Page 59 - Electrical Properties of Materials

42 The electron

back by the potential barrier at z =0. If E > V 2 , the electron slows down but
carries on regardless.
How should we formulate the equivalent quantum mechanical problem? We
should represent our electron by a wave packet centred in space on –z 0 and
should describe its momentum with an uncertainty p. We should use the
wave function obtained as initial condition at t = 0 and should solve the time-
dependent Schrödinger equation. This would be a very illuminating exercise,
alas much too difficult mathematically.
We have to be satisfied by solving a related problem. We shall give our
electron a definite energy, that is a definite momentum, and we shall put up with
the concomitant uncertainty in position. We shall not be able to say anything
about the electron’s progress towards the potential barrier, but we shall have
a statistical solution which will give the probability of finding the electron on
either side of this potential barrier.
Specifying the momentum and not caring about the position of the electron
is not so unphysical as you might think. The conditions stated may be approx-
∗
∗ imated in practice by shooting a sufficiently sparse electron beam towards the
So that the interaction between the
electrons can be neglected. potential barrier with a well-defined velocity. We are not concerned then with
the positions of individual electrons, only with their spatial distribution on the
average, which we call the macroscopic charge density. Hence we may identify
2
e|ψ| with the charge density.
2
This is not true in general. What is always true is that |ψ| gives the prob-
2
The charge of the electron is not ability of an electron being found at z. Be careful, |ψ(z)| does not give the
smoothed out. When the electron fraction of the electron’s charge residing at z. If, however, a large number
2
is found, the whole electron is of electrons behave identically, then |ψ| may be justifiably regarded to be
there. proportional to the charge density.
Let us proceed now to the mathematical solution. In region 1 where V 1 =0
the solution is already available in eqns (3.18) and (3.19),

E
1 =exp –i t {A exp(ik 1 z)+ B exp(–ik 1 z)}, (3.21)

and
2mE
2
k = . (3.22)
1
2
In region 2 the equation to be solved is as follows (the time-dependent part
of the solution remains the same because E is specified)
2
2
∂ ψ
+(E – V 2 )ψ = 0, (3.23)
2m ∂z 2
with the general solution

E
2 =exp –i t {C exp(ik 2 z)+ D exp(–ik 2 z)}, (3.24)

where
2m
2
k = (E – V 2 ). (3.25)
2
2