ABSTRACT
So far, we have only considered systems whose energy levels are discrete, such as the harmonic oscillator or the spin 1/2 magnet. The energy levels of a particle in free space, on the other hand, are continuous, as are those of many other systems, and we now extend our treatment to cover the continuum case. We shall hnd that we have to replace the level degeneracy (that is, the number of quantum states in any one energy level) with a new variable, the density of states, which is dehned as the number of states per unit volume per unit energy interval, and which is in general a function of the kinetic energy of the particle. We use a trick that is common in quantum mechanics, to pretend that
the particle is in a large but hnite box, in which the potential is zero so that the energy E is simply the kinetic energy. The energy levels in this box are discrete, although close together. At the end of the calculation, we allow the size of the box to become inhnite, so that the separation of the levels tends to zero. For a given energy E, we shall hnd that the number of states whose energy is less than E is proportional to volume in a three-dimensional (not
enthalpy G) as the number of states per unit volume with energy less than E. In a one-or two-dimensional system, G(E) is dehned as the number of states with energy less than E per unit length or area respectively. The density of states is then the derivative g(E) # dG
dE , the number of states
per unit volume in the energy range E to E + dE being g(E) dE. Note that while the degeneracy g of a discrete state is a dimensionless number, the dimensions of g(E) in d dimensions are (energy)1(length)d. The number of states between E and E+dE in volume V is V g(E) dE,
and from the canonical distribution (Equation (5.4)) the probability that one of these states is occupied is Z1e#E . Hence, the probability that the particle has an energy between E and E + dE is
P (E) dE = Z1V g(E)e#E dE. (6.1)
As before, Z is the partition function, and P (E) is a probability density, with dimension (energy)1. Since
$ 0 P (E) dE = 1, the partition function is
Z = V
g(E)e#E dE. (6.2)
Comparing Equations (6.2) and (5.8) we see that when we go from a sum over a discrete set of energy levels to an integral over a continuum, we must replace the level degeneracy g with the number of states in the interval dE; that is, V g(E) dE. As a simple example of this limiting process, consider the partition func-
tion at high temperature of a harmonic oscillator of frequency " enclosed in a box of unit volume. The quantized energy levels are equally spaced and nondegenerate, being given by Ej = (j +
1 2 )h", j = 0 to . If kBT ( h",
the Boltzmann factor e#Ej changes by very little when j increases by 1, so that the energy E can be regarded as a continuous variable. The sum
j gje #Ej in Equation (5.8) for Z can then be approximated by an inte-
gral, with gj replaced by g(E) dE as in Equation (6.2). Since the separation of the energy levels is h", there are (h")1 states in unit energy interval, so that g(E) = (h")1, and
Z (h")1 8
e#E dE = (#E")1 = kBT
h" . (6.3)
Hence, from Equation (5.9)
EX = ( lnZ (#
= #1 = kBT. (6.4)
For 3NA, oscillators U = 3NAEX = 3R0T , in agreement with the high
We now calculate the density of states for a nonrelativistic free particle of mass m, for which the dispersion relation is (see Equation (B.1))
E = h2
2m 2 = W2
2m k2,
where is the de Broglie wavelength and k = 2$/ . The derivation begins with a calculation of the normal modes of waves in a box with perfectly re4ecting walls. This part of the calculation is valid for any dispersion relation. We start with a one-dimensional box of length L, and calculate the
density of states per unit length as L . We put the origin at one end of the box and let the amplitude of the wave at a distance x from the origin at time t be (x, t). Since the walls are assumed to be impenetrable, the boundary conditions are the same as for a string hxed at both ends: (0, t) = (L, t) = 0. We look for the normal modes, which are simply the standing wave solutions of the wave equation with the given boundary conditions. This problem is elementary, and we know that the solutions have the
form (see Appendix B)
(x, t) = A sinx et (6.5)
where A is a normalizing constant, = 2$" (where " is the frequency of the mode), and the boundary condition (L, t) = 0 requires that = n$/L, where n is any positive nonzero integer. Some low normal modes (that is, the solutions for |(x, t)| consistent with the boundary conditions) are illustrated in Figure 6.1. We now ask how many modes there are with less than a certain value,
say k. We call the number of these modes N(k). This is the number of integers n less than kL$ . Since L is very large,
kL $ is a large number for
all but the very lowest k, so that its fractional part is negligible and we can write N(k) kL$ . Hence, the number of modes with < k, per unit length, is
G(k) = N(k)
L = k
$ . (6.6)
From the dispersion relation (Equation (B.1)) given above, k = (2mE) 1/2
W . Substituting for k in Equation (6.6) we hnd for a spinless particle (which has only one quantum state for each normal mode)
G(E) = (2mE)1/2
$W . (6.7)
Figure 6.1. Normal modes in one dimension: is plotted vertically (with the zero of each mode displaced for clarity) and x horizontally.
