According to this page of "Quan­tum Me­chan­ics for En­gi­neers" by Leon van Dom­me­len, for one particle the integral over all space:

$$I=\int (\nabla\psi)^2 + V\psi^2 \mathrm{d}^3\mathbf{r}\tag{1}$$ with $\psi$ a real one-particle wavefunction and $V$ a real potential, is the same as:

$$I'=\int (\nabla|\psi|)^2 + V|\psi|^2 \mathrm{d}^3\mathbf{r}.\tag{2}$$

What happens to points where the wavefunction changes sign?

I assume also, that the potential does not include spin and velocity terms, but I'm not sure.

Why should these integrals be equal? This argument is from a series of statements to prove that the ground state can be real, positive and unique. It seems trivial but I require this integral for a more serious problem.

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    $\begingroup$ To answer "what happens to points where the wavefunction changes sign?", the article shows here ww2.eng.famu.fsu.edu/~dommelen/quantum/style_a/… that the sign cannot changes, and thus it's always the same. $\endgroup$ – rnels12 May 18 '18 at 11:12
  • $\begingroup$ Yeah @mels12 but that requires the above. $\endgroup$ – Mauricio May 18 '18 at 11:36

Leon van Dom­me­len is making the claim that for a differentiable real wave function $\psi:\mathbb{R}^3\to \mathbb{R}$, we have$^1$ $$ \nabla |\psi| ~=~{\rm sgn}(\psi)\nabla \psi, \tag{A}$$ so that the kinetic energy term in eqs. (1) & (2) is unchanged by replacing $\psi\to|\psi| $. He concludes that if a real wavefunction $\psi$ minimizes the expectation value $\frac{\langle \psi |\hat{H}|\psi\rangle}{\langle \psi |\psi\rangle}$ of the Hamiltonian operator, so does $|\psi|$.


$^1$ The absolute value $|\psi|$ is not necessarily differentiable, but eq. (A) still make sense in distribution theory.

  • $\begingroup$ Ok that's why I thought but that's seems fishy. If the argument is to use $|\psi|$ as a wavefunction, then its derivative can be easily discontinuous and then not a valid wavefunction. $\endgroup$ – Mauricio May 18 '18 at 19:20
  • $\begingroup$ Leon van Dom­me­len is apparently not assuming at this stage that the TISE is satisfied. $\endgroup$ – Qmechanic May 18 '18 at 19:41

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