# Photon density of states: Polarization/Helicity degree of freedom?

Sakurai's "Advanced Quantum Mechanics" states in Eq. (2.116) that the density of states of a single photon with $\vec k$ vector pointing into the solid angle $d\Omega$ is given by $$\rho_{d\Omega}(E) = \frac{V\omega^2}{(2\pi)^3}\frac{d\Omega}{\hbar c^3}. \qquad (1)$$ This formula seems to be correct, I also found it in two other books.

I found a derivation of this formula in the book "Quantum Mechanics in Chemistry" by Schatz/Ratner, which I sketch in the following:

• We assume that the photon is enclosed in a large cube of side length $L$ such that its volume is $V=L^3$.

• The "wavefunction" of the photon is a plane wave $e^{i\vec k \cdot \vec r}$.

• Because the wavefunction has to be zero at the boundaries, we have a quantization of the wave vector: $\vec k = \frac{2\pi}{L}\vec n$ with $n_{x,y,z} = 0, \pm 1, \pm 2, \dots$

• This means that the number of states per unit "volume" in $\vec k$ space is given by $(\frac{L}{2\pi})^3 = \frac{V}{(2\pi)^3}$.

• Since the relation between energy and wavenumber is $E=\hbar c k$, the volume in $\vec k$ space of all states with energy between $E$ and $E+dE$ and wave vector pointing in the solid angle $d\Omega$ is given by $\frac{dE}{\hbar c} \times k^2 d\Omega$.

Multiplying the $\vec k$ space volume with the number of states per unit volume, we arrive at Eq. (1).

My question is: In the derivation we completely neglected the polarization/helicity degree of freedom of the photon. Shouldn't the DOS be twice as big?

• Yes, you need to multiple for the different polarizations. – Ben S Jan 20 '17 at 17:10