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Suppose that a Quantum System can be described by the wavefunction $\psi(\vec{x},t)$, but due to the occurence of chaotic noise within the Quantum System, only the "filtered" wavefunction $$\phi(\vec{x},t)= \int_V d^3y \int_{-T}^T d \tau\psi(\vec{x}+\vec{y},t+\tau)$$ with the cubical fluctuation volume $V$ and characteristic fluctuation time $T$ (These functions can also be space-time dependent!) can be described with well-known Quantum mechanical equations.

It holds $$i \hbar \frac{\partial}{\partial t} \phi=H \phi$$ with the well-determined Hamiltonian $H$ of the physical System. In this question I consider $$H = - \frac{\hbar^2}{2m} \Delta + V.$$ Now I can Substitute the Ansatz for $\phi$ and I obtain by using the Leibnitz rule:

$$\int_V d^3y \int_{-T}^T d \tau \left(i \hbar \frac{\partial}{\partial t} + \frac{\hbar^2}{2m} \Delta - V\right)\psi(\vec{x}+\vec{y},t+\tau)= -\left(i \hbar \frac{\partial}{\partial t} + \frac{\hbar^2}{2m} \Delta\right) \left(\int_V d^3y \int_{-T}^T d \tau\right) \psi(\vec{x}+\vec{y},t+\tau)=....$$

The right Hand side is not easy to expand out; there occur a lot of integrals and differences in $\psi$. The resulting equation will be linear, but it is an integro-algebro-differential-equation that is difficult to solve. I have no idea how to solve this equation and also I cannot find a Propagator $$\Pi(\vec{x}-\vec{x'},t-t')$$ which determines Transition rules like $$\psi(\vec{x},t)=\int d^3x' \int dt' \Pi(\vec{x}-\vec{x'},t-t')\psi(\vec{x'},t').$$ Can anybody give hints how I can solve the equation for $\psi$ (it is enough to Show how I can find a Propagator similar to Feynman propagator)?

Every helpful contribution will be greatly appreciated.

P.S.: It might be easier to introduce rectangle functions for the integrals, i.e.: $\int_{-T}^T d \tau (...) = \int_{- \infty}^\infty d \tau \,{\rm rect}_{2T}(\tau) (...)$. Here, the subscript $2T$ denotes that the rectangle function has the broadness $2T$.

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  • $\begingroup$ Have you considered filtering the function in spectral space? $\endgroup$ – Nick P Mar 1 '15 at 18:30
  • $\begingroup$ No, I haven't considered yet. $\endgroup$ – kryomaxim Mar 1 '15 at 18:31

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