3.3 Solutions
205
P(x) =
1
π
√
a 2 − x 2
(5)
3.57 Schrodinger’s equation in one dimension is
−
2
2m
d
2
ψ
dx 2 + V (x)ψ = Eψ
(1)
Given
ψ = exp(−
1
2
ax
2 )
( 2 )
Differentiating twice,
we get
d
2
ψ
dx 2 exp(−
1
2
ax
2 )(a
2 x
2
− a)
( 3 )
Inserting (2) and (3) in (1), we get
V (x) = E +
2
2m
(a
2 x
2
− a)
( 4 )
Minimum value of V (x) is determined from
dV
dx
=
2 a
2 x
m
= 0
Minimum of V (x) occurs at x = 0
From (4) we find 0 = E −
2 a
2m
(a) Or the eigen value E =
2 a
2m
(b) V (x) =
2 a
2m
+
2
2m
(a
2 x
2
− a) =
2 a
2 x
2
2m
3.58 < V > n =
E n
2
<
P
2
2m
> n =< H > n − < V > n =
1
/ 2 E n
< P
2
> n = m E n
Also < x > n = 0; < P > n = 0
< (Δx)
2
>=< x
2
> n − < x
2
> n
<
ΔP
2
>=< P
2
> n − < P n >
2
=< P
2
> n = m E n
But < x
2
> n =
∞
−∞ u
∗
n (x)x
2 u n (x)dx; ξ = αx