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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.

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You have a t typo in the first equation, it should be an i. This question is badly phrased. – Ron Maimon Oct 17 '11 at 8:14

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 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:

$$ \frac{\partial j(x,t)}{\partial x} = \frac{\hbar}{2mi} \left[\psi^* \frac{\partial ^2 \psi}{\partial x^2} - \psi \frac{\partial^2 \psi^*}{\partial x^2}\right] = \frac{\hbar}{2mi} \frac{\partial}{\partial x} \left[ \psi^* \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^*}{\partial x} \right ] (3) $$

$$ \mbox{(2) and (3)} \Leftrightarrow \frac{\partial P(x,t)}{dt} + \frac{\partial j(x,t)}{\partial x} = 0 $$

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I got to refresh my latex ability by writing this. However, I can't seem to figure out how to write the correct Psi letter with Mathjax. In my previous (three years old) LaTeX documents, I simply used \psi without any special package. Please edit my post to correct that, thanks! – ChrisR Oct 16 '11 at 16:40
Dear @ChrisR, what's exactly wrong with \psi? Your answer seems perfect. MathJax is just a particular implementation of TeX and LaTeX into web pages (which I also use on my blog and recommend to others). If you don't like the shape of \psi, maybe you wanted the capital \Psi ? – Luboš Motl Oct 16 '11 at 18:20
Dear @LubošMotl, thank you for asking. In my LaTex documents, the lowercase \psi looks like this one . It is the same letter, just slightly printed differently. – ChrisR Oct 16 '11 at 19:09
Also there seems to be several minus sign bugs in your answer (and on the site you linked!) – wsc Oct 17 '11 at 5:27
@wsc Thank you for pointing that out. I think I have corrected them now, after redoing the math on paper. – ChrisR Oct 17 '11 at 22:41

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