I would like some assistance working with commutators

[hmparticle9]

I am starting to realise that I am struggling with commutators, and I want to address this before moving on. This particular example involves the angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$.

$$[L_x, L_y] = [y p_z - z p_y, z p_x - x p_z] = [y p_z, z p_x] - [y p_z, x p_z] - [z p_y, z p_x] + [z p_y, x p_z]$$
This is an application (twice) of $[A+B, C] = [A,C] + [B,C]$. Good so far.

My book says the following: "from the canonical commutation relations the two middle terms drop out". If I expand the two middle terms and apply the canonical commutation relations a few times to "shift" things about. For instance:
$$[y p_z, x p_z] = y p_z x p_z - x p_z y p_z = yx p_z p_z - xy p_z p_z = 0$$
Then (only then) can I say that the middle terms drop out. But seeing that they drop out directly from the canonical commutation relations, I can't see it. So that is my first problem. I might be reading into things a bit too much, and the author might mean that I need to use the canonical commutation relations a few times for things to drop out.

My second issue is that it is stated that
$$ [y p _z, z p_x] = y p_x [p_z, z], \text {and } [z p_y, x p_z] = x p_y [z, p_z] $$

Again, repeated use of canonical commutation relations we can say
$$[z p_y, x p_z] = zp_y x p_z - x p_z z p_y = p_y x(zp_z - p_z z) = x p_y [z, p_z]$$

I just want to make sure I am following things correctly, and that I'm not missing a trick

 

[TSny]

For instance:
$$[y p_z, x p_z] = y p_z x p_z - x p_z y p_z = yx p_z p_z - xy p_z p_z = 0$$
Then (only then) can I say that the middle terms drop out. But seeing that they drop out directly from the canonical commutation relations, I can't see it.

Your method is fine. To shortcut it, you can note that all factors in $[y p_z, x p_z]$ commute among themselves. Thus, $x$ commutes with $y$ and $p_z$ , $y$ commutes with $x$ and $p_z$ , and $p_z$ commutes with itself and $x$ and $y$. So, the result of the commutator is zero.

Again, repeated use of canonical commutation relations we can say
$$[z p_y, x p_z] = zp_y x p_z - x p_z z p_y = p_y x(zp_z - p_z z) = x p_y [z, p_z]$$

Good.

Since $x$ commutes with $z$ , $p_y$ , and $p_z$ , $x$ can treated as a factor that can be pulled out of the commutation. Likewise $p_y$ commutes with $x$, $z$, and $p_z$. So, $p_y$ can be pulled out. It doesn't matter whether you "factor out" the $x$ to the left side or the right side. Same for $p_y$. And it doesn't matter the order in which you pull out the $x$ and $p_y$. Thus, you can go immediately from $[z p_y, x p_z]$ to $x p_y [z, p_z]$. Be sure to respect the order of $z$ and $p_z$.

While getting familiar with this sort of manipulation, if you have any doubt it's a good idea to write things out as you have done.