# Is there a multicolor single photon?

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fxdung
We know that we can not cut a single photon into many photons. So that there must be a single color for a single photon?Because if a single photon has many color then it will be dispersed through prism(so a single photon would be cut into many photons(?))

• A. Neumaier

Gold Member
We know that we can not cut a single photon into many photons. So that there must be a single color for a single photon?Because if a single photon has many color then it will be dispersed through prism(so a single photon would be cut into many photons(?))

A single photon has a single frequency (what you might call "color"). However, it is possible to split a photon into 2 photons using "parametric down conversion" (PDC) using a nonlinear crystal (although the proper description of that process is complicated). When PDC occurs, there is conservation of momentum, energy, etc. as would be expected. I.e. total energy of 1 photon in equals total energy of 2 photons out.

Mentor
A single photon has a single frequency
Actually, I think even this is not always true. AFAIK the particle number operator and the frequency operator do not commute, so they have no joint eigenstates; so a "single photon" that is a number eigenstate will not have a well-defined frequency. A "single photon" that is actually a coherent state will, but then it is not an eigenstate of the number operator and the term "single photon" has to be interpreted differently when applied to it (in terms of the expectation value of photon number).

• DrChinese and gentzen
Mentor
We know that we can not cut a single photon into many photons.
No, we don't know that, because the term "a single photon" in quantum field theory does not mean what you are thinking it means. In fact, as my post #3 just now notes, the term does not even have a single meaning; it can mean different things for different types of quantum field states.

fxdung
What is mathematical formula of frequency operator?

Last edited:
Gold Member
2022 Award
Actually, I think even this is not always true. AFAIK the particle number operator and the frequency operator do not commute, so they have no joint eigenstates; so a "single photon" that is a number eigenstate will not have a well-defined frequency. A "single photon" that is actually a coherent state will, but then it is not an eigenstate of the number operator and the term "single photon" has to be interpreted differently when applied to it (in terms of the expectation value of photon number).
What do you mean by "frequency operator"? I guess you mean the Hamiltonian. For a single-photon mode with momentum ##\vec{k}## and helicity ##h## the energy eigenvalues are ##E=|\vec{k}|## (using natural units with ##\hbar=c=1##), and ##E=\omega##.

This single-photon Fock state of course is not a proper state vector representing a photon, because as any plane wave, it's not normalizable to one. They are generalized eigenstates "normalizable to a ##\delta##-distribution". A true single-photon state is always a "wave packet", i.e., something like
$$|\psi \rangle \propto \int_{\mathbb{R}^3} \mathrm{d}^3 k A(\vec{k}) \hat{a}^{\dagger}(\vec{k},h) |\Omega \rangle,$$
where ##A(\vec{k})## is square integrable. I've written down a one-photon state with helicity ##h##. You can of course have arbitrary superpositions (i.e., arbitrary polarization states). Since ##A(\vec{k})## has a finite width, indeed such a state is not a monochromatic one (although you can make it arbitrarily close to one; you only have to make ##A(\vec{k})## sharply peaked around some momentum ##\vec{k}_0##); correctly normalized ##|A(\vec{k})|^2## gives the probability distribution for detecting a photon with momentum ##\vec{k}## (and thus energy/frequency ##|\vec{k}|##). It's an eigenstate of the photon-number operator with eigenvalue 1, i.e., you have prepared precisely one photon.

A coherent state is not a single-photon state. It is not even an eigen state of the photon-number operator but an eigen state of an annihilation operator. The photon number is Poisson distributed in such a state, which represents (if of not too low intensity) rather a classical em. wave than "photons". Sometimes popular-science sources mix up single-photon states with low-intensity coherent states. You can make the expectation value of the photon number arbitrarily small for such a coherent state, even smaller than 1. Then this coherent state is pretty close to the vacuum state, and if you detect something it's very likely to be the response to a single photon, but the probability to detect 2 or more photons is also not exactly 0.

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