# Determine numerical infinity for Schrodinger equation $−\psi''(x) + x^ 2 \psi(x) = E\psi(x)$

Consider the following Schrodinger equation for the harmonic oscillator with real $x$: $$−ψ''(x) + x^ 2 ψ(x) = Eψ(x).$$

I solve the last equation using shooting method and implicit Runge-Kutta integration. For the eigenvalue $E=1$, the corresponding eigenfunction looks like: where I set $x=4$ as $+∞$ and $x=-4$ as $-∞$ . From the Fig, we see that the eigenfunction exponentially decay at $x=\pm 4$ .

However, when eigenvalue $E=17$, the corresponding eigenfunction is too complicated so that it won't exponentially decays at $x=\pm 4$ as shown on the following Fig. In fact, the eigenfunction decays at $x=\pm 6$ as shown on the following Fig. My Question

The numerical infinity should be large (like $x=6$) for large eigenvalue $E$, and small (like $x=4$) for small eigenvalue. If independent from the eigenvalue, I uniformly set numerical infinity as a very large number, it would waste computational resource. So, is there any formula to determine the exact numerical infinity based on different inputted eigenvalue please?

• What do you call "decaying"? What is the right threshold for it? 1/10, 1/100, 1/1000... The exact numerical infinity is at inifinity, because that is the domain of the problem. On the other hand, how do you know the eigenvalues beforehand? – nicoguaro Apr 18 '16 at 17:28
• The eigenvalue can be analytically calculated with results: $E = 1, 3, 5, 7...$. I am sorry that due to low reputation, I only upload two figures. Now from the 3rd Fig, the eigenfunction is decayed in the sense that the derivative of the eigenfunction with respect to $x$ at the numerical infinity is almost zero. – pig Apr 18 '16 at 20:54
• The eigenvectors can be analytically calculated as well. – nicoguaro Apr 18 '16 at 20:55
• Please don't cross post. – nicoguaro Apr 18 '16 at 21:59
• Crossposting is discouraged. See this discussion. I am not a mathematician, but a lot of people in SciComp are... so I think that your claim is quite incorrect. – nicoguaro Apr 18 '16 at 22:15

You look at the question the wrong way. First, let us consider the exact problem and its solution: $$-\psi''(x) + x^2 \psi(x) = E\psi(x),$$ with boundary values $\psi(\pm\infty)=0$. Now you want to consider truncating your domain somewhere, i.e., you want to consider a different problem that reads $$-\tilde\psi''(x) + x^2 \tilde\psi(x) = E\tilde\psi(x),$$ with boundary values $\tilde\psi(\pm x_\infty)=0$ where $x_\infty$ is a finite termination point of your domain.

Clearly, these two problems will have different answers, $\psi$ and $\tilde\psi$. The question you are asking is, in essence, how large you have to choose $x_\infty$. But this cannot be answered without also saying how small you want the difference $\psi-\tilde\psi$ to be!

If you choose $x_\infty$ small, then you can solve the problem of course, but $\psi-\tilde\psi$ will be large. On the other hand, if you want the difference to be small, you will likely have to choose $x_\infty$ to be large. In general, if you require that $\|\psi-\tilde\psi\|\le\varepsilon$, then there will be some value $\bar x(\varepsilon)$ so that any $x_\infty\ge \bar x(\varepsilon)$ will be a valid choice of the domain size.

• If I understand your answer correctly, you suggest that I need numerically solve two equations simultaneously - the original equation and the complex conjugate of the original equation - afterwards, I compare the difference $\psi(x) - \bar \psi(x)$ from the two solutions at the chosen numerical infinity $x_\infty$. If the difference is not vanished, then I have to increase the chosen value for $x_\infty$ and numerically solve the two equations simultaneously again. Then do the comparison again. I think that your method may work, but numerically it is expensive. Anyway, I give you +1. – pig Apr 23 '16 at 6:25
• No, no, you misunderstand. $\bar\psi$ isn't the complex conjugate (I should have chosen a different notation, maybe $\tilde\psi$). It's just the solution of a different problem, on a finite domain, whereas $\psi$ is the solution on the infinite domain. Of course, numerically, you can't solve for $\psi$ because -- well -- the domain is infinite. – Wolfgang Bangerth Apr 23 '16 at 13:01
• I've edited the answer to use $\tilde\psi$ instead of $\bar\psi$. – Wolfgang Bangerth Apr 23 '16 at 13:03
• Thank you for your clarification! Looks like I need to do the following things: After numerically calculating $\tilde \psi(x)$, then due to exponentially decay (i.e. $\psi'(\infty) = 0$), $\tilde \psi'(x_\infty)$ must be vanished as well. Numerically, I can set $\tilde \psi'(x_\infty) < \epsilon$. If the last inequality is not satisfied, then I must increase $x_\infty$, then solve the ODE again and find the new solution $\tilde \psi(x)$ until the last inequality is satisfied. All looks well, except that I may never find $\tilde \psi(x)$ at first hand due to incorrect selection of $x_\infty$. – pig Apr 23 '16 at 22:19
• The Runge-Kutta integration may not guarantee to converge if $x_\infty$ is not appropriately chosen. If it's not converged, then $\tilde \psi(x)$ can not be obtained, consequently there is no way to do the following test $\tilde \psi'(x_\infty) < \epsilon$. But thank you for your answer! – pig Apr 23 '16 at 22:26

You should maybe look at the analytical form of these solutions (here's the wikipedia page), this would probably help you to derive a relation between the energy level $n$ and the typical length of your state (I would look at the $H_n$ functions given in the link above).

• It would be much better to determine the length of my state without knowing the analytical form. However, I give you +1, since your answer implies that I should look at asymptotic solution for the eigenfunctions. The asymptotic form must provide information where $\psi'(x)$ and $\psi(x)$ vanishes on the $x$. – pig Apr 18 '16 at 22:02