# Partial derivative of probability density (squared modulus of the wave function) wrt to position (1D)?

Here's a snippet from Introduction to Quantum Mechanics by David Griffiths (Sec 1.5): I understand how we used the Schrodinger equation to go from a partial in time to a double partial in position. But why is there a $-$ sign in the parentheses? Shouldn't there be a $+$ instead, since

$\Large \frac{\partial |\Psi|^2}{\partial x} = \frac{\partial (\Psi^*\Psi)}{\partial x} = \Psi^*\frac{\partial\Psi}{\partial x}\ +\ \frac{\partial\Psi^*}{\partial x}\Psi$

You should try solving this on a piece of paper. I don't think that you understand correctly how the author passes from $\partial/\partial t$ to $\partial^2/\partial^2 x$. You can not use Schrodinger's equation for $|\Psi|^2$, it is only valid for $\Psi(x, t)$:

$$i \hbar \frac{\partial}{\partial t} \Psi(x, t) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \Psi(x, t).$$

For $\Psi^*$ another equation holds, which can be obtained by conjugating both sides of the Schrodinger's equation:

$$- i \hbar \frac{\partial}{\partial t} \Psi^*(x, t) = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} \Psi^*(x, t).$$

Note the minus sign on the l.h.s. It comes from conjugating an imaginary unit ($i^* = -i$). This minus sign is responsible for the minus sign in your answer.

Now I trust you to carefully expand both parts of your equation and see for yourself that they are indeed equal.

• Oh I see! So we first expand the partial in time using the product rule, and then use Schrodinger's equation, yes? Sep 8, 2016 at 10:45
• @TusharRakheja Exactly. Did you prove your formula yet? :) I suggest you finish this, it would obviously help you in the future. Sep 8, 2016 at 10:56
• Yes, I was able to finish the proof! Thank you very much! :D Sep 8, 2016 at 11:07