I found this great question about the solution of the Schrodinger equation for a particle in a constant gravitational field, but the solution they wanted is to the time independent Schrodinger equation.

$$E \psi=\frac{-\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}+mgx\psi$$

Wave function of a particle in a gravitational field

I want a solution for the time dependent Schrodinger equation for a particle in a constant gravitational field, one with dispersion, where the energy is not exactly known. How do I get it? Basically I am trying to get a solution to this equation

$$i\hbar \frac{\partial\psi}{\partial t}=\frac{-\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}+mgx\psi$$


The time-dependent SE in PDE shorthand:

$$i\hbar \psi_t=-\frac{\hbar^2}{2m}\psi_{xx}+mgx\psi$$

To solve it, we use separation of variables, by assuming the Ansatz: $$\psi(x,t)=\Psi(x)\phi(t)$$ Inserting into the PDE: $$i\hbar \Psi(x)\phi'(t)=-\frac{\hbar^2}{2m}T(t)\Psi''(x)+mgx\Psi(x)T(t)$$ Dividing both sides by $\psi(x,t)$ we get: $$i\hbar\frac{\phi'}{\phi}=-\frac{\hbar^2}{2m}\frac{\Psi''}{\Psi}+mgx=E$$ Where $E$ is a separation constant.

So we obtain two separate DEs, one in $t$ and one in $x$: $$i\hbar\frac{\phi'}{\phi}=E\tag{1}$$ $$\Rightarrow \phi(t)=e^{-\frac{{ E i t}}{\hbar}}$$ $$-\frac{\hbar^2}{2m}\frac{\Psi''}{\Psi}+mgx=E\tag{2}$$ $(2)$ is basically the DE you'll find in your link.

  • $\begingroup$ I thought of that but it doesn't look like the kind of solution I want. I want a solution where the expectation value of the position is $x_0+v_0t-1/2gt^2$ and where there is finite uncertainty in all observables $\endgroup$ Feb 12 '21 at 22:01
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    $\begingroup$ $(2)$ inevitably leads to an Airy function. I don't see how you can 'force' a classical solution like the one in your comment onto this DE and its Airy solution. $\endgroup$
    – Gert
    Feb 12 '21 at 22:05
  • $\begingroup$ What if I use time dependent perturbation theory? $\endgroup$ Sep 23 '21 at 6:56
  • $\begingroup$ Worth a try I suppose but I'm not well versed in perturbation theory. Sorry. $\endgroup$
    – Gert
    Sep 23 '21 at 16:14
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    $\begingroup$ @RyanParikh if you want a solution that resembles the classical form, you could try working in the Heisenberg picture, in which you calculate time dependent operators that act on a fixed initial wavefunction. I have not looked at what happens in this case, but it would not surprise me if was solvable and gave you something like what you are after $\endgroup$ Sep 23 '21 at 16:23

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