7
$\begingroup$

Consider the (quantum) Hamiltonian on the real line $$H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x).$$

Let us assume that the potential $V$ is an even smooth functions with exactly two non-degenerate minima, and $\lim_{|x|\to +\infty}V(x)=+\infty$. Such a $V$ is called double well potential.

Let $E_1,E_2$ be the first two minimal eigenvalues of $H$. It is known in physics literature (see problem 3 after $\S$ 50 in Landau-Lifshitz, vol. 3) that under some extra assumption which are not quite specified there one has $$E_2-E_1 \approx \frac{\omega\hbar}{\pi} \exp(-\frac{1}{\hbar} C), $$ where $\omega, C$ are positive constant which can be written down explicitly, and the result is understood asymptotically as $\hbar\to 0$.

I am looking for a mathematically more rigorous discussion of this result where, in particular, the assumptions are formulated more explicitly.

$\endgroup$
6
  • 2
    $\begingroup$ I take it you mean $\lim_{|x|\to\infty}V(x)=+\infty$ in your definition of a double-well potential, right? $\endgroup$
    – gmvh
    Apr 25, 2021 at 10:42
  • $\begingroup$ Oops... Corrected. Thanks. $\endgroup$
    – asv
    Apr 25, 2021 at 11:07
  • $\begingroup$ I presume you mean the WKB approximation for the level splitting? $\endgroup$ Apr 25, 2021 at 13:22
  • 2
    $\begingroup$ I don't understand the term "nondegenerate". That usually means that there are two things that are not equal (possibly the values of $V$ at the two minima in the present context). Since $V$ is an even function, I suspect it's more likely that something is degenerate here as opposed to nondegenerate? $\endgroup$ Apr 25, 2021 at 15:01
  • 2
    $\begingroup$ @MichaelEngelhardt : By non-degenerate minima I mean that the second derivatives at the minima is positive. $\endgroup$
    – asv
    Apr 25, 2021 at 18:02

2 Answers 2

9
$\begingroup$

This is either

Helffer-Sjostrand

https://www.tandfonline.com/doi/abs/10.1080/03605308408820335

or Barry Simon

https://www.jstor.org/stable/pdf/2007072.pdf?refreqid=excelsior%3A258084917fff9e0c10088abbb2679c55

PS: I cannot resist pointing out that the last paper is in Annals of Mathematics proving, as you notice yourself, an exercise in Landau/Lifshitz.

$\endgroup$
4
$\begingroup$

Double wells, by Evans M. Harrell, Comm. Math. Phys. 75, 239 (1980), should be sufficiently rigorous. The result for the tunnel splitting, theorem 2.22, applies to a symmetric double-well potential in arbitrary number $n$ of dimensions, constructed as the sum $V(x-f)+V(x+f)$ of two single-well potentials $V(x)$, bounded and of compact support, after translation by $\pm f$.

The tunnel splitting is given for large $f$ in terms of the eigenfunction $\Phi(x)$ for the lowest eigenstate in the potential $V(x)$, $$E_1-E_2=\left(\int\Phi(x-f)[V(x-f)+V(x+f)]\Phi(x+f)\,dx\right)(1+o(f^{-n})).$$

$\endgroup$
3
  • $\begingroup$ In my question the potential is symmetric (even). By non-degenerate I meant that the second derivative at the minima is positive. $\endgroup$
    – asv
    Apr 25, 2021 at 18:06
  • 2
    $\begingroup$ I think from a formal point of view, the problem here is that $h\to 0$ in the OP doesn't in an obvious way correspond to pushing the wells apart by sending $f\to\infty$. Rather, it seems equivalent to a rescaling $V(x)\to V(hx)$ of a potential (which does push them out to infinity, but also changes their shape). $\endgroup$ Apr 25, 2021 at 22:12
  • $\begingroup$ @ChristianRemling --- true, thanks for pointing this out; perhaps from the physics point of view a scaling with the separation of the wells is somewhat more natural than a scaling with Planck's constant... $\endgroup$ Apr 26, 2021 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.