Page 62 - Electrical Properties of Materials

The electron in a potential well 45

3.8 The electron in a potential well
In our previous examples, the electron was free to roam in the one-dimensional
space. Now we shall make an attempt to trap it by presenting it with a region of
low potential energy which is commonly called a potential well. The potential L is the width of the well.
profile assumed is shown in Fig. 3.4. If E > V 1 , the solutions are very similar
to those discussed before, but when E < V 1 , a new situation arises.
We have by now sufficient experience in solving Schrödinger’s equation
for a constant potential, so we shall write down the solutions without further
discussions.
In region 3 there is only an exponentially decaying solution
ψ 3 = C exp(–γ z), (3.29)

where
2m
2
γ = (V 1 – E). (3.30)
2
In region 2 the potential is zero. The solution is either symmetric or
∗
antisymmetric. Accordingly, ∗ This is something we have not proved.
It is true (the proof can be obtained fairly
ψ 2s = A cos kz (3.31) easily from Schrödinger’s equation) in
general that if the potential function is
or symmetric the solution must be either
symmetric or antisymmetric.
ψ 2a = A sin kz (3.32)
where
2m
2
k = E. (3.33)
2
In region 1 the solution must decay again, this time towards negative infinity.
If we wish to satisfy the symmetry requirement as well, the wave function must
look like
ψ 1 = ± C exp γ z. (3.34)
† There are a number of different proofs
Let us investigate the symmetric solution first. We have then eqns (3.29) and
that these conditions are physically reas-
(3.31), and eqn (3.34) with the positive sign. The conditions to be satisfied are onable. For example, direct integration
†
the continuity of ψ and ∂ψ/∂z at L/2 and –L/2, but owing to the symmetry of Schrödinger’s equation over an inter-
it is sufficient to do the matching at (say) L/2. From the continuity of the wave val involving a boundary can yield the
condition that ∂ψ/∂z must generally be
function
continuous. One exception is when the
L L potential step at the boundary becomes
A cos k – C exp –γ = 0. (3.35) infinite, for example when an electron is
2 2
confined in an infinite potential well. In
this case, ∂ψ/∂z is not continuous.
1 2 3
V V
1 V = V
3 1
V = 0
2
L 0 L z Fig. 3.4
–
2 2 An electron in a potential well.