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
\begin{equation}
\rho_{d\Omega}(E) = \frac{V\omega^2}{(2\pi)^3}\frac{d\Omega}{\hbar c^3}. \qquad (1)
\end{equation}
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:
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We assume that the photon is enclosed in a large cube of side length $L$ such that its volume is $V=L^3$.
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The "wavefunction" of the photon is a plane wave $e^{i\vec k \cdot \vec r}$.
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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 $n_{x,y,z} = 0, \pm 1, \pm 2, \dots$
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This means that the number of states per unit "volume" in $\vec k$ space is given by $(\frac{L}{2\pi})^3 = \frac{V}{(2\pi)^3}$.
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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?
Best Answer
It seems like the problem is just a misconception. OP's Eq. (1) really is not the density of one-particle states, but instead the density of one-particle modes. To obtain the real density of states, one indeed has to multiply the formula by 2. In rate calculations using this density in Fermi's golden rule, many common textbooks (like Sakurai's "Advanced Quantum Mechanics") take care of this factor by summing up results for two different polarizations in the end of the calculation.