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 ${\underline k}$ is less than some given value $k$. Since the length of the wave number vector is the distance from the origin in wave number state, the points with ${\underline k}$ $\raisebox{.3pt}{$<$}$ $k$ form an octant of a sphere with radius $k$. 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)

\mbox{octant volume:}\quad{\textstyle\frac{1}{8}} {\textstyle\frac{4}{3}} \pi k^3

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

\mbox{single state volume:}\quad \Delta k_x \Delta k_y \Delta k_z

where $\Delta{k}_x$, $\Delta{k}_y$, and $\Delta{k}_z$ are the spacings between the points in the $x$, $y$, and $z$ directions respectively. To find the approximate number of points inside the octant of the sphere, take the ratio of the two volumes:

\mbox{number of spatial states inside:}\quad
\frac{\pi k^3}{6\Delta k_x \Delta k_y \Delta k_z}

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

\Delta k_x = \frac{\pi}{\ell_x} \qquad
\Delta k_y = \frac{\pi}{\ell_y} \qquad
\Delta k_z = \frac{\pi}{\ell_z}

Plug this into the expression for the number of points in the octant to get:
\mbox{number of spatial states inside:}
\quad \frac{{\cal V}}{6\pi^2} k^3 %
\end{displaymath} (D.12)

where ${\cal V}$ is the (physical) volume of the box $\ell_x\ell_y\ell_z$. Each wave number point corresponds to one spatial state, but if the spin of the particles is $s$ then each spatial state still has $2s+1$ different spin values. Therefore multiply by $2s+1$ to get the number of states.

To get the density of states on a wave number basis, take the derivative with respect to $k$. The number of states ${\rm d}{N}$ in a small wave number range ${\rm d}{k}$ is then:

{\rm d}N = {\cal V}{\cal D}_k {\,\rm d}k \qquad {\cal D}_k = \frac{2s+1}{2\pi^2} k^2

The factor ${\cal D}_k$ is the density of states on a wave number basis.

To get the density of states on an energy basis, simply eliminate $k$ in terms of the single-particle energy ${\vphantom' E}^{\rm p}$ using ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hbar}k^2$$\raisebox{.5pt}{$/$}$$2m$. That gives:

{\rm d}N = {\cal V}{\cal D}{\,\rm d}{\vphantom' E}^{\rm p}...
...\frac{2m}{\hbar^2}\right)^{3/2} \sqrt{{\vphantom' E}^{\rm p}}

The used expression for the kinetic energy ${\vphantom' E}^{\rm p}$ 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 $\Delta{k}_x\Delta{k}_y\Delta{k}_z$. The number of states with wave number ${\underline k}$ less than some given value $k$ was found by computing how many such little block volumes are contained within the octant of the sphere of radius $k$.

The problem is that a wave number ${\underline k}$ 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 $y$-​dimension $\ell_y$ of the box containing the particles is small, then $\Delta{k}_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$$\raisebox{.5pt}{$/$}$$\ell_y$ is large. That is illustrated in figure 6.12. In this case, there are no states inside the sphere at all if $k$ is less than $\Delta{k}_y$. Regardless of what (D.12) claims. In the range $\Delta{k}_y$ $\raisebox{.3pt}{$<$}$ $k$ $\raisebox{.3pt}{$<$}$ $2\Delta{k}_y$, illustrated by the red sphere in figure 6.12, the red sphere gobbles up a number of states from the plate ${\underline k}_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\Delta{k}_y$. This number of states can be estimated as

\frac{\frac14\pi(k_x^2+k_z^2)}{\Delta k_x \Delta k_z}

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

k_x^2+k_z^2=k^2-k_y^2 = k^2 - \left(n_y\Delta{k}_y\right)^2

with $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 for the lowest plate. Substituting for $\Delta{k}_x$, $\Delta{k}_y$, and $\Delta{k}_z$ in terms of the box dimensions then gives
\mbox{spatial states per plate:} \quad \frac{A}{4\pi}
\qquad \mbox{if}\quad \bigg[\ldots\bigg] > 0 %
\end{displaymath} (D.13)

Here $A$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x\ell_z$ is the area of the quantum well and $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is the plate number. For nonrelativistic speeds $k^2$ is proportional to the energy ${\vphantom' E}^{\rm p}$. Therefore the density of states, which is the derivative of the number of states with respect to energy, is constant.

In the range $2\pi$$\raisebox{.5pt}{$/$}$$\ell_y$ $\raisebox{.3pt}{$<$}$ $k$ $\raisebox{.3pt}{$<$}$ $3\pi$$\raisebox{.5pt}{$/$}$$\ell_y$ a second quarter circle of states gets added. To get the number of additional states in that circle, use $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 for the plate number in (D.13). For still larger values of $k$, just keep summing plates as long as the expression between the square brackets in (D.13) remains positive.

If the $z$-​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 $k$ hits the line that is closest to the origin, having quantum numbers $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. Beyond that value of $k$, the number of states on the line that is within the sphere is

\frac{\sqrt{k^2- (n_y\Delta k_y)^2 - (n_z\Delta k_z)^2}}{\Delta k_x}

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
\mbox{spatial states per line:} \quad
\frac{\ell}{\pi} \...
\qquad \mbox{if}\quad \bigg[\ldots\bigg] > 0 %
\end{displaymath} (D.14)

with $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x$ the length of the quantum wire. For still larger values of $k$ sum over all values of $n_y$ and $n_z$ for which the argument of the square root remains positive.

For nonrelativistic speeds, $k^2$ 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:

\mbox{spatial states per point:} \quad 1
\qquad \bigg[k^...
- \left(n_z\frac{\pi}{\ell_z}\right)^2 \bigg] > 0 %
\end{displaymath} (D.15)

In other words, to get the total number of states inside, simply add a 1 for each set of natural numbers $n_x$, $n_y$, and $n_z$ for which the expression in brackets is positive. The derivative with respect to energy, the density of states, becomes a series of delta functions at the energies at which the states are hit.