I I would like some assistance working with commutators

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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
 
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hmparticle9 said:
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.

hmparticle9 said:
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.
 
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