# I How can the photon wave function be described?

#### orisomech

Summary
Fermions such as the electron and proton can be described by wave function in momentum and in space. what about photons?
Fermions such as the electron and proton can be described by wave function in momentum and in position, and it is possible to get the momentum wavefunction from space wave function and vice versa by Fourier Transform.
what about photons? can photons be described by position wave function?
If they cannot be described by position wave function, what is the meaning of their location?
the uncertainty principle of Heisenberg for fermions follows directly from Fourier transform of the space wave function. what about the case of photons?
For example : the double slit experiment.
For both fermions and photons there is an interference pattern, depending on the size of the slit compared to the wavelength of the particle.
It is easy to understand it for fermions because you can describe them in both position and momentum wave function , but for photons it's not the same.

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#### A. Neumaier

Science Advisor
It depends on what the wave function is required to satisfy.

For example, the Silberstein vector is a valid 1-photon wave function in the same sense as the solutions of the Dirac equation with positive energy Fourier transform are wave functions for the electron. See this paper by Tamburini and Vicino. But its squared norm does not give a position probability density.

You may wish to browse my theoretical physics FAQ with several related articles.

#### vanhees71

Science Advisor
Gold Member
This is highly misleading. A single-particle wave-function interpretation for interacting is at best valid in a restricted regime, where no creation and annihilation processes take place. For massless quanta, like photons, which are in no way restricted to be created and annihilated all the time, it's completely obsolete, no matter how you formulate it. The Riemann-Silberstein vector is a nice calculational tool but not the way QED is usually formulated, or is there a working QFT formulation with the RS vector only, i.e., without using the four-potential + gauge invariance? At best it should be simply equivalent to usual QED, which is the most accurate description of nature we are aware of today.

#### A. Neumaier

Science Advisor
The Riemann-Silberstein vector is a nice calculational tool but not the way QED is usually formulated
The description by the Riemann-Silberstein vector is fully equivalent to the textbook description of the 1-photon sector of QED, by a unitary transform. Just as the position and momentum representation for a nonrelativistic particle.

#### vanhees71

Science Advisor
Gold Member
That's for non-interacting photons. That's of course mathematically clear, but non-interacting quanta are not so interesting, since they are not observable (an exception may be cases like a possible sterile neutrino, which is observable in a very weak sense only).

#### A. Neumaier

Science Advisor
That's for non-interacting photons. That's of course mathematically clear, but non-interacting quanta are not so interesting, since they are not observable (an exception may be cases like a possible sterile neutrino, which is observable in a very weak sense only).
Single photons are always described in the noninteracting Hilbert space, just like single electrons. Photons interacting with electrons can be discussed by considering an approximate Hamiltonian in the tensor product of these Hilbert spaces. The wave function at a fixed time is independent of this; it sees nothing of the interactions.

#### orisomech

This is highly misleading. A single-particle wave-function interpretation for interacting is at best valid in a restricted regime, where no creation and annihilation processes take place. For massless quanta, like photons, which are in no way restricted to be created and annihilated all the time, it's completely obsolete, no matter how you formulate it. The Riemann-Silberstein vector is a nice calculational tool but not the way QED is usually formulated, or is there a working QFT formulation with the RS vector only, i.e., without using the four-potential + gauge invariance? At best it should be simply equivalent to usual QED, which is the most accurate description of nature we are aware of today.
Sorry, I'm kind of lost you. From what I learned in QED photons are described in momentum space, so a photon here or in the other side of the universe are equivalent if they have the same momentum, which is obviously wrong.
How is QED describe the photon different than I described?

#### vanhees71

Science Advisor
Gold Member
Sorry, I'm kind of lost you. From what I learned in QED photons are described in momentum space, so a photon here or in the other side of the universe are equivalent if they have the same momentum, which is obviously wrong.
How is QED describe the photon different than I described?
Well, the usual description of relativistic QT is quantum field theory. A photon wave function makes not too much sense, at least not when it comes to interacting photons, and all we know about photons is through interactions with matter.

