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I was reading Pilot wave theory and De Broglie–Bohm theory pages on Wikipedia that I found how similar they are comparing to classical physics and I wondered what happens if we just replace the density with a delta function,

$$\rho=q \delta^3(\pmb r-\pmb r') =\frac{-q}{4\pi} \nabla.(\nabla \frac{1}{\vert\pmb r-\pmb r'\vert} ) $$

On the wave function:

$$\Psi=\sqrt \rho \mathit exp (iS/\hbar)$$

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  • $\begingroup$ FWIW, a square root of a Dirac delta function does not make sense. $\endgroup$
    – Qmechanic ♦
    Commented Dec 9, 2020 at 21:23
  • $\begingroup$ @Qmechanic Maybe you are right but I think we can define the square root of a Laplacian. $\endgroup$
    – Martin Spinoza
    Commented Dec 10, 2020 at 5:56
  • $\begingroup$ @MartinSpinoza, you can, but it has nothing to do with $\sqrt{\rho}$. The square root of the Laplacian is taken in the sense of composition of linear operators. $\endgroup$
    – fqq
    Commented Dec 10, 2020 at 17:45

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