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I assume it's not too difficult but I don't know how to show the equivalence between two different expressions of the guidance equation in Bohmian mechanics. The following is the form I already encountered quite often:

$$\dot{\textbf{x}}(t) = \frac{\hbar}{2im|\psi|^2}(\psi^*\nabla\psi - \psi\nabla\psi^*)$$

I recently read "de Broglie-Bohm, delayed-choice double-slit experiment and density matrix" by John Bell where he illustrates shortly the basics of Bohmian Mechanics and mentions that "the particle rides along on the wave at some position $\textbf{x}(t)$ with velocity:

$$\dot{\textbf{x}}(t) = \frac{1}{m} \frac{\partial}{\partial\textbf{r}} \text{Im} \log\psi(t,\textbf{r})\bigg\rvert_\textbf{r=x}$$

He doesn't call this equation the guidance equation but I think they're supposed to be the same - does anyone know how to show the equivalence of these two? Where does the $\text{Im} \log$ come from?

Thanks in advance!

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Just write out the derivatve in your second expression. Remember that $$ \partial_x( \ln \psi)= \frac{1}{\psi} \partial_x \psi, $$ and ${\rm Im} f= (f-f^*)/2i$ and there you are.

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  • $\begingroup$ It was indeed easier than expected - thank you!! $\endgroup$ – Juri V Jan 19 at 7:20

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