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We know that the energy of the photon is given by $\hbar\omega$ and it so happens that this exact $\omega$ is the frequency we would obtain from the many photon EM wave. How does one relate the two?

Is the energy spacing between states of a harmonic oscillator related to the coherent state formed by the system? If so, how does one extend this to the case of EM waves?

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    $\begingroup$ I'm not sure I completely understand your question. We have a quantum mechanical description of the EM field in second quantization, which is built upon photons. We can write the expectation value of the electric and magnetic field, and (as expected from Ehrenfest's theorem), they show the classical EM wave behavior. This matches the two aspects quite nicely, I think, and relate the frequency of the EM wave to the frequency of the photon. Would that be an answer in the direction you are looking for? (I can elaborate in a full answer if necessary) $\endgroup$
    – user245141
    Commented Jan 15, 2020 at 13:41
  • $\begingroup$ Yes. This is in the direction of what I am looking for. Can you please elaborate? $\endgroup$
    – Superfast Jellyfish
    Commented Jan 15, 2020 at 14:00

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There is a standard way to quantize (non-relativisticly) the EM field. Based on the classical energy density $$H = \frac{1}{8\pi}\int\! d^3r \left[|\vec{E}(\vec{r})|^2+|\vec{B}(\vec{r})|^2\right]$$ We write everything in terms of the vector potential $$\vec{A}(\vec{r},t)$$ which we expand in plain waves $\vec{q}_{\vec{k}}(t) e^{i\vec{k} \vec{r}}$. Note that we do not make any assumptions about the time-dependent, or the frequency. That will come out naturally. Then from the Hamiltonian is $$ H = \sum_{\vec{k}} |\dot{\vec{q}}_{\vec{k}}|^2 + \omega_{\vec{k}} |\vec{q}_{\vec{k}}|^2$$ with $\omega_{\vec{k}} = c|\vec{k}|$. Now we quantize these modes, with $p=\dot{q}$ the conjugate momenta of $q$, and you see that the EM field is described as sum of Harmonic oscillators. The creation and annihilation operators of these "Harmonic oscillators" add or remove a quanta of $\hbar\omega_{\vec{k}}$ from the field, and these are the photons.

You can work out how $\vec{E}$ and $\vec{B}$ look in terms of these fields, and you get that $\vec{E}$ has expansion in modes that propagate like the calssical fields, that is with an exponent of $i(\vec{k}\vec{r}-\omega_\vec{k} t)$.

A nice point here is that for a state of the field with a well-defined number of photons, the expectation value of both the electric and magnetic field is zero (just like that the momenta and position of the Harmonic oscillator is zero for a state with well defined $n$). You need a coherent state in order to describe the classical limit. In fact - this is the origin of the term "coherent state", as it was born in quantum optics!

More detailed equations can be found in the wikipedia page on the subject.

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  • $\begingroup$ I would like to add a comment to this already excellent answer, now that I have studied this a bit. The harmonic evolution in time of the fields come directly from the fact that energy goes as $\sim n\hbar\omega$. That means that the time evolution of the state goes as $e^{-in\omega t}$. The reason these phases don’t cancel out during expectation of fields because the energy eigenstates are not field eigenstates. In fact the phase factors of the field’s expectation value works out to be exactly $e^{-i\omega t}$ $\endgroup$
    – Superfast Jellyfish
    Commented Sep 13, 2020 at 8:38
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This question has no answer at present as we do not have a mechanical model of a single photon.

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