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3 Quantum Mechanics – II
This is the continuity equation where the probability current J =
1
2im
(ϕ
∗
∇ϕ−
ϕ∇ϕ
∗ )
And probability density
ρ =
i
2m
ϕ
∗ ∂ϕ
∂t
− ϕ
∂ϕ
∗
∂t
For a force free particle the solution of the Klein – Gordan equation is ϕ =
A e
i(p.x−Et)
The probability density is
ρ =
i
2m
A
∗ e
−i(p.x−Et)
(i AE) e
−i(p.x−Et)
−
Ae
−i(p.x−Et)
i A
∗ E
e
−i(p.x−Et)
=
i
2m
A
∗ A (−i E) − A A
∗ (i E)
=
|A|
2
2m
[E + E] = E
|A|
2
m
As E can have positive and negative values, the probability density could
then be negative
3.6 (a) Class I: Refer to Problem 3.25
ψ 1 = Ae
βx (−∞ < x < −a)
ψ 2 = D cos αx (−a < x < +a)
ψ 3 = Ae
−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
αxdx +
∞
a
A
2 e
−2βx dx = 1
A
2 e
−2βa
/2β + D
2 [a + sin(2αa)/2α] + A
2 e
−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 α a = a e
−βa
(2)
Combining (1) and (2) gives
D =
a +
1
β
−1
A = e
βa cosαa
a +
1
β
−1