6.3 Den­sity of States

Up to this point, this book has pre­sented en­ergy lev­els in the form of an en­ergy spec­trum. In these spec­tra, each sin­gle-par­ti­cle en­ergy was shown as a tick mark along the en­ergy axis. The sin­gle-par­ti­cle states with that en­ergy were usu­ally listed next to the tick marks. One ex­am­ple was the en­ergy spec­trum of the elec­tron in a hy­dro­gen atom as shown in fig­ure 4.8.

How­ever, the num­ber of states in­volved in a typ­i­cal macro­scopic sys­tem can eas­ily be of the or­der of 10$\POW9,{20}$ or more. There is no way to show any­where near that many en­ergy lev­els in a graph. Even if print­ing tech­nol­ogy was up to it, and it can only dream about it, your eyes would have only about 7 10$\POW9,{6}$ cones and 1.3 10$\POW9,{8}$ rods to see them.

For al­most all prac­ti­cal pur­poses, the en­ergy lev­els of a macro­scopic sys­tem of non­in­ter­act­ing par­ti­cles in a box form a con­tin­uum. That is schemat­i­cally in­di­cated by the hatch­ing in the en­ergy spec­trum to the right in fig­ure 6.1. The spac­ing be­tween en­ergy lev­els is how­ever very many or­ders of mag­ni­tude tighter than the hatch­ing can in­di­cate.

Ta­ble 6.1: En­ergy of the low­est sin­gle-par­ti­cle state in a cube with 1 cm sides.

It can also nor­mally be as­sumed that the low­est en­ergy is zero for non­in­ter­act­ing par­ti­cles in a box. While the low­est sin­gle par­ti­cle en­ergy is strictly speak­ing some­what greater than zero, it is ex­tremely small. That is nu­mer­i­cally il­lus­trated by the val­ues for a 1 cm$\POW9,{3}$ cu­bic box in ta­ble 6.1. The ta­ble gives the low­est en­ergy as com­puted us­ing the for­mu­lae given in the pre­vi­ous sec­tion. The low­est en­ergy oc­curs for the state $\pp111////$ with $n_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. As is com­mon for sin­gle-par­ti­cle en­er­gies, the en­ergy has been ex­pressed in terms of elec­tron volts, one eV be­ing about 1.6 10$\POW9,{-19}$ J. The ta­ble also shows the same en­ergy in terms of an equiv­a­lent tem­per­a­ture, found by di­vid­ing it by 1.5 times the Boltz­mann con­stant. These tem­per­a­tures show that at room tem­per­a­ture, for all prac­ti­cal pur­poses the low­est en­ergy is zero. How­ever, at very low cryo­genic tem­per­a­tures, pho­tons in the low­est en­ergy state, or ground state, may have a rel­a­tively more sig­nif­i­cant en­ergy.

The spac­ing be­tween the low­est and sec­ond low­est en­ergy is com­pa­ra­ble to the low­est en­ergy, and sim­i­larly neg­li­gi­ble. It should be noted, how­ever, that in Bose-Ein­stein con­den­sa­tion, which is dis­cussed later, there is a macro­scopic ef­fect of the fi­nite spac­ing be­tween the low­est and sec­ond-low­est en­ergy states, minis­cule as it might be.

The next ques­tion is why quan­tum me­chan­ics is needed here at all. Clas­si­cal non­quan­tum physics too would pre­dict a con­tin­uum of en­er­gies for the par­ti­cles. And it too would pre­dict the en­ergy to start from zero. The en­ergy of a non­in­ter­act­ing par­ti­cle is all ki­netic en­ergy; clas­si­cal physics has that zero if the par­ti­cle is at rest and pos­i­tive oth­er­wise.

Still, the (anti) sym­metriza­tion re­quire­ments can­not be ac­com­mo­dated us­ing clas­si­cal physics. And there is at least one other im­por­tant quan­tum ef­fect. Quan­tum me­chan­ics pre­dicts that there are more sin­gle-par­ti­cle states in a given en­ergy range at high en­ergy than at low en­ergy.

To ex­press that more pre­cisely, physi­cists de­fine the “den­sity of states” as the num­ber of sin­gle-par­ti­cle states per unit en­ergy range. For par­ti­cles in a box, the den­sity of states is not that hard to find. First, the num­ber ${\rm d}{N}$ of sin­gle-par­ti­cle states in a small wave num­ber range from $k$ to $k+{\rm d}{k}$ is given by, {D.26},

$${\rm d}N = {\cal V}{\cal D}_k { \rm d}k \qquad {\cal D}_k = \frac{2s+1}{2\pi^2} k^2 %$$ (6.5)

Here ${\cal V}$ is the vol­ume of the box that holds the par­ti­cles. As you would ex­pect, the big­ger the box, the more par­ti­cles it can hold, all else be­ing the same. Sim­i­larly, the larger the wave num­ber range ${\rm d}{k}$, the larger the num­ber of states in it. The fac­tor ${\cal D}_k$ is the den­sity of states on a wave num­ber ba­sis. It de­pends on the spin $s$ of the par­ti­cles; that re­flects that there are $2s+1$ pos­si­ble val­ues of the spin for every given spa­tial state.

