Probability current Conservation of probability: Suppose a wavefunction has ${\partial \mathbb P \over \partial t} = -t f(x,t)$ and ${\partial j \over \partial x} = i f(x,t)$. How does it follow that ${\partial \mathbb P \over \partial t} = {-\partial j \over \partial x}$? Thanks.
 A: Are you looking for a proof?
If so, this link (which has some sign errors as pointed out in the comments) proves it as follows (without the sign errors):
We start by differentiating the definition of the probability density with respect to time only:
$$
\frac{\partial P(x,t)}{\partial t} = \frac{\partial}{\partial t}\left (\psi^*(x,t) \psi(x,t)\right) = \left[ \frac{\partial\psi^*}{\partial t}\psi + \psi^*\frac{\partial\psi}{\partial t} \right] (1)
$$
We now exploit Schrödinger's Equation and its complex conjugate:
$$
-\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi =  i \hbar \frac{\partial \psi}{\partial t}
$$
$$
-\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* =  -i \hbar \frac{\partial \psi^*}{\partial t}
$$
And inject them into (1):
$$
\frac{\partial P(x,t)}{\partial t} = \frac 1 {i\hbar}\left[ \frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2}\psi - V(x)\psi^*\psi -\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2}\psi^* + V(x)\psi\psi^* \right]
$$
Which corresponds to:
$$
\frac{\partial P(x,t)}{\partial t} = \frac 1 {i\hbar} \frac{\hbar^2}{2m} \left[ \frac{\partial^2\psi^*}{\partial x^2}\psi - \psi^*\frac{\partial^2\psi}{\partial x^2}\right] = \frac{\hbar}{2mi}\frac{\partial}{\partial x} \left[ \frac{\partial\psi^*}{\partial x}\psi - \psi^*\frac{\partial\psi}{\partial x} \right] (2)
$$
The probability current is defined as follows:
$$
j(x,t) = \frac{\hbar}{2mi} \left[\psi^* \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^*}{\partial x} \right]
$$
Hence, its differential with respect to the $x$ axis is the following:
$$
\frac{\partial j(x,t)}{\partial x} = \frac{\hbar}{2mi} \left[\psi^* \frac{\partial ^2 \psi}{\partial x^2} - \psi \frac{\partial^2 \psi^*}{\partial x^2}\right] = \frac{\hbar}{2mi} \frac{\partial}{\partial x} \left[ \psi^* \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^*}{\partial x} \right ] (3)
$$
$$
\mbox{(2) and (3)} \Leftrightarrow \frac{\partial P(x,t)}{dt} + \frac{\partial j(x,t)}{\partial x} = 0
$$
