# Potential energy of particle in delta function potential

What is the potential energy of a particle in the single bound state $$\psi_b(x)=\frac{\sqrt{m\alpha}}{\hbar}e^{-\frac{m\alpha}{\hbar^2}|x|}$$ of the Dirac-delta potential well $$V(x) = -\alpha \delta(x)$$ which has total energy $$\langle H\rangle =E=-\frac{m\alpha^2}{2\hbar^2}$$? Wouldn't the potential energy become infinite?

As always in Quantum Mechanics, we discuss expected values of observable quantities, so it only makes sense to look at $$\langle V\rangle$$. We have: $$\langle V\rangle= \langle \psi_b\mid \hat V\psi_b\rangle = \int^{+\infty}_{-\infty} \overline\psi_b(-\alpha\delta(x))\psi_bdx = -\alpha\int^{+\infty}_{-\infty}|\psi_b(x)|^2\delta(x)dx = -\alpha |\psi_b(0)|^2=-\frac{m\alpha^2}{\hbar^2}$$ This also implies that the expected kinetic energy $$T$$ is: $$\langle T\rangle = \langle H - V\rangle = -\frac{m\alpha^2}{2\hbar^2}+\frac{m\alpha^2}{\hbar^2} = \frac{m\alpha^2}{2\hbar^2}$$ which turns out to be $$-E$$.