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Let's say we have an arrangement of spins in 2D space (as given in the below picture).

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Assume that the $z$-axis is out of the plane and a spin (circled in red) makes an angle $\theta$ with the $x$-axis. This spin precesses in its direction. If it precesses of course it has some component in the $z$-direction. One can construct the $x$-(or $y$-) direction component of spin by measuring the angle $\theta$. For example, $\hat S_x = \hbar \cos \theta$ and $\hat S_y = \hbar \sin \theta$. This kind of decomposition is readily available from the already known direction of spin. But I can't see how can we measure the $z$-component of spin in this formalism. Is there any way to measure the $z$-component of spin for this kind of spin arrangement?

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  • $\begingroup$ "If it precesses of course it has some component in the 𝑧-direction." Is this true? If I hold my arms out horizontally and spin about a point on the floor, must my hands point in a direction that has a vertical component? $\endgroup$
    – Marius LadegĂĄrd Meyer
    Commented May 15, 2021 at 14:16
  • $\begingroup$ If the red arrow was linear momentum and it were precessing then I would agree that there is a z-component of angular momentum, but that is something else entirely. $\endgroup$
    – Marius LadegĂĄrd Meyer
    Commented May 15, 2021 at 14:18
  • $\begingroup$ @MariusLadegĂĄrdMeyer let me be more specific, I am actually measuring magnonic spin current $J$ in a 2D non-collinear magnet (the one given above). To measure x-component of spin current in y-direction, I use operator $\hat J_y^x=\hat S_x \hat v_y$. And for y-component $\hat J_y^y=\hat S_y \hat v_y$. I believe that when one particle (magnon) moves, this magnon, even though it's travelling in $xy$ plane, has some $z$-component of spin. I want a current operator for this component. I thought, if I can make an operator for $z$-component of spin, I can make current operator as $ J_y^z= S_z v_y$ $\endgroup$
    – Luqman Saleem
    Commented May 15, 2021 at 15:23
  • $\begingroup$ @MariusLadegĂĄrdMeyer can't I use commutation relation $\{\hat S_x,\hat S_y\}=i\hbar \hat S_z$ to get $\hat S_z$? $\endgroup$
    – Luqman Saleem
    Commented May 15, 2021 at 15:25

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