Eigenvalues of n dot L, QM

1. Jun 30, 2014

fluidistic

1. The problem statement, all variables and given/known data
I've been told that the operator $\hat n \cdot \hat {\vec L}$ has the same eigenvalues as $L_z$. Later I've been told that it has the same eigenvalues as any component of $\hat {\vec L}$. But I am a bit confused, as far as I understand the eigenvalues of $L_x$, $L_y$ and $L_z$ cannot be the same at the same time. Hmm now that I think about it, if $L_z$ has eigenvalues $\hbar m$, then both $L_x$ and $L_y$ have no eigenvalues? This is impossible... I am missing something.
I ask this because I found a problem that more or less states that if $|l,m>$ is an eigenvector of both $\hat L$ and $\hat L_z$, then find the coefficients $c_i$ in function of $\theta$ and $\phi$ such that $|\psi>=c_1 |1,1>+c_0 |1,0>+c_{-1}|1,-1>$ is an eigenvector of $\hat n \cdot \hat {\vec L}$.

2. Relevant equations
$\hat A \psi =\lambda \psi$.

3. The attempt at a solution
I think I know a way how to tackle the problem. I'd write $\hat n =\cos \phi \sin \theta \hat i + \cos \phi \cos \theta \hat j + \cos \theta \hat k$. Then write $\hat {\vec L}=L_x \hat i + L_y \hat j + L_z \hat k$. Then write $L_x$ and $L_y$ in terms of $L_{\pm}$ because I know how they act on $|1,m>$
With all this in mind I could write the left side of the following equation: $(\hat n \cdot \hat {\vec L}) |\psi >= \lambda |\psi>$.
What I did not know is that apparently I could take $\lambda = 0$, $\hbar$ or $-\hbar$.

2. Jun 30, 2014

dauto

No body said they have the same eigenvalues at the same time. they have the same set of eigenvalues but at any one given time they may have different eigenvalues taken from the common set Lx and Ly always have eigenvalues which may sometimes be zero. Having a zero eigenvalue is not the same thing as having no eigenvalue. Even when Lz is m at least one of Lx and Ly must have a non-zero eigenvalue (except if the particle's spin is zero).

3. Jun 30, 2014

fluidistic

Ah ok I see now. Thank you.

I still don't understand why $\hat n \cdot \hat {\vec L}$ and $\hat L_z$ share the same eigenvalues.
It's not like $\hat n$ is arbitrary since it must satisfy that $(\hat n \cdot \hat {\vec L}) |\psi >= \lambda |\psi>$ and I can't just assume that $\hat n = \hat z$ for instance.

4. Jun 30, 2014

Oxvillian

Sure they do!

$L_z$ and $L_x$ are just operators, they have the same eigenvalues no matter what time of day it is. Indeed, they have the same set of eigenvalues as each other - to claim otherwise would be to claim that God prefers the $x$-direction to the $z$-direction, or maybe vice-versa.

Probably what you mean is that a given state vector can't be both an eigenstate of $L_z$ and of $L_x$. (?)

5. Jul 1, 2014

king vitamin

I think your confusion lies in the fact that $\hat n \cdot \hat {\vec L}$ and $\hat L_z$ always have different eigenvectors as long as $\hat n \neq \pm \hat z$. You need to solve for the eigenvalues/eigenvectors in the different cases, and it turns out that they do always have the same eigenvalues, even though the eigenvectors are different for a given basis. The reason this makes sense is because your choice of a basis (which direction $\hat{z}$ points in) is arbitrary; you could consider another coordinate system (and basis for spin states) where $\hat{n} = \hat{z}$ and you should get the same eigenvalues.

6. Jul 1, 2014

vanhees71

Take a rotation that puts the $z$ axis of an arbitrary Cartesian basis to the unit vector $\vec{n}$. This is represented by a unitary tranformation in the Hilbert space of states and thus, $\hat{J}_z$ and $\vec{n} \cdot \hat{\vec{J}}$ has the same eigenvalues. The eigenstates of the latter operator are given by the unitary transformation of the eigenstates of the former. It's a good exercise to write this down in formulas!