Physics 6210/Spring 2007/Lecture 27
Lecture 27
Relevant sections in text: §3.6, 3.7
Angular momentum addition in general
We can generalize our previous discussion of 2 spin 1/2 systems as follows. Suppose we
are given two angular momenta J~1 and J~2 (e.g., two spins, or a spin and an orbital angular
momentum, or a pair of orbital angular momenta). We can discuss both angular momenta
at once using the direct product space as before, with a product basis |j1 , m1 i ⊗ |j2 , m2 i.
We represent the operators on product vectors as
~
J~1 (|αi ⊗ |βi) = (J|αi)
⊗ |βi,
and
~
J~2 (|αi ⊗ |βi) = |αi ⊗ (J|βi),
and extend to general vectors by linearity. The product basis |j1 , m1 i ⊗ |j2 , m2 i is the
basis corresponding to the commuting observables provided by (J12 , J22 , J1z , J2z )).
The total angular momentum is defined by
~J = J~1 + J~2 .
A set of commuting observables that includes the total angular momentum is provided
by the operators (J12 , J22 , J 2 , Jz ). Note that both bases are eigenvectors of J12 and J22
since these commute with all components of the individual and total angular momentum
(exercise). We also note that product eigenvectors |j1 , j2 , m1 , m2 i are in fact eigenvectors
of Jz with eigenvalues given by m = m1 + m2 since
Jz |j1 , j2 , m1 , m2 i = (J1z + J2z )|j1 , j2 , m1 , m2 i = (m1 + m2 )h̄|j1 , j2 , m1 , m2 i.
But we will have to take linear combinations of product basis vectors to get total angular
momentum vectors – eigenvectors of J 2 .
The basis of total angular momentum eigenvectors are denoted |j1 , j2 , j, mi. For given
values of j1 and j2 , it can be shown that (see the text)
j = |j1 − j2 |, |j1 − j2 | + 1, . . . , j1 + j2 − 1, j1 + j2
with (as usual)
m = −j, −j + 1, . . . j − 1, j.
The two sets of commuting observables defining each kind of basis are not all compatible.
In particular, J1z and J2z do not commute with J 2 . So the set of total angular momentum
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Physics 6210/Spring 2007/Lecture 27
eigenvectors will be distinct from the eigenvectors of the individual angular momenta. Total
angular momentum eigenvectors, |j1 , j2 , j, mi can be expressed as linear combinations of
|j1 , j2 , m1 , m2 i where the superposition will go over various m1 , m2 values. Of course one
can also expand |j1 , j2 , m1 , m2 i in terms of |j1 , j2 , j, mi where the superposition will be
over various j, m values. The coefficients in these superpositions are known as the ClebschGordan coefficients. We have worked out a very simple example of all this in the case of
a pair of spin 1/2 systems. There is a general theory of Clebsch-Gordan coefficients which
we shall not have time to explore. Instead we will briefly visit another, relatively simple,
and relatively important example.
Example: A particle in 3-d with spin 1/2
We now apply these ideas and obtain some formulas describing the angular momentum
of a particle moving in 3-d with spin 1/2. Of course this is a model of a non-relativistic
electron so it is extremely important.
To begin, we specify the Hilbert space of states. Using the tensor product construction,
we can view it as the space of formal linear combinations of the product basis:
|~x, ±i = |~xi ⊗ |Sz , ±i.
A general state is of the form
Z
|ψi =
d3 x (a+ (~x)|~x, +i + a− (~x)|~x, −i) .
As usual, |a± |(~x)|2 is the probability density for finding the particle at ~x with spin up/down
along the z axis. Note that the normalization condition is
Z
d3 x |a+ |2 + |a− |2 = 1.
Alternatively, we can characterize the state in terms of its components in the basis defined
