# Pulling out partial derivatives?

1. May 18, 2013

### Cogswell

I'm reading through the book Quantum Mechanics (Second Edition) by David J. Griffiths and it got to the part about proving that if you normalise a wave function, it stays normalised (Page 13).

That part that I don't get is how they say:

$\dfrac{i \hbar}{2m} \left( \Psi^* \dfrac{\partial^2 \Psi}{\partial x^2} - \dfrac{\partial^2 \Psi^*}{\partial x^2} \Psi \right) = \dfrac{\partial}{\partial x} \left[\dfrac{i \hbar}{2m} \left( \Psi^* \dfrac{\partial \Psi}{\partial x} - \dfrac{\partial \Psi^*}{\partial x} \Psi \right) \right]$

How can they just pull out a partial operator like that?
Because if you expand it out again it would give you:

$\dfrac{i \hbar}{2m} \left( \dfrac{\partial}{\partial x} \left[ \Psi^* \dfrac{\partial \Psi}{\partial x}\right] - \dfrac{\partial}{\partial x} \left[ \dfrac{\partial \Psi^*}{\partial x} \Psi \right] \right)$

The operator will be applied to the wrong $\Psi$ and also won't you need to apply the product rule to is as well?

2. May 18, 2013

### wotanub

If you do just what you said, you'll see that the extra terms cancel and you'll be back at the original expression. Take out a scrap of paper.

3. May 18, 2013

### lugita15

If you apply the product rule to the first term of $\dfrac{i \hbar}{2m} \left( \dfrac{\partial}{\partial x} \left[ \Psi^* \dfrac{\partial \Psi}{\partial x}\right] - \dfrac{\partial}{\partial x} \left[ \dfrac{\partial \Psi^*}{\partial x} \Psi \right] \right)$, you'll get a term $\dfrac{\partial \Psi}{\partial x}\dfrac{\partial \Psi^*}{\partial x}$, which cancels out with a term $- \dfrac{\partial \Psi}{\partial x}\dfrac{\partial \Psi^*}{\partial x}$ you'll get if you apply the product rule to the second term.

4. May 18, 2013

### Cogswell

Haha right, thank you. I thought there was a special property of partial derivatives that I didn't know.