By definition, the wave function can be obtained by acting the position eigenstate to a state of the system, e.g., $\langle x\vert \psi \rangle$. For the wave function of an electron travelling in one-dimensional space, we can calculate the wave function of an electron by the way mentioned above, $\psi(x)=\langle x \vert \psi\rangle$. I'm okay with this.

Q1) However, what about the case of photon? Can we also define the wave function of photon in the same way above?

Q2) What about the following? Let consider a single photon state of light, which encoded to a left-propagating electromagnetic wave in one-dimensional space. Then, we place the infinite number of detectors along the space and then repeat the detection measurement for photon's position so many times at different times. Would what will be measured over space be almost same as the square of the wave function of photon, i.e., $\vert \psi(x)\vert^{2}=\vert \langle x\vert {\rm single~photon}\rangle\vert^{2}$? Can we say this way?

Q3) Would $\vert \psi(x)\vert^{2}$ in Q2 be equivalent to the electromagnetic wave?


2 Answers 2


The Quantum Mechanics people use to deal with electrons is a non-relativistic theory. It is, in great part, based on the quantization of the Hamiltonian (there's also the spin part which is added by hand in that setting),

$$H = \dfrac{P^2}{2m}+V$$

And the resulting theory has proved to be not Lorentz invariant, so, in disagreement with the Special Theory of Relativity.

This lead people to develop the Relativistic Quantum Mechanics based on the equations of Klein-Gordon and Dirac. Despite producing wave-function equations that are indeed Lorentz invariant, these approaches come with some issues.

I believe that from one intuition standpoint, the best way to think about this, is that by virtue of the existence of the rest energy $E = mc^2$, mass can be converted into energy and energy can be converted into mass. Traditional Quantum Mechanics can't deal with this, because the way it is constructed makes it into a theory of a fixed number of particles.

To really deal with relativistic phenomena in the context of Quantum Mechanics one needs to go into Quantum Field Theory.

Notice, for example, in the case of the electron, that the Relativistic Quantum Mechanics approach develops the Dirac equation as the equation for a wave-function, hence a classical field, whereas the Quantum Field Theory approach develops the equation as the equation for a quantum field.

So the answer to your question: the problem in treating the photon like one treats the electron in Quantum Mechanics, is that photons are always relativistic particles. It is not possible to deal with photons outside the context of special relativity.

For more into the quantum treatment of photons, I suggesting reading Merzbacher's Quantum Mechanics chapters on the matter. The chapter on "identical particles" introduces the general framework used (and as a matter of fact, it is the framework for free quantum fields), and a later chapter applies the general ideas to describe photons and hence, to provide a quantum description of electromagnetism.


Not a true answer, but too long for a comment. A 1-particle probability theory runs into certain difficulties when we add relativity into quantum mechanics, but we can't help but do that for a speed-$c$, zero-rest-mass particle such as a photon.

A non-relativistic quantum model of one particle has a probability density $\rho=\psi^\ast\psi$ and a probability $3$-current $\mathbf{j}$ with continuity equation $\dot{\rho}+\boldsymbol{\nabla}\cdot\mathbf{j}=0$ encoding the conservation of the total probability $1$. You can use the Schrödinger equation to obtain an expression for $\mathbf{j}$ consistent with this condition.

This current admits an easy relativistic generalisation, a $4$-current $j^\mu$. For solutions of the Klein-Gordon equation (a relativistic cousin of the Schrödinger equation with a similar motivation), $\partial_\mu j^\mu=0$ in flat spacetime (or $\nabla_\mu j^\mu=0$ in curved spacetime). In fact, this is just the same equation again if you define $\rho:=j^0,\,\mathbf{j}^i:=j^i$.

However, $\int d^3\mathbf{x}j^0$ can be positive, negative or zero for solutions of the Klein-Gordon equation. This ultimately scuppers any attempt to interpret a relativistic wavefunction as a 1-particle probability amplitude.

  • $\begingroup$ Just adding to this, a photon is by definition relativistic always, so position is undefined for it. So, the answer to your question is "no". To go more deeply into the reasoning, compare the postulates of QM, which you seem to try to apply to a photon to the ones of QFT and QED, which was actually developed in order to describe the photon. There is no position operator in QFT and QED whatsoever. There is also a legit explanation on why relativistic QM does not describe a photon, which you might want to look at. $\endgroup$
    – MsTais
    Oct 21, 2016 at 16:38
  • $\begingroup$ @MsTais;J.G. Thanks for your comments. Can we define the position for photons in an operational way which relies on measurements or operations? For example, we can measure a position for photon, can't we? Let say, we send a single photon wave packet which is localized in time and space in one-dimensional waveguide structure. Then we place an infinite number of APD along the waveguide, by which we measure the position of light going through in different times. We repeat this experiment so many times for the same source. What we measure is the square of the wave function, isn't it? $\endgroup$
    – Veteran
    Oct 23, 2016 at 17:44
  • $\begingroup$ @user58226 You can have an empirical approximation for how much light strikes a given region in a time period. The physical explanation is that the photon field $A^\mu$ manifests itself empirically as particles and/or waves. However, we don't have a theory of 1 photon's probability of being somewhere, and cannot for the reasons discussed here. (The closest you can get is, "the photon this atom just emitted must have had about this much momentum, so has a certain chance of being elsewhere later".) $\endgroup$
    – J.G.
    Oct 23, 2016 at 18:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.