(It should be noted that for the above ex­pres­sion for ${\cal D}_k$ to be valid, the wave num­ber range ${\rm d}{k}$ should be small. How­ever, ${\rm d}{k}$ should still be large enough that there are a lot of states in the range ${\rm d}{k}$; oth­er­wise ${\cal D}_k$ can­not be ap­prox­i­mated by a sim­ple con­tin­u­ous func­tion. If the spac­ing ${\rm d}{k}$ truly be­comes zero, ${\cal D}_k$ turns into a dis­tri­b­u­tion of in­fi­nite spikes.)

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

$$\fbox{$\displaystyle {\rm d}N = {\cal V}{\cal D}{ \rm d}{... ...{2m}{\hbar^2}\right)^{3/2} \sqrt{{\vphantom' E}^{\rm p}} $} %$$ (6.6)

The re­quire­ments on the en­ergy range ${\rm d}{\vphantom' E}^{\rm p}$ are like those on ${\rm d}{k}$.

The fac­tor ${\cal D}$ is what is con­ven­tion­ally de­fined as the den­sity of states; it is on a unit en­ergy range and unit vol­ume ba­sis. In the spec­trum to the right in fig­ure 6.1, the den­sity of states is in­di­cated by means of the width of the spec­trum.

Note that the den­sity of states grows like $\sqrt{{\vphantom' E}^{\rm p}}$: quickly at first, more slowly later, but it con­tin­ues to grow. There are more states per unit en­ergy range at higher en­ergy than at lower en­ergy. And that means that at nonzero en­er­gies, the en­ergy states are spaced many times tighter to­gether still than the ground state spac­ing of ta­ble 6.1 in­di­cates. As­sum­ing that the en­er­gies form a con­tin­uum is an ex­tremely ac­cu­rate ap­prox­i­ma­tion in most cases.

The given ex­pres­sion for the den­sity of states is not valid if the par­ti­cle speed be­comes com­pa­ra­ble to the speed of light. In par­tic­u­lar for pho­tons the Planck-Ein­stein ex­pres­sion for the en­ergy must be used, ${\vphantom' E}^{\rm p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\omega$, where the elec­tro­mag­netic fre­quency is $\omega$ $\vphantom0\raisebox{1.5pt}{$=$}$ $ck$ with $c$ the speed of light. In ad­di­tion, as men­tioned in sec­tion 6.2, pho­tons have only two in­de­pen­dent spin states, even though their spin is 1.

It is con­ven­tional to ex­press the den­sity of states for pho­tons on a fre­quency ba­sis in­stead of an en­ergy ba­sis. Re­plac­ing $k$ with $\omega$$\raisebox{.5pt}{$/$}$​$c$ in (6.5) and $2s+1$ by 2 gives

$$\fbox{$\displaystyle {\rm d}N = {\cal V}{\cal D}_\omega {\... ...ga \qquad {\cal D}_\omega = \frac{1}{\pi^2c^3} \omega^2 $} %$$ (6.7)

The fac­tor ${\cal D}_\omega$ is com­monly called the “den­sity of modes” in­stead of den­sity of states on a fre­quency ba­sis.

Key Points

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The spec­trum of a macro­scopic num­ber of non­in­ter­act­ing par­ti­cles in a box is prac­ti­cally speak­ing con­tin­u­ous.

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The low­est sin­gle-par­ti­cle en­ergy can al­most al­ways be taken to be zero.

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The den­sity of states ${\cal D}$ is the num­ber of sin­gle-par­ti­cle states per unit en­ergy range and unit vol­ume.

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More pre­cisely, the num­ber of states in an en­ergy range ${\rm d}{\vphantom' E}^{\rm p}$ is ${\cal V}{\cal D}{ \rm d}{\vphantom' E}^{\rm p}$.

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To use this ex­pres­sion, the en­ergy range ${\rm d}{\vphantom' E}^{\rm p}$ should be small. How­ever, ${\rm d}{\vphantom' E}^{\rm p}$ should still be large enough that there are a lot of states in the range.

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For pho­tons, use the den­sity of modes.