Angular momentum operator commutation relation

[Bernard]

I am reading a proof of why

$$\left[ \hat{L}_x, \hat{L}_y \right ] = i \hbar \hat{L}_z$$

Given a wavefunction $\psi$,
$$\hat{L}_x, \hat{L}_y \psi = \left( -i\hbar \right)^2 \left( y \frac{\partial}{\partial z} - z \frac {\partial}{\partial y} \right ) \left (z \frac{\partial \psi}{\partial x} - x \frac{\partial \psi}{\partial z } \right )$$

$$= -\hbar ^2 \left ( yz \frac{\partial ^2 \psi}{\partial z \partial x} + y \frac{\partial \psi}{\partial x} - yx \frac{\partial ^2 \psi}{\partial z^2} - z^2 \frac{\partial ^2 \psi}{\partial y \partial x} + zx \frac{\partial ^2 \psi}{\partial y \partial z} \right )$$

This is a simple expansion of the brackets. I don't understand however, where the second term in the brackets of the last equation comes from?

 

[blue_leaf77]

Use the product rule to calculate $y \frac{\partial}{\partial z} \left(z\frac{\partial \psi}{\partial x}\right)$ because both $z$ and $\frac{\partial \psi}{\partial x}$ are a function of $z$.

 

[fresh_42]

The product rule:
$$
y\cdot \frac{\partial}{\partial z}(z \cdot \frac{\partial \psi}{\partial x})= y\cdot \frac{\partial}{\partial z}(z)\cdot \frac{\partial \psi}{\partial x} + y\cdot z\cdot \frac{\partial}{\partial z}(\frac{\partial \psi}{\partial x})=y \cdot 1 \cdot \frac{\partial \psi}{\partial x} + y\cdot z \cdot \frac{\partial^2 \psi}{\partial z \partial x}
$$

 

[Bernard]

The product rule:
$$
y\cdot \frac{\partial}{\partial z}(z \cdot \frac{\partial \psi}{\partial x})= y\cdot \frac{\partial}{\partial z}(z)\cdot \frac{\partial \psi}{\partial x} + y\cdot z\cdot \frac{\partial}{\partial z}(\frac{\partial \psi}{\partial x})=y \cdot 1 \cdot \frac{\partial \psi}{\partial x} + y\cdot z \cdot \frac{\partial^2 \psi}{\partial z \partial x}
$$

yes I understand that but why not, by the same reasoning, this is not applied on the other terms? (or maybe it is and I am not seeing it) i.e. why not

$$y \frac{\partial}{\partial z} \left( -x \frac {\partial \psi}{\partial z} \right ) = y \frac{\partial(-x)}{\partial z} \frac{\partial \psi}{\partial z} - xy \frac{ \partial ^2 \psi}{\partial z^2}$$

but just $$y \frac{\partial}{\partial z} \left( -x \frac {\partial \psi}{\partial z} \right ) = - xy \frac{ \partial ^2 \psi}{\partial z^2}$$

 

[blue_leaf77]

What is $y \frac{\partial(-x)}{\partial z}$ equal to? Recall that $x$ and $z$ are independent variables.

 

[Bernard]

What is $y \frac{\partial(-x)}{\partial z}$ equal to? Recall that $x$ and $z$ are independent variables.

yep that is zero. it just slipped my mind. Thanks