Semiclassical volume of a single photon An electromagnetic plane wave propagating in the +z direction with  $\vec{E} = E_0cos(\vec{k}\cdot\vec{r}-wt)\hat{x}$ and $ \vec{B} = \frac{E_0}{c}cos(\vec{k}\cdot\vec{r}-wt)\hat{y}$. Let us now model the plane wave as an orthorhombic lattice of closely spaced photons with side lengths $l_x, l_y,l_z$. The Poynting vector is $\vec{S} = \frac{1}{u_0}\vec{E} \times{\vec{B}} = \frac{1}{u_0c}E_0^2cos^2(\vec{k}\cdot\vec{r}-wt)\hat{z} $ and the average power going through the cross-sectional area $l_xl_y$ is $P = \frac{E_0^2l_xl_y}{2u_0c}$. Since a single photon crosses over in a time $\frac{l_z}{c}$, we get the energy for the photon as $E = \frac{E_0^2l_xl_yl_z}{2u_0c^2}$, which when combined with $ E = hf $ gives us the volume  
$V = \frac{2hf}{\epsilon_0E_0^2}$
Is there is a meaningful sense in which the above volume of a photon is indeed correct, or at-least correct in the context of plane waves? I understand that in QM the picture is more complex and that there is a sense in which a photon is a point particle with no volume. Is there perhaps a quantum mechanical analog of this argument that gives a similar result, or is this semiclassical volume completely off the mark? 
 A: I don't have a single, unified answer to your question, but I do have a list of observations:


*

*Your equation for V is determined, up to the unitless constant 2 in front, by dimensional analysis. Your other reasoning, such as the stuff about the orthorhombic lattice, doesn't seem at all convincing to me, but in any case can only affect the unitless constant, which is not especially interesting.

*There is a sense in which all the fundamental particles of physics are pointlike. On the other hand, we can also say that the "size" of a particle is quantifiable by its $\Delta x$ as in the Heisenberg uncertainty principle, and this can be arbitrarily large. These aren't really contradictory statements, just different ways of defining size.

*Photons don't really have a wavefunction in the sense of the Schrodinger equation, like a $\Psi(x,t)$. The hand-wavy argument to this effect is that if we had eigenstates of position, they would have infinite uncertainty in energy, but then this energy would allow for the creation of all kinds of particle-antiparticle pairs, and therefore we wouldn't have just "the" photon.

*Nevertheless there are certainly contexts where it makes sense to talk about the kind of thing you're discussing. For example, if an atom is in an excited state, then a photon it emits is going to have a coherence length of $\sim ct$, where $t$ is the lifetime, and this implies a certain volume.
