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Here's a homework problem I'm working on. I am not asking for the answer, but any guidance or comments on the approach are appreciated.

Given that a measurement of $L^2$ for a free particle has resulted in the value $6\hbar^2$, what are the possible results for a measurement of $L_x$?

I've been following the approach given on pp. 378-380 in Liboff (4th ed.). I've made it as far as expressing the eigenfunctions of $L_x$ as sums of the eigenfunctions of $L^2$... Here's what I've got:

$$\begin{align*} X_1 &= \frac{1}{2}\left(|2,2\rangle - |2,1\rangle + |2,-1\rangle - |2,-2\rangle\right)\\ X_2 = -X1 &= -\frac{1}{2}\left(|2,2\rangle - |2,1\rangle + |2,-1\rangle - |2,-2\rangle\right)\\ X_3 &= \frac{1}{2}\left(\sqrt{\frac{3}{2}}|2,2\rangle - |2,0\rangle + \sqrt{\frac{3}{2}}|2,-2\rangle\right)\\ X_4 = -X3 &= -\frac{1}{2}\left(\sqrt{\frac{3}{2}}|2,2\rangle - |2,0\rangle + \sqrt{\frac{3}{2}}|2,-2\rangle\right)\\ X_5 &= \frac{1}{2}(|2,2\rangle + |2,1\rangle - |2,-1\rangle - |2,-2\rangle)\\ X_6 = -X_5 &= -\frac{1}{2}\left(|2,2\rangle + |2,1\rangle - |2,-1\rangle - |2,-2\rangle\right) \end{align*}$$

The next step appears to be to express $L^2$ as sums of the eigenfunctions of $L_x$, which does not seem possible.

It was a lot of algebra, matrices, determinants, etc. to get to this point so it is certainly possible that I've miscalculated something accidentally, but does anyone see anything fundamentally wrong here with this approach? And/or is there a simpler way?

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@KyleKanos Actually, eqnarray is outdated and it's better to use align. –  Wouter Dec 2 '13 at 16:02
Oh, and is this question about a hydrogen atom? That would be important to mention. –  David Z Dec 2 '13 at 16:05
@Wouter: completely unawares of deprecation of eqnarray. Is that something that happened like 10 years ago and no one told me? –  Kyle Kanos Dec 2 '13 at 16:07
$L^2$ has a dimension of $\hbar^2$, so you have an error in the formulation of the problem. –  user23660 Dec 2 '13 at 16:12
I may be missing something, but from the problem setup, (which I assume you mean $\hat{L}^2=6\hbar^2$), then quantum number $l=2$. The operator $\hat{L}_x$ can yield the value $-l\hbar$ to $+l\hbar$ in integer steps. I don't there's enough information to specify which of these five are or aren't possible for the problem unless you give more info. –  BMS Dec 2 '13 at 16:17

2 Answers 2

If $L^2$ is $6\hslash^2$ this means that the quantum number $l$ is $2$, hence the quantum number $m$ goes from $-2$ to $+2$, and the possible values of $L_x$ are $-2\hslash, -\hslash, 0, \hslash, 2\hslash$

Conventionally $L_z$ is used instead than $L_x$, but what applies to $L_z$ must apply also to $L_x$ and $L_y$: there are no preferential directions in space.

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Thanks, MiMo. I understand why the possible values of $L_z$ are $-2\hbar, -\hbar, 0, \hbar, 2\hbar$, and Liboff does a good job explaining this in his text. It looks like I am going to have to convince myself why this is also the case for $L_x$...and presumably $L_y$. –  Ramblin Wreck Dec 2 '13 at 17:24
There are no preferential direction in space, what applies to one axis must apply to the others... –  MiMo Dec 2 '13 at 17:33

Since the operator $L^2$ yielded $6\hbar^2$, one can deduce that $l=2$ after measurement. In general, the only possible measurements of one of the $L_i$ components are from $-l\hbar$ to $+l\hbar$ in integer steps. If you're comfortable with this for $L_Z$, note that none of the angular momentum operators $L_i$ are fundamentally more special than the other two, so the same will be true for $L_x$. Based on the (lack of) information given in the problem, I don't think there's enough to say anything more than this. Perhaps if we knew what was done prior to to the $L^2$ measurement we could say more.

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