# Commutator relations

1. Jan 16, 2014

### Observer Two

I'm doing something horribly wrong in something that should be very easy. I want to show that:

$[L^2, x^2] = 0$

So:

$[L^2, x x] = [L^2, x] x + x [L^2, x]$

$L^2 = L_x^2 + L_y^2 + L_z^2$

Therefore: $[L^2, x] = [L_x^2 + L_y^2 + L_z^2, x] = [L_x^2, x] + [L_y^2, x] + [L_z^2, x]$

= $L_y [L_y, x] + [L_y, x] L_y + L_z [L_z, x] + [L_z, x] L_z$

$= -i h L_y z - ih z L_y + i h L_z y + i h y L_z$

So

$[L^2, x x] = -i h L_y z x - ih z L_y x + i h L_z y x + i h y L_z x + -i h x L_y z - ih x z L_y + i h x L_z y + i h x y L_z$

And now?

This ought to be a lot easier.

2. Jan 16, 2014

### strangerep

Where did this exercise come from? I suspect you misunderstand what $x^2$ means here, but I can't be sure without seeing the original unedited question.

(BTW, you're supposed to use the homework template when posting in this forum -- else you risk attracting the Wrath of the Mentors.)

3. Jan 16, 2014

### Observer Two

Ughy. You are right. I looked it up again and it's not actually x^2 as in "position operator squared" but X^2 = x^2 + y^2 + z^2.

I didn't sleep for 29 hours now, working for some exams. I'm slightly confused. My bad. Next time I'll use the exact template.

4. Jan 16, 2014

### strangerep

This is a really bad technique. You are probably more confused than you realize, maybe even approaching a delirium state. Non-sleeping is counter-productive. Things that take ages to understand or perform while tired tend to be much quicker when you're fresh.

Go and sleep (even if just for a few hours -- set the alarm clock accordingly).