neutral particles such as the neutron also have
magnetic moments, which you can think of as arising from some sort of internal
charge distribution or current. (But don’t think too hard.)
In the presence of a magnetic field B(t), a magnetic moment µ feels an applied
torque µ×B, and hence possesses a potential energy −µ·B. Because of this torque,
the magnetic moment (and hence the particle) will precess about the field. (Recall
that torque is equal to the time rate of change of angular momentum, and hence a
non-zero torque means that the angular momentum vector of the particle must be
changing.) Let us see what we can learn about this precession.
We restrict consideration to particles essentially at rest (i.e., with no angular
momentum) in a uniform magnetic field B(t), and write µs = γS where γ = gq/2mc
is called the gyromagnetic ratio. (If the field isn’t uniform, there will be a force
equal to the negative gradient of the potential energy, and the particle will move in
space. This is how the Stern-Gerlach experiment works, as we will see below.) In
the case of a spin one-half particle, the Hamiltonian given by
Hs = −µs · B(t) = −
gq~
gq
S · B(t) = −
σ · B(t) .
2mc
4mc
(4.13)
To understand how spin behaves, we look at this motion from the point of view of
the Heisenberg picture, which we now describe. (See Griffiths, Example 4.3 for a
somewhat different approach.)
The formulation of quantum mechanics that we have used so far is called the
Schrödinger picture (SP). In this formulation, the operators are independent of
time, but the states (wave functions) evolve in time. The stationary state solutions
to the Schrödinger equation
H|Ψ(t)i = i~
∂
|Ψ(t)i
∂t
(where H is independent of time) are given by
|Ψ(t)i = e−iEt/~ |ψi = e−iHt/~ |ψi
32
(4.14)
where H|ψi = E|ψi and |ψi is independent of time.
In the Heisenberg picture (HP), the states are independent of time, and the
operators evolve in time according to an equation of motion. To derive the equations
of motion for the operators in this picture, we freeze out the states at time t = 0:
|ψH i := |Ψ(0)i = |ψi = e+iHt/~ |Ψ(t)i .
(4.15a)
In the SP, the expectation value of an observable O (possibly time dependent) is
given by
hOi = hΨ(t)|O|Ψ(t)i = hψ|eiHt/~ Oe−iHt/~ |ψi .
In the HP, we want the same measurable result for hOi, so we have
hOi = hψH |OH |ψH i = hψ|OH |ψi .
Equating both versions of these expectation values, we conclude that
OH = eiHt/~ Oe−iHt/~
(4.15b)
where OH = OH (t) is the representation of the observable O in the HP. Note in
particular that the Hamiltonian is the same in both pictures because
HH = eiHt/~ He−iHt/~ = eiHt/~ e−iHt/~ H = H .
Now take the time derivative of (4.15b) and use the fact that H commutes with
e±iHt/~ :
dOH
i
i
∂O −iHt/~
= HeiHt/~ Oe−iHt/~ − eiHt/~ Oe−iHt/~ H + eiHt/~
e
dt
~
~
∂t
=
i
∂O −iHt/~
[H, OH ] + eiHt/~
e
.
~
∂t
If O has no explicit time dependence (for example, the operator pt has explicit time
dependence), then this simplifies to the Heisenberg equation of motion
i
dOH
= [H, OH ] .
dt
~
(4.16)
Also note that if H is independent of time, then so is HH = H, and (4.16) then
shows that H = const and energy is conserved.
Returning to our problem, we use the Hamiltonian (4.13) in the equation of
motion (4.16) to write
i
i gq
i gq
dSi
= [Hs , Si ] = −
Bj [Sj , Si ] = −
Bj i~εjik Sk
dt
~
~ 2mc
~ 2mc
gq
gq
=+
εikj Sk Bj =
(S × B)i
2mc
2mc
or
gq
dS(t)
=
S(t) × B(t) = µs (t) × B(t) .
dt
2mc
33
(4.17)
This is just the operator version of the classical equation that says the time rate of
change of angular momentum is equal to the applied torque. From this equation we
see that the spin of a positively charged particle (q > 0) will precess in the negative
sense about B, while the spin of a negatively charged particle (q < 0) will precess
in the positive sense.
