# Why do we not include contributions from $\delta(x)$ in the energy eigenvalue of the delta potential bound state?

Consider a delta function potential $$V(x)=-\alpha \delta(x)$$. There is a bound state

$$\psi(x) = \frac{\sqrt{m\alpha}}{\hbar} \exp\left(-\frac{m\alpha|x|}{\hbar^2}\right),$$ with energy eigenvalue $$E=-\frac{m\alpha^2}{2\hbar^2}.$$

However, if we plug the wavefunction into $$E=\langle\psi|H|\psi\rangle,$$ then we get $$E=\left\langle\psi\middle|\frac{p^2}{2m}\middle|\psi\right\rangle-\alpha\langle\psi|\delta(x)|\psi\rangle\\ = -\frac{m\alpha^2}{2\hbar} - \frac{m\alpha}{\hbar^2}\int \exp\left(-\frac{2m\alpha|x|}{\hbar^2}\right)\delta(x)\, dx\\ =-\frac{3m\alpha^2}{2\hbar^2},$$

which is not equal to the eigenvalue. So how do we get around not including the contribution from $$\delta(x)$$ in $$\langle\psi|H|\psi\rangle$$?

• How is $\langle p^2/2m\rangle$ negative? Mar 30, 2021 at 9:46
• exactly. In the final formula, your first term should be positive, so in the end you get $-m\alpha^2/2\hbar^2$. Mar 30, 2021 at 9:49
• By my calculations, the first term is positive as already pointed out by previous comments. Remember that the second derivative of $|x|$ at the origin is proportional to $\delta(x)$ so you get an additional contribution from there. Mar 30, 2021 at 9:52
• To clarify, the integrals in the regions $x>0$ and $x<0$ separately give $- \frac{m \alpha^2}{4\hbar^2}$ each which is what probably gave you the $-\frac{m \alpha^2}{2\hbar^2}$ that you have. The additional contribution from the Dirac delta function at the origin is $\frac{m \alpha^2}{\hbar^2}$ for a total of $+\frac{m \alpha^2}{2\hbar^2}$ Mar 30, 2021 at 9:56

The derivative of $$\left|x\right|$$ is not $$1$$, as seems to be required for the calculation of your eigenvalue to be $$-\frac{m\alpha^{2}}{2\hbar^{2}}$$
Instead, the right derivative in the sense of distributions, is, $$\frac{d}{dx}\left|x\right|=\theta\left(x\right)-\theta\left(-x\right)$$ with $$\theta$$ being Heaviside's step function. When differentiating again, you get, $$\frac{d^{2}}{dx^{2}}\left|x\right|=\delta\left(x\right)+\delta\left(-x\right)$$ You should be alright if you redo the calculation with these prescriptions.