What is the density and energy of a photon? As I understand, photons are considered mass-less, which is a necessary condition for moving at the speed of light. However, does that mean their density is 0, as they will occupy some volume. If their density is zero, that means there is no matter inside a photon. Thus, shouldn't a photon be able to pass through matter instead of colliding with it? As $E = mc^2$, shouldn't a photon have zero energy, as it has zero mass?
 A: The complete energy relation is
$$E^2 = m^2c^4 + \lvert \vec{p} \rvert^2 c^2$$
The photon has $m=0$, so we are left with
$$E = \lvert \vec{p} \rvert c$$
(we don't care about the negative solution here). But according to De Broglie, it is $\vec{p} = \hbar \vec{k}$, so we have
$$E = \hbar c \lvert \vec{k} \rvert$$
but $k=\frac{2\pi}{\lambda}$ and $\hbar=\frac{h}{2\pi}$. Therefore
$$E=h\frac{c}{2\lambda} = h \nu$$
where $\nu$ is the frequency of the photon.
A photon is not a matter particle (fermion), but a force carrier particle (boson). A boson is not subject to the Pauli exclusion principle. Any number of bosons can occupy the same quantum state. So, if you want to call it like that, one could say the photon does not "occupy some volume". But it still interacts with matter, as it couples to electric charges. The simplest example is Compton scattering.
A: It does not matter how many photons comprise a wave, photons are massless thus so is the wave. Minimum density maximum momentum. The amount of energy density in light is equivalent to the amount of energy density of space.
