Knowing all spin components at the same time [duplicate]

You can't know all spin components simultaneously due to the commutation relation (& Heisenberg's uncertainty principle):

$$[S_x, S_y] = i\hbar S_z$$

But what if you know that $$S_z=0$$? Then that implies that $$S_x$$ and $$S_y$$ commute and thus you know all components at the same time.

Note: that $$S^2$$ doesn't have to be equal to 0, since $$S_x$$ and $$S_y$$ can be non-zero. This is what I want to stress, as opposed to the answer is another linked post in the comments, which assumes that $$s=0$$.

Note: I am referring to the eigenvalues of the operators. So with $$S_z = 0$$ I mean that $$m_s = 0$$, i.e. an eigenstate of $$S_z$$ with an eigenvalue of 0.

Edit:

I believe that the conclusion is that the only simultaneous eigenstate of Sx, Sy and Sz are if the eigenvalue is 0. Is this correct?

• The operator is not the same as the value. The commutation relations hold for any value. Commented May 19 at 20:18
• But if two operators commute that implies that they have simultaneous eigenstates, which means that you have an eigenstate where the eigenvalues corresponding to $S_x, S_y, S_z$ are all known. Commented May 19 at 20:19
• If there is a common eigenstate of Sx, Sy and Sz, doesn't that imply that all of their eigenvalues are defined simultaneously? (since you know that Sz = 0 and then Sx and Sy commute) Commented May 19 at 20:22
• I don't understand from that post specifically why l = 0 is necessary. Why can't you have a state | 2 0> for example with l=2 and m=0? This state has Lz = 0, but a nonzero L^2, while at the same time having defined value for Lx and Ly which should have non zero eigenvalues Commented May 19 at 20:28
• The logic is that the commutation relations imply that if you have an eigenvector of both $L_x,L_y$, then it must be an eigenvector of $L_z$ with eigenvalue $0$. Using the other commutation relations, obtained by cyclic interchanging the operators, shows that the eigenvalues of $L_x,L_y$ must be $0$, too. Then it is trivial to see that the eigenvalue of $L^2$ is also $0$. This is what the answer is elaborating. Where is the problem with the argument? Commented May 19 at 20:46

Why use spin, when you can work it for orbital angular momentum. The analog to a vector boson is the $$P$$ state, with eigenfunctions:

$$|1, m\rangle = Y_1^m(\Omega)$$

with the $$m=0$$ state looking like:

$$Y_1^0 \propto \cos\theta = z/r$$

For simplicity, I'll work on the unit sphere:

$$|1, 0\rangle \propto z$$

It's an eigenstate of:

$$L_z = x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x }$$

since $$z$$ doesn't depend on $$x$$ or $$y$$.

If I apply:

$$L_x = y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y }$$

I get:

$$L_x z = y$$

and likewise for $$L_x$$.

So it is clearly not an eigenstate of $$L_x$$ or $$L_y$$.

• Is the conclusion that the only simultaneous eigenstate of the three operators are if all of their eigenvalues are 0? Commented May 19 at 20:52