B
S
µs
Figure 1: Precession for q < 0
Let us specialize to the case where B is both uniform and independent of time,
and write B = B0 ẑ. Then the three components of (4.17) become
dSx (t)
gqB0
=
Sy (t)
dt
2mc
Defining
dSy (t)
gqB0
=−
Sx (t)
dt
2mc
dSz (t)
= 0 . (4.18)
dt
gqB0
2mc
we can combine the Sx and Sy equations to obtain
ω0 =
(4.19)
d2 Sx (t)
= −ω02 Sx (t)
dt2
and exactly the same equation for Sy (t). These equations have the solutions
Sx (t) = a cos ω0 t + b sin ω0 t
Sy (t) = c cos ω0 t + d sin ω0 t
Sz (t) = Sz (0) .
Clearly, Sx (0) = a and Sy (0) = c. Also, from the equations of motion (4.18) we
have ω0 Sy (0) = (dSx /dt)(0) = ω0 b and −ω0 Sx (0) = (dSy /dt)(0) = ω0 d so that our
solutions are
Sx (t) = Sx (0) cos ω0 t + Sy (0) sin ω0 t
Sy (t) = Sy (0) cos ω0 t − Sx (0) sin ω0 t
Sz (t) = Sz (0) .
34
(4.20)
These can be written in the form Sx (t) = A cos(ω0 t+δx ) and Sy (t) = A sin(ω0 t+δy )
where A2 = Sx (0)2 +Sy (0)2 . Since these are the parametric equations of a circle, we
see that S precesses about the B field, with a constant projection along the z-axis.
For example, suppose the spin starts out at t = 0 in the xy-plane as an eigenstate
of Sx with eigenvalue +~/2. This means that
hSx (0)i = hx̂ + |Sx (0)|x̂ +i =
~
2
hSy (0)i = hSz (0)i = 0 .
Taking the expectation value of equations (4.20) we see that
hSx (t)i =
~
cos ω0 t
2
~
hSy (t)i = − sin ω0 t
2
hSz (t)i = 0 .
(4.21)
This clearly shows that the spin stays in the xy-plane, and precesses about the zaxis. The direction of precession depends on the sign of ω0 , which in turn depends
on the sign of q as defined by (4.19).
How does this look from the point of view of the SP? From (4.14), the spin state
evolves according to
|χ(t)i = e−iHt/~ |χ(0)i = ei(gqBt/2mc)Sz /~ |χ(0)i = eiω0 tSz /~ |χ(0)i .
Comparing this with (4.10a), we see that in the SP the spin precesses about −ẑ
with angular velocity ω0 .
Now let us consider the motion of an electron in an inhomogeneous magnetic
field. Since for the electron we have q = −e < 0, we see from (4.12) that µs is antiparallel to the spin S. The potential energy is Hs = −µs · B, and as a consequence,
the force on the electron is
F = −∇Hs = ∇(µs · B) .
Suppose we have a Stern-Gerlach apparatus set up with the inhomogeneous B
field in the z direction (i.e., pointing up):
In our previous discussion with a uniform B field, there would be no translational
force because ∇Bi = 0. But now this is no longer true, and as a consequence, a
particle with a magnetic moment will be deflected. The original S-G experiment
was done with silver atoms that have a single valence electron, but we can consider
this as simply an isolated electron.
It is easy to see in general what is going to happen. Suppose that the electron is
in an eigenstate of Sz with eigenvalue +~/2. Since the spin points up, the magnetic
35
moment µs is anti-parallel to B, and hence Hs = −µs · B > 0. Since the electron
wants to minimize its energy, it will travel towards a region of smaller B, which is
up. Obviously, if the spin points down, then µs is parallel to B and Hs < 0 so the
electron will decrease its energy by moving to a region of larger B, which is down.
Another way to look at this is from the force relation. If µs is anti-parallel to
B (i.e., spin up), then since the gradient of B is negative (the B field is decreasing
in the positive direction), the force on the electron will be in the positive direction,
and hence the electron will be deflected upwards. Similarly, if the spin is down,
then the force will be negative and the electron is deflected downwards.
