Page 58 - Electrical Properties of Materials

The electron meeting a potential barrier 41

The solution of this differential equation is a wave in space. Hence the general
solution of Schrödinger’s equation for the present problem is

E

=exp –i t {A exp(ikz)+ B exp(–ikz)}. (3.18)

In this example we have chosen the potential energy of the electron as zero, A and B are constants representing
thus eqn (3.19) must represent the kinetic energy. Hence, we may conclude the amplitudes of the forward and
that k must be equal to the momentum of the electron. We have come to this backward travelling waves, and k
conclusion before, heuristically, on the basis of the wave picture, but now we is related to E by
have the full authority of Schrödinger’s equation behind us.
2 2
You may notice too that p = k is an alternative expression of de Broglie’s k
E = . (3.19)
relationship, thus we have obtained from Schrödinger’s equation both the wave 2m
behaviour and the correct wavelength.
What can we say about the position of the electron? Take B =0 for sim-
plicity, then we have a forward travelling wave with a definite value for k.The
probability of finding the electron at any particular point is given by |ψ(z)| 2
which according to eqn (3.18) is unity, independently of z. This means physic-
ally that there is an equal probability of the electron being at any point on the
z-axis. The electron can be anywhere; that is, the uncertainty of the electron’s
position is infinite. This is only to be expected. If the value of k is given then
the momentum is known, so the uncertainty in the momentum of the electron
is zero; hence the uncertainty in position must be infinitely great.

3.5 The electron as a particle
Equation (3.17) is a linear differential equation, hence the sum of the solutions
is still a solution. We are therefore permitted to add up as many waves as we
like; that is, a wave packet (as constructed in Chapter 2) is also a solution of
Schrödinger’s equation.
We can now be a little more rigorous than before. A wave packet represents ∗ The constant K may be determined by
2
an electron because | (z)| is appreciably different from zero only within the the normalization condition
packet. With the choice a(k) = 1 in the interval k and zero elsewhere, it ∞ 2
follows from eqns (2.16) and (2.17) that the probability of finding the electron |ψ(z)| dz =1.
–∞
is given by ∗

2
1
sin ( kz)
2
|ψ(z)| = K 2 . (3.20) 1 2
1 ( kz)
2 V
3.6 The electron meeting a potential barrier Electron
Consider again a problem where the motion of the electron is constrained in
V 2
one dimension, and the potential energy is assumed to take the form shown in V = 0
1
Fig. 3.2.
0 z
You are familiar with the classical problem, where the electron starts some-
where on the negative z-axis (say at –z 0 ) in the positive direction with a definite Fig. 3.2
velocity. The solution may be obtained purely from energetic considerations. An electron incident upon a potential
If the kinetic energy of the electron E is smaller than V 2 , the electron is turned barrier.