3.3 Solutions
169
Or
d
2
ψ(x)
dx 2 +
2m E
2
ψ(x) = 0
( 2 )
Writing
α
2
=
2m E
2
(3)
Equation (2) becomes
d
2
ψ
dx 2 + α
2
ψ = 0
( 4 )
which has the solution
ψ(x) = A sin αx + B cos αx
(5)
where A and B are constants of integration. Take the origin at the left
corner, Fig 3.5.
Fig. 3.5 Square potential
well of infinite depth
Boundary condition:
ψ(0) = 0; ψ(a) = 0
The first one gives B = 0. We are left with
ψ = A sin αx
(6)
The second one gives
αa = nπ, n = 1, 2, 3 . . .
(7)
n = 0 is excluded as it would give a trivial solution.
Using the value of α in (6)
ψ n (x) = Asin
nπ x
a
(8)
This is an unnormalized solution. The constant A is determined from
normalization condition.
169
Or
d
2
ψ(x)
dx 2 +
2m E
2
ψ(x) = 0
( 2 )
Writing
α
2
=
2m E
2
(3)
Equation (2) becomes
d
2
ψ
dx 2 + α
2
ψ = 0
( 4 )
which has the solution
ψ(x) = A sin αx + B cos αx
(5)
where A and B are constants of integration. Take the origin at the left
corner, Fig 3.5.
Fig. 3.5 Square potential
well of infinite depth
Boundary condition:
ψ(0) = 0; ψ(a) = 0
The first one gives B = 0. We are left with
ψ = A sin αx
(6)
The second one gives
αa = nπ, n = 1, 2, 3 . . .
(7)
n = 0 is excluded as it would give a trivial solution.
Using the value of α in (6)
ψ n (x) = Asin
nπ x
a
(8)
This is an unnormalized solution. The constant A is determined from
normalization condition.