It is worth pointing out that a particle with angular momentum l will split into
2l + 1 distinct components. So if the magnetic moment is due to orbital angular
momentum, then there will necessarily be an odd number of beams coming out of
a S-G apparatus. Therefore, the fact that an electron beam splits into two beams
shows that its angular momentum can not be due to orbital motion, and hence is a
good experimental verification of the existence of spin.
Suppose that the particle enters the S-G apparatus at t = 0 with its spin aligned
along x̂, and emerges at time t = T . If the magnetic field were uniform, then the
expectation value of S at t = T would be given by equations (4.21) as
=
hSiunif
T
~
~
(cos ω0 T )x̂ − (sin ω0 T )ŷ .
2
2
(4.22)
Interestingly, for the inhomogeneous field this expectation value turns out to be
zero.
To see this, let the particle enter the apparatus with its spin along the x̂ direction,
and let its initial spatial wave function be the normalized wave packet ψ0 (r, t). In
terms of the eigenstates of Sz , the total wave function at t = 0 is then (from (2.9a))
"
#
ψ0 (r, 0)
1
Ψ(r, 0) = √
.
(4.23a)
2 ψ0 (r, 0)
When the particle emerges at t = T , it is a superposition of localized spin up wave
packets ψ+ (r, T ) and spin down wave packets ψ− (r, T ):
"
#
ψ+ (r, T )
1
.
(4.23b)
Ψ(r, T ) = √
2 ψ− (r, T )
Since the spin up and spin down states are well separated, there is no overlap
between these wave packets, and hence we have
ψ+ (r, T ) ψ− (r, T ) ≈ 0 .
(4.23c)
Since the B field is essentially in the z direction there is no torque on the
particle in the z direction, and from (4.17) it follows that the z component of spin
is conserved. In other words, the total probability of finding the particle with spin
36
up at t = T is the same as it is at t = 0. The same applies to the spin down states,
so we have
Z
Z
2 3
2
|ψ+ (r, T )| d r = |ψ0 (r, 0)| d3 r
and
and therefore
Z
Z
2
|ψ− (r, T )| d3 r =
2
|ψ+ (r, T )| d3 r =
Z
Z
2
|ψ0 (r, 0)| d3 r
2
|ψ− (r, T )| d3 r .
(4.24)
From (4.23a), the expectation value of S at r and t = 0 is given by
~
hSir,0 = Ψ† (r, 0) σΨ(r, 0)
2
~
= Ψ† (r, 0) (σx x̂ + σy ŷ + σz ẑ)Ψ(r, 0)
2
=
~
|ψ0 (r, 0)|2 x̂
2
and hence the net expectation value at t = 0 is (since ψ0 is normalized)
Z
~
hSi0 = hSir,0 d3 r = x̂ .
2
(This should have been expected since the particle entered the apparatus with its
spin in the x̂ direction.) And from (4.23b), using (4.23c) we have for the exiting
particle
~
hSir,T = Ψ† (r, T ) σΨ(r, T )
2
~
= Ψ† (r, T ) (σx x̂ + σy ŷ + σz ẑ)Ψ(r, T )
2
n
o
~
2
2
=
|ψ+ (r, T )| − |ψ− (r, T )| ẑ .
2
Integrating over all space, we see from (4.24) that
hSiinhom
=0
T
(4.25)
as claimed.
Why does the uniform field give a non-zero value for hSiT whereas the inhomogeneous field gives zero? The answer lies with (4.19). Observe that the precessional
frequency ω0 depends on the field B. In the Stern-Gerlach apparatus, the B field is
weaker up high and stronger down low. This means that particles lower in the beam
precess at a faster rate than particles higher in the beam. Since (4.25) is essentially
37
an average of (4.22), we see that the different phases due to the different values of
ω0 will cause the integral to vanish.
Of course, this assumes that there is sufficient variation ∆ω0 in precessional
frequencies that the trigonometric functions average to zero. This will be true as
long as
(∆ω0 )T ≥ 2π .
