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.


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?

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    $\begingroup$ The operator is not the same as the value. The commutation relations hold for any value. $\endgroup$ Commented May 19 at 20:18
  • $\begingroup$ 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. $\endgroup$
    – Stallmp
    Commented May 19 at 20:19
  • $\begingroup$ 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) $\endgroup$
    – Stallmp
    Commented May 19 at 20:22
  • $\begingroup$ 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 $\endgroup$
    – Stallmp
    Commented May 19 at 20:28
  • 1
    $\begingroup$ 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? $\endgroup$ Commented May 19 at 20:46

1 Answer 1


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 $L_y$.

  • 1
    $\begingroup$ Is the conclusion that the only simultaneous eigenstate of the three operators are if all of their eigenvalues are 0? $\endgroup$
    – Stallmp
    Commented May 19 at 20:52

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