Commutation of (L^2)op and (Lz)op

1. Nov 19, 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.

2. Relevant equations

Relevant equations are included in the question

3. The attempt at a solution

My attempted solution is attached (Scan0004.jpg). I work it out, but I'm going wrong somewhere as I'm finding that they do not commute as they should.

Thank you in advance for any help!

Andrew

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2. Nov 19, 2009

Ben Niehoff

This is a rather convoluted way to do this, so I'm not going to try to pick apart all of it. But I can tell you that you have a significant error in the very first three lines. You have attempted to set

$$L_z = L_x L_y - L_y L_x$$

when the correct expression is

$$L_z = \frac{1}{i \hbar} (L_x L_y - L_y L_x)$$

Now, if you like, you can set $\hbar = 1$, as this amounts to just choosing a special system of units. However, you cannot set $i = 1$. The imaginary unit i is crucial to getting the cancellation you need.