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.

  • 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. $\endgroup$
    – makt
    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. $\endgroup$
    – makt
    Apr 25, 2021 at 18:02

2 Answers 2


This is either



or Barry Simon


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.


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 $V(x)$, $$E_1-E_2=\left(\int\Phi(x-f)[V(x-f)+V(x+f)]\Phi(x+f)\,dx\right)(1+o(f^{-n})).$$

  • $\begingroup$ In my question the potential is symmetric (even). By non-degenerate I meant that the second derivative at the minima is positive. $\endgroup$
    – makt
    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, privacy policy and cookie policy

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