1000 Solved Problems in Modern Physics

(Romina) #1

216 3 Quantum Mechanics – II


∴(Sp+Sn)·(Sp+Sn)=S·S
Sp^2 +Sn^2 +2Sp·Sn=S^2 = 0

(^1) / 2 (1/ 2 +1)+ (^1) / 2 (1/ 2 +1)+2Sp·Sn= 0
Or Sp·Sn=− 3 /4. Orσp·σn=− 3
(ii) For triplet state S= 1
3 / 4 + 3 / 4 +2Sp·Sn=1(1+1)
∴Sp·Sn= 1 / 4
But Sp=^1 / 2 σpand Sn=^1 / 2 σn
∴σp·σn= 1
3.80 From the definition of angular momentum
L=r×p, we can write


L=





∣∣

ijk
xyz
pxpypz





∣∣

=i(ypz−zpy)+j(zpx−xpz)
+k(xpy−ypx)
=iLx+jLy+kLz

Fig. 3.26Cartesian and polar
coordinates


Lx=ypz−zpy=−i

(

y


∂z

−z


∂y

)

Ly=zpx−xpz=−i

(

z


∂x

−x


∂z

)

(1)

Lz=xpy−ypx=−i

(

x


∂y

−y


∂x

)

Ifθ is the polar angle,φthe azimuthal angle andr the radial distance,
(Fig. 3.26). Then
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