Biomedical Engineering Reference
In-Depth Information
which can be rewritten
d
2
ψ
d
x
2
8π
2
m
(ε
−
U
0
)
+
ψ
=
0
(11.9)
h
2
The standard solution to this second-order differential equation is
A
sin
8π
2
m
(ε
B
cos
8π
2
m
(ε
−
U
0
)
−
U
0
)
ψ
=
x
+
x
h
2
h
2
where
A
and
B
are constants of integration. We know that ψ
=
0 in regions I and III so the
wavefunction must also be zero at
x
=
0 and at
x
=
L
in order to be continuous at these
points. Substituting ψ
=
0at
x
=
0 shows that
B
=
0. If we substitute ψ
=
0at
x
=
L
we get
A
sin
8π
2
m
(ε
−
U
0
)
0
=
L
h
2
This means that either
A
0 or the argument of sin must be an integral multiple of π
(written
n
).
A
cannot be 0, for otherwise the wavefunction would be 0 everywhere, and so,
on rearranging
=
n
2
h
2
8
mL
2
ε
n
=
U
0
+
(11.10)
A
sin
n
π
L
x
ψ
n
=
The energy is quantized, and so I have added a subscript to the ε for emphasis. It generally
happens that the imposition of boundary conditions leads to energy quantization. It is often
necessary when tackling such problems to match up both the wavefunction and its gradient
at the boundaries, but in this particular case the solution appeared just by considering the
wavefunction. (In fact, the infinities in this particular potential mean that the first derivative
is not continuous at the boundaries; 'real' potentials do not contain such sudden jumps to
infinity.)
Parameter
n
is the
quantum number
and at first sight the allowed values are
n
=
−
...
2,
−
=
0 is not allowed for the same reason that
A
cannot be 0. We
can deduce the constant of integration
A
by recourse to the Born interpretation of quantum
mechanics; the probability of finding the particle between
x
and
x
1, 0, 1, 2, ...The value
n
+
d
x
is
ψ
n
(
x
) d
x
Probabilities have to sum to 1 and so we require
L
ψ
n
(
x
) d
x
=
1
0
from which we deduce
2
L
sin
n
π
x
ψ
n
=±
L
