Are there analytic solutions to the time-Dependent Schrödinger equation, or is the equation too non-linear to solve non-numerically?

Specifically - are there solutions to time-Dependent Schrödinger wave function for an Infinite Potential Step, both time dependent and time inpendent cases?

I have looked, but everyone seems to focus on the time-Independent Schrödinger equation.

  • $\begingroup$ Comment to the post (v2): In the TDSE, does the Hamiltonian have explicit time-dependence or not? $\endgroup$
    – Qmechanic
    Sep 7, 2017 at 10:56

4 Answers 4


The complete solution for the time dependent equation with an infinite potential step is found by the method of images. Given any initial wavefunction

$$ \psi_0(x) $$

for x<0, you write down the antisymmetric extension of the wavefunction

$$ \psi_0(x) = \psi_0(x) - \psi_0(-x) $$

And you solve the free Schrodinger equation. So any solution of the free Schrodinger equation gives a solution for the infinite potential step. This is not completely trivial to make, because the solutions do not vanish in any region. But, for example, the spreading delta-function

$$ \psi(x,t) = {1\over \sqrt{2\pi it}} e^{-(x-x_0)^2\over it } $$

Turns into the spreading, reflecting, delta function

$$ \psi(x,t) = {1\over \sqrt{2\pi it}} e^{-(x-x_0)^2\over it } - {1\over \sqrt{2\pi it}} e^{-(x+x_0)^2\over it } $$

You can do the same thing with the spreading Gaussian wavepacket, just subtract the solution translated to +x from the solution translated to -x. In this case, normalizing the wavefunction is hard when the wavefunction start out close to the reflection wall.

Time independent infinite potential wall

The solution to the time independent problem of the infinite potential wall are all wavefunctions of the form

$$ \sin(kx) $$

for all k>0. Superposing these solutions gives all antisymmetric functions on the real line.

To find this solution, note that the time independent problem (eigenvalue problem) for the Schrodinger equation is solved by sinusoidal waves of the form $e^{ikx}$, and you need to superposes these so that they are zero at the origin, to obey the reflection condition. This requires that you add two k-waves up with opposite signs of k and opposite sign coefficients.

The opposite sign of k just means that the wave bounces off the wall (so that k changes sign), while the opposite sign of the coefficient means that the phase is opposite upon reflection, so that the wave at the wall cancels.

General solution

The time dependent problem for a time independent potnetial is just the sum of the solutions to the time independent problem with coefficients that vary in time sinusoidally.

If the eigenfunctions $\psi_n$ are known, and their energies $E_n$ are known, and the potential doesn't change in time, then the,

$$ \psi(t) = \sum_n C_n e^{-iE_n t} \psi_n(x) $$

is the general solution of the time dependent problem. This is so well known that generally people don't bother saying they solved the time-dependent problem once they have solved the eigenvalue problem.

The general solution of the time-dependent Schrodinger equation for time dependent potentials doesn't reduce to an eigenvalue problem, so it is a different sort of thing. this is generally what people understand when you say solving the time-dependent equation, and this reflects the other answers you are getting. I don't think this was the intent of your question, you just wanted to know how to solve the time dependent equation for a time independent potential, in particular, for an infinite reflecting potential wall. This is just the bouncing solution described above.

  • $\begingroup$ Full marks Ron, this was just what I was trying to understand. Ironic that its so trivial no one mentions it. Can you recommend a text or source for this? $\endgroup$
    – metzgeer
    Dec 1, 2011 at 1:09
  • $\begingroup$ I don't think there is any more to say about it that what I said above, also, look here: physics.stackexchange.com/questions/12611/… . The grandaddy of all exact solution for a time independent problem is the Gaussian wavepacket, which you can work out most easily from knowing the stochastic version. I worked it out here: en.wikipedia.org/wiki/User:Likebox/Schrodinger , it used to be on Wikipedia, before that project degenrated. I can't recommend any other literature on elementary stuff, unfortunately. $\endgroup$
    – Ron Maimon
    Dec 1, 2011 at 1:41
  • $\begingroup$ @metzgeer: It doesn't contain this problem specifically (I think, it may since have been updated), but I highly recommend Introduction to Quantum Mechanics by Griffiths. It contains some introductory examples and problems to work though which are very helpful for solidifying the methods used in quantum mechanics at this level. It is also in a conversational tone that will probably appeal to you if you enjoy using SE. :) $\endgroup$
    – qubyte
    Dec 1, 2011 at 4:15
  • $\begingroup$ @MarkS.Everitt: Oh thank Saint Albertus Magnus the patron saint of scientists, for physics books at a conversational tone - or I would never make it :) $\endgroup$
    – metzgeer
    Dec 1, 2011 at 10:22
  • $\begingroup$ @metzgeer: Eek, sorry I hope I didn't cause any offence. I recommend it because it's my favourite! $\endgroup$
    – qubyte
    Dec 1, 2011 at 10:32

Indeed, analytic solutions to the time-dependent Schrodinger equation are rare, much rarer than that for the time-independent Schrodinger equation.

