In the 3rd volume of the Landau-Lifschitz text book ("Quantum mechanics") problem 3 after section 50 studies the Schroedinger equation on the line with symmetric double well potential, e.g. $U(x)=g(x^2-a^2)^2$ with $g,a>0$. In particular they estimate the difference between the first two lowest energy levels using the method of quasi-classical approximation. (The final result is not essential to my question, nevertheless see p. 184 in the 3rd edition of the book.)

As far as I understand, the essential point in the argument was that the potential is symmetric, i.e. $U(-x)=U(x)$. Apparently it is even more important that the potential $U$ has the same value in the two local minima (which are also the global minima).

I am wondering what happens in the case of the non-symmetric double well potential which has two local minima and assumes at them different values. For example I would be interested to see a proof of the following result I heard: if the wave function of a particle is localized near the local minimum where $U$ is larger (called a fake vacuum or metastable state) then after a long time the particle will be localized near the other local minimum (the true vacuum) due to tunneling.

I heard that metastable states are studied sometimes using eigen-functions of the Hamiltonian with non-real eigenvalues. A discussion of that would be of particular interest to me as well as a comparison with the standard approach using the usual spectrum of the Hamiltonian.


Cute and interesting question.

What really happens is that the stable points are the global minima (if more than one) of the Hamiltonian, i.e. the states having the lowest eigenvalues.

The effect of tunneling can be seen by using the WKBJ method (just as is done when estimating the life time of alpha decays).

Thanks for the question, I hope my comment has been useful.

  • $\begingroup$ Do you have a reference for more details? Thanks. $\endgroup$
    – MKO
    May 18 '18 at 13:40

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.