0
$\begingroup$

We know that the number of bound states for an attractive delta potential is one. If so what will the number of bound states for a particle in a repulsive delta potential? If $V(x)= +a \cdot \delta(x)$?

$\endgroup$
  • 8
    $\begingroup$ How could a repulsive force bind anything? $\endgroup$ – G. Smith Aug 15 '19 at 17:49
3
$\begingroup$

There is no bound state for the positive Dirac potential. Look at the derivation of the bound state in e.g. these lecture notes. Formula (11.15) for the wave function of the bound state $$\Phi(x,t)=\frac{\sqrt{am}}{\hbar}e^{-\frac{am|x|}{\hbar^2} + i \frac{a^2 m t}{2\hbar^3} }$$ is derived without assuming $a>0$. But if a negative $a$ is inserted in the formula it becomes un-normalisable because of the first spatial term inside the exponential is now positive, implying a growing amplitude away from the spike. Hence demanding normalisation implies no bound state in this case.

$\endgroup$
  • $\begingroup$ Thank you sir. Answer is really helpful $\endgroup$ – Abin Philip Aug 16 '19 at 12:16

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.