In fact, this places a constraint on the minimum amount of inhomogeneity that the
S-G apparatus must have in order to split the beam. From (4.19) we have
∆ω0 = −
e
e ∂B0
∆B0 ≈ −
∆z
me c
me c ∂z
where ∆z is the distance between the poles of the magnet. Therefore, in order for
the apparatus to work we must have
e ∂B0
2π
.
T ≥
me c ∂z
∆z
5
The Addition of Angular Momentum
The basic idea is simply to express the eigenstates of the total angular momentum
operator J of two particles in terms of the eigenstates of the individual angular
momentum operators J1 and J2 , where J = J1 + J2 . (Alternatively, this could be
the addition of spin and orbital angular momenta of a single particle, etc.) Each
particle’s angular momentum operator satisfies (where the subscript a labels the
particle and where we take ~ = 1)
[Jai , Jaj ] = i
3
X
εijk Jak
k=1
with corresponding eigenstate |ja ma i. And since we assume that [J1 , J2 ] = 0, it
follows that
[Ji , Jj ] = [J1i + J2i , J1j + J2j ] = [J1i , J1j ] + [J2i , J2j ]
=i
=i
3
X
k=1
3
X
εijk J1k + i
3
X
εijk J2k = i
k=1
3
X
k=1
εijk J1k + J2k
εijk Jk
k=1
and hence J is just another angular momentum operator with eigenstates |jmi.
Next we want to describe simultaneous eigenstates of the total angular momentum. From [J1 , J2 ] = 0, it is clear that [J1 2 , J2 2 ] = 0, and since [Ja 2 , Jaz ] = 0, we
see that we can choose our states to be simultaneous eigenstates of J1 2 , J2 2 , J1z
and J2z . We denote these states by |j1 j2 m1 m2 i.
38
Alternatively, we can choose our states to be simultaneous eigenstates of J 2 ,
J1 , J2 2 and Jz . That [J 2 , Jz ] = 0 follows directly because J is just an angular
momentum operator. And the fact that [Ja 2 , Jz ] = 0 follows because Ja is an
angular momentum operator, Jz = J1z + J2z and [J1 , J2 ] = 0. Finally, to show that
[J 2 , Ja 2 ] = 0, we simply observe that
2
J 2 = J1 + J2
2
= J1 2 + J2 2 + 2J1 · J2
= J1 2 + J2 2 + 2(J1x J2x + J1y J2y + J1z J2z ) .
It is now easy to see that [J 2 , Ja 2 ] = 0 because [J1 , J2 ] = 0 and [Ja 2 , Jai ] = 0 for
i = x, y, z. We denote these simultaneous eigenstates by |j1 j2 jmi.
However, let me emphasize that even though [J 2 , Jz ] = 0, it is not true that
2
[J , Jaz ] = 0. This means that we can not specify J 2 in the states |j1 j2 m1 m2 i, and
we can not specify either J1z or J2z in the states |j1 j2 jmi.
We know that the angular momentum operators and states satisfy (with ~ = 1)
J 2 |jmi = j(j + 1)|jmi
(5.1a)
Jz |jmi = m|jmi
p
J± |jmi = j(j + 1) − m(m ± 1)|jm ± 1i .
(5.1b)
(5.1c)
Let us denote the two individual angular momentum states by |j1 m1 i and |j2 m2 i.
Then the two-particle basis states are denoted by the various forms (see Part I,
Section 1 in the handout Supplementary Notes on Mathematics)
|j1 j2 m1 m2 i = |j1 m1 i|j2 m2 i = |j1 m1 i ⊗ |j2 m2 i .
When we write hj1 j2 m1 m2 |j1′ j2′ m′1 m′2 i we really mean
(hj1 m1 | ⊗ hj2 m2 |)(|j1′ m′1 i ⊗ |j2′ m′2 i) = hj1 m1 |j1′ m′1 ihj2 m2 |j2′ m′2 i .
