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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.

Qmechanic
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Michael
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2 Answers2

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For example, if a red traffic light is shining at 700nm, does each photon have 'quatum wavelength' of 700nm

yes

Does this mean that (for a given instant in time) the probability of detecting said photon rises and falls every 700nm?

Yes, if the wave is plane polarized, so that the classical energy density has this oscillation. The probability density has to scale the same way as the classical energy density, so that the average rate of energy deposition on a detector is in accord with the correspondence principle.

user280073
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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?)

nosuchthingasmagic
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