3.3 Solutions
215
3.3.6 Angular Momentum
3.77 L =
i j k
x y z
p x p y p z
= i(yp z − zp y ) − j(x p z − zp x ) + k(x p y − yp x )
= i L x + j L y + k L z
Thus, L x = yp z − zp y , L y = zp x − x p z , L z = x p y − yp x
(1)
[L x , L y ] = L x L y − L y L x
= (yp z − zp y )(zp x − x p z ) − (zp x − x p z )(yp z − zp y )
= yp z zp x − yp z x p z − zp y zp x + zp y x p z − zp x yp z + zp x zp y
+ x p z yp z − x p z zp y
(2)
But [ p x , p y ] = [x, p y ] = 0, etc
(3)
(2) becomes
[L x , L y ] = yp x p z z − yxp
2
z − z
2 p y p x + xp y zp y − yp x zp z + z
2 p x p y
+ yxp
2
z − xp y p z z = [z, p z ](xp y − yp z )
( 4 )
But [z, p z ] = [z, −i
∂
∂z
] = −i
z,
∂
∂z
(5)
Further,
z,
∂
∂z
= −1
( 6 )
So
[z, p z ] = i
(7)
Combining (1), (4) and (7) we get
[L x , L y ] = iL z
(8)
3.78 Given spin state is a singlet state, that is S = 0
S 1 + S 2 = S
Form scalar product by itself
S 1 · S 1 + S 1 · S 2 + S 2 · S 1 + S 2 · S 2 = S · S
S 1
2
+ 2 S 1 · S 2 + S 2
2
= S
2
= 0
Now, S 1
2
= S 2
2
= (1/2)(1/2 + 1) = 3/4
Therefore S 1 · S 2 = −(3/4)
2
3.79 For the n – p system
S p + S n = S
and S p
2
= S n
2
= s(s + 1) with s = 1/2
(i) For singlet state, S = 0