Since these two-particle states form a complete set, we can write the total combined angular momentum state of both particles as
X
(5.2)
|j1′ j2′ m1 m2 ihj1′ j2′ m1 m2 |j1 j2 jmi .
|j1 j2 jmi =
j1′ j2′
m1 m2
However, we now show that many of the matrix elements hj1′ j2′ m1 m2 |j1 j2 jmi vanish,
and hence the sum will be greatly simplified.
For our two-particle operators we have
Jz = J1z + J2z
and
39
J± = J1± + J2± .
While we won’t really make any use of it, let me point out the correct way to write
operators when dealing with tensor (or direct) product states. In this case we should
properly write operators in the form
J = J1 ⊗ 12 + 11 ⊗ J2 .
Then acting on the two-particle states we have, for example,
Jz (|j1 m1 i ⊗ |j2 m2 i) = J1z ⊗ 1 + 1 ⊗ J2z (|j1 m1 i ⊗ |j2 m2 i)
= J1z |j1 m1 i ⊗ |j2 m2 i + |j1 m1 i ⊗ J2z |j2 m2 i
= (m1 + m2 )(|j1 m1 i ⊗ |j2 m2 i) .
As I said, while we won’t generally write out operators and states in this form, you
should keep in mind what is really going on.
Now, for our two-particle states we have
Ja 2 |j1 j2 m1 m2 i = ja (ja + 1)|j1 j2 m1 m2 i
Jaz |j1 j2 m1 m2 i = ma |j1 j2 m1 m2 i
where a = 1, 2. Taking the matrix element of J1 2 acting to both the left and right
we see that
hj1′ j2′ m1 m2 |J1 2 |j1 j2 jmi = j1′ (j1′ + 1)hj1′ j2′ m1 m2 |j1 j2 jmi
= hj1′ j2′ m1 m2 |j1 j2 jmij1 (j1 + 1)
or
[j1′ (j1′ + 1) − j1 (j1 + 1)]hj1′ j2′ m1 m2 |j1 j2 jmi = 0 .
Since this result clearly applies to J2 2 as well, we must have
hj1′ j2′ m1 m2 |j1 j2 jmi = 0
if j1′ 6= j1 or j2′ 6= j2 .
In other words, equation (5.2) has simplified to
X
|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 jmi .
|j1 j2 jmi =
m1 m2
Next, from Jz = J1z + J2z we can let J1z + J2z act to the left and Jz act to the
right so that
hj1 j2 m1 m2 |Jz |j1 j2 jmi = (m1 + m2 )hj1 j2 m1 m2 |j1 j2 jmi
= hj1 j2 m1 m2 |j1 j2 jmim .
This shows that
hj1 j2 m1 m2 |j1 j2 jmi = 0
40
unless m = m1 + m2
and hence equation (5.2) has become
X
|j1 j2 m1 m2 ihj1 j2 m1 m2 |j1 j2 jmi .
|j1 j2 jmi =
m1 m2
m1 +m2 =m
Finally, we regard the values of j1 and j2 as fixed and understood, and we simply
write the total angular momentum state as
X
|jmi =
|m1 m2 ihm1 m2 |jmi .
(5.3)
m1 m2
m1 +m2 =m
The complex numbers hm1 m2 |jmi are called Clebsch-Gordan coefficients.
They are really nothing more than the elements of the unitary transition matrix that
takes us from the {|m1 m2 i} basis to the {|jmi} basis. (This is just the statement
from linear algebra that given two bases {ei } and P
{ēi } of a vector space V , there is
a nonsingular transition matrix P such that ēi = j ej pji . Moreover, ēi is just the
ith column of P . If both bases are orthonormal, then P is in fact a unitary matrix.
See Theorem 15 in Supplementary Notes on Mathematics.)
Since (normalized) eigenfunctions corresponding to distinct eigenvalues of a hermitian operator are orthonormal, we know that
hj ′ m′ |jmi = δjj ′ δmm′
hm′1 m′2 |m1 m2 i = δm1 m′1 δm2 m′2 .
and
Therefore, taking the inner product of equation (5.3) with itself we see that
X
|hm1 m2 |jmi|2 = 1 .
(5.4)
m1 m2
m1 +m2 =m
Given j1 and j2 , this holds for any resulting values of j and m. (This is really
just another way of saying that the columns of the unitary transition matrix are
normalized.)
