I have been thinking about this for a while. I think I misunderstood something about the basics of quantum waves.

Let's look at light diffracted in conditions similar to the double slit experiment. The intensity pattern we can observe on the screen is due to the wavelike propagation of photons. We can then construct a wavefunction, whose square will give us the probability of a photon being measured at a certain point in space. The maxima of this probablity distribution will resemble the intensity maxima on our screen. The intensity of light is the square of the amplitude of electric wave associated with it.

So it seems to me that these two things tell us the same thing. The quantum wavefunction will give us the probability of a photon being at a point in space, which will be proportional to the intensity of our light (more probability means more photons over time, which means more energy). We would get the same results from calculating the electromagnetic wave at each point, right? So what is different in these two things? Does the electromagnetic wave resemble the quantum wavefunction? Is it okay to think of the EM wave as a result of the probability distribution calculated from QM? (the stronger the EM wave, the more photons can be measured, which means bigger probability... etc)

I have been reading forums and I can't find satisfying answers to this question.


1 Answer 1


Photons don't have an associated wave function. You either use the (fully classical) Maxwell's equations for the electromagnetic field (without photons) or Quantum Electrodynamics (which doesn't work with wave functions at all).

For more details, see What equation describes the wavefunction of a single photon?.

  • $\begingroup$ Is it fair to say, then, that the double slit experiment with photons is not a quantum-mechanical effect? While for e.g. electrons it is a quantum mechanical one? $\endgroup$ May 28, 2019 at 14:01
  • 1
    $\begingroup$ @doublefelix No, it is not fair to say that. "Photons" is an intrinsically quantum-mechanical concept, so no phenomenon involving them may said to be classical. The fact that it makes no sense to associate a wavefunction to photons does not mean they are classical; quite the opposite: it means that you need to use the whole machinery of QFT to describe them. On the other hand, it is perfectly fair to say that the DSE with light is a classical effect (in that it can be described using classical field theory). $\endgroup$ May 28, 2019 at 14:19
  • $\begingroup$ Neither absense of particle conservation, nor absense of an equation (including Shroedinger one) imply absense of wave function. Hence an answer can't be accepted. $\endgroup$
    – Dims
    Feb 20 at 18:43

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