Is one photon the quanta of excitation between modes or Energy eigenstates?

Question

When we quantize a EM field with appropriate boundary conditions (say in a waveguide/cavity) we get modes denoted by the $vec{k}$ and we know that $omega = c|vec{k}|$ so do modes represent different frequencies of the EM wave ($omega_k$) and higher modes mean higher frequencies?
Also Fock states $|nrangle$ are photon number states and energy eigenstates of the QHO ( quantized EM wave) . Since $$E_n=hbaromega (n+frac{1}{2})$$ higher energy means greater frequency means more photons. Photons are excitations between the energy eigenstates, and since $omega propto |vec{k}|$ , does more photons means a higher mode ? Do photons serve as excitations between different modes ?
I'm mostly confused about how we can have many photons in a single mode and not change the frequency(energy) and is it  $$E_n=hbaromega_k (n+frac{1}{2}) { } \OR\ E_n=hbaromega (n_k+frac{1}{2})$$ and the difference between them.

Emilio Pisanty · Answer

higher energy means greater frequency means more photons

No.
The energy in the mode is the product of the frequency of the mode and the number of photons in that mode,
$$
E_k = hbar omega_k (n_k+tfrac12).
$$
Thus:

A higher frequency (at the same number of photons) raises the energy.
A higher number of photons (at the same frequency) raises the energy.

Both of these are independent of each other, and if all you know is that the energy went up, you cannot tell which mechanism caused it.
It is also important to note that increasing the number of photons in the mode does not change the mode, but changing the frequency does. Thus, at a hand-wavy level, you get "higher modes" when you increase the frequency (but be careful with that term, which is only loosely defined).

Roger Vadim · Answer

Every mode is a separate harmonic oscillator. In the context of electromagnetic field a mode is essentially the spatial structure of oscillations corresponding to a given frequency and polarization. Thus, the energy in mode $k$ is given by the number of quanta/photons in this mode, $n_k$:
$$E_{k,n_k} = hbaromega_kleft(n_k+frac{1}{2}right),$$
whereas the enregy of the electromagnetic field is obtained by summing over all the modes:
$$E_{tot} = sum_k E_{k,n_k} = sum_khbaromega_kleft(n_k+frac{1}{2}right).$$
Note further that $k$ really stands for all the indices characterizing a mode, typically the three-dimensional wave vector and the polarization index $k=(mathbf{k},nu)$.

Is one photon the quanta of excitation between modes or Energy eigenstates?

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