Now, we know that −j1 ≤ m1 ≤ j1 and −j2 ≤ m2 ≤ j2 . Therefore, from
m = m1 + m2 , the maximum value of m must be j1 + j2 . And since −j ≤ m ≤ j,
it follows that the maximum value of j is jmax = j1 + j2 . Corresponding to this
maximum value j1 + j2 of j, we have a multiplet of 2(j1 + j2 ) + 1 values of m, i.e.,
−(j1 + j2 ) ≤ m ≤ j1 + j2 . On the other hand, since there are 2j1 + 1 possible
values of m1 , and 2j2 + 1 possible values of m2 , the total number of (not necessarily
distinct) possible m1 + m2 = m values is (2j1 + 1)(2j2 + 1).
The next highest possible value of m is j1 + j2 − 1 so that there is a j state with
j = j1 + j2 − 1 and a corresponding multiplet of 2(j1 + j2 − 1) + 1 = 2(j1 + j2 ) − 1
possible m values. We continue lowering the j values by one, and for each such
value j = j1 + j2 − k there are 2(j1 + j2 − k) + 1 possible m values. However, the
total number of m values in all multiplets must equal (2j1 + 1)(2j2 + 1).
41
Example 5.1. Consider the case j1 = 1, j2 = 2. Then the total number of possible
m values is
(2j1 + 1)(2j2 + 1) = 3 · 5 = 15 .
We can arrange these values of m = m1 + m2 in a table with m1 across the top and
m2 down the side:
m1
−1
0
1
−2 −3 −2 −1
−1 −2 −1
0
m2
0
1
0 −1
1
0
1
2
1
2
3
2
The distribution of m values is as follows:
value of m = m1 + m2 : 3
number of occurances: 1
2 1
2 3
0 −1 −2 −3
3
3
2
1
We have mmax = 1 + 2 = 3 so that jmax = j1 + j2 = 3, and there are seven values
of m in this multiplet: 3, 2, 1, 0, −1, −2, −3. If we eliminate these, then the next
highest value of m is 2, so there must exist a j = 2 state with the five m values
2, 1, 0, −1, −2. Eliminating these, the highest remaining m value is 1, so there is
a j = 1 state with the three m values 1, 0, −1. This exhausts all of the m values.
Note that summing the number of m values in each multiplet we have 7 + 5 + 3 = 15
as it must.
The point of this example is to illustrate the origin of the two ways of getting
an m value of j1 + j2 − 1: one way is the multiplet corresponding to j = j1 + j2 ,
and the second way is the multiplet corresponding to j = j1 + j2 − 1. Similarly, the
case m = j1 + j2 − 2 enters into three multiplets: j = j1 + j2 , j = j1 + j2 − 1 and
j = j1 + j2 − 2.
In any case, for each nonnegative integer k, the state with j = j1 +j2 −k contains
2(j1 + j2 − k) + 1 values of m, subject to the constraint
b
X
k=0
[2(j1 + j2 − k) + 1] = (2j1 + 1)(2j2 + 1) .
(5.5)
The minimum value jmin = j1 + j2 − b is defined by the integer b which is to be
determined. Recalling the formula
m
X
k=n
k=
m−n+1
(m + n)
2
42
we have
b
X
[2(j1 + j2 − k) + 1] = 2(j1 + j2 )(b + 1) − 2
k=0
b
X
k + (b + 1)
k=0
= 2(j1 + j2 )(b + 1) − (b + 1)b + (b + 1)
= (b + 1)[2(j1 + j2 ) + 1 − b] .
Then equation (5.5) becomes
(b + 1)[2(j1 + j2 ) + 1 − b] = (2j1 + 1)(2j2 + 1)
or
b2 − 2(j1 + j2 )b + 4j1 j2 = 0 .
