Angular momentum commutators

1. Feb 17, 2016

whatisreality

1. The problem statement, all variables and given/known data
Show that $<L_x^2> = <L_y^2>$ using the commutation relations. The system is in the eigenstate |l,m> of $L^2$ and $L_z$.

2. Relevant equations
$[L_x, L_y] = i \hbar L_z$
$[L_y, L_z] = i \hbar L_x$
$[L_z, L_x] = i \hbar L_y$
$[L_x, L^2] = 0$
$[L_y, L^2] = 0$
$[L_z, L^2] = 0$

3. The attempt at a solution
I wish I could attempt this, but I have no idea how I'm supposed to get any sort of expression involving either $<L_x^2>$ or $<L_y^2>$. From the commutators, anyway. With ladder operators I could get somewhere! I had a vague notion they were both meant to be zero, but the second part asks you to calculate $<L_x>$ and gives lots of marks for it. So I don't think it's zero.

How do I get to the squares of the operators from the commutators??

2. Feb 17, 2016

drvrm

To work out this commutator, an identity is useful. For any 3 operators , and try for getting the relationship

3. Feb 17, 2016

DrDu

Just a guess: What happens if you multiply the second eq with Ly and the third with Lx?

4. Feb 17, 2016

whatisreality

I get really really close!
$[L_y, L_z] L_y = L_y L_z L_y - L_zL_y^2 = i\hbar L_x L_y$

$[L_Z, L_x] L_x = -[L_x, L_z] L_x = -(L_xL-z-L_zL_x)L_x$
$=L_zL_x^2 - L_xL_zL_x = -i\hbar L_yL_x$

Rearranging,
$L_zL_y^2 = L_yL_zL_y - i\hbar L_xL_y$
And $L_zL_x^2 = L_xL_zL_x - i\hbar L_yL_x$
If $L_x L_y = L_y L_x$, then $L_x = L_y$, I think, but not sure that's true.

Due to a limited knowledge of operators, I'm also unclear what this tells me. If $L_x = L_y$, does that mean $<L_x> = <L_y>$?
And does that mean $<L_x^2> = <L_y^2>$?

5. Feb 17, 2016

whatisreality

So an alternative approach to using the operators already written above would be to find expressions for the commutators
$[L_x^2, L_z]$ and $[L_y^2, L_z]$ ?

6. Feb 17, 2016

drvrm

I guess that you have to find expectation values of square of angular momentum components so somebody should use square relations like L^2 =........... then take expectation value of the system in a state say I l.m> then only the measurement can be done- there may be problems on the way but let us see!

7. Feb 17, 2016

drvrm

to make calculations more quick raising and lowering operators can be used when you get expressions for square of the ang. momentum operators see a book by Powell or Schiff on quantum mechanics-there is a very easy book by Anderson

8. Feb 17, 2016

vela

Staff Emeritus
I don't see your logic here, but $L_x L_y \ne L_y L_x$. The operators don't commute since $[L_x,L_y] \ne 0$.

Of course. You can always do the same things to both side of an equation.

If $L_x = L_y$, then yes, of course, because anywhere $L_x$ appears, you can replace it with $L_y$. But $L_x \ne L_y$, so you haven't proved $<L_x^2> = <L_y^2>$.

9. Feb 17, 2016

PeroK

Here's some logic. You are given three things:
1) the commutator relations;
2) $L^2 = L_x^2 + L_y^2 + L_z^2$;
3) $<L^2> = \hbar^2 l(l+1)$ and $<L_z> = \hbar m$

I suggest you will need all of this information to prove that (under these circumstances) $<L_x^2> = <L_y^2>$. I doubt you will prove anything more fundamental such as $L_x^2 = L_y^2$

In particular, you may make progress initially manipulating expressions for the operators, but sooner or later you will need to work with the expected value.

... you may also need to assume that all these operators are self-adjoint!

Last edited: Feb 17, 2016
10. Feb 17, 2016

drvrm

additional hint; use the expression for L+ and L- to get Lx ^2 and Ly^2 added together and use a state say Il.m> to calculate the expectation value!

11. Feb 17, 2016

DrDu

No, that's at variance with your first commutator.

12. Feb 17, 2016

whatisreality

What does at variance with mean?

Is that a good point to stop working with operators then, and start using expectation values?

13. Feb 17, 2016

whatisreality

Is $<L^2 > = < L_x^2 + L_y^2 + L_z^2 > = < L_x^2 > + <L_y^2> + <L_z^2>$? Can you split the expectation value like that?

And is $<L_zL_y^2> = <L_z> <L_y^2>$?

From the definition of the expectation value, I think the first one is true. Don't think the second one is...

Last edited: Feb 17, 2016
14. Feb 17, 2016

DrDu

15. Feb 17, 2016

DrDu

The first splitting is always true. the second statement <L_zL_y^2> = <L_z> <L_y^2> holds only true for the expectation values of Eigenfunctions of L_z as <L_z=hbar m < .

16. Feb 17, 2016

whatisreality

So because the system is in the normalised eigenstate $|l,m>$ that would be true?

17. Feb 17, 2016

DrDu

Yes

18. Feb 17, 2016

whatisreality

OK. Here's where I've got to:
$<L_zL_y^2> = <L_z><L_x^2> = \hbar m <L_y^2>$

$<L_zL_x^2> = \hbar m <L_x^2>$ by the same reasoning as above.
And what I tried to do with it then was silly, so that's as far as I've got.
Is that along the right lines? I haven't actually used any commutator relations there, and that doesn't show the required relation yet. If that's right, where do I go from there?

Last edited: Feb 17, 2016
19. Feb 17, 2016

Oxvillian

whatisreality - I fear you're not going in the correct direction here

What do you have against ladder operators? If you need to use $L_x$ and $L_y$ in the $L_z$ basis, expressing them in terms of ladder operators is usually the way to go. And I promise you it will work!

20. Feb 18, 2016

Oxvillian

Oh I see, you're supposed to "use the commutation relations"

Well, I guess the properties of the ladder operators follow from the angular momentum commutation relations, so you're kind of using them indirectly...

Or you could "use the commutation relations" to show that

$$L_y \; = \; e^{-i\frac{\pi}{2}L_z} \; L_x \; e^{i\frac{\pi}{2}L_z}$$

and then use that to get the required result.