As massless quanta photons are utmost relativistic, and there's nothing that can hinder them from being produced and annihilated whatsoever, i.e., since they are massless, there's no energy gap, i.e., the slightest acceleration of a charged particle produces soft photons. That's why there's no limit, in which a single-particle wave description for photons interacting with charged matter could be valid.

So the right description is a theory, where photons can be created and annihilated through interactions with charged particles/matter enabling us to describe, how they are observed, and the most convenient (in fact also the only way making physical sense) is relativistic quantum field theory.

You start with quantizing the electromagnetic field coupled to charged particles. Full QED uses some (qantized) relativistic field for the charged particles too. The first and most simple physically relevant case is a system of fields, describing electrons and positrons interacting with the em. field (photons). You can as well also look at the approximation, where the matter particles are described by non-relativistic QT (in both 1st and 2nd quantized form).

You'll soon realize that there is some problem due to the gauge invariance of electromagnetism when trying to canonically quantize the em. field, described by the four-potential. The most simple way is to fix the gauge using the Coulomb-gauge condition $\vec{\nabla} \cdot \vec{A}=0$ and realizing that the temporal component $A^0$ is not an independent field but given by the Poisson equation,
$$\Delta A^0 = -\rho,$$
with the solution
$$A^0(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(t,\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
Then you can just quantize the field components $\vec{A}$ in the canonical way, taking into account the gauge constraint.

You end up with a Hamiltonian consisting of a free-field part, a matter part (including the static Coulomb interaction), and an interaction part, with which you can start to do perturbation theory in the interaction picture (carefully avoiding trouble related to Haag's theorem by first working in a finite volume with periodic boundary conditions for the fields).

The free em. field has an interpretation in terms of photons. It has two polarization-degrees of freedom (as can be immediately seen from the Coulomb-gauge condition), and you can expand the quantized em. field in terms of annihilation (associated with the positive-frequency modes) and creation (associated with the negative-frequency modes) operators, annihilating and creating a single photon in a given momentum-polarization state. The Hilbert space is realized as a socalled Fock space. The demand of microcausality, i.e., that the local observable operators should commute at space-like separation of their arguments, and of the existence of a ground state of lowest energy forces you to quantize the em. field as bosons (which is of course in accordance with observations). This holds true for any field with an integer spin, but photons are special, because they are massless; they have not three spin-degrees of freedom as expected for a massive particle with spin 1 but only two polarization states, which can be chosen as the helicity eigenstates; helicity is the projection of total (sic!) angular momentum of the quantum to the direction of its momentum, and there are only two values left, namely $\lambda=\pm 1$, referring to the left- and right-circular polarized em. plane wave modes.

Then you can also quantize the (massive) charged particle part too. In this case it also contains the Coulomb interactions, giving rise to bound states. If you have a proton and an electron state, they form hydrogen atoms, which are described in the non-relativistic limit and neglecting spin as you know it from the QM 1 lecture, but there may also be scattering of an electron and a proton.

Then the interaction of such a system of charged matter particles with the em. radiation field leads to the description of all kinds of radiative processes, including spontaneous and stimulated emission as well as absorption of photons on a hydrogen atom leading to transitions from one to another bound state. In lowest order that's all a nice exercise in time-dependent perturbation theory.

Then there are also higher-order corrections, which are very tricky, because they give rise of infinite integrals. It took the physicists quite a while to resolve this issue by perturbative pertubation theory. In fact to get the issue resolved, the famous measurement of the Lamb shift, as a radiative correction of this kind, in the hydrogen spectrum triggered the theorists to finally resolve this issue with the divergences and to invent renormalization theory (among them Bethe, who got the issue solved first within the non-relativistic theory for hydrogen, Feynman, Schwinger, and Tomonaga, for which work the latter 3 got the Nobel prize in physics; Bethe got his Nobel a bit later for the explanation how stars shine).

