6
$\begingroup$

Given a certain Hamiltonian, for example,

$$ H = -\frac{1}{2}\frac{\partial^2}{\partial x^2 } + x^4 . $$ , what methods can we use to approximate the $n$th eigenvalue, for very large $n$?

For small $n$ -- the ground state and the first few low-energy states -- the eigen-energies are easy to calculate, say, with numerical methods, perturbation theory, or even the variational method. But the standard literature of QM doesn't cover the calculation of very high eigen-energies.

Is there any reliable method to approximate $E_n$ for large $n$, preferably with some control over the error?

$\endgroup$
  • $\begingroup$ Hi John, I edited your (interesting) question with the hope that it gets reopened (I changed the wording so that it doesn't sound like a HW question). If you don't like the edit, you can roll it back by clicking on the edit button. $\endgroup$ – AccidentalFourierTransform Dec 18 '16 at 12:58
  • $\begingroup$ @AccidentalFourierTransform You changed the question in quite ignorant way and someone also deleted my comments - for large energy level number WKB usually works very well and it IS mentioned in most standard courses. If I remember correctly, exactly for $x^4$ it gives error of order $0.01$% already at $n=10$ or something like that! I would convert it into proper answer when I'll get opportunity to go through literature for error estimates. $\endgroup$ – OON Dec 18 '16 at 14:49
  • $\begingroup$ @OON You cannot convert that into a proper answer because the question is closed, as you can see. My edit, while rather drastic, was intended to get the question reopened so that you, or anyone else, can actually answer it. $\endgroup$ – AccidentalFourierTransform Dec 18 '16 at 14:58
  • $\begingroup$ @AccidentalFourierTransform I understand that and by all means welcome anything to reopen it. Sorry that my comment sounded harsh. However I don't get why comments were deleted and wanted to note that "But the standard literature of QM doesn't cover the calculation of very high eigen-energies." is not correct. $\endgroup$ – OON Dec 18 '16 at 15:03
  • $\begingroup$ @OON Comments are meant to be transient, and are routinely deleted once they have served their purpose. Don't put information of long-term value in a comment and expect it to stay there. $\endgroup$ – David Z Dec 21 '16 at 20:21
3
$\begingroup$

As @OON mentioned, the WKB is the archetypal tool in the cottage industry studying this system, starting with Bender & Woo 1969 and going on to Cornwall & Tiktopoulos 1993 and on and on...

To leading order, the energy in WKB is well-known to simply amount to the Bohr-Sommerfeld-Wilson quantization condition. You correctly infer that $\hbar$ is superfluous and absorbable in the variables to 1, but let's keep it here, $$ \left (-\frac{\hbar^2}{2} \partial_x^2 +x^4-E_n\right ) \psi_n(x)=0, $$ to connect to 101 yr old formulas, namely, the semiclassical phase integral with turning points at $x=\pm E_n^{1/4}$, $$ \int_{-E_n^{1/4}}^{E_n^{1/4}}dx \sqrt{2(E_n-x^4)}= (n+1/2)\pi \hbar, $$ for n=0,1,2,3,...,10000, ... (The 1/2 is the so-called Maslov correction, cf Hall ch 15.2 .)

This is immediately solved to $$ E_n=\left ( \frac{(n+1/2)\pi \hbar}{\sqrt{2} C} \right )^{4/3} ~, $$ where $C\equiv \int_{-1} ^1 dy \sqrt{1-y^4}= B(1/4,3/2)/2= \sqrt{\pi}~ \Gamma (5/4)/\Gamma (7/4) \sim 1.748$, so might skip the 1/2 for your chosen n, for which the "many wavelengths in the well" part of the approximation works all too well.

In fact, if you pursued the references, you'd see that this system is the prototype laboratory for WKB asymptotics methods.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.