1000 Solved Problems in Modern Physics

172 3 Quantum Mechanics – II

The solutions are

u 1 (r)=Asinkr+Bcoskr;r<R (11)

u 2 (r)=Ce−γr+Deγr;r>R (12)

Boundary conditions: asr→ 0 ,u 1 → 0

and asr→∞,u 2 must be finite. This means thatB=D=0.

Therefore the physically accepted solutions are

u 1 =Asinkr (13)

u 2 =Ce−γr (14)

At the boundary,r=R,u 1 =u 2 and their first derivatives

(

du 1

dr

)

r=R

=

(

du 2

dr

)

r=R

These lead to

AsinkR=Ce−γr (15)

AkcoskR=−γCe−γr (16)

Dividing the two equations

kcotkR=−γ (17)

Or

cotkR=−

γ

k

(18)

NowV 0  W, so cotkRis a small negative quantity. ThereforekR ≈

π/ 2 k^2 R^2 =

(π

2

) 2

Or

M(V 0 −W)R^2

^2

=

π^2

4

Again neglectingWcompared toV 0

V 0 R^2 ≈

π^2 ^2

4 M

3.20 The inside wave function is of the formu=Asinkr. BecauseV(r)=0for
r>R, we need to consider contribution tofrom within the well alone.

=

∫R

0

u∗(−V 0 )udr=−V 0 A^2

∫R

0

sin^2 krdr

=

(

−

V 0 A^2

2

)∫ R

0

(1−cos 2kr)dr

=−V 0 A^2

[

R

2

−

sin 2kR

4 k

]