1000 Solved Problems in Modern Physics

3.3 Solutions 215

3.3.6 Angular Momentum ................................

3.77 L=

∣ ∣ ∣ ∣ ∣ ∣

ijk

xyz

pxpypz

∣ ∣ ∣ ∣ ∣ ∣

=i(ypz−zpy)−j(xpz−zpx)+k(xpy−ypx)

=iLx+jLy+kLz

Thus,Lx=ypz−zpy,Ly=zpx−xpz,Lz=xpy−ypx (1)

[Lx,Ly]=LxLy−LyLx

=(ypz−zpy)(zpx−xpz)−(zpx−xpz)(ypz−zpy)

=ypzzpx−ypzxpz−zpyzpx+zpyxpz−zpxypz+zpxzpy

+xpzypz−xpzzpy (2)

But [px,py]=[x,py]= 0 ,etc (3)

(2) becomes

[Lx,Ly]=ypxpzz−yxp^2 z−z^2 pypx+xpyzpy−ypxzpz+z^2 pxpy

+yxp^2 z−xpypzz=[z,pz](xpy−ypz)(4)

But [z,pz]=[z,−i

∂

∂z

]=−i

[

z,

∂

∂z

]

(5)

Further,

[

z,

∂

∂z

]

=−1(6)

So

[z,pz]=i (7)

Combining (1), (4) and (7) we get

[Lx,Ly]=iLz (8)

3.78 Given spin state is a singlet state, that isS= 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 12 +2S 1 ·S 2 +S 22 =S^2 = 0

Now, S 12 =S 22 =(1/2)(1/ 2 +1)= 3 / 4

Therefore S 1 ·S 2 =−(3/4)^2

3.79 For the n – p system

Sp+Sn=S

and Sp^2 =Sn^2 =s(s+1) withs= 1 / 2

(i) For singlet state, S= 0