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This calculation is for a double slit experiment setup which is experiencing a far field radiation from an extended monochromatic thermal source. I assume the source is 1-D and it's length is $b$. Here is the schematics.

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I think a single photon state emitted from such source is: $$|\Psi\rangle = \int^{\frac{b}{2}}_{-\frac{b}{2}} db \sum_n c_n \int d\omega f_n(\omega)\hat{a}^\dagger_n(\omega) e^{- i \omega F(z_n, b)} |0\rangle$$. But I am little lost what should be $F(z_n, b)$?

I think a coherent state representation of the same source is: $$|{\alpha}\rangle = \int^{\frac{b}{2}}_{-\frac{b}{2}} db \prod_{\bf k} |\alpha_{\bf k} e^{- i \omega F'(z_n, b, k)} \rangle$$. But I am little lost what should be $F'(z_n, b, k)$?

The value of n could be either 1 or 2 corresponding to the first or second slit respectively. Each $\omega$ is a mode coming from the point sub source. Are my expressions correct?

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1 Answer 1

I don't think this is a quantum optics problem. Just look up the van-Cittert-Zernike theorem . The (complex) visibility is the Fourier transform of the mutual coherence function of the source.

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