It's well-known that hydrogen atom described by time-independent Schrödinger equation (neglecting any relativistic effects) is completely solvable analytically.

But are any initial value problems for time-dependent Schrödinger equation for hydrogen solvable analytically - maybe with infinite nuclear mass approximation, if it simplifies anything? For example, an evolution of some electron wave packet in nuclear electrostatic field.

  • $\begingroup$ What do you mean by "analytically"? You probably don't mean the math definition, which is that the function converges to its Taylor series. If you mean "involving simple functions" the you should know there's no qualitative difference between numerical integration and special functions. In fact many common special functions are evaluated by your computer via the differential equation they satisfy. $\endgroup$ – user10851 Feb 23 '14 at 22:44
  • $\begingroup$ @ChrisWhite I mean explicit solution, in terms of such functions, which don't require to set up a dense spatial grid and propagate the solution in small temporal steps to find the value at a given point in spacetime with required precision. $\endgroup$ – Ruslan Feb 24 '14 at 4:15
  • $\begingroup$ Assuming infinite mass of the nucleus, we basically have an electron wave-packet in a central 1/r potential. One thing that comes to mind is the scattering on a Coulomb potential problem. I can't say off the top of my head if it's integrable, but it seems like a reasonable starting point. $\endgroup$ – LLlAMnYP Dec 23 '15 at 11:54
  • $\begingroup$ For a discussion of a nicely-looking quasiclassical wave packet as initial condition, see this answer. $\endgroup$ – Ruslan May 20 '18 at 20:25

Of course there are. That's why we solve the time-independent version of the Schrödinger equation to begin with: because given any eigenfunction $\psi_0$ of the hamiltonian with eigenvalue $E$, the phase-evolved combination $$ \psi(t) = e^{-iEt/\hbar}\psi_0 $$ is a solution of the time-dependent Schrödinger equation, and, moreover, any linear combination of such solutions is still a solution.

There is, of course, an understandable prejudice against taking a stationary state as an initial condition for the TDSE (but it is just a human prejudice with no real meat to back it up). If that really bothers you, then you can just take a nontrivial linear combination, like, say, $$ \psi = \frac{\psi_{100}+\psi_{210}}{\sqrt{2}}, $$ and it will then show oscillations in both the position-space and momentum-space probability distributions. To borrow from my answer to Is there oscillating charge in a hydrogen atom?, the explicit wavefunction is given by

$$ \psi(\mathbf r,t) = \frac{\psi_{100}(\mathbf r,t) + \psi_{210}(\mathbf r,t)}{\sqrt{2}} = \frac{1}{\sqrt{2\pi a_0^3}} e^{-iE_{100}t/\hbar} \left( e^{-r/a_0} + e^{-i\omega t} \frac{z}{a_0} \frac{ e^{-r/2a_0} }{ 4\sqrt{2} } \right) , $$ and this goes directly into the oscillating density: $$ |\psi(\mathbf r,t)|^2 = \frac{1}{2\pi a_0^3} \left[ e^{-2r/a_0} + \frac{z^2}{a_0^2} \frac{ e^{-r/a_0} }{ 32 } + z \cos(\omega t) \, \frac{e^{-3r/2a_0}}{2\sqrt{2}a_0} \right] . $$

Taking a slice through the $x,z$ plane, this density looks as follows:

This combination gives you an explicit analytical solution of the time-dependent Schrödinger equation. Now, again, it is understandable to dismiss this as somehow "not being a real wavepacket", partly because from some perspectives it might feel "too easy", but all of those are human prejudices with very little support on well-defined and truly meaningful mathematical criteria on the initial conditions or the corresponding solution. This is an honest-to-goodness electronic wavepacket moving under the influence of a point-charge nucleus.

  • $\begingroup$ That was quite stupid of me to fail to rule out such trivial initial conditions as superposition of finite number of eigenstates in the question body (I did mention this in a comment to another answer though). $\endgroup$ – Ruslan Nov 4 '17 at 20:26
  • $\begingroup$ @Ruslan The question you want to ask is ill-posed, because there is no well-defined "trivial-o-meter" for initial conditions. The only thing that's stopping you from seeing these solutions as legitimate wavepackets are your own personal prejudices. They're understandable, but they're still prejudices and they're still exclusively human. $\endgroup$ – Emilio Pisanty Nov 4 '17 at 20:29
  • $\begingroup$ I do agree with you (now, 3 years after I posted the question). I'm not sure now how to get out of the trap: the answer about finite eigenstate superposition is obvious, and there are two such answers posted here. They don't answer what I wanted to ask, but they still do have a point. Maybe I should just accept one... $\endgroup$ – Ruslan Nov 4 '17 at 20:34

What you do have available is an explicit knowledge of the eigenvalues and eigenvectors (also for the continuous spectrum). By expanding your initial wavepacket in terms of the eigenvectors you then obtain its value for later times as a sum (or integral for continuous spectrum) with added weight factors exp[-i$\lambda$t], where $\lambda$ is the eigenvalue associated with the corresponding eigenvector.

  • $\begingroup$ The problem with this approach is that one needs to do lots of numeric 3D integrations to find projections of initial state on eigenstates. And if the initial state has too high localization so that much of continuous spectrum states have contribution to initial state, then one would have to do even more integrals. This approach doesn't seem as a very good one in general, this is why I asked about direct solution of time-dependent equation. $\endgroup$ – Ruslan Jan 28 '14 at 12:34
  • $\begingroup$ Yes, I see your problem. Did you consider a Green's function approach? $\endgroup$ – Urgje Jan 29 '14 at 11:31
  • $\begingroup$ I don't know much about Green's functions. Could you elaborate on how to apply them to this problem (and what to read to understand your explanation better)? $\endgroup$ – Ruslan Jan 29 '14 at 11:38
  • $\begingroup$ The Green's function of the time-independent case is known, both in coordinate and momentum representation. Whether or not that helps in your case I do not know. $\endgroup$ – Urgje Jan 30 '14 at 16:08

Solution of an initial value problem can be written as integral of the initial function $\psi_0$ multiplied by the propagator of the Schr. equation. Depending on the function $\psi_0$, the integral may or may not be calculable in terms of simple functions. I do not know of any initial function $\psi_0$ and potential $A(t)$ that would admit simple exact solution; the equation with time-dependent term is difficult to solve. More rewarding way seems to be to find the solution with a computer. The real problem is I think elsewhere - how do we find appropriate function $\psi_0$ to describe real atoms? Often the first eigenfunction of the Hamiltonian is used, but I do not think this is particularly well motivated.

  • 1
    $\begingroup$ My question was "are there any initial value problems", not "what is the definite solution". Of course, by initial conditions I mean not such trivial ones as superposition of finite eigenfunctions, but some form of wave packets. $\endgroup$ – Ruslan Mar 25 '14 at 19:06
  • $\begingroup$ OK, I've edited my answer accordingly. Sorry I can't be of more help. $\endgroup$ – Ján Lalinský Mar 26 '14 at 21:07

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