A.10 Density of states, periodic boundary conditions and black-body radiation

In previous thermal and statistical physics courses we have tended to consider a particle in a box (side lengths $L_x$, $L_y$ $L_z$), with boundary condition that the wavefunction must vanish at the wall. Then the energy eigenfunctions are of the form

for positive integers $l,m,n$, that is, the allowed values of the wavenumber $k_i$ are quantised, though very closely spaced for a macroscopic box. The density of states gives the number of states in the vicinity of a given momentum or energy. In momentum space, the number of states with $k\left(\equiv \left|k\right|\right)$ in the range $k\to k+\text{d}k$, (where $\text{d}k$ is small, but much bigger than the spacing between states) is

See here for details of the derivation. Note that as $k_x$, $k_y$ and $k_z$ all have to be positive, the vector $k$, which isn’t quite a momentum becuase we are dealing with standing waves, has to lie in the positive octant.

In quantum mechanics we prefer not to deal with standing waves, but with momentum eigenstates which are travelling waves. But we still want the advantage of a finite box so that the states remain countable. The solution is to use periodic boundary conditions in which, when a particle reaches a wall at, say, $\left(x,y,L_z\right)$ it leaves the box and reappears, with the same momentum, at $\left(x,y,0\right)$. This may sound artificial but we get the same expression for $D\left(k\right)$; the advantage is that we can usefully talk about $D\left(k\right)$ as well.

In this case the boundary condition is $\psi \left(x,y,L_z\right)=\psi \left(x,y,0\right)$ and the wavefunction is

noting now that $k_x=2\pi l∕L_x$ etc since a whole number of wavelengths have to fit into the box. We have fixed the normalisation so that there is one particle in the box; this differs from the $\delta$-function normalisation used elsewhere. We can now talk about the number of states with $k$ in the range $k_x\to k_x+\text{d}{k}_x$, $k_y\to k_y+\text{d}{k}_y$ and $k_z\to k_z+\text{d}{k}_z$ (where $\text{d}{k}_i$ is small, but much bigger than the spacing between states)

where $\text{d}{\Omega }_k=sin{\theta }_k\text{d}{\theta }_k\text{d}{\varphi }_k$. We have obtained the same expression for $D\left(k\right)$, as advertised. This time though we integrated over all values of ${\theta }_k$ and ${\varphi }_k$, not just the positive octant.

The density of states can be defined with respect to energy, or to frequency, as well. In each case the number of states remains the same: $D\left(k\right)\text{d}k=D\left(E\right)\text{d}E=D\left(\omega \right)\text{d}\omega$ so the density of states will change by the inverse of the factor which relates $k$ and the new variable: $D\left(x\right)=D\left(k\right)\left(\text{d}k∕\text{d}x\right)$.

For non-relativistic particles $k=\sqrt{2mE}∕\hslash$, so $D\left(E\right)=\left(Vm\sqrt{2mE}\right)∕\left(2{\pi }^2{\hslash }^3\right)$. For photons though, $k=E∕\hslash c$, so $D\left(E\right)=\left(V{E}^2\right)∕\left(2{\pi }^2{\hslash }^3{c}^3\right)$ and $D\left(\omega \right)=\left(V{\omega }^2\right)∕\left(2{\pi }^2{c}^3\right)$. In the notes we define $D\left(\omega \widehat{k}\right)$ which is akin to $D\left(k\right)$; the angle integral hasn’t yet been done, but the switch to frequency has, bringing in a factor of $1∕c^3$.

If a particle has more than one spin state we need to multiply by the degeneracy factor which is 2 for spin-$\frac{1}{2}$ electrons and for photons.

Bose-Einstein statisitics gives the average number of photons $n\left(\omega ,T\right)$ in a mode of frequency $\omega$ at temperature $T$. Hence we obtain the Planck law for the energy density in space at frequency $\omega$,

Note the dimesions, which are energy/frequency/length${}^3$. It’s a “double density” - per unit volume, but also per unit $\omega$. To get the full energy density $U\left(T\right)$ we integrate over $\omega$.

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