In just about every text I read (online or in paper), when they handle the time-dependent Schrödinger Equation, I see something along the lines of "we always assume the potential is independent of time." Why is this? Are there not plenty of circumstances when this isn't valid? Aren't most experiments done with varying potentials (NMR for example, the magnetic field, which affects the potential, is changing in time)? Is this assumption made in textbooks just for pedagogical reasons, to make life easier?

If we don't make this assumption, then it seems to me that the Schrödinger equation is no longer separable and we can no longer just apply the time-evolution operator as is usually done (and the time-independent equation is no longer valid).

Perhaps tangential to the main question but: Also, if we want to solve it numerically, it seems to me we also can't simplify using split-step Fourier transforms or into a form handled by Runge-Kutta. Is this correct? I'm especially interested in exploring the numerical analysis, but I guess I should post that question in the scientific computing SE.

Of course, when I say "potential" I mean $V\left(\vec r, t\right)$ in the equation \begin{equation} i\hbar\frac{\partial}{\partial t} \Psi\left(\vec r, t\right) = \left[\frac{-\hbar^2}{2m}\nabla^2+V(\vec r, t)\right]\Psi\left(\vec r, t\right) \end{equation} and the assumption whose justification I don't understand is $V\left(\vec r, t\right)=V\left(\vec r\right)$.

  • $\begingroup$ Out of curiosity, what does "tangential to another question" mean? Is it just a fancy way to say it is related, or does it mean the relation between the two questions is specific and if so how? $\endgroup$
    – Winston
    Commented May 12, 2019 at 5:34
  • $\begingroup$ I just mean the question about the numerical analysis is related to my main question superficially. I wanted to ask it but I'm not sure this is the right place -- I would expect this is the right place for the rest of the questions, though. $\endgroup$ Commented May 12, 2019 at 5:38
  • $\begingroup$ Related: en.wikipedia.org/wiki/Dyson_series $\endgroup$
    – user87745
    Commented May 12, 2019 at 5:44
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    $\begingroup$ Related (maybe even duplicate): physics.stackexchange.com/q/17768 $\endgroup$
    – user191954
    Commented May 12, 2019 at 6:03

1 Answer 1


There are plenty of situations where the potential depends on time. The core reason you haven't seen them is likely that you haven't been looking in the right places.

However, that said, there is indeed a clear separation between the static and the time-dependent components of the potential. For the vast majority of experiments where we use a time-dependent probe to interact with the system, the probe is extremely weak (by several orders of magnitude) when compared to the natural hamiltonian of the system. This means that it is best treated using perturbation theory, so that the best strategy is to solve the time-independent Schrödinger equation for the dominating structural part of the hamiltonian (which generally doesn't depend on time) and then worry about the probe.

Moreover, a huge number of experiments are done, for various reasons, using oscillating potentials which are very close to monochromatic. For those potentials, it is often possible to move to a 'rotating frame' in which the interaction hamiltonian effectively becomes static, which makes the analysis much simpler.

Still, there's plenty of situations where none of this is valid, particularly if the probe is strong enough to get out of the perturbative regime. But even then, it is still important to have the structure of the system (i.e. the eigenstates of the interaction-free hamiltonian) at hand, as they are generally important parts of the analysis, even when they no longer play an explicit role in solving the TDSE.

If you want a deeper exploration of these themes, I recommend David Tannor's Quantum Mechanics: A Time-Dependent Perspective.

And finally,

Also, if we want to solve it numerically, it seems to me we also can't simplify using split-step Fourier transforms or into a form handled by Runge-Kutta. Is this correct?

No, it's not. Time-dependent potentials are perfectly solvable using the standard numerical methods. They might need a small bit of fine-tuning, but nothing more.

  • $\begingroup$ Right, of course I had forgotten about studying time-dependent perturbations. Could you name some examples where the probe would be out of the perturbative regime (or just a system that might be studied without perturbation theory)? As for numerical methods, I do not see how to use e.g. Runge-Kutta, since my understanding is that RK4 solves equations of the form $\partial_t \Psi = f(x,\Psi)$ but now we have $f(x,t,\Psi)$ since $V$ depends on $t$ in addition to $x$. $\endgroup$ Commented May 12, 2019 at 19:10
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    $\begingroup$ Good examples from my neck of the woods are high-order harmonic generation and above-threshold ionization in the tunnelling regime. Doubtless there are others. $\endgroup$ Commented May 12, 2019 at 19:17
  • $\begingroup$ Regarding numerical methods: are you seriously doubting that the TDSE can be solved numerically? If you've only been shown a restricted class of Runge-Kutta solvers, then go look for a text that deals with broader variants of the method. This google search is a good starting point - the zoo of methods for time-dependent QM is far too broad to mention here. Pretty much every method here, other than eigenvalue methods, can be used for time-dependent problems. $\endgroup$ Commented May 12, 2019 at 19:23
  • $\begingroup$ No, of course I am not doubting it can be solved numerically; just don't understand how the standard methods apply, which to me amounts to RK (I have very limited numerical PDE experience). Thanks for the resources. $\endgroup$ Commented May 12, 2019 at 19:38

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