Page 48 - Electrical Properties of Materials

Applications to electrons 31

We may now define two velocities. One is ω 0 /k 0 , which corresponds to
the previously defined phase velocity, and is the velocity with which the central
components propagate. The other velocity may be defined by looking at the
expression for A. Since A represents the envelope of the wave, we may say that
the envelope has the same shape whenever
(ω – ω 0 )t –(k – k 0 )z = constant. (2.24)

Hence, we may define a velocity,
∂z ω – ω 0
v g = = , (2.25) v g is called the group velocity be-
∂t k – k 0 cause it gives the velocity of the
which, for sufficiently small δk, reduces to wave packet.
∂ω

v g = . (2.26)
∂k
k=k 0
2.4 Applications to electrons

We have discussed some properties of waves. It has been an exercise in math-
ematics. Now we take a deep plunge and will try to apply these properties to
the particular case of the electrons. The first step is to identify the wave packet
with an electron in your mental picture. This is not unreasonable. We are say-
ing, in fact, that where the ripples are, there must be the electron. If the ripples
are uniformly distributed in space, as is the case for a single frequency wave,
the electron can be anywhere. If the ripples are concentrated in space in the
form of a wave packet, the presence of an electron is indicated. Having identi-
fied the wave packet with an electron, we may identify the velocity of the wave
packet with the electron velocity.
What can we say about the energy of the electron? We know that a photon
of frequency ω has an energy
E = hf = ω, (2.27)

where f is the frequency of the electromagnetic wave and = h/2π. Ana-
logously, it may be suggested that the energy of an electron in a wave packet
centred at the frequency ω is given by the same formula. Hence, we may write
down the energy of the electron, taking the potential energy as zero, in the form

1
2
ω = mv . (2.28)
2 g
We can differentiate this partially with respect to k to get
∂ω ∂v g
= mv g (2.29)
∂k ∂k
which, with the aid of eqn (2.26), reduces to
∂v g
= m . (2.30)
∂k
Integrating, and taking the integration constant as zero, we get
k = mv g , (2.31)