Showing Commutator Relations for [L^2, x^2]

[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.

 

[strangerep]

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

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

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.)

 

[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.

 

[strangerep]

I didn't sleep for 29 hours now, working for some exams. I'm slightly confused.

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).

 

[paranormal]

I can understand your frustration with trying to prove this commutator relation. It may seem like a simple task, but sometimes even the simplest equations can be difficult to solve. However, it is important to approach this problem systematically and logically.

First, let's review the definition of a commutator. The commutator of two operators A and B is defined as [A, B] = AB - BA. In this case, we have A = L^2 and B = x^2. So the commutator [L^2, x^2] would be L^2x^2 - x^2L^2.

Next, let's expand L^2x^2 using the definition of the angular momentum operator L^2 = L_x^2 + L_y^2 + L_z^2. This gives us:

L^2x^2 = (L_x^2 + L_y^2 + L_z^2)x^2

= L_x^2x^2 + L_y^2x^2 + L_z^2x^2

Now, using the commutator definition, we can rewrite L_x^2x^2 as [L_x^2, x^2] + x^2L_x^2. Similarly, L_y^2x^2 can be rewritten as [L_y^2, x^2] + x^2L_y^2 and L_z^2x^2 as [L_z^2, x^2] + x^2L_z^2.

Substituting these back into our original equation, we get:

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

Now, let's focus on the commutators [L_x^2, x^2], [L_y^2, x^2], and [L_z^2, x^2]. Using the commutator definition, we can rewrite them as:

[L_x^2, x^2] = L_x^2x^2 - x^2L_x^2

[L_y^2, x^2] = L_y^2