In Griffith's book, the following equation is derived based on schroedinger's equation $\frac \partial {\partial t} \Psi = \frac {i h}{2m} \frac {\partial^2 \Psi} {\partial x^2} – h^{-1}V\Psi$:
$$\frac \partial {\partial t} |\Psi|^2 = \frac {i h}{2m} \frac \partial {\partial x}\big (\frac {\partial \Psi}{\partial x}\Psi^* – \frac {\partial \Psi^*}{\partial x} \Psi \big )$$
I understand the derivation, but it is intuitively very strange to me, because the potential function doesn't appear anywhere. Surely the potential applied to a particle at some moment has an influence over where that particle is likely to be measured the next moment. How can I intuitively make sense of this?
Edit: For example, one implication of this is that the derivative of expected position $\frac {\partial \langle x\rangle}{\partial t}$ doesn't depend on the potential applied to the particle (as it isn't, in griffith's equation 1.29).
Best Answer
In order to derive the continuity equation for the probability density current, it is necessary that the wavefunction $\Psi$ satisfies Schrödinger's equation for a given Hamiltonian (containing $V$). Therefore, the potential is implicitly present even if it is not explicit in the formula. Maybe, one could write $\Psi_V$, to make explicit such a point.