above. In this case we organize the information into a 2-component column vector whose
entries are complex valued functions. This gadget is known as a spinor field:
h~x, +|ψi
a+ (~x)
=
.
h~x, +|ψi
a− (~x)
The position, momentum and spin operators are defined on the product basis as follows:
~ xi ⊗ |Sz , ±i) = ~x|~x, ±i,
X(|~
P~ (|~xi ⊗ |Sz , ±i) = (P~ |~xi ⊗ |Sz , ±i,
~ xi ⊗ |Sz , ±i) = |~xi ⊗ (S|S
~ z , ±i).
S(|~
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Physics 6210/Spring 2007/Lecture 27
This implies that, on the spinor fields the position and momentum operators do their usual
~ multiplies, P~ differentiates) on each component function while the spin operators
thing (X
do their usual thing via 2 × 2 matrices. For example, the orbital angular momentum is
~ =X
~ × P~ and acts as
L
!
h̄ ~
x
×
∇a
(~
x
)
a
(~
x
)
+
+
~ × P~
X
= h̄i
,
a− (~x)
x × ∇a− (~x)
i~
and
Sx
a+ (~x)
a− (~x)
h̄
=
2
a− (~x)
a+ (~x)
.
We can now define the total angular momentum of the system as the operator
~ + S.
~
J~ = L
~ and S
~ commute and satisfy the angular momentum commutation
As usual, because L
relations, we have
[Ja , Jb ] = ih̄abc Jc ,
so the total angular momentum has all the general properties we deduced previously. For
~ only generates rotations
example, J~ generates rotations of the system as a whole, while L
~ only generates rotations of the spin. We can only
of position and momentum, and while S
~ and
simultaneously diagonalize L2 , S 2 , J 2 and one component, say, Jz . Setting J~1 = L
~ we have j1 = l = 0, 1, 2, . . . and j2 = s = 1 . For a state specified by |l, s = 1/2, j, mi
J~2 = S
2
we then have that the possible values for j are
1
1
j = l − ,l + .
2
2
How are the total angular momentum eigenvectors related to the original product
eigenvectors (eigenvectors of Lz and Sz )? We will sketch the construction. Begin with the
eigenvectors of total angular momentum with the maximum value of angular momentum:
1
1
1
|j1 = l, j2 = , j = l + , m = l + i.
2
2
2
Given j1 and j2 , there is only one (linearly independent) product eigenvector which has
the appropriate eigenvalue for Jz , namely
1
1
|j1 = l, j2 = , m1 = l, m2 = i,
2
2
so we have
1
1
1
1
1
|j1 = l, j2 = , j = l + , m = l + i = |j1 = l, j2 = , m1 = l, m2 = i.
2
2
2
2
2
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Physics 6210/Spring 2007/Lecture 27
To get the lower eigenvalues of Jz we can apply the ladder operator
J− = L− + S−
to this ket and normalize it. The details can be found in the text. Similarly, we can also
start with the minimum values for j = l − 12 and m = −l + 21 and construct the states with
the other m values using the raising operator. In this way we can express all the total
angular momentum eigenvectors in terms of the product eigenenvectors. The result is not
too complicated. We have, for j = l ± 21 ,
r
h r
1
1
1
1
1
1
1
± l ± m + |l, m− i⊗|+i+ l ∓ m + |l, m+ i⊗|−i.
|j1 = l, j2 = , j = l± , mi = √
2
2
2
2
2
2
2l + 1
In terms of spinor fields we get the total angular momentum spinors Yj,m (θ, φ), representing the total angular momentum eigenvectors by taking components in the product
basis:

 q
1
l
(θ, φ)
± l ± m + 2Y
1
ml =m− 21

 q
Yj,m = √
.

1
l
2l + 1
l ∓ m + 2Y
(θ,
φ
1
ml =m+ 2
The upper (lower) component of this column vector determines the probability density for
finding a particle at various angular locations and with a spin up (down) along z, given
that its total angular momentum is known with certainty to have the values specified by
j = l ± 21 and m.
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