Addition of Angular Momenta
Consider a system of two spin-1 /2 particles (whose orbital degrees of freedom we ignore).
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which obey (15.1.2a) (i=l,2)
( 15.1.2b)
Since s; = l /2, and m; = ± 1/2 has freedom only in sign, let us use the compact notation 1++), 1+-), 1-+), 1--) to denote the states. For instance, (15.1.3)
and so on. These four vectors form the product basis. They represent states that have well-defined values for the magnitude and z component of the individual spins. Suppose now that we choose not to look at the individual spins but the system as a whole. What are the possibk values for the magnitude and z component of the system spin, and what are the s~ates that go with these values? This is a problem in addition of angular momenta, which is the topic of this chapter. ; In terms of the operators s['> and J®s?'.
Si 2 > which act on the one-particle
spaces, S, =S[ 11 ®Im and S,=
403
404
Consider the operator
CHAPTER 15
(15.1.4)
which we call the total angular momentum operator. That Sis indeed the total angular momentum operator is supported by (1) our intuition; (2) the fact that it is the generator of rotation for the product kets, i.e., rotations of the whole system; (3) the fact that it obeys the commutation rules expected of a generator of rotations, namely, (15. 1.5)
as may be readily verified. Our problem is to find the eigenvalues and eigenvectors of S 2 and S=. Consider first (15.1.6)
which commutes with si' sL S~:. and s2z· We expect it to be diagonal in the product basis. This is readily verified:
.n
!i)
S=I++)=(S1=+ S2JI++)= (2+2 I++) 1
Szf+
)=01+-)
(15.1.7)
S=l-+)=01-+) Szf--)=-nl -) Thus the allowed values for the total z component are By the method of images (or any other method) ++
s _____. 1i [ -
prod~ct basts
0I 0 0
n, 0, and -fi.
,- -+ 0 0 0 0
0 0 0 0
JJ
(15.1.8)
Note that the eigenvalue .s·= = 0 is twofold degenerate, and the eigenspace is spanned by the vectors i +-) and -+ ). If we form some linear combination, af·t-----)+131-+) , we still get an eigenstate with s==O, but this state will not have definite values for s~., and 82, (unless a or J3 = 0). Consider next the operator (15.1.9)
Although S 2 commutes with Si and S~, it does not commute with S1: and Szz because of the S 1 ·S 2 tenn, which has S 1x, S 1y, etc. in it. By explicit computation,
++ s2---->
n.z [
product basis
20 0
O
+-
·-+
0
0
1 0
1 0
n
(15.1.10)
Thus we see that although I++) and 1---) are eigenstates of S 2 [s(s+ 1)=2], the states of zero Sz, namely, 1+-) and 1-+), are not. However, the following linear combinations are:
1+-)+1·····•·> 1+-)-1·-·+) 21/2
(s= 1)
(15.1.11) (s=O)
Exercise 15.1.1. *Derive Eqs. (15.1.10) and (15.1.11). It might help to use (15.1.12)
This completes the solution to the problem we undertook. The allowed values for total spin are s =I and 0, while the allowed values of s= are 11., 0, and -11.. The corresponding eigenstates in the product basis are
Is= 1m= 1,
S1
=
1/2 sz= 1/2)=1++)
Is= 1 m=O,
Sj
=
1/2 s2= 1/2) = T
112
[1 +- >+ 1-+ >l
Is= 1m= -1, s1 = 1/2 s2 = 1/2)= 1--) ls=Om=O,
sl=l/2sz=l/2)=2
112
(15.1.13)
[1+-)-l-+)]
These vectors represent
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