144
Chapter 3. Wave optics
Following the analysis of the preceding section, we define a state
function
4
Ψ(x, t) =
a k ψ k (x, t)
k
4
−iH k t/¯ h
=
a k u k (x) e
k
4
1
i(k·x−ω k t)
= √
a k e
,
(3.79)
V k
where the summation over k represents a summation over all possible values of (n x , n y , n z ). As previously, the probability that a
single precise measurement of the total energy yields a specific
value H k is given by |a k |
2 . Following the procedure of the earlier
analysis, we find
1
iω k t
−ik·x
a k = e
√
d
3 x Ψ(x, t) e
,
(3.80)
V V
where the integral is over the cubic volume V .
Next we define a function Φ(k, t) which satisfies
1
dV k
√
a k =
Φ(k, t).
(3.81)
V
(2π) 3/2 dN
The reason for this precise definition will become clear shortly.
Substituting this into (3.79) and making use of (3.68), the state
function is
1
4 dV k
i(k·x−iω k t)
Ψ(x, t) =
Φ(k, t) e
.
(3.82)
(2π) 3/2
k dN
We now consider the limiting case where the cubic volume V is
taken to be very large. According to the preceding arguments, the
lattice of eigenstates in k-space becomes very dense. In this case
the sum can be represented by the integral
i(k·x−ω k t)
Ψ(x, t) =
1
d
3 k Φ(k, t) e
,
(3.83)
(2π) 3/2