# 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!

• There’s no reason not to consider the photon a particle that travels in a straight line and can fit through a slit that’s not too small. Mar 28, 2019 at 23:27
• I don't understand exactly what you mean by that. Mar 29, 2019 at 0:57
• a photon can have an energy density within it’s local space which is small enough to fit through a very small slit. There is no need for it to be spread out over space. Mar 29, 2019 at 1:05
• A photon can not be described by a particle traveling in a straight line. It is completely delocalized. Mar 31, 2019 at 16:36

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.

• You write: suppose the position of some object with energy density ρ is in an equal superposition of x1 or some distant position x2. What do you mean exactly? What will be $T^{00}(x)$ for a photon with an energy $E=\hbar {f}$? Mar 28, 2019 at 18:49
• @descheleschilder Of course, you noticed that I swapped the one photon for a simpler situation! I did this because the situation for a photon is much harder, because it's embedded in a full quantum field theory. The operator $T^{00}(x)$ is an infinite-dimensional one with eigenvalues ranging from zero to (in principle) infinity, reflecting the fact that you could potentially see any of these results, depending on how we did the measurement. It's so horrible that we almost never speak of it. Mar 28, 2019 at 19:05

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.

• I didn't mean the density of a big number of photons but of a single photon. Why can't the accompanying wavefunction be expressed in the position representation? Mar 28, 2019 at 18:25

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.

• I gave you a +1 but I have some questions. Why is $\rho$ in the first equation a charge density when the argument of the delta function is the position? Hasn't $\rho$, in this case, the unit $(Cm)$ instead of $\frac{C}{m}$? Secondly, why has a photon by definition a single wavelength? Thirdly, why can't the last wavefunction be normalized? Then, what do you mean by **in some sense **? And last but not least, where is the answer to my question? Mar 28, 2019 at 20:55
• question 1: The unit of the delta function is the reciprocal of length in one dimension, reciprocal of area in 2, and reciprocal of volume in 3 dimensions. Multiply by charge q, you have C/m^3,units of charge density. The units are independent of the argument (mostly). A vector argument often implies you are dealing with 3 dimensions which indicates you use the correct dimensions for 3D space. In general, the dirac delta has the units needed for the final integral to have the right units. Consequently, the units are reciprocal wave numbers in one dimensional momentum space. Mar 28, 2019 at 22:28
• Question 2: If a photon has a single, well defined energy by definition, right? I am fuzzy there. But given that, E=hf mandates that a specific Energy implies a specific frequency. All photons travel at the speed of light, c=wavelength * frequency=angular frequency/wave number. So a specific energy and a specific speed implies a specific momentum/wave number allowing for the use of the diract delta in momentum space. Mar 28, 2019 at 22:33
• question 3: The resulting wave function can't be normalized because it is not square-integrable. It's modulus is 1. The integral of the modulus goes to infinity. You can only normalize it if the integral of the modulus has a finite value. Apparently there are ways to get around this difficulty but I'm unfamiliar with second quantization and such. Mar 28, 2019 at 22:37
• "in some sense" illustrates my misgivings about associating a speed with a Plane Wave. in 3D is precisely a plane wave traveling at speed c. some strange phenomenon happen that move over c but don't exchange information at speeds faster than c. For example, its possible to arrange a space ship, a light source, and a distant screen in such away that the shadow traverses the screen faster than light. All but the subluminal space ship moving at the speed of light and yet the shadow moves faster. I'm pretty sure something similar can be arranged with a few photon wave functions. Mar 28, 2019 at 22:47