TITLE: What is the relationship between the classical wavelength of light and the quantum mechanical wavefunction of a photon?
QUESTION [4 upvotes]: It is common to speak of the 'wavelength' of light. For instance visible light has a wavelength of around 400 to 700 nm. A single photon can also have a wavelength, given apparently by $\lambda = \frac{hc}{E} $. My understanding is that a photon's wavelength is not related to the physical size of a photon, which is a actually a point particle. Instead the wavelength of a photon is related to the probability of finding a photon at a particular coordinate in spacetime.
So is the wavelength of a photon from the quantum perspective the same value as the wavelength of a classical EM wave? For example, if a red traffic light is shining at 700nm, does each photon have 'quatum wavelength' of 700nm? Does this mean that (for a given instant in time) the probability of detecting said photon rises and falls every 700nm?
It would also be nice to see whavetever equations are relevant.

REPLY [0 votes]: The question asks about the wavelength and the wavefunction. To clarify, those are two distinct things. The wavelength is a property (e.g., 700 nm), while the wavefunction is a function that can vary over space and time and gives the probability of detecting the photon in a certain state. While there are different theories, a photon has just one wavelength.
I remember being really impressed when I was in school that the probabilistic nature of light usually associated with quantum mechanics is already apparent from classical electricity and magnetism. To recap, in classical E&M, what is "waving" is the electromagnetic field. According to classical E&M, the density per unit volume stored in an electromagnetic field is proportional to
$$\frac{1}{2} \left(\epsilon_0 E^2 + \frac{1}{\mu_0}B^2\right).$$
For a monochromatic plane wave, the electric and magnetic contributions are equal, so it is proportional to $E^2$. (You can find details in any standard source on E&M.)
Now if you accept that the field represents a group of photons in which each photon has energy $\hbar\omega$ (where $\omega$ is the frequency of the electromagnetic wave and is related to the wavelength $\lambda$ by $\lambda= \frac{2\pi c}{\omega}$), the density of photons in the field is proportional $\frac{E^2}{\hbar \omega}$, which you can think of as the probability density for finding a photon. With the $E^2$, you already get the interference effects of quantum mechanics.
I should also note that since photons are relativistic particles, you need to go beyond non-relativistic quantum mechanics for an accurate description (see, e.g.: What equation describes the wavefunction of a single photon?)
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