On the PDE $i\tau\psi_t(x,t) + e^{-\alpha t}\psi_{xx}(x,t) = x^2e^{\alpha t}\psi(x,t)$.

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I would like to solve or at least further understand the solution of the following PDE:

$$i\tau\psi_t(x,t) + e^{-\alpha t}\psi_{xx}(x,t) = x^2e^{\alpha t}\psi(x,t)$$

such that $\int_{-\infty}^\infty\psi^*(x,t)\psi(x,t)dx = 1$ for all $t > 0$.

This arises from Quantum Mechanics when trying to solve the Schrödinger equation for an explicitly time-dependent Hamiltonian.

I know that seperation of variables doesn't work here, because of the exponential factors that involve time. I've also thought about maybe using a Fourier Transform, but that doesn't really help in the spatial variable, because then the $x^2$ becomes a second derivative.

I would not be surprised if no analytical solution is possible here and I am forced to resort to numerical methods, but I have not given up hope yet that somebody might have an idea for a clever ansatz or a clever coordinate transform that reduces this to something more manageable.

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Well, it is separable, in the suitable variables, and reducible to a nasty eigenvalue problem for one of them.

That is, split your linear operator, $$ 0=\left ( i\tau \partial_t + e^{-\alpha t} \partial_x^2 -x^2 e^{\alpha t} \right )\psi \\ = \left ( i\tau (\partial_t -\frac{\alpha x}{2}) +(\frac{i\tau \alpha}{2}x\partial_x + e^{-\alpha t} \partial_x^2 -x^2 e^{\alpha t} ) \right )\psi \\ \equiv (L_1 +L_2)\psi, $$ and set $\psi= e^{-iEt/\tau} \phi(x ~e^{\alpha t/2}) $ for some eigenvalue constants E and associated functions $\phi$.

Since $L_1 \phi(x ~e^{\alpha t/2}) =0$ and $(L_1-E) e^{-iEt/\tau} =0 $, one has $$ -E\phi (x ~e^{\alpha t/2}) =L_2 ~\phi (x ~e^{\alpha t/2}) = \left (\frac{i\tau \alpha}{2}x\partial_x + e^{-\alpha t} \partial_x^2 -x^2 e^{\alpha t} \right ) \phi (x ~e^{\alpha t/2}), $$ so that $$ -E\phi (y) = \left (\frac{i\tau \alpha}{2}y\partial_y + \partial_y^2 -y^2 \right ) \phi (y), $$ a potentially messy eigenvalue problem you are invited to analyze for your solutions. (It is covered in E Kamke's monumental "phone book", though...)