# Time dependent wave function of a particle in a gravitational field

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

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

Hi I am coming back to edit this question in hopes that I can direct students in the right direction. The reason I asked this question in February 2021 was due to a fundamental misunderstanding of quantum mechanics and the Schrodinger equation. I did not understand the role of the time independent Schrodinger equation and I did not see the use in decomposing the wave function into a sum of stationary states. Here is a resource that greatly helped me understand why you would want to do it and HOW to do it. I have linked to the specific page that made things click for me. https://farside.ph.utexas.edu/teaching/qmech/Quantum/node101.html

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.

• 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 Commented Feb 12, 2021 at 22:01
• $(2)$ 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.
– Gert
Commented Feb 12, 2021 at 22:05
• What if I use time dependent perturbation theory? Commented Sep 23, 2021 at 6:56
• Worth a try I suppose but I'm not well versed in perturbation theory. Sorry.
– Gert
Commented Sep 23, 2021 at 16:14
• @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 Commented Sep 23, 2021 at 16:23