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In Yanhua Shih's book on quantum optics, the coherence functions are expressed in terms of effective wave function. Here are the expressions for single photon wave packets.

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To derive the coherence functions we have used Heisenberg picture where the field operators, not the wave functions, are time dependent.

We can also take a look at the effective wave function for thermal source.

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So, it is confusing why we have a wave function with time as a parameter. What is the physical significance of effective wave function?

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Why would it be odd that time is a parameter? –  Dan Jun 11 '13 at 0:27
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In Heisenberg picture states (kets) do not depend on time. But the so-called effective wave function here must clearly depend on time, seeing how it is defined.

Think of nonrelatvistic QM, where the value $\psi\left(\mathbf{x},t\right)$ of a wave function at some point in space and time does not depend on whether you're working in Schrödinger or Heisenberg picture (or Dirac picture, or any picture you want). This is so because it is equal to a scalar product. But when you change pictures, kets are multiplied by an operator $\hat{U}$ and bras by its inverse operator, leaving the scalar product unchanged. So if the wave function is time-dependent in Schrödinger picture, it also is in Heisenberg picture.

To get back to the effective photon wave function, more information about the states $\mid\!\Psi_n\rangle$ would be needed to tell you what the wave function really means. Show me how these states are defined and I might be able to tell you more.

In the meantime, I hope this helps.

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