2
$\begingroup$

I have read this question:

Why doesn't there exist a wave function for a photon whereas it exists for an electron?

where annav says:

Here is the [wavefunction of the photon,][1] which is a solution of a quantized maxwell's equation:

Now write the complex wave function as the sum of real imaginary parts $\bar E_{\tau}(\bar r)$ and $\bar B_{\tau}(\bar r),$ $$\bar \psi_{\tau}(\bar r,t)=2^{-1/2}\left(\bar E_{\tau}(\bar r,t)+i\bar B_{\tau}(\bar r,t)\right).$$

where Chiral Anomaly says:

In this generalized sense, a single photon can have a wavefunction. When physicists say that a photon doesn't have a wavefunction, they mean that it doesn't have a wavefunction that is a function of the eigenvalues of position observables, and that's because it doesn't have any strict position observables.

Can we define a wave function of photon like a wave function of an electron?

where J.G. says:

This ultimately scuppers any attempt to interpret a relativistic wavefunction as a 1-particle probability amplitude.

where MS Tais in a comment says:

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".

photon wave function, double slit, single photon source

where Punk_Physicist says:

There's an old argument by Newton and Wigner, that the photon as a massless particle can't have a position operator and therefore no position space wave function.

Now for clarification, the question is about more then just context, and using the phrase wave function in different ways. Photons do not have strict position observables. Though, they are QM objects, and I do believe that all elementary particles in the SM are QM objects, and should have a wave function. Now for photons the Fourier transform of $f(\nu)$ is sometimes considered as wave function, but this is for photons meaningless, as time oscillations are fast, but the observation process is too slow. Our eyes observe the frequency distribution, not the Fourier transform. Because of the transversal nature of photons, the Fourier transform can never be interpreted as something in space (only along planes perpendicular to momentum).

Wave function of a photon?

Question:

  1. Does a photon have a wave function or not?
$\endgroup$
  • 3
    $\begingroup$ I think this is pretty well answered by Chiral Anomaly's answer, which you linked. People mean different things when they say "has a wavefunction". Depending on what exactly you mean by that phrase, the answer is either yes or no. $\endgroup$ – knzhou Apr 10 at 19:40
  • 1
    $\begingroup$ I also asked a related question (with another excellent answer by Chiral Anomaly) here. $\endgroup$ – knzhou Apr 10 at 19:41
  • 1
    $\begingroup$ see Hawton, Margaret. "Photon wave functions in a localized coordinate space basis." Physical Review A 59.5 (1999): 3223. $\endgroup$ – ZeroTheHero Apr 10 at 21:28
  • $\begingroup$ @ZeroTheHero thank you I am trying, but I cannot access the full version. journals.aps.org/pra/abstract/10.1103/PhysRevA.59.3223 $\endgroup$ – Árpád Szendrei Apr 10 at 21:35
  • 1
    $\begingroup$ Does this answer your question? Why doesn't there exist a wave function for a photon whereas it exists for an electron? $\endgroup$ – WillO Apr 11 at 11:36
2
$\begingroup$

The difference is that in classical physics electron is a particle, while electromagnetic field is a wave. When they are quantized, the particle-wave duality kicks in: the electron acquires wave-like properties - a wave function which can undergo interference and produce a discrete spectrum when confined, whereas the electromagnetic field acquires particle-like properties - it now has a momentum and it is created in discrete units (photons). Thus, properly speaking, the modes of the electromagnetic field are its wave functions. (This hardly comes as a surprise, if you read about quantizing electromagnetic field, which nearly always involves modes expansion and then imposing the commutation relations on the expansion coefficients.)

Things become really odd when we perform second quantization for electrons, which is mathematically identical with the first quantization for the electromagnetic field.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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