D.26 Density of states

This note derives the density of states for particles in a box.

Consider the wave number space, as shown to the left in figure 6.1. Each point represents one spatial state. The first question is how many points have a wave number vector whose length is less than some given value . Since the length of the wave number vector is the distance from the origin in wave number state, the points with form an octant of a sphere with radius . In fact, you can think of this problem as finding the number of red points in figure 6.11.

Now the octant of the sphere has a volume

(in wave
number space, not a physical volume)

Conversely, every wave number point is the top-left front corner of a little block of

volume

where , , and are the spacings between the points in the , , and directions respectively. To find the approximate number of points inside the octant of the sphere, take the ratio of the two

volumes:

Now the spacings between the points are given in terms of the sides
, , and of the box containing the
particles as, (6.3),

Plug this into the expression for the number of points in the octant to get:

where is the (physical) volume of the box . Each wave number point corresponds to one spatial state, but if the spin of the particles is then each spatial state still has different spin values. Therefore multiply by to get the number of states.

To get the density of states on a wave number basis, take the
derivative with respect to . The number of states
in a small wave number range is then:

The factor is the density of states on a wave number basis.

To get the density of states on an energy basis, simply eliminate
in terms of the single-particle energy using
. That gives:

The used expression for the kinetic energy is only valid for nonrelativistic speeds.

The above arguments fail in the presence of confinement. Recall that each state is the top-left front corner of a little block in wave number space of volume . The number of states with wave number less than some given value was found by computing how many such little block volumes are contained within the octant of the sphere of radius .

The problem is that a wave number is only inside the sphere octant if all of its little block is inside. Even if 99% of its block is inside, the state itself will still be outside, not 99% in. That makes no difference if the states are densely spaced in wave number space, like in figure 6.11. In that case almost all little blocks are fully inside the sphere. Only a thin layer of blocks near the surface of the sphere are partially outside it.

However, confinement in a given direction makes the corresponding spacing in wave number space large. And that changes things.

In particular, if the -dimension of the box containing the
particles is small, then is large.
That is illustrated in figure 6.12. In this case, there are
no states inside the sphere at all if is less than
. Regardless of what (D.12) claims. In
the range ,
illustrated by the red sphere in figure 6.12, the red sphere
gobbles up a number of states from the plate
. This number of states can be estimated as

since the top of this ratio is the area of the quarter circle of states and the bottom is the rectangular area occupied per state.

This expression can be cleaned up by noting that

with 1 for the lowest plate. Substituting for , , and in terms of the box dimensions then gives

Here is the area of the quantum well and 1 is the plate number. For nonrelativistic speeds is proportional to the energy . Therefore the density of states, which is the derivative of the number of states with respect to energy, is constant.

In the range a second quarter circle of states gets added. To get the number of additional states in that circle, use 2 for the plate number in (D.13). For still larger values of , just keep summing plates as long as the expression between the square brackets in (D.13) remains positive.

If the -dimension of the box is also small, like in a quantum
wire, the states in wave number space separate into individual lines,
figure 6.13. There are now no states until the sphere of
radius hits the line that is closest to the origin, having quantum
numbers 1. Beyond that value of ,
the number of states on the line that is within the sphere is

since the top is the length of the line inside the sphere and the bottom the spacing of the states on the line. Cleaning up, that gives

with the length of the quantum wire. For still larger values of sum over all values of and for which the argument of the square root remains positive.

For nonrelativistic speeds, is proportional to the energy. Therefore the above number of states is proportional to the square root of the amount of energy above the one at which the line of states is first hit. Differentiating to get the density of states, the square root becomes an reciprocal square root.

If the box is small in all three directions, figure 6.14,
the number of states simply becomes the number of points inside the
sphere: