Wavefunction of a photon

Does anyone have an explicit closed-form expression for the wavefunction of a single photon from a multipolar source propagating through free space? Any basis is acceptable as long as it is a single photon state.

A reference would also be appreciated, but not essential.

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A possible duplicate has been suggested: Does a photon have a wave function or not?

But this question primarily concerns the existence of the wavefunction and is not what I am looking for. None of the answers provide an explicit expression for the wave function, and neither the question nor the answers discuss a multipole source. The multipole source, in particular, is central to my question.

• physics.stackexchange.com/questions/542943/… Commented Apr 20, 2020 at 2:25
• Thanks, I added a clarification that any basis is ok. I am not tied to the position basis if that is problematic.
– Dale
Commented Apr 20, 2020 at 2:31
• There is a formula for the wavefunction in that question, but I am not sure that is what you are looking for. Commented Apr 20, 2020 at 2:35
• You might find cft.edu.pl/~birula/publ/APPPwf.pdf useful. Commented Apr 20, 2020 at 3:35
• Does this answer your question? Does a photon have a wave function or not? Commented Apr 20, 2020 at 5:53

I think you are asking for the wavefunction of a single photon in a state of sharp total angular momentum $$j$$. This is, $$$$|jmk\lambda\rangle=\frac{(2j+1)}{4\pi}\int\sin{\beta}d\alpha d\beta \{D^{j}(\alpha,\beta,0)^{m}_{\ \lambda}\}^{*}|\vec{k},\lambda\rangle$$$$ In this equation the 3-momentum of the photon is the vector $$\vec{k}$$. The magnitude of $$\vec{k}$$ is $$k$$. The vector $$\vec{k}$$ is defined as being rotated by Euler angles $$\alpha,\beta$$ from a fiducial momentum $$\vec{k}_{0}$$ along the z-axis. The actual 3-momentum is, $$\vec{k}=R(\alpha,\beta,0)\vec{k}_{0}$$ where $$R(\alpha,\beta,\gamma)$$ is the rotation matrix and $$\alpha,\beta,\gamma$$ are Euler angles. The Euler angle convention is $$\alpha$$ is rotate about z, $$\beta$$ is rotate about resultant y, $$\gamma$$ is rotate about resultant z. The matrix $$D^{j}(\alpha,\beta,0)^{m}_{\ \lambda}$$ is Wigner's D-matrix. $$\lambda=\pm 1$$ is the photon's helicity. The states $$|\vec{k},\lambda\rangle$$ are the linear-momentum-helicity eigenvectors. The states $$|jmk\lambda\rangle$$ are angular-momentum-helicity eigenvectors. This result is equation (8.7.2), page 147 of thge book "Group Theory in Physics" by Wu-Ki Tung. It is also equation (28.35) on page 218 of the book "Relativistic Theory of Reactions", by J. Werle.

• Thank you! Is there any clear way to relate the angular or other parameters to a multipolar source, or is the multipole source simply not relevant somehow?
– Dale
Commented Apr 20, 2020 at 11:09
• @Dale : You've now asked the question that I was interested in, "What is the relation between a source (atom or nucleus) that undergoes a transition to produce these single photon states?" I don't know the answer because I've got distracted on other things. However, Wu-Ki Tung page 150 says a little about it. However, I think one really has to study chapter V on Radiation in Quantum Electrodynamics (Volume 4 2nd Edition Course of Theoretical Physics by Landau and Lifshitz).
– user7154
Commented Apr 20, 2020 at 13:43
• Doing a little more reading, I think that you are right that multipolar solutions do indeed correspond to states with sharp total angular momentum. Since the states of the atom differ by angular momentum, that difference must correspond to the angular momentum of the resulting photon. That makes the atom itself a multipole source, although I was originally thinking of generic multipolar sources
– Dale
Commented Apr 20, 2020 at 15:07

I posted an answer to a similar question here. Admittedly, many different things may be implied when talking about the wave function of a photon. However, one should keep in mind is that, unlike electrons, photons are classically waves. Quantization neither adds nor subtracts from their wave-like properties, but injects discreteness (i.e. makes them from an electromagnetic field into countable photons). Their wave modes remain the same - plain waves of the electromagnetic field.

