# Comutation of (Lz)op and (L^2)op

1. Nov 24, 2009

### metgt4

1. The problem statement, all variables and given/known data

It has been shown that the operators (Lx)op and (Ly)op do not commute but satisfy the following equation:

(Lx)op(Ly)op - (Ly)op(Lx)op = i(hbar)(Lz)op

(a) Use this relation and the two similar equations obtained by cycling the coordinate labels to show that (L2)op(Lz)op = (Lz)op(L2)op, that is, these two operators commute. [Hint: You do not need to introduce the differential formulas for the operators. Use the fact that (AB)C = A(BC) where A, B, and C are operators]

The question continues, but this is the part I am having trouble with.

I already attempted a solution (scan is attached), and my prof gave me the following hint:

"Hi Andrew - one suggestion: instead of writing every operator as a commutator,
consider instead moving operators from right to left using commutation relations.
For example, on the LHS you have X^2*Z=X(XZ) = X *(ZX) + stuff (using a commutator
relation), which in turn is equal to (XZ)*X = Z*X^2 + more stuff (using a commutator
relation). But then you have something that's on the RHS, ie Z*X^2."

My problem here is that I have not been able to find a commutator relation that allows me to move the components of the equation around like that, nor have I been able to work a relation out myself.

I'd like to work out as much of this problem as I can, so hints would be preferred!

Andrew

2. Relevant equations

3. The attempt at a solution

2. Nov 24, 2009

### Feldoh

I think your professor is trying to say something like this:

If we have $$[x,y p_x] = y[x,p_x]$$ because y commutes can just be factored out.

3. Nov 24, 2009

### jdwood983

You're looking to see that $[L^2,L_z]=0$ right? And since you know that $L^2=L_x^2+L_y^2+L_z^2$, you can say

$$[L_x^2+L_y^2+L_z^2,L_z]=something$$

This should be a good push in the right direction.

4. Nov 24, 2009

### jdwood983

Normally you have that $[A,B]=-[B,A]$ unless $A$ and $B$ commute, which case $[A,B]=[B,A]=0$. So if you need to prove that $[L^2,L_z]=[L_z,L^2]$, you just need to show that

$$[L_x^2+L_y^2+L_z^2,L_z]=something$$

and

$$[L_z,L_x^2+L_y^2+L_z^2]=something$$

where both $something$'s are the same thing, zero.

EDIT: That's weird...I got an email saying you responded already, by it's not here after I posted this...hmmm....a glitch in the Matrix??

5. Nov 24, 2009

### metgt4

Must have been a glitch. I was in class/the library for the last few hours! Thanks for the help there though, I'll keep working on it and hopefully won't run into any more trouble!

6. Nov 26, 2009

### metgt4

I decided to go back and plug in the actual values of each operator. My work is in attached. Is my physics/math correct, or did I oversimplify things?

#### Attached Files:

• ###### scan0006.jpg
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7. Nov 26, 2009

### Ben Niehoff

Your professor is trying to explain to you a useful manipulation. All you do is take the definition of a commutator

$$[a,b] = ab - ba$$

$$ab = ba + [a,b]$$

Now you have a means of getting one operator to "move through" another, and you can work with just the commutator algebra, rather than having to use any particular representation of the operators. For example:

$$a^2b = aab = aba + a[a,b] = ba^2 + [a,b]a + a[a,b]$$

Using this idea, you should be able to solve your problem using only the commutation relations alone; you don't need the representation in terms of differential operators.

8. Nov 26, 2009

### diazona

Yes, that's exactly what I was about to say myself.

9. Nov 26, 2009

### jdwood983

That seems a strange way of going about it. I would have just used the fact that $[AB,C]=[A,C]B+A[B,C]$ and come up with

$$[L_x^2+L_y^2+L_z^2,L_z]=[L_x^2,L_z]+[L_y^2,L_z]+[L_z^2,L_z]=[L_x,L_z]L_x+L_x[L_x,L_z]+[L_y,L_z]L_y+L_y[L_y,L_z]=onward$$

and likewise for the other part.