I Why “If Lz has a well-defined value, then Lx and Ly do not”？

[hilily]

On Griffiths's book page 166 first paragraph he said "
It's not merely that you don't know all three components of L; there simply aren't three components—a particle just cannot have a determinate angular momentum vector, any more than it can simultaneously have a determinate position and momentum. If Lz has a well- defined value, then Lx and Ly do not."

 

[PeterDonis]

This is because angular momentum operators in different directions do not commute, so it is impossible for there to be any states that are eigenstates of more than one of them. The example you quote is a state which is an eigenstate of $L_z$, but therefore cannot be an eigenstate of any other angular momentum operator, including $L_x$ and $L_y$.

 

[hilbert2]

There's a single exception to this, the case where the squared angular momentum $L^2$ is zero and all of $L_x ,L_y ,L_z$ are zero as well.

 

[PeroK]

This is because angular momentum operators in different directions do not commute, so it is impossible for there to be any states that are eigenstates of more than one of them. The example you quote is a state which is an eigenstate of $L_z$, but therefore cannot be an eigenstate of any other angular momentum operator, including $L_x$ and $L_y$.

In general, if two operators do not commute then there cannot be a complete set of common eigenstates. But, there can be some states that are common eigenstates of the non-commuting operators.

 

[hilbert2]

If I have two noncommuting $2\times 2$ matrices $\mathbf{A}$ and $\mathbf{B}$, and two diagonal $n\times n$ matrices $\mathbf{P}$ and $\mathbf{Q}$, I can form the block matrices

$\begin{bmatrix}\mathbf{A} & 0 \\ 0 & \mathbf{P}\end{bmatrix}$

and

$\begin{bmatrix}\mathbf{B} & 0 \\ 0 & \mathbf{Q}\end{bmatrix}$,

which have an arbitrary number $n$ of linearly independent common eigenstates, but a couple of non-common ones. So actually it's possible that the non-common eigenstates are a minority, even though this doesn't happen often unless you deliberately construct this kind of a case.

 

[vanhees71]

On Griffiths's book page 166 first paragraph he said "
It's not merely that you don't know all three components of L; there simply aren't three components—a particle just cannot have a determinate angular momentum vector, any more than it can simultaneously have a determinate position and momentum. If Lz has a well- defined value, then Lx and Ly do not."

This is very misleading since a quantum system always has all the observables you can define for it, i.e., no matter in which state the system is defined you can always measure any of its observables. Of course, it depends on the state the system is prepared in, which observables take determined values. Griffiths's QT book seems to be quite sloppy in its formulations!

The other postings in this thread are of course correct concerning the question, when an observable takes a determined value (neglecting the possibility of degeneracy).

 

[Birrabenzina]

On Griffiths's book page 166 first paragraph he said "
It's not merely that you don't know all three components of L; there simply aren't three components—a particle just cannot have a determinate angular momentum vector, any more than it can simultaneously have a determinate position and momentum. If Lz has a well- defined value, then Lx and Ly do not."

Since you're studying quantum mechanics there, you need to take the basic laws and apply those to your case. I can give you two suggestions, you have to work this out yourself.
Use these relations, and the answer will be yours:
\begin{align}
&\left[x_i,p_j\right]=i\hbar\delta_{ij}\\
&L_i=\epsilon_{ijk}x_jp_k
\end{align}