1000 Solved Problems in Modern Physics

(Tina Meador) #1

3.3 Solutions 215


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


3.77 L=

∣ ∣ ∣ ∣ ∣ ∣

ijk
xyz
px pypz

∣ ∣ ∣ ∣ ∣ ∣

=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
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