172
3 Quantum Mechanics – II
The solutions are
u 1 (r ) = A sin kr + B cos kr; r < R
(11)
u 2 (r ) = Ce
−γ r
+ De
γ r ; r > R
(12)
Boundary conditions: as r → 0, u 1 → 0
and as r → ∞, u 2 must be finite. This means that B = D = 0.
Therefore the physically accepted solutions are
u 1 = A sin kr
(13)
u 2 = Ce
−γ r
(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
A sin k R = Ce
−γ r
(15)
Ak cos k R = −γ Ce
−γ r
(16)
Dividing the two equations
k cot k R = −γ
(17)
Or
cot k R = −
γ
k
(18)
Now V 0 W , so cot kR is a small negative quantity. Therefore kR ≈
π/2 k
2 R
2
=
π
2
2
Or
M(V 0 − W )R
2
2
=
π
2
4
Again neglecting W compared to V 0
V 0 R
2
≈
π
2
2
4M
3.20 The inside wave function is of the form u = A sin kr . Because V (r ) = 0 for
r > R, we need to consider contribution to < V > from within the well alone.
< V >=
R
0
u
∗ (−V 0 )u dr = −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 2k R
4k
3 Quantum Mechanics – II
The solutions are
u 1 (r ) = A sin kr + B cos kr; r < R
(11)
u 2 (r ) = Ce
−γ r
+ De
γ r ; r > R
(12)
Boundary conditions: as r → 0, u 1 → 0
and as r → ∞, u 2 must be finite. This means that B = D = 0.
Therefore the physically accepted solutions are
u 1 = A sin kr
(13)
u 2 = Ce
−γ r
(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
A sin k R = Ce
−γ r
(15)
Ak cos k R = −γ Ce
−γ r
(16)
Dividing the two equations
k cot k R = −γ
(17)
Or
cot k R = −
γ
k
(18)
Now V 0 W , so cot kR is a small negative quantity. Therefore kR ≈
π/2 k
2 R
2
=
π
2
2
Or
M(V 0 − W )R
2
2
=
π
2
4
Again neglecting W compared to V 0
V 0 R
2
≈
π
2
2
4M
3.20 The inside wave function is of the form u = A sin kr . Because V (r ) = 0 for
r > R, we need to consider contribution to < V > from within the well alone.
< V >=
R
0
u
∗ (−V 0 )u dr = −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 2k R
4k