# Why photon wave function does not exist?

by exponent137
Tags: exist, function, photon, wave
 P: 287 I hear some reasons that photon exist in 4D space-time, wave function not and so on. But, an electron can be described with de Broglie waving and we can use wave function to describe electron. Frequency of the wave function is the same as energy/\hbar and k of wave function is the same as that of de Broglie k. But why this is not appropriate for the photon. Maybe because photon does not have rest energy? What is connection between wave function and de Broglie waving is described in http://iopscience.iop.org/0031-9120/43/4/013
P: 533
 What is connection between wave function and de Broglie waving is described in http://iopscience.iop.org/0031-9120/43/4/013

 Sci Advisor P: 3,549 Why photon wave function does not exist? I am not quite sure about the meaning of the statement either: Even accepting that there is no position operator for photons, I do not quite see why this precludes writing down a wavefunction although maybe only in momentum space. In QFT for a free photon field you have a Fock space with a vacuum vector |0> and hence also all kind of n-particle vectors which can be obtained by acting with various ##a_{k\sigma}^+## on it. I don't see any reason why not to call these vectors "momentum space wavefunctions".
P: 1,020
there is an argument by dirac about integral spin particles which states that momentum representation is sufficient for integral spin particles such as photon and there is no coordinate representation of these particles.
 The answer is to be found in a hidden assumption in our work. Our argument is valid only provided the position of the particle is an observable. If this assumption holds, the particle must have a spin angular momentum of half a quantum. For those particles that have a different spin the assumption must be false and any dynamical variables x1, x2, x3 that may be introduced to describe the position of the particle cannot be observables in accordance with our general theory. For such particles there is no true Schrödinger representation. One might be able to introduce a quasi wave function involving the dynamical variables x1, x2, x3, but it would not have the correct physical interpretation of a wave function—that the square of its modulus gives the probability density. For such particles there is still a momentum representation, which is sufficient for practical purposes.
He was talking about dirac eqn or I should say his own eqn.
P: 4,568
 Quote by DrDu I am not quite sure about the meaning of the statement either: Even accepting that there is no position operator for photons, I do not quite see why this precludes writing down a wavefunction although maybe only in space. In QFT for a free photon field you have a Fock space with a vacuum vector |0> and hence also all kind of n-particle vectors which can be obtained by acting with various ##a_{k\sigma}^+## on it. I don't see any reason why not to call these vectors "momentum space wavefunctions".
You are right, it is fully justified to call it momentum-space wave function. You can also take the Fourier transform of it and get something like a position-space "wave function", but physically it has a somewhat different interpretation than wave function in non-relativistic QM.
 P: 24 What do you mean by the word wavefunction? I guess you can always go to the classical limit, where you have classical fields for photons.
 Sci Advisor Thanks P: 2,305 Usually the wave function of a particle is the position representation of a pure quantum state. Of course for particles with non-vanishing spin this wave function has also a spin (or polarization) index. The physical interpretation of such wave functions is clear in the non-relativistic case via Born's rule: The modules squared of the wave function is the probability density to find a particle at the point given by the spatial coordinates with the spin given by the spin index (usually meaning that the spin-z component is $\sigma$ with $\sigma \in \{-s,-s+1,\ldots,s \}$. This interpretation fails in the relativistic case. For photons, i.e., massless particles with spin 1, you cannot even define such a wave function formally since there is no position operator for photons. Thus the notion of a wave function for photons is not even defineable!
P: 3,549
 Quote by vanhees71 This interpretation fails in the relativistic case. For photons, i.e., massless particles with spin 1, you cannot even define such a wave function formally since there is no position operator for photons. Thus the notion of a wave function for photons is not even defineable!
Yes, but this argument does not exclude the possibility to define a wavefunction in momentum representation?
P: 1,020
 Quote by DrDu Yes, but this argument does not exclude the possibility to define a wavefunction in momentum representation?
that is what is necessarily dirac's argument I have provided.
P: 87
 Quote by vanhees71 For photons, i.e., massless particles with spin 1, you cannot even define such a wave function formally since there is no position operator for photons. Thus the notion of a wave function for photons is not even defineable!
Photons can be detected at a specific location (for instance on a CCD pixel or rods/cones of a retina). Is that detected position an eigenvalue of some operator? That operator would be a position operator, but at the same time it is said that there is no position operator for photon and that position of a photon is not an observable.

Do I understand something wrong?
P: 4,568
 Quote by mpv_plate Photons can be detected at a specific location (for instance on a CCD pixel or rods/cones of a retina). Is that detected position an eigenvalue of some operator? That operator would be a position operator, but at the same time it is said that there is no position operator for photon and that position of a photon is not an observable. Do I understand something wrong?
In modern literature, measurements which can be described in terms of collapse into an eigenstate of an operator are referred to as projective measurements. There are also more general measurements described in terms of positive operator valued measures (POVM), which are not projective measurements. Measurement of position of the photon is an example of such a more general measurement.

See e.g.
http://arxiv.org/abs/1007.0460