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} \bi...
...2 \bigg]^{1/2}
\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^2 ...
- \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.