Correct me if I'm wrong, but equations in QM are quite always obtained by looking at the energy dependance of the problem of interest. For example, for Schrodinger's equation one just uses

$E = \frac{p^2}{2m}$

which, when translated into the language of operators, gives the known formula (for a free particle). The same happens with Klein-Gordon. One starts with

$E^2 = m^2c^4 + c^2 p^2$

and then, again, by translating to operators arrives at the KG equation (for a free particle, again).

Here comes the question, When using De Broglie relations

$E = \hbar \omega$

$p = \hbar k$

One can found the dispersion relation of the solutions easily by solving the equations or by replacing this identities in the definition of energy. What happens in the presence of a potential? Does the dispersion relation changes? Can we make, using the right potential, solutions for the Schrodinger equation that are not dispersive?

  • $\begingroup$ This is mostly beyond me, but if you look up Kramers-Kronig related material on wikipedia, it might be of help. $\endgroup$
    – user108787
    Oct 19, 2016 at 12:31

1 Answer 1


When you have a position dependent potential $V$, you have no simple wave solutions to the Schrödinger equation anymore and, in general, the de Broglie relations do not hold. However, if you have a potential that varies slowly with position, there exists the so called Wentzel-Kramers-Brillouin (WKB) approximation for the solution of the Schrödinger equation, which leads to quasi-sinusoidal wave functions with wavelengths and amplitudes that change slowly with position. This is frequently used for obtaining approximate analytical (quasi-classical) solutions, e.g., for tunneling probabilities. More about the WKB approximation, you will find here https://en.wikipedia.org/wiki/WKB_approximation .


Your Answer

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

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