1000 Solved Problems in Modern Physics

(Grace) #1

228 3 Quantum Mechanics – II


(b) First we show that the wavefunction is normalized.

|ψ|^2 dr=

1

4 π

∫∞

0

|g(r)|^2 r^2 dr

∫π

0


∫ 2 π

0

(1+cosφsin 2θ)sinθdφ

=

1

2

∫π

0

sinθdθ= 1

The probability for the occurrence ofLz=is
∫ [√
1
3

(



2

)

Y 11

] 2

dΩ=(2/3)


(3/ 8 π)sin^2 θ. 2 πsinθdθ

=(1/2)

∫+ 1

− 1

(1−cos^2 θ)dcosθ= 2 / 3

The probability for the occurrence ofLz=0is
∫ [√
1
3

Y 10

] 2

dΩ=

∫+ 1

− 1

(

3

4 π

)

. 2 πcos^2 θdcosθ= 1 / 3


3.96 (a)[Jz,J+]=JzJ+−J+Jz=Jz(Jx+iJy)−(Jx+iJy)Jz
=JzJx−JxJz+i(JzJy−JyJz)
=[Jz,Jx]+i[Jz,Jy]=iJy−iiJx
=iJy+Jx=(Jx+iJy)=J+
=J+,in units of.
(b) From (a),JzJ+=J+Jz+J+
JzJ+|jm>=J+Jz|jm>+J+|jm>
=J+m|jm>+J+|jm>
=(m+1)J+|jm>
J+|jm>is nothing but|j,m+ 1 >apart from a possible normalization
constant. Thus
J+|jm>=Cjm+|j,m+ 1 >
Given a state|jm>, the state|j,m+ 1 >must exist unlessCjm+van-
ishes for that particularm. Sincejis the maximum value of m by definition.
There can not be a state|j,j+ 1 >, i.e.Cjj+must vanish.J+is known as
the ladder operator. Similarly,J−lowers m by one unit.
(c)J+=Jx+iJy
J−=Jx−iJy


Therefore,Jx=

1

2

(J++J−)=



01 /


20

1 /


201 /


2

01 /


20



Jy=

1

2 i

(J+−J−)=



0 −i/


20

i/


20 −i/


2

0 i/


20



[Jx,Jy]=JxJy−JyJx=i



10 0

00 0

00 − 1


⎠=iJz
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