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