Solving for b using the quadratic formula yields
p
b = (j1 + j2 ) ± (j1 + j2 )2 − 4j1 j2
= (j1 + j2 ) ± (j1 − j2 )
so that b = 2j1 or b = 2j2 . Since jmin = j1 + j2 − b, we see that either jmin = j2 − j1
or jmin = j1 − j2 . If j1 6= j2 then one of these is negative and must be rejected, and
hence we conclude that jmin = |j1 − j2 |. Thus we arrive at jmin ≤ j ≤ jmax or
|j1 − j2 | ≤ j ≤ j1 + j2 .
(5.6)
Equations (5.1c), (5.3), (5.4) and (5.6) are all we need to calculate the ClebschGordan coefficients. The procedure is best illustrated by a specific example.
Example 5.2. Let us consider the case j1 = j2 = 1/2 because you (should) already
know the answer. In this case we have m1 = m2 = ±1/2, so according to equation
(5.6) we know that 0 ≤ j ≤ 1.
First we have jmax = j1 + j2 = 1, and the maximum value of m is m1 + m2 = 1
also. From equation (5.3) we have the single term
E
E
1 1 ED 1 1
11 =
11
2 2 2 2
and by equation (5.4) we must have
E
D1 1
11
2 2
2
= 1.
This specifies the topmost state up to a phase factor. Choosing this phase to be 1
(the Condon-Shortley convention), we are left with
D1 1
E
1 1 = +1 .
2 2
43
Hence equation (5.3) becomes
E
1 1E
11 =
2 2
(5.7a)
where the state on the left-hand side of this equation is the |jmi state, and the
state on the right-hand side is the |m1 m2 i state.
To construct the next lowest state, we act on the left-hand side of this last result
with J− , and on the right-hand side with J− = J1− + J2− . Using equation (5.1c),
we obtain
E r3 1 √
1
1 1E
1E
210 =
+
−
+
−
4 4
2 2
2
2
or
E
1 1E
1
1E
1
−
+
.
(5.7b)
−
10 = √
2 2
2
2
2
Since (by equation (5.3))
E
E
E
1
1 1 ED 1 1
1 ED 1
1
−
10 =
10
−
− 10 + −
2
2 2
2
2 2
2 2
we see that
or
D1
E D 1 1
E
1
1
10 = −
10 = √ .
2
2
2 2
2
We now act on equation (5.7b) with J− = J1− + J2− again to obtain
"r
#
E
√
1
1
1E
1E
1
3 1
2 1 −1 = √
− −
+
+ − −
4 4
2
2
2
2
2
and hence
−
D
E
1E
1
1 −1 = − −
2
2
(5.7c)
E
1
1
− 1 −1 = 1.
2
2
This completes the j = 1 multiplet which, for obvious reasons, is called the triplet
state.
Now for the j = 0 state. We know from equation (5.3) that
E
E
E
1
1 ED 1
1
1 1 ED 1 1
00 =
−
− 00 + −
−
00
2
2 2
2
2 2
2 2
E
E
1 1
1
1
a+ −
b.
−
:=
2
2
2 2
−
To solve for a and b, we first see that from equation (5.4) we have a2 + b2 = 1.
Next, we note that the |jmi = |1 0i state is orthogonal to the |jmi = |0 0i state,
and therefore
1
0 = h1 0|0 0i = √ (a + b)
2
44
√
so that a = −b. Therefore 2a2 = 1 so that a = −b = 1/ 2 and we have
E
1
1 1E
1E
1
− −
−
00 = √
2
2 2
2 2
(5.7d)
which is called the singlet state.
We have thus constructed the well-known wavefunctions for two electrons:
E
E


1 1 = 21 12





n
E
Eo
E
triplet symmetric states:
− 12 12 + 21 − 21
1 0 = √12



E
E


 1 −1 = − 1 − 1
2
2
singlet antisymmetric state:
E
1
1 1E
1E
1
− −
.
−
00 = √
2
2 2
2 2
While we will soon return to discuss the symmetry of wave functions in detail,
it should be fairly clear that exchanging electrons 1 and 2 in any of the triplet
states leaves the wave function unchanged, and hence the states are referred to as
symmetric. Similarly, exchanging electrons 1 and 2 in the singlet state changes
the overall sign of the state, and hence this is called an antisymmetric state.
45