I came across this confusion when trying to numerically solve the time-dependent Schrodinger equation.

The short version of my question is: when solving for time-dependent wavefunctions $\psi(t,x)$ on paper, why do we not need boundary conditions in addition to initial condition $\psi(0,x)=$some function of $x$?

My thoughts: we are told that in order to solve a time-dependent Schrodinger equation we only need to specify a normalized initial wavefunction. However the Schrodinger equation looks extremely like a heat equation, and yet when solving the heat equation we do need to specify both initial temperature distribution as well as some boundary conditions. So what is causing us to treat these two equations differently? Is it the fact that in the Schrodinger case we have a normalization condition for the wavefunction, which is like a non-local boundary condition? Another way of saying this is that we are imposing boundary conditions at infinity? Or is there something I am misunderstanding in my logic?

A follow up question: does this mean that naive implementation of NDSolve[] in Mathematica is not enough to deal with the Schrodinger equation as you will always get an error due to missing boundary conditions?

The problems I have in mind are wavepacket evolution in scattering and quantum tunneling. In these cases, I prepare an initial wavepacket for $\psi(0,t)$ (I am not restricting to any wavefunction ansatz) and just evolve numerically to obtain $\psi(t,x)$. So it seems that we need not specify any boundary conditions, or am I missing something?

  • 1
    $\begingroup$ Who says you don't need boundary conditions? $\endgroup$ Mar 28, 2022 at 4:22
  • $\begingroup$ I am just really confused where it is hidden. For example in a bound state wavepacket evolution (e.g. initial coherent state in a harmonic potential) we might argue that the boundary conditions are imposed at infinity. But what about a wavepacket real-time scattering problem? Where do you impose boundary conditions if we don’t know what happens at future infinity (which is what we are trying to predict in the first place)? $\endgroup$ Mar 28, 2022 at 16:17
  • $\begingroup$ You're trying to find solutions for the given boundary conditions. If you don't know what the boundary conditions are then you would have to introduce that as parameters in your model as well. But when you're solving an instance of the equation you're still going to be assuming boundary conditions. Is this what you are getting at? $\endgroup$ Mar 28, 2022 at 18:59

1 Answer 1


The boundary conditions are imposed when solving the time independent Schrodinger equation, depending on what potential the particle exists in. Take the 1D particle in a box - the solutions for $\psi(x)$ are well known and depend on the boundary conditions of the box: $$\psi(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$

Now for the time dependent Schrodinger equation, we can simply tack on a temporal function ($e^{-iE_n t/\hbar}$) to our $\psi(x)$, remembering that a full solution is a superposition of states. This gives $$\Psi(x,t) = \sum_{n=0}^\infty a_n \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)e^{-iE_n t/\hbar}$$

where the $a_n$ for the 1D well can be found from the normalization condition $$\int_{-\infty}^\infty \left|\Psi(x,t)\right|^2 dx = \sum_{n=1}^\infty| a_n|^2 = 1$$

We must have an initial state (usually $\Psi(x,0)$) to determine time evolution.

Not sure if it will be helpful or not, but I've attached an image of some mathematica code regarding the above and the TDSE for a 1D well.

enter image description here

  • $\begingroup$ A appreciate you writing such a long answer. I am aware that TISE implements boundary conditions. However I am trying to understand why we do not seem to implement boundary conditions for arbitrary potentials, specifically in the scattering case. For example, one could prepare an initial coherent state wave-packet (which is normalizable, contrary to plain waves), let it go through a lump shaped potential and ask what is the shape of the wave function at some time. In this case, you wouldn't have the luxury to use analytic eigenstates. However, you only need to specify a $\psi(0,x)$. $\endgroup$ Mar 28, 2022 at 4:05
  • $\begingroup$ I think that given an initial state, and a known potential, this will determine your $\psi(x)$ and thusly the $\Psi(x,t)$. It is all dependent on the potential in question. When discussing arbitrary potentials, or time dependence outside of a known closed system, perhaps we can turn to time-dependent perturbation theory, which is how most systems are tackled which do not have an analytic solution. $\endgroup$
    – bleuofblue
    Mar 28, 2022 at 4:44
  • $\begingroup$ Yes but then we are back to something like the s-matrix program which is not very suitable for describing real-time evolution or non-perturbative effects like quantum tunneling. I have a hunch that in this case boundary conditions are imposed at infinity or maybe the classical turning point. However, how would we impose boundary conditions like $\psi(t,\infty)$ without knowing the future is where I’m not sure about. $\endgroup$ Mar 28, 2022 at 16:27

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