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In the second quantization time-independent field operator can be expanded in the orthonormal basis: $$\hat{\Psi}(\mathbf{x}) = \sum\limits_{i}\hat{a}_{i}\ \phi_{i}(\mathbf{x})$$

Time evolution of the field operator is given by: $$\hat{\Psi}(\mathbf{r},t) = e^{it/\hbar \hat{H}}\hat{\Psi}(\mathbf{x})e^{-it/\hbar \hat{H}} = \sum\limits_{i}\hat{a}_{i}(t)\ \phi_{i}(\mathbf{x})$$ Time dependence is transferred to the operators. Can we have time-dependent basis states rather than operators: $$\hat{\Psi}(\mathbf{r},t) = \sum\limits_{i}\hat{a}_{i}\ \phi_{i}(\mathbf{x},t)$$ ? How would the Fock state look like ?

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  • $\begingroup$ you are using the Heisenberg picture, where the time dependence is absorbed into the the operators.. If you work in the Schrodinger picture then your basis states will have time dependence... $\endgroup$
    – Bruce Lee
    Commented Feb 22, 2016 at 14:08
  • $\begingroup$ I can't find second quantization in Schrodinger picture. Everyone is doing it in Heisenberg picture. $\endgroup$
    – WoofDoggy
    Commented Feb 22, 2016 at 14:15
  • $\begingroup$ actually it can be done...why the second quantisation is done in Heisenberg picture (or in interaction picture) is because it is more convenient to do so... $\endgroup$
    – Bruce Lee
    Commented Feb 22, 2016 at 14:25
  • $\begingroup$ So now the question is: where I can find a good reference about second quantization in Schrodinger picture? $\endgroup$
    – WoofDoggy
    Commented Feb 22, 2016 at 14:29
  • $\begingroup$ I'am afraid I don't know... :)...you can edit and re-ask it for a resource recomendation...or ask about the problems in second quantization in Schrodinger picture... $\endgroup$
    – Bruce Lee
    Commented Feb 22, 2016 at 14:32

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The answer is yes only if $\hat{H}$ is quadratic in the creation and annihilation operators (with suitable additional regularity assumptions). In other words, it is possible only if the map is a so-called Bogol'ubov transformation.

There is in fact a theorem called Shale's theorem (for bosons) and Shale-Stinespring (for fermions) giving the conditions for a quadratic operator to be the generator of a Bogol'ubov transformation.

If $\hat{H}$ is more than quadratic, it is easy to see (heuristically at least) that it cannot preserve the linear form of the field operator, and so OP's last expression cannot be true.

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  • $\begingroup$ So I can also take it as a kind of approximation. I could see a few papers that assumed this kind of expansion when dealing with Bose-Einstein condensates. $\endgroup$
    – WoofDoggy
    Commented Feb 22, 2016 at 14:53
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    $\begingroup$ In some sense yes. As far as I know, you get a Bogoliubov hamiltonian (i.e. transformation) as the evolution operator for quantum fluctuations around some suitable states in the mean field limit of many-body bosonic theories (and that is indeed related to condensation). For a mathematical take on the subject see e.g. this paper and this one. $\endgroup$
    – yuggib
    Commented Feb 22, 2016 at 15:18

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