A paradigm is the time-dependent harmonic oscillator. Then you have the Landau-Zener-Stueckelberg problem, the Demkov-Osherov problem.

Another case I can recall is the Stey-Gibberd problem, which is introduced pedagogically in the paper. This model is simple and beautiful, yet unfortunately little known.

An artificial model (Luttinger-like) realizable in the one-dimensional tight binding model is also exactly solvable.

But overall, the list is short.


The equation is analytically solveable if you allow the potential to vary in magnitude but keep the borders fixed. Then, you can assume the form $\sum_{n} A_{n}(t) \sin\left(\frac{\pi n x}{L}\right)$ for the wavefunction.

Substituting this form of the wavefunction into $\frac{\hbar}{i}\frac{d\psi}{dt}=-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}+V(t)\psi$ results in:

$$0=\sum_{n} \sin\left(\frac{\pi n x}{L}\right)\left[ \frac{\hbar \dot A_{n}}{i} - \frac{\hbar^{2}A_{n}\pi^{2}n^{2}}{2mL} +V A_{n} \right]$$

Since each sine term has independent nodes and antinodes, each of the enclosed factors must be independently zero. The solution integrates to:

$$A_{n}=a_{n}\exp\left(\frac{\pi^{2}\hbar n^{2} i t}{2mL^{2}}\right)\exp\left(\frac{i}{\hbar}\int V(t) dt\right)$$

Where the $a_{n}$ are constants. Note that this differs from the standard infinite square well solution only by the second factor involving the integral of the potential energy. Also, note that there is nothing that depends on $n$ in this term, so this solution can be pulled out of the sum entirely, and thus simply multiplies the old wavefunction by an overall phase, and generates a physically identical wavefunction to our old, constant-$V(t)$ solution.

I'm almost certain that exact anlaytic solutions exist for less trivial solutions, but varying the potential on the infinite square well doesn't do much, in the end.

  • $\begingroup$ This is for a infinite potential well isn't it, I'm trying to understand an infinite potential step, do I simply increase the width of the well to let L approach infinity? $\endgroup$
    – metzgeer
    Nov 30, 2011 at 4:08
  • $\begingroup$ @metzgeer: I don't understand what an infinite potential step is--do you mean a potential that is $V=V(t)$ for $x > 0$, and $\infty$ otherwise? $\endgroup$ Nov 30, 2011 at 16:46
  • $\begingroup$ you're quite right jerry. I was thinking of a time varying $\psi(x,t)$ variable, not in terms of V(x,t) I should have said V(x) = $\infty$ for all time. Mea culpa. $\endgroup$
    – metzgeer
    Dec 1, 2011 at 1:00
  • $\begingroup$ @metzgeer: in that case, the time-independent schrodinger solutions, with each basis function multiplied by $\exp(iEt)$ gives you the correct time-dependent wavefunction. This will be true for all potentials that don't depend on the time. $\endgroup$ Dec 1, 2011 at 12:18

The universe has a sense of humour, so I cannot resist supplementing a question
which has the word « analytic » in its title... in a minute, you will see why I would have preferred you to use the phrase « closed-form », with the following observation--answer.

For any time-independent potential $V(x)$, let $H = -{\partial^2\over\partial x^2} + V(x)$. (What I am about to say works for any time-independent $H$, and if a system is isolated, the Hamiltonian is always time-independent even though Schroedinger's equation is time-dependent.) Suppose the system starts in the initial state $\psi_o$, a wave function of $x$ of course. Then the following analytic function of time gives the solution to the time-dependent Schroedinger equation:

$$\psi(x,t) = e^{itH}\cdot\psi_o.$$

This is similar to the last formula given by Mr. Maimon which told you, as he explained, how to get the solution to the time-dependent Schroedinger equation once you have solved the time-independent equation for all its eigenvalues and eigenstates. The difference is that in that formula, $E$ was one of the many eigenvalues of the operator $H$, but here we can simply plug in $H$ as an operator into the holomorphic (analytic) function $e^z$ (one way is by using the power series of this analytic function).

Hence the universes' pun between analytic as a synonym for closed-form expression but also as a synonym for holomorphic which means it can be expressed as a convergent power series and extended as a function to the entire complex number plane, which is done in some approaches to Quantum Field Theory, e.g., by Streater and Wightman, and in some approaches to path integrals.

This approach is less practical for your specific situation than the answer by Mr. Maimon which is well-adapted to your specific problem...but it gives a closed-form formula, generalises well even to potentials $V$ with singularities, infinities, etc., and sometimes can help you think about the physics of the problem without getting lost in the gory details of calculating the answer.

I have often wondered whether it can be extended to time-varying potentials...I suspect it could be...

  • 1
    $\begingroup$ It can only be generalized as $\psi(x,t) = \exp (-i \int H(t') dt' ) \psi(x,0)$ if the Hamiltonian commutes with itself at different times, $[H(t),H(t')] = 0$. $\endgroup$
    – perplexity
    Jan 19, 2012 at 13:56

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