Suppose a particle in free space given by:

$$\psi(x,t) = Ae^{ik(x-\frac{\hbar k}{2m}t)} + Be^{-ik(x-\frac{\hbar k}{2m}t)}.$$

Why are numerical solutions necessary in order to plot this? Why can't values for $x$ be plugged in for each value of $t$ and plotted exactly for the real and imaginary components? Implementations I've found seem to use Euler or Runge-Kutta methods to approximately integrate from a differential rather than a solution. I'm missing something in my understanding.

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    $\begingroup$ In general you would need to numerically solve the SE. For this case you of course would not need to. More information would be needed here though. This is certainly a solution, but it is not the solution, as you have not specified initial conditions. Perhaps the implementations you have looked at are involving some additional assumptions you were unaware of. $\endgroup$ – Aaron Stevens Oct 15 at 0:14
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    $\begingroup$ It's completely unecessary to use numerics to solve for $\psi$ as you have written it. In fact, by virtue of you having written $\psi(x,t)$ in a closed form expression, there is no need for numerics. However, the Schrodinger equation is applicable to far more scenarios than just free space. If there is some external potential energy $V(c)$ than the eigenfunctions will be complicated and $\psi(x,t)$ might not have such a nice closed form expression. $\endgroup$ – user1379857 Oct 15 at 0:24
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    $\begingroup$ “Implementations I have found ...” Could you provide an example for some context? $\endgroup$ – J. Murray Oct 15 at 0:31
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    $\begingroup$ @JacksonCapper In that link you aren't looking at a free particle. There is a potential barrier that exists. Since the program has the option of modifying the barrier, it makes sense that it would implement a numerical solver in order to handle the various allowed cases. $\endgroup$ – Aaron Stevens Oct 15 at 0:59
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    $\begingroup$ Analytic solutions aren't possible for every ODE/PDE, so numerical methods are used $\endgroup$ – Kyle Kanos Oct 15 at 0:59

Why are numerical solutions necessary in order to plot this?

They aren't. Numerical solutions of the TDSE are only necessary when no analytical solution is known.

The only real-world reason for implementing a numerical solution of the Schrödinger equation when there are analytical solutions available that makes sense is when you're implementing a multi-purpose solver that needs to work robustly and consistently over a range of situations that includes potentials for which no analytical solution is known; it might then be impractical to switch back-and-forth from numerical to analytical solutions. (This is the case for the solver you've linked in the comments.)

Of course, one might also choose to solve numerically for situations where analytical solutions are known, as training for writing numerical solvers, or as testing for one of those, but those are not "necessary" in any sense.

If you have examples in front of you that don't seem to fit either of these patterns, it is on you to produce those examples.

  • $\begingroup$ I'd argue that testing your numerical routine with analytic solutions is necessary. But otherwise I'd agree with the rest of the post. $\endgroup$ – Kyle Kanos Oct 20 at 15:43

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