# What equation describes the wavefunction of a single photon?

The Schrödinger equation describes the quantum mechanics of a single massive non-relativistic particle. The Dirac equation governs a single massive relativistic spin-½ particle. The photon is a massless, relativistic spin-1 particle.

What is the equivalent equation giving the quantum mechanics of a single photon?

• – user4552 Jun 3 '13 at 15:41

There is no quantum mechanics of a photon, only a quantum field theory of electromagnetic radiation. The reason is that photons are never non-relativistic and they can be freely emitted and absorbed, hence no photon number conservation.

Still, there exists a direction of research where people try to reinterpret certain quantities of electromagnetic field in terms of the photon wave function, see for example this paper.

• You can also say that the wavefunction of a photon is defined as long as the photon is not emitted or absorbed. The wavefunction of a single photons is used in single-photon interferometry, for example. In a sense, it is not much different from the electron, where the wave-function start to be problematic when electrons start to be created or annihilated... – Frédéric Grosshans Nov 17 '10 at 10:19
• I agree. For the electrons there is a possibility to slow them down to non-relativistic speeds, but there is no such possibility for photons. I would also add that there is an interesting discussion about photons and electrons in the Peierls's book "Surprises in theoretical physics". – Igor Ivanov Nov 18 '10 at 21:46
• Igor, I can't reconcile your wording with Frédéric's comment. Yes, there is no possibility for photons to slow down relativistically, but so what? Unless I misunderstand, there is still a spatial wavefunction (complexly valued over R^3) for the photon which obeys a relativistic Schrodinger equation. Yes, we have to assume the photon is not emitted or absorbed, but the same is true with electrons! The description of the latter in terms of a spatial wavefunction also breaks down when they are emitted or absorbed. – Jess Riedel Jun 3 '13 at 20:36
• You should google W.E. Lamb's anti-photon paper. – Jan Bos Sep 29 '16 at 13:52
• The field does still (in the Schrodinger picture) have a state which is a vector in a Hilbert space that evolves according to the Schrodinger equation, though. It's just that said state isn't usefully interpreted in terms of number-conserved "photons". – AGML May 29 '17 at 22:40

There is a slight confusion in this question. In quantum field theory, the Dirac equation and the Schrödinger equation have very different roles. The Dirac equation is an equation for the field, which is not a particle. The time evolution of a particle, ie, a quantum state, is always given by the Schrödinger equation. The hamiltonian for this time evolution is written in terms of fields which obey a certain equation themselves. So, the proper answer is: Schrödinger equation with a hamiltonian given in terms of a massless vector field whose equation is nothing else but Maxwell's equation.

The maxwell equations, just like in classical electrodynamics. You'll need to use quantum field theory to work with them though.

While the answers above are great, I Felt it was lacking what The question asked regarding an equation analogous to the Schrodinger (or Dirac) equation.

There is a quantity called the Riemann-Silberstein vector ( https://en.wikipedia.org/wiki/Riemann%E2%80%93Silberstein_vector#Photon_wave_function ), First used by infamous Bernhardt Riemann to demonstrate a concise formulation of Maxwell's equations.

This “vector” has the form:

$$\vec{F}=\vec{E}+ic\vec{B}$$

A quick search online, demonstrates that classical electrodynamics written in this form can be quite useful in solving problems.

In the quantum realm, a quantity analogous to the wavefunction can be written for a single photon. Such a quantity has the form:

$$i\hbar\partial_{t}\vec{F}=c\left(\vec{S}\cdot\frac{\hbar}{i}\vec{\nabla}\right)\vec{F}$$ Which can be written simply in the form:

$$i\hbar\partial_{t}\vec{F}=c\left(\vec{S}\cdot\hat{P}\right)\vec{F}$$

This can be a useful quantity for examining the properties of a single photon. Start with the Wikipedia page, it's actually quite an interesting and useful quantity.

The general concept of quantum mechanics is that particles are waves. On of hand-waveing "derivations" of quantum mechanics is assumption that phase of particles behaves in the same way as phase of light $\exp( i \vec{k}\cdot \vec{x} - iE t / \hbar)$ (see Feynman Lectues on Physics, Volume 3, Chapter 7-2).

For light that is monochromatic (or almost monochromatic), just take Maxwell Equations plus add assumption that one photon can't be partially absorbed. Most of the time it suffices to use the paraxial approximation, or even - plane wave approximation. It works for standard quantum mechanics setups like Elitzur–Vaidman bomb-tester.

For nonmonochronatic light its much more complicated. More on nature of quantum mechanics of one photon: Iwo Bialynicki-Birula, On the Wave Function of the Photon, Acta Physica Polonica 86, 97-116 (1994).

A single photon is described quantum mechanically by the Maxwell equations, where the solutions are taken to be complex. The Maxwell equations can be written in the form of the matrix Dirac equation, where the Pauli two-component matrices, corresponding to spin 1/2 electrons, are replaced by analogous three-component matrices, corresponding to spin 1 photons. Since the Dirac equation and corresponding Maxwell equation are fully relativistic, there is no problem with the mass of the photon being zero, as there would be for a Schroedinger-like equation. See http://www.nist.gov/pml/div684/fcdc/upload/preprint.pdf.

According to Wigner's analysis, the single photon Hilbert space is spanned by a basis parameterized by energy-momenta on the forward light cone boundary, and a helicity of $\pm 1$.

However, a manifestly Lorentz covariant description in position space has to include a fictitious longitudinal photon with a helicity of 0. This degree of freedom is pure gauge, and decouples. Interestingly enough, the state norm is now positive semidefinite, instead of positive definite, with the transverse modes having positive norm and the longitudinal ones having zero norm.

There are several different waves associated with a photon. In QED the photon is associated with a classical solution of the (4-)vector potential. The vector potential contains features that are not physical, as a change of gauge is not reflected in any change of physical properties. Thus its role as a wave function might be somewhat questionable.But still, there must be a wave which explains the well known interference and diffraction patterns.

When we see a screen illuminated by laser light that have passed through a double slit, our eyes receives photons scattered by the atoms on the surface of the screen. Atoms absorb and emit photons as quantum electric dipole antennas. This implies that the atoms are sensitive to the electric field. From the vector field associated with the photon an electric field can be calculated. This field is gauge independent thus a physical field. This field is a solution to Maxwell's equations and will describes the usual interference and diffraction patterns.

My answer is more of comment on other correct answers: you cannot build a delta-function for the photon in 3D becase the longitudinal component of a massless vector field is missing. But that does not mean there is no useful and meaningful concept of a wave function in the single-photon sector. This is just a peculiar fact about free electromagnetic field, you basically cannot localize light to a region smaller that the characteristic wave-length. Maxwell equations for the source-less (solenoidal) component of the vector potential field $\bf{A}$ play the role of the Schrodinger equation.

I recommend the book of Rodney Loudon "The Quantum Theory of Light" as good resource to really understand the quantum level of desription for light.