# "Probability flux": Imaginary part

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

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 $$i^2= -1$$ $$\implies$$ $${-\frac{\hbar}{m}\textrm{div}\big(\Im(\bar{ψ}\nablaψ)\big)}$$.

• Physically is it possible for $\partial |\psi|^2/\partial t$ to be negative, and certainly $(-i)(-i)=-1$. Commented Aug 27, 2023 at 11:28

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