Recently, I study quantum optics and deal with quantization of EM field in a cavity. We know we can express/quantize vector potential in terms of $\hat{a},\hat{a}^{\dagger}$ to get a quantized EM field in a cavity.
$$
\vec{A}(\vec{r},t)=\sum_{n,\sigma}\sqrt{\frac{\hbar}{2\epsilon_0\omega_n V}}\vec{e}_{n,\sigma}\Big[\hat{a}_{n,\sigma}e^{i(\vec{k}_n\cdot\vec{r}-\omega_nt)}+\hat{a}_{n,\sigma}^{\dagger}e^{-i(\vec{k}_n\cdot\vec{r}-\omega_nt)}\Big]
$$
The quantized Hamiltonian is:
$$
\hat{H}=\sum_{k}\hbar\omega_k(\hat{n}_k+\frac{1}{2})
$$
The eigenstate of quantized Hamiltonian is: $\left| n_1,n_2,n_3,… \right>=\left|n_1\right>\otimes\left|n_2\right>\otimes\left|n_3\right>…$ The state means there are $n_1$ photons in the first mode and $n_2$ photons in second mode and so on…
So every mode has it own number of photons and photons in the different modes are not at the same frequencies. But why do we take photons as indistinguishable particles?
Best Answer
Calling photons indistinguishable particles is an artefact of thinking of them as fermion-like particles, rather than excitation quanta of electromagnetic field - the phrase is used to make the point to those (yet) unfamiliar with second quantization (see also this answer). If you take a specific mode of the field, with occupation number $n_k$, there is no way of distinguishing its $m_k$-th quantum from $m_k+1$-th or any other. But, since your already know how to quantize the field, this even does not come into question.
The photons in different modes (with different k-vectors, frequencies and polarization) are of course distinguishable.