Is there no coordinate conjugate to $L_x$?

Question

The conjugate momentum corresponding to $\phi$ (azimuthal angle in sp. polar coordinate) is $L_z$ (sometimes written $L_\phi$) which is frequently used in quantum mechanics. Why is there no coordinate conjugate to $L_x$ (or $L_y$)?

Answer 1

In spherical polar coordinates $\phi$ is the azimuthal angle for one possible orientation of the reference directions in space.

It is important to keep in mind that space is isotropic and there is nothing that makes $\hat{z}$ a special direction when it comes to saying

"I'll define my polar coordinates like this."

So of course there is a coordinate that is conjugate to $L_x$. I think I'll call it $\phi_x$ and it is the azimuthal angle for a set of spherical polar coordinates where the polar angle is measured away from $+\hat{x}$.

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The relationship of $\phi_x$ to to the unit vectors of the more usual spherical polar coordinate system is, however, complicated and location dependent.

Answer 2

My answer will take a tack different from @dmckee. One possible definition of conjugate variables is that their Poisson bracket $\{A,B\}=1$. The bracket in spherical coordinates is $$ \{f,g\}=\frac{1}{L\sin\theta}\left(\frac{\partial f}{\partial \phi}\frac{\partial g}{\partial\theta}- \frac{\partial g}{\partial \phi}\frac{\partial f}{\partial\theta}\, . \right) $$ so that, with the standard definitions $$ L_z=L\cos\theta\, ,\qquad L_x=L\sin\theta\cos\phi\, ,\qquad L_y=L\sin\theta\sin\phi $$ one reproduces the usual Poisson bracket between angular momentum operators. Here, $\theta$ is the polar angle and $\phi$ the azimuthal angle.

With this indeed $L_z=L\cos\theta$ is conjugate to $\phi$ in the sense that $\{L_z,\phi\}=1$.

Continuing this way you can get a partial differential equation to be satisfied by the function $f_x(\theta,\phi)$ that is to be conjugate to $L_x$. Alternatively, you can easy verify that $L_\theta= L \phi \sin\theta + B(\theta)$ is conjugate to $\theta$ for arbitrary $B$.

An obvious problem of course is that there are only two spherical angles so you cannot expect that two "coordinate angles" will be conjugate to three momenta.

If you want the QM version replace the classical bracket with the quantum commutation so that $i\hbar \{f,g\}\to [\hat f,\hat g]$ although there are problems in defining an angle operator, as discussed elsewhere, and there are other issues as well.