• "Quantization neither adds nor subtracts from their wave-like properties" this is wrong, see this one photon at a time experiment sps.ch/en/articles/progresses/… . P hotons in mainstream physics are point particles in the standard model. en.wikipedia.org/wiki/Standard_Model it is probability that has wave properties in QM Commented Apr 20, 2020 at 7:50
• @annav I think your concept of the photon is the particle physicist's view. In the standard model they are treated as point particles. But that does not mean that the are point particles. The mainstream optical physicist thinks of the photon as an excitation of an EM mode. Which is right? Well ... they both work for their purposes. Like seemingly everything related to photons there are two ways of looking at it. Commented Apr 20, 2020 at 11:43
• @annav If what I remember about the standard model is correct, the Hamiltonian is written in terms of raising and lowering operators acting on a field for each type of particle. It seems that for some purposes the mainstream approach to the standard model thinks of particles as particles, e.g. Feynman diagrams. For other purposes as excitations of a wave, e.g. the Hamiltonian. Commented Apr 20, 2020 at 12:00
• @garyp there is no contradiction. Field theory uses the plane wave wave functions of the particles in the table as a kind of "coordinate" system on which creation and annihilation operators act. After all Feynman diagrams directly use field theory. Because plane waves go from - to + infinity , real particles (as tracks in the bubble chamber for example) have to be represented by wavepackets in this mathematcal description, where the HUP is automatically obeyed. Commented Apr 20, 2020 at 12:30
• @garyp One cannot be in the quantum mechanical frame and talk of the photon as a wave. There is a probability that has wave characteristics, look at the one photon at a time experiment sps.ch/en/articles/progresses/… . It is confusing classical and quantum frames to talk of the photon as a wave.. The excitation is again a one particle excitation . Commented Apr 20, 2020 at 12:48

I define the photon wave function in a covariant formulation which has four polarisation states, two of which are not observable. Some authors use only transverse states, but the other two states would appear on Lorentz transformation, and they appear to be necessary to derive the classical correspondence correctly.

For momentum $$p=(P^0,\mathbf p)$$, define a longitudinal unit 3-vector, $$\mathbf w(\mathbf p,3) = \frac {\mathbf p} {|\mathbf p |}$$ and orthogonal transverse unit 3-vectors $$\mathbf w(\mathbf p,1),\mathbf w(\mathbf p,2)$$ such that for $$r,s= 1,2,3$$ $$\mathbf w(\mathbf p,r) \cdot \mathbf w(\mathbf p,s) = \delta_{rs}$$ Define normalised spin vectors, $$\mathbf w(\mathbf p,0) = (1,\mathbf 0)$$ $$\mathbf w(\mathbf p,r) = (0,\mathbf w(\mathbf p,r))$$ For momentum $$p$$ a photon plane wave state is given by the wave function, $$\langle x|p,r\rangle = \lambda(|\mathbf p |,r) w(\mathbf p,r) e^{-ix \cdot p}$$ where $$p^2 = 0$$ and $$\lambda$$ is determined by relativistic considerations $$\lambda(|\mathbf p|, r) = \frac 1{(2\pi)^{3/2}} \frac 1{\sqrt{2p^0}}$$

You can then express a photon wave function as an integral

$$f^a(x) = \frac 1{(2\pi)^{3/2}} = \sum\limits_{r=0}^3 \int \frac{d^3\mathbf p}{\sqrt{2p^0}} \mathbf w(\mathbf p,r) e^{-ix \cdot p} \langle \mathbf p, r|f\rangle$$

I took this from lecture notes at Cambridge, and I am not sure which books do things much the same way (there are some normalisation choices, as well as choice of gauge). I have given more detail in A Construction of Full QED Using Finite Dimensional Hilbert Space and in The Mathematics of Gravity and Quanta

• Thank you, but I am not looking for plane waves, I am looking for the wavefunction for a photon from a multipolar source. Is there a relationship there with what you wrote?
– Dale
Commented Apr 20, 2020 at 11:12
• Of course. Plane waves are a basis. As in any wave problem, transform your initial condition to find the coefficients in the momentum basis, then you have the complete solution. Commented Apr 20, 2020 at 12:04
• @CharlesFrancis Isn't $\langle x \vert p \rangle = e^{i p \cdot x}$?
– Sidd
Commented Apr 7 at 23:23

A wave function for a free k-mode photon is defined in a recent paper: https://hal-cea.archives-ouvertes.fr/cea-04021491/document You can find a detailed description in the above reference in HAL or in the Journal of Modern Physics: https://www.scirp.org/journal/paperinformation.aspx?paperid=123267

• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review Commented Jun 15, 2023 at 18:34