Let's say, that we are considering the basic Heisenberg model with only two spin-particles. So our Hamiltonian can be written as follows:
$$ H = \sum_{\langle i,i' \rangle} S^{(x)}_i S^{(x)}_{i'} + S^{(y)}_i S^{(y)}_{i'} + S^{(z)}_i S^{(z)}_{i'}. $$
In the case of two spins with $S = \frac{1}{2} $, it can also be represented as a matrix:
$$ H = \frac{1}{4} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. $$
The eigenvectors for this matrix are as follows:
$$ |0\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, |1\rangle = \begin{pmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{pmatrix}, |2\rangle = \begin{pmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{pmatrix}, |3\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} $$
(I didn't have any better idea for naming those vectors).
Now, when we want to calculate the expected value of spin in given axis for only one particle (e.g. the first one), we can use matrices in the following form:
$$ S_j^{(1)} = \sigma_j \otimes \mathbb{1}, $$ where $j$ can take values corresponding to axes of our interest, i.e. $j \in \{x,y,z\}$ and $\sigma_j$ are spin matrices (so, these are Pauli matrices multiplied by some factor). Let's write down two of them:
$$ S_x^{(1)} = \frac{1}{2} \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}, \\ S_z^{(1)} = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}. $$
Finally, if we calculate the expected value of spin in the $z$ axis on the first site for all of the eigenvectors, we get:
$$ \langle 0 | S_z^{(1)} |0 \rangle = \frac{1}{2}, \langle 1 | S_z^{(1)} |1 \rangle = 0, \langle 2 | S_z^{(1)} |2 \rangle = 0, \langle 3 | S_z^{(1)} |3 \rangle = -\frac{1}{2}, $$ which makes sense for me. But, when I want to calculate expected values in the $x$ axis, I always get 0:
$$ \langle 0 | S_x^{(1)} |0 \rangle = \langle 1 | S_x^{(1)} |1 \rangle = \langle 2 | S_x^{(1)} |2 \rangle = \langle 3 | S_x^{(1)} |3 \rangle = 0. $$
Why is that so? I don't think, that these results say, that spin never lies in the $XY$ plane, because when I switch from $S_x$ to $S_x^2$ operators, I don't get 0's anymore. Does it have something to do with the basis, in which those operators are written (I belive, that all of them are usually written in the basis of the $S_x$ operator)?
Edit 1: Below I give results of all expected values for each eigenvector of the Hamiltonian.
$|0\rangle$ (eigenvalue = $\frac{1}{4}$ )
$$\langle S_x \rangle = 0, \langle S_y \rangle = 0, \langle S_z \rangle = \frac{1}{2} $$ $$\langle S_x^2 \rangle = \frac{1}{4}, \langle S_y^2 \rangle = \frac{1}{4}, \langle S_z^2 \rangle = \frac{1}{4} $$ $$ \langle S^2 \rangle = \frac{3}{4} $$
$|1\rangle$ (eigenvalue = $\frac{1}{4}$ )
$$\langle S_x \rangle = 0, \langle S_y \rangle = 0, \langle S_z \rangle = 0 $$ $$\langle S_x^2 \rangle = \frac{1}{4}, \langle S_y^2 \rangle = \frac{1}{4}, \langle S_z^2 \rangle = \frac{1}{4} $$ $$ \langle S^2 \rangle = \frac{3}{4} $$
$|2\rangle$ (eigenvalue = $-\frac{3}{4}$ )
$$\langle S_x \rangle = 0, \langle S_y \rangle = 0, \langle S_z \rangle = 0 $$ $$\langle S_x^2 \rangle = \frac{1}{4}, \langle S_y^2 \rangle = \frac{1}{4}, \langle S_z^2 \rangle = \frac{1}{4} $$ $$ \langle S^2 \rangle = \frac{3}{4} $$
$|3\rangle$ (eigenvalue = $\frac{1}{4}$ )
$$\langle S_x \rangle = 0, \langle S_y \rangle = 0, \langle S_z \rangle = -\frac{1}{2} $$ $$\langle S_x^2 \rangle = \frac{1}{4}, \langle S_y^2 \rangle = \frac{1}{4}, \langle S_z^2 \rangle = \frac{1}{4} $$ $$ \langle S^2 \rangle = \frac{3}{4} $$
For the Pauli matrices we have $\sigma_x^2 = \sigma_y^2 = \sigma_z^2 = \mathbf{1}$, but for higher values of spin (where e.g. $S_x^2 \neq \mathbf{1}$) we would get analogical results.
Edit 2: Because $[H, S_x] \neq 0$ can we understand this, that when spin does not align in the $Z$ axis, it has equal probability of aligning in any possible direction in the $XY$ plane, and because of this we get expected values equal to 0?
Something like holding a pen vertically on a table and then releasing it. From its initial upright position ($Z$) it can fell in any possible direction on the table (our $XY$ plane).
