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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.

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The time dependent equation for a time independent potential is solved by superpositions of the solutions to the time-independent problem, with the coefficients varying as exp(iEt). The equation is completely linear. – Ron Maimon Nov 30 '11 at 4:20
@RonMaimon: Not so simple. If this were the case, there'd be no reason to ever make a rotating wave approximation. Time dependence is particularly troublesome when there is no interaction picture which removes it. In an example case of a sinusoidal drive on a low dimensional system, there is no rotating frame without time dependence, and in general no analytic solution. Having said that, even in the time independent case you are fortunate if your ODE has an attainable analytic solution beyond a hand-wavey "it's just a matrix exponential". – qubyte Nov 30 '11 at 17:09
@Mark: Yes it is so simple, your examples are completely irrelevant. No need muddy this simple question--- it's just asking about the time-dependent equation for a time independent potential. A sinusoidal drive is not time independent. Rotating frames produce time independent forces in the rotating frame, a global centrifugal potential and a coriolis magnetic-type force, so they can be made time independent. Lots and lots of problems have analytic solutions if you start with an known ground state and vary coefficients. – Ron Maimon Nov 30 '11 at 18:08
@RonMaimon: Calm down there. I misread the start of your comment. In any case I'm not "muddying the question" since the comment was clearly directed at you. Anyway, in my experience it's only the special cases that can be solved. Throw a few small systems together and you've got a many body problem (unless you have a lot of symmetry). As we rarely deal with just a few two level systems, that pretty much describes everything unless you're willing to approximate. But I digress... – qubyte Nov 30 '11 at 18:22
@Everitt: I agree. I didn't mean to sound abrasive, just haven't had my coffee yet. – Ron Maimon Nov 30 '11 at 18:35

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.

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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? – metzgeer Dec 1 '11 at 1:09
I don't think there is any more to say about it that what I said above, also, look here:… . 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: , it used to be on Wikipedia, before that project degenrated. I can't recommend any other literature on elementary stuff, unfortunately. – Ron Maimon Dec 1 '11 at 1:41
@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. :) – qubyte Dec 1 '11 at 4:15
@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 :) – metzgeer Dec 1 '11 at 10:22
@metzgeer: Eek, sorry I hope I didn't cause any offence. I recommend it because it's my favourite! – qubyte Dec 1 '11 at 10:32

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.

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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? – metzgeer Nov 30 '11 at 4:08
@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? – Jerry Schirmer Nov 30 '11 at 16:46
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. – metzgeer Dec 1 '11 at 1:00
@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. – Jerry Schirmer Dec 1 '11 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...

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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, $[H(t),H(t')] = 0$. – perplexity Jan 19 '12 at 13:56

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