Can a single photon have an energy density? In one question, which is further irrelevant for thís question, the comment was made that a single photon can have an energy density.
I didn't agree. Off course the wavefunction is spread out in space, which seems to suggest that the energy is spread out in space also, giving the photon an energy density. 
The question is though if the energy is really spread out over space. I think not so because if we look at the photon (without disturbing it, so this can happen only in our minds), the photon erratically (the wavefunction is related to the probability density to find it at some infinitesimal interval) jumps from one place to another, within the confines of its wavefunction and without us knowing it. Because of this, a photon density isn't a well-defined concept. At least, according to me. I hope there is someone who disagrees!
 A: First of all, the wave function of a photon can't be represented in the position representation. So, you can't define a probability density in the same manner as electrons. Instead, second quantazation language is used to represent the wave function of a photon, which basically counts the number of photons in a certain state (wavenumber). 
Secondly, the notion of a photon comes from the fact that if we measure light, we only measure quantized 'ticks'. Each tick represents a photon that is absorbed. If we want to measure the intensity/energy density of ligth, the number of ticks per second are measured.
If we can measure the number of photons per second, we can convert this to a photon density by realizing the photons travel at the speed of light. So, the photon density is a well defined concept, but a photon itself is completely delocalized. 
A: I'm not sure about a photon, but its possible to associate a charge density with an infinitesimally small source of charge. 
$\rho=q\delta(\vec{x'}-\vec{u}t);$
This represents the charge density, $\rho$, in terms of the diract delta function. The delta function is roughly an infinitely high, infinitely narrow spike when its argument is zero. This argument corresponds to a point charge moving at constant velocity $\vec{u}$. 
By definition, a photon has a single energy, an therefor a single frequency and consequently wavelength. 
$E=\hbar\omega$
$c=\omega/\kappa$
So the wave number is:  $k=E/\hbar c$
We can use the dirac delta function again to represent the photon in momentum space:
$$\phi(k)=\delta(k-E/\hbar c)$$
The most generic one dimensional wave function in the space domain is :
$\psi(x)=\int_{-\infty}^{\infty} \phi(k)e^{i(kx-Et/\hbar)}dk$
In this specific case:
$$\psi(x)=\int_{-\infty}^\infty \delta(k-E/\hbar c)e^{i(kx-Et/\hbar)}dk=e^{\frac{iE}{\hbar c}(x-ct)}$$
This cannot be normalized but does hold kinematic information of the photon, e.g. $i\hbar\partial\psi/\partial t=E\psi$. 
It in some sense moves at the speed of light. 
So the 3D analog might be the best you can do in undergraduate quantum mechanics.
By the uncertainty principle, an infinitely precise momentum/energy implies infinitely imprecise knowledge of its location. 
You don't know where the photon is so you don't know the density. So it seems your intuition is correct. 
A: You can define the energy density $T^{00}(x)$ for any field configuration, including a single photon. But it's crucial to remember that the energy density is a quantum object, just like everything else in quantum mechanics. The energy density at a given point can be in superposition.
For example, suppose the position of some object with energy density $\rho$ is in an equal superposition of $x_1$ or some distant position $x_2$. The energy density at $x_1$ is not $\rho/2$, it is an equal quantum superposition of $\rho$ and $0$. Just like standard quantum mechanics doesn't tell you where the particle "really" is, it doesn't say where the energy "really" is. As such, energy density doesn't introduce any new conceptual difficulties beyond those that were already there to begin with.
