One of the conditions for canonical transformations is that all momentum variables should commute. But $(L_x ,L_y)=L_z \neq 0$. Does that mean these are not canonical co ordinates? But aren't point transformations a special case of canonical transformations? How can these transformations be point but not canonical?

EDIT- Spherical co-ordinates are $(r, \theta, \phi)$. The momenta corresponding to $\theta$ and $\phi$ are $L_x$ and $L_z$. They don't commute as I already wrote.

By point transformations, I mean the co ordinate transformations in the Lagrangian formalism. But even disregarding that, we use spherical co-ordinates all the time in Hamiltonian and Quantum mechanics. They can't be non-canonical, right?

  • $\begingroup$ How did you find the momentum corresponding to $\theta$? Your conclusion that it's $L_x$ looks wrong. $\endgroup$
    – Ruslan
    Commented Feb 3, 2022 at 10:13
  • $\begingroup$ @Ruslan I wrote them backwards sorry. $\theta$ has $L_z$ as it rotates about the $z$ axis. $\endgroup$
    – Ryder Rude
    Commented Feb 3, 2022 at 11:12
  • $\begingroup$ Regardless of naming, only one angle has a well-defined axis: the one that rotates around $\hat e_z$. The other angle is that of a standing spherical wave along (and into) the $z$-axis, and if you find corresponding momentum, it won't coincide with any of Cartesian components of angular momentum. $\endgroup$
    – Ruslan
    Commented Feb 3, 2022 at 11:17
  • $\begingroup$ @Ruslan Do you know of the other momentum? Why doesn't it have a popular name? $\endgroup$
    – Ryder Rude
    Commented Feb 3, 2022 at 11:21
  • $\begingroup$ I don't know what it's called. I suppose it doesn't have much use because the interpretation of the variable is not so enlightening, nor are such states with $|m|\ll \ell$ very relevant to the classical limit. $\endgroup$
    – Ruslan
    Commented Feb 3, 2022 at 11:31

1 Answer 1


Spherical co-ordinates are $(r,\theta,\phi)$. The momenta corresponding to $\theta$ and $\phi$ are $L_x$ and $L_z$.

No they aren't. It's true that $p_\phi=L_z$, but $p_\theta= \cos(\phi) L_y - \sin(\phi) L_x$.

The free-particle Lagrangian in spherical coordinates is $$L = \frac{1}{2} m\big(\dot r^2 + r^2\dot \theta^2 + r^2\sin^2(\theta) \dot \phi^2\big)$$ $$\implies \pmatrix{p_\theta \equiv \frac{\partial L}{\partial \dot\theta} = mr^2\dot \theta\\p_\phi \equiv \frac{\partial L}{\partial \dot \phi} = mr^2\sin^2(\theta) \dot \phi}$$

Recall that $$L_x = m(y \dot z - z \dot y) = m r^2\big(-\dot \theta \sin^2(\theta)\sin(\phi)- \dot \phi\sin(\theta)\cos(\phi)\cos(\theta) - \dot \theta \cos^2(\theta)\sin(\phi)\big)$$ $$= -mr^2\big(\dot \theta \sin(\phi) + \dot \phi \sin(\theta)\cos(\theta)\cos(\phi)\big)$$ $$L_y= m(z\dot x - x\dot z)= mr^2(\dot \theta \cos^2(\theta) \cos(\phi) - \dot \phi \sin(\theta)\cos(\theta)\sin(\phi) + \dot \theta\sin^2(\theta)\cos(\phi)\big)$$ $$= mr^2\big(\dot \theta \cos(\phi) -\dot \phi \sin(\theta)\cos(\theta) \sin(\phi)\big)$$ $$\implies -\sin(\phi)L_x + \cos(\phi)L_y = mr^2\dot\theta = p_\theta$$ as promised.

The Poisson bracket of the two angular momentum coordinates is then

$$\{p_\theta,p_\phi\} = \{\cos(\phi)L_y,L_z\} - \{\sin(\phi) L_x,L_z\}$$ Noting that $\{AB,C\}=A\{B,C\}+\{A,C\}B$ and that $\{f(\phi),L_z\} = \{f(\phi),p_\phi\} = f'(\phi)$, we quickly find that $\{p_\theta,p_\phi\}=0$ as expected.


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