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Can a system be in an eigenstate of the total angular momentum operator $L^2$ without being in an eigenstate of any of $L_x$, $L_y$, and $L_z$? What would such a state look like?

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Can a system be in an eigenstate of the total angular momentum operator $L^2$ without being in an eigenstate of any of $L_x$, $L_y$, and $L_z$?

Yes.

First consider the three states $|\ell, m\rangle$ with $\ell=1$ and $m=\pm 1$ and $m=0$. These are all eigenstates of $L^2$ and $L_z$.

Now consider an arbitrary linear combination $$ |\Psi\rangle = \sum_{m=-1}^1 c_m|1, m\rangle\;, \tag{A} $$ for arbitrary $c_m$.

The state in Eq. (A) above is still an eigenstate of $L^2$, but it is not generally an eigenstate of any of the individual components $L_x$ $L_y$ or $L_z$.

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  • $\begingroup$ Thanks! Does this hold for any arbitrary choice of x, y, and z axis? That is, would it be possible that the angular momentum state (or spin state) of a particle is not an eigenstate of $L_i$ for any arbitrary axis $i$ and yet it is an eigenstate of $L^2$? $\endgroup$
    – Lory
    Commented Apr 21, 2023 at 4:15
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    $\begingroup$ If think the answer is yes depending on what exactly you mean. For example, if you start by fixing the axes, then whatever axes you choose you can always call one of them "z" and setup eigenstates wrt to that axis. Then the whole argument goes through as before. But if you are asking something else, then I'm not sure. $\endgroup$
    – hft
    Commented Apr 21, 2023 at 5:23
  • $\begingroup$ Thanks, that helps. $\endgroup$
    – Lory
    Commented Apr 21, 2023 at 5:45
  • $\begingroup$ I would have chosen for the system to be an eigenstate of $L^2$ and $( L_x + L_y )/\sqrt2$, and that will obviously satisfy the conditions too. $\endgroup$ Commented Apr 21, 2023 at 5:49

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