In all this you nowhere need the idea of a position for a photon. The theory, however, of course also describes the detection of photons. One way is to use the photoelectric effect. In the most simple form the photon scatters on the condactance electron of a metal, hitting it out of the metal (i.e., a transition from a bound state of the electron in the metal to a scattering state of this electron). This "photoelectrons" can then be measured. Seen as a measurement of the "position" of an electron, it's just defined by the location of the detector, which as a massive object of course has a position observable.

This is quite analogous to the observation of visible light: What you precept is just the interaction of a classical electromagnetic wave with the retina in your eye, producing an electric pulse, which is later processed in your brain. What you call "intensity of the light" physically is given by the energy density of the electric field, and this can be easily defined by the field operators of this field.

Also what I've just called "a classical em. field" is quantum-field-theoretically described by a socalled coherent state, which is not a state of definite photon number like the Fock states used to define the Hilbert space for the free em. field but a superposition of such states with any number of photons, defining a state of well-defined phase rather than of well-defined photon number.

#### orisomech

In all this you nowhere need the idea of a position for a photon. The theory, however, of course also describes the detection of photons. One way is to use the photoelectric effect. In the most simple form the photon scatters on the condactance electron of a metal, hitting it out of the metal (i.e., a transition from a bound state of the electron in the metal to a scattering state of this electron)
It doesn't make any sense. if you don't need the position of the photon, the electrons hitting the metal in photoelectric effect can be equally be photons that are in the other side of the universe. clearly something is missing in this description, don't you agree? what you are saying is that all photons in the universe that have the same momentum are same.

#### PrashantGokaraju

The coordinates of a photon are meaningful only in cases where the characteristic dimensions of the problem are large in comparision with the wavelength.

#### Demystifier

Science Advisor
2018 Award
A photon wave function makes not too much sense, at least not when it comes to interacting photons, and all we know about photons is through interactions with matter.
We could say the same also for the electron, or any other particle, that all we know about it is through the interactions with matter. So the interactions are not the reason why the photon wave function is problematic.

The photon wave function is problematic if we want it to describe the probability density of photon positions in a Lorentz covariant way. Otherwise, if that's not what we want from the wave function of the photon, then it is a perfectly well defined entity.

#### orisomech

The coordinates of a photon are meaningful only in cases where the characteristic dimensions of the problem are large in comparision with the wavelength.
I guess your'e right, but this is just an estimation. what if you do experiments with length dimension of the size of the photon wavelength than cannot use that.
it seems that there is no perfect description of the photon nature.

#### vanhees71

Science Advisor
Gold Member
It doesn't make any sense. if you don't need the position of the photon, the electrons hitting the metal in photoelectric effect can be equally be photons that are in the other side of the universe. clearly something is missing in this description, don't you agree? what you are saying is that all photons in the universe that have the same momentum are same.
No, I don't agree, because what you say is disproven by all experiments done so far, and the experiments involving photons are among the most accurate ones possible. It's called quantum optics!

#### A. Neumaier

Science Advisor
The photon wave function is problematic if we want it to describe the probability density of photon positions in a Lorentz covariant way.
No; this is not the cause of the problem. The probability density of an electron is also not defined in a Lorentz covariant way.

#### vanhees71

Science Advisor
Gold Member
We could say the same also for the electron, or any other particle, that all we know about it is through the interactions with matter. So the interactions are not the reason why the photon wave function is problematic.

The photon wave function is problematic if we want it to describe the probability density of photon positions in a Lorentz covariant way. Otherwise, if that's not what we want from the wave function of the photon, then it is a perfectly well defined entity.
The difference is that a massive quantum has a valid non-relativistic limit, for which position observables are well defined.

