Page 56 - Electrical Properties of Materials

Solutions of Schrödinger’s equation 39

The physical content of eqn (3.4) will be clearer when we shall treat
more practical problems, but there is one thing we can say immediately.
Schrödinger’s equation does not tell us the position of the electron, only the
probability that it will be found in the vicinity of a certain point.
The description of the electron’s behaviour is statistical, but there is nothing
particularly new in this. After all, you have met statistical descriptions before,
in gas dynamics for example, and there was considerably less fuss about it.
The main difference is that in classical mechanics, we use statistical meth-
ods in order to simplify the calculations. We are too lazy to write up 10 27
differential equations to describe the motion of all the gas molecules in a
vessel, so we rely instead on a few macroscopic quantities like pressure, tem-
perature, average velocity, etc. We use statistical methods because we elect
to do so. It is merely a question of convenience. This is not so in quantum
mechanics. The statistical description of the electron is inherent in quantum
theory. That is the best we can do. We cannot say much about an electron at
a given time. We can only say what happens on the average when we make
many observations on one system, or we can predict the statistical outcome
of simultaneous measurements on identical systems. It may be sufficient to
make one single measurement (specific heat or electrical conductivity) when
the phenomenon is caused by the collective interaction of a large number of
electrons.
We cannot even say how an electron moves as a function of time. We can-
not say this because the position and the momentum of an electron cannot be
simultaneously determined. The limiting accuracy is given by the uncertainty
relationship, eqn (2.33).

3.3 Solutions of Schrödinger’s equation
Let us separate the variables and attempt a solution in the following form
(r, t)= ψ(r)w(t). (3.7)

Substituting eqn (3.7) into eqn (3.4), and dividing by ψw we get Now r represents all the spatial
variables.
2
2
∇ ψ 1 ∂w
– + V =i . (3.8)
2m ψ w ∂t
Since the left-hand side is a function of r and the right-hand side is a function
of t, they can be equal only if they are both separately equal to a constant which
we shall call E, that is we obtain two differential equations as follows:
∂w
i = Ew (3.9)
∂t
and
2 2
– ∇ ψ + Vψ = Eψ. (3.10)
2m
The solution of eqn (3.9) is simple enough. We can immediately integrate
and get
E

w =exp –i t , (3.11)