184
3 Quantum Mechanics – II
Fig. 3.13
For (a) the inside and outside wave functions are as in the deuteron
Problem 3.19. For (b) the inside wave function is similar but the outside
function becomes constant (W 1 = 0) and is a horizontal line.
3.36 (a) Class I: Refer to Problem 3.25
ψ 1 = Ae
βx (−∞ < x < −a)
ψ 2 = D cos ax(−a < x < +a)
ψ 3 = A e
−βx (a < x < ∞)
Normalization implies that
−a
−∞
|ψ 1 |
2 dx +
a
−a
|ψ 2 |
2 dx +
∞
a
|ψ 3 |
2 dx = 1
−a
−∞
A
2 e
2βx dx +
a
−a
D
2 cos
2
αx dx +
∞
a
A
2 e
−2βx dx = 1
A
2 e
−2βa
2β
+ D
2
a +
sin(2αa)
2α
+
A
2e
−2βa
2β
= 1
Or
A
2 e
−2βa
/β + D
2 (a + sin(2αa)/2α) = 1
( 1 )
Boundary condition at x = a gives
D cos αa = ae
−βa
(2)
Combining (1) and (2) gives
D =
a +
1
β
−1
A = e
βa cos αa
a +
1
β
−1