The Wiener-Khinchin theorem says that the Fourier transform of the degree of first order coherence is equal to the power spectral density normalized (Loudon, Quantum theory of light, pag. 102):
$F(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}g^{(1)}(\tau)exp(i\omega\tau)d\tau$,
Where $F(\omega)$ is the normalized spectrum and $g^{(1)}(\tau)$ is the first order degree of coherence.
Let's take a multimode state of light like multimode coherent light, in which we consider only wave vector parallel to the z axis direction of the beam:
$|\{\alpha\}\rangle = |\alpha_{k_1} \rangle|\alpha_{k_2} \rangle|\alpha_{k_3} \rangle... $ ,
This state is for definition eigenstate of the positive-frequency part of the electric field operator. From this follow that for multimode coherent light, like in single mode: $|g^{(1)}(z_1, t_1; z_2, t_2)| = g^{(2)}(z_1, t_1; z_2, t_2) = 1$.
If it's stil possible to use the Wiener-Khinchin theorem, and knowing that from $|g^{(1)}(z_1, t_1; z_2, t_2)| = 1$ follow $g^{(1)}(z_1, t_1; z_2, t_2) = exp(i\phi)$, we have:
$F(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}exp(i(\omega\tau + \phi) )d\tau$,
For single mode coherent light we have:
$g^{(1)} = exp(-i\omega_0\tau)$
Then:
$F(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}exp(i(\omega - \omega_0)\tau)d\tau = \delta(\omega - \omega_0)$.
It's reasonable because single mode have one frequency, so the spectral distribution is just a Dirac delta function.
In multimode i immagine a sum of Dirac's delta centered in each frequency:
$F(\omega) = \frac{1}{N}\sum_{i}^{N}\delta(\omega - \omega_i)$.
If we take the inverse Fourier transform, a part maybe for some constant:
$\int_{-\infty}^{\infty}F(\omega) exp(-i\omega\tau)d\omega = \frac{1}{N}\sum_{i}^{N}\int_{-\infty}^{\infty}\delta(\omega - \omega_i)exp(-i\omega\tau)d\omega = \frac{1}{N}\sum_{i}^{N}exp(-i\omega_i\tau)$,
$g^{(1)}(\tau) = \frac{1}{N}\sum_{i}^{N}exp(-i\omega_i\tau)$.
Now, for multimode coherent light we know $g^{(1)}(z_1, t_1; z_2, t_2) = exp(i\phi)$ because the modulus is one.
How can be $exp(i\phi) = \frac{1}{N}\sum_{i}^{N}exp(-i\omega_i\tau)$? Or equivalently $|\frac{1}{N}\sum_{i}^{N}exp(-i\omega_i\tau)| = 1$?
This is the starting point:
$\frac{1}{N}|\sum_{i}^{N}exp(-i\omega_i\tau)| = \frac{1}{N}\sqrt{[N + \sum_{i < j }^{N}2cos((\omega_i - \omega_j)\tau)]} $.
Because we are in a cavity of lenght L, we have $\omega_n = n\frac{c2\pi}{L}$, so $\omega_i - \omega_j = (i - j)\frac{c2\pi}{L}$ ,
Then:
$|\frac{1}{N}\sum_{i}^{N}exp(-i\omega_i\tau)| = \frac{1}{N}\sqrt{[ N + \sum_{i < j }^{N}2cos((i - j) (\frac{2\pi}{L})c\tau)]}$
The summation under the square root clearly must be equal to $N(N-1)$ in order to have $|g^{(1)}(\tau)| = 1$. For $\tau = 0$ works well, how can be possible with this term: $\frac{c\tau}{L}$? The formula doesn't seem to be independent of $\tau$, but the final result have to be. If $c\tau = nL$, with n any integer, it will works.
Another question: in general we know that the bandwidth is inversely proportional to the coherence time. Here the coherence time is infinity, so we have to be zero bandwidth. But multimode light can have an arbitrary number of different frequencies. This doesn't represent bandwidth? Maybe because it's discrete, not continuous?