In the standard quantum-optics description the "probability density of photon positions" is rather described as the detection probability of a photon detector with some good (but always of course finite) position resolution, like a photo plate or a CCD camera. The "probability density of photon positions" turns out to be proportional to the energy density of the electromagnetic field, which is in complete accordance also with how the "intensity of light" is described in classical electrodynamics. The latter is of course also an approximation of the quantum description in cases, where this approximation is applicable (e.g., for coherent states of large intensity (aka a laser), thermal radiation (aka a light bulb)).

#### Demystifier

Science Advisor
2018 Award
The difference is that a massive quantum has a valid non-relativistic limit, for which position observables are well defined.
The position observable in non-relativistic QM does not depend on the mass, so one can define it even for a massless particle. See the link in post #15 above.

#### vanhees71

Science Advisor
Gold Member
The point is that for a massless particle there's no threshold for creation of new particles (for photons bremsstrahlung has no threshold, and you must even correct the wrong treatment of the asymptotic free states of charged particles to get rid of IR and collinear divergences). That's why for massless particles it's at least very problematic to use naive 1st-quantization pictures.

For massive particles like electrons it makes some sense in the non-relativistic regime. That's why to solve the Kepler problem for the Dirac equation to get the hdyrogen fine structure as if were a 1st-quantization "wave function" leads to pretty good results. Of course, this is finally explained by considering the full QFT case within standard perturbation theory (now, as far as I know, at the 5-6-loop level for the radiation corrections).

#### A. Neumaier

Science Advisor
The position observable in non-relativistic QM does not depend on the mass, so one can define it even for a massless particle. See the link in post #15 above.
You didn't define there any operator!

#### vanhees71

Science Advisor
Gold Member
How so? Massless particles do not lead to physically interpretible representations of the Galilei group. Massless particles in non-relativistic physics simply don't make sense.

#### PeterDonis

Mentor
can photons be described by position wave function?
Strictly speaking, no, because, unlike massive particles, photons have no non-relativistic approximation, and the relativistic method of obtaining a localized wave function for a particle, called Newton-Wigner localization, does not work for massless particles like photons.

if you don't need the position of the photon, the electrons hitting the metal in photoelectric effect can be equally be photons that are in the other side of the universe
No, they can't, because the metal is not on the other side of the universe, it's in your lab. (Btw, I assume you mean photons hitting the metal, not electrons--the electrons are in the metal and get knocked out of atoms in the metal by the incoming photons and cause a current.) The photons causing the photoelectric current have to be in the lab, otherwise they wouldn't hit the metal.

The difference here is that, when you measure the photoelectric effect, you are not measuring the "photon wave function". You are measuring some observable that has a reasonably localized position, and which tells you about the photons coming in. It is perfectly possible for an observable connected with photons to be localized, even though, as I said above, there is no valid way to construct a localized photon wave function.

#### olgerm

Gold Member
It depends on what the wave function is required to satisfy.
I think the requirement is that wavefunction square equals to probability density of finding photons with given proterties(such as location) or expected value of photons with such propeties.

this paper http://www.rrp.infim.ro/2014_66_2/A7.pdf claims that there is a photon wavefunction.

#### A. Neumaier

Science Advisor
I think the requirement is that wavefunction square equals to probability density of finding photons with given properties(such as location) or expected value of photons with such propeties.

this paper http://www.rrp.infim.ro/2014_66_2/A7.pdf claims that there is a photon wavefunction.
But this wave function has not the probability interpretation that you require.

#### olgerm

Gold Member
But this wave function has not the probability interpretation that you require.
I think they claim, that it has. Look page 6. Especially equation 32. If by photon_density they mean probility_density .

#### A. Neumaier

Science Advisor
I think they claim, that it has. Look page 6. Especially equation 32. If by photon_density they mean probility_density .
I don't believe their claim. This would imply that there is an associated position operator with commuting components for the photons, which does not exist unless rotation invariance is broken, which is physically unacceptable.

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