"Probability flux": Imaginary part

Question

I was confronted with this problem from "Mathematik für Chemiker", 8th edition (Ansgar Jüngel, Hans G. Zachmann):

$$ \begin{align} \partial_t |\psi|^2 &= \partial_t (\bar\psi\psi) \\ &= \partial_t\bar\psi\cdot\psi+\bar\psi\cdot\partial_t \psi \\ &= {-\frac{i\hbar}{2m}\Delta\bar\psi\psi+\frac{i\hbar}{2m}\bar\psi\Delta\psi} \\ &= -\frac{i\hbar}{2m}\textrm{div}\big(\nabla\bar\psi\psi-\bar\psi\nabla\psi\big) \\ &= \frac{\hbar}{m}\textrm{div}\big(\Im(\bar\psi\nabla\psi)\big) \quad\quad\quad (13.7) \end{align} $$

I calculated it over and over, but eventually got the same sign error:

$-\frac\hbar m\mathrm{div}\big(\Im(\bar\psi\nabla\psi)\big)$ instead of $\frac\hbar m\mathrm{div}\big(\Im(\bar\psi\nabla\psi)\big)$

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Here is my working: $$ \begin{align} \partial_t|\psi|^2 &= \partial _t(\psi\bar\psi) \\ &= (\partial_t\psi)\bar\psi + \psi\partial_t\bar\psi \\ &= \psi\partial_t\bar\psi + (\partial_t\psi)\bar\psi \\ &= \Big(-\frac{i\hbar}{2m}\Delta\bar\psi\Big)\psi + \Big(\frac{i\hbar}{2m}\Delta\psi\Big)\bar\psi \\ &= -\frac{i\hbar}{2m}\big((\Delta\bar\psi)\psi - (\Delta\psi)\bar\psi\big) \\ &= -\frac{i\hbar}{2m}\Big( \frac{\partial^2\bar\psi}{\partial x^2}\cdot\psi - \frac{\partial^2\psi}{\partial x^2}\cdot\bar\psi \Big) \\ &= -\frac{i\hbar}{2m}\frac{\partial}{\partial x}\Big( \frac{\partial\bar\psi}{\partial x}\cdot\psi - \frac{\partial\psi}{\partial x}\cdot\bar\psi \Big) \\ &= -\frac{i\hbar}{2m} \textrm{div}\big((\nabla\bar\psi)\psi - (\nabla\psi)\bar\psi\big)\\ &= -\frac{i\hbar}{2m}\mathrm{div}\Big(-2i\Im\big((\nabla\psi)\bar\psi\big)\Big)\\ &= -\frac\hbar m\mathrm{div}\Big(\Im\big((\nabla\psi)\bar\psi)\big)\Big)\\ &= -\frac\hbar m\mathrm{div}\big(\Im(\bar\psi\nabla\psi)\big) \end{align} $$

I used the $\bar{z}-z = -2ib = -2i\Im(z)$ approach. In my understanding this should result in the cancellation of $-2$ and $-\frac12$, leaving us with $\frac{i\hbar}m\textrm{div}\big(i\Im(\bar\psi\nabla\psi)\big)$. Using $i^2= -1$ $\implies$ ${-\frac{\hbar}{m}\textrm{div}\big(\Im(\bar{ψ}\nablaψ)\big)}$.

Answer 1

Your result and calculation are correct, and you reproduce properly the well known continuity equation for a free particle,

$ \partial_t \rho = - \vec \nabla \cdot \vec j $

with $\rho = |\psi|^2$ and $\vec j= \frac{\hbar }{m} \ \Im \{\bar \psi\nabla \psi \}$ such that,

$ \partial_t|\psi|^2 = - \nabla \cdot \frac{\hbar}{m}\Im \{ \bar \psi\nabla \psi \} $

There is a sign error in the last expression of equation 13.7, which should contain a minus sign.