Consider the potential $V(x)=\frac{2}{x^2}$ and let $\frac{\hbar^2}{2m}=1$ for convenience. Now consider the function $\psi(x)=\delta(x)$. According to Griffiths (electrodynamics book) problem 1.45(a), $$x\delta'(x)=-\delta(x)\tag{1}.$$ I'm not sure if I can do this but if I write



The Schrodinger equation now looks like

\begin{align} &-\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E \psi(x)\tag{4}\\ &-\frac{2\delta(x)}{x^2}+\frac{2}{x^2}\delta(x)=0 \delta(x).\tag{5} \end{align}

So it looks like $\delta(x)$ is an eigenstate with eigenvalue zero. But this goes against my intuition and is probably wrong but I'm not sure where the fault lies. Is it the derivative of the delta function? Can energy (eigenvalue of Hamiltonian) be zero? Potential is maximum at $0$, so how can the probability be maximum at $0$?

  • $\begingroup$ Your trial wavefunction is not normalizable. To find your bearings before getting lost in distributions, you might consider a narrow Gaussian. $\endgroup$ Apr 3, 2019 at 19:06
  • $\begingroup$ Something strange but I cannot tell it precisely. $\delta(x)$ is the eigenstate of position operator. But it cannot simultaneously be the eigenstate of the Hamiltonian since $[\hat{H}, \hat{x}] \neq 0$. $\endgroup$
    – K_inverse
    Apr 11, 2019 at 8:39

1 Answer 1

  1. OP's eq. (1) is a well-known identity in distribution theory. However, the expression $$\frac{\delta(x)}{x}$$ in OP's eq. (2) is mathematically ill-defined.

  2. If one considers a function times a Dirac delta distributions $$f(x)\delta(x),$$ then (as a minimum$^1$ requirement) the function $f$ should have a limit from left and right at $x\!=\!0$, so that it makes sense to replace $f(x)\delta(x)$ with $$\frac{f(0^+)+f(0^-)}{2}\delta(x).$$


$^1$ Mathematics textbooks on distribution theory typically assumes that $f\in C^{\infty}(\mathbb{R})$ is smooth, but that is too restrictive for many physics applications.


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