Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutation Rules for L and J

  1. Apr 10, 2015 #1
    So the total angular momentum operator J commutes with any scalar operator S. The argument for this is that J is the generator of 'turntable rotations' (by this I mean we rotate the whole object about an axis, along with its orientation) and the expectation value of any scalar operator has to be invariant under such a rotation. This tells us that S commutes with the rotation operator and thus its generator J.

