Can commutation relations prove <Lx^2> = <Ly^2> in an eigenstate of L^2 and Lz?

[whatisreality]

Homework Statement

Show that $<L_x^2> = <L_y^2>$ using the commutation relations. The system is in the eigenstate |l,m> of $L^2$ and $L_z$.

Homework Equations

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

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??

 

[drvrm]

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

To work out this commutator, an identity is useful. For any 3 operators

,

and

try for getting the relationship

 

[DrDu]

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

 

[whatisreality]

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

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 $<L_x^2> = <L_y^2>$?

 

[whatisreality]

To work out this commutator, an identity is useful. For any 3 operators

,

and

try for getting the relationship

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 $[L_y^2, L_z]$ ?

 

[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!

 

[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

 

[vela]

If $L_x L_y = L_y L_x$, then $L_x = L_y$, I think, but not sure that's true.

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

If $L_x = L_y$, does that mean $<L_x> = <L_y>$?

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

And does that mean $<L_x^2> = <L_y^2>$?

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 $<L_x^2> = <L_y^2>$.

 

[PeroK]

I get really really close!

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 $<L_x^2> = <L_y^2>$?

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!

 

[drvrm]

So an alternative approach to using the operators already written above would be to find expressions for the commutators
[L2x,Lz][Lx2,Lz][L_x^2, L_z] and [L2y,Lz][Ly2,Lz][L_y^2, L_z] ?

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!

 

[DrDu]

If $L_x L_y = L_y L_x$, then $L_x = L_y$, I think, but not sure that's true.

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

 

[whatisreality]

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

What does at variance with mean?

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

 

[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...

 

[DrDu]

What does at variance with mean?

It is a contradiction.

 

[DrDu]

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...

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 < .

 

[whatisreality]

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 < .

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

 

[DrDu]

Yes

 

[whatisreality]

Yes

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?

 

[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!

 

[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.

 

[DrDu]

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 $<L_x^2> = <L_y^2>$?

You are quite close. However you have to look at the results of multiplication by e.g. Lx one time from the left and one time from the right. Then you can get rid of terms like LxLzLx comparing the two equations.

 

[PeroK]

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...

The result, I believe, does not follow from the commutation properties alone. It requires the other properties of these particular AM operators. The properties of the ladder operators follow directly from the properties of the AM operators, whether you use them explicitly or not. You could do the whole thing without defining $L_{\pm}$ explicitly but it would be just the same proof essentially, just more complicated.

In particular, I believe the result requires that $L_z$ is self-adjoint and hence eigenfunctions corresponding to distinct eigenfunctions are orthogonal. Some key properties of the ladder operators, hence $L_x$ and $L_y$, hinge on this.

(I've just seem DrDru's latest post. Perhaps I'm wrong and everything follows from the commutation relations alone. I don't see it, I must confess.)

 

[whatisreality]

The result, I believe, does not follow from the commutation properties alone. It requires the other properties of these particular AM operators. The properties of the ladder operators follow directly from the properties of the AM operators, whether you use them explicitly or not. You could do the whole thing without defining $L_{\pm}$ explicitly but it would be just the same proof essentially, just more complicated.

If using ladder operators counts as using commutation relations, I can prove it. I hope that's the case, but I'll ask my lecturer when I get the chance!

You are quite close. However you have to look at the results of multiplication by e.g. Lx one time from the left and one time from the right. Then you can get rid of terms like LxLzLx comparing the two equations.

OK, I'll give that a go. Thanks :)

Does $<L_x^2L_y^2> = <L_y^2L_x^2>$ imply $<L_x^2> = <L_y^2>$ in some way? I don't see how, but if so I've proved it!

 

[Oxvillian]

The result, I believe, does not follow from the commutation properties alone.

Well, the result simply isn't true for a general state, so it won't be possible to get it from the commutation properties of the operators alone.

One way or another, the result follows from the rotational symmetry of the $L_z$ eigenstates. Roughly speaking, the $L_z$ eigenstates don't care what you call the $x$-axis and what you call the $y$-axis. The challenge (I guess) is to explain this in a language that includes commutation relations somewhere along the line.

 

[TSny]

Consider $\left< L_y L_z L_x \right>$ and reduce it in two different ways using the commutation relations. Reduce it one way to get something involving $\left< L_x^2 \right>$. Reduce it another way to get something involving $\left< L_y^2 \right>$

 

[drvrm]

I wish I could attempt this, but I have no idea how I'm supposed to get any sort of expression involving either <L2x><Lx2> or <L2y><Ly2>. 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 <Lx><Lx> 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??

After such a long sharing i think some more hints can be dropped so that your confusion gets cleared
1. express square of the two components x . y added together as L^2- Lz^2
express L+ and L- raising and lowering operators as sum and difference of Lx and i.Ly ; add the two to get 2.Lx similarly get Ly square the expression to get Lx^2 and similarly Ly^2
2. take a state Ilm> and calculate the expectation values ;you should get the results as <L+^2> and similarly L-^2 expectation will vanish .pl. do this exercise ;use property of ladder operator L+I lm> = I l, m+1> and similarly for L- but it will lower m value.

 

[whatisreality]

Thank you everyone for your help! I got there in the end! :) But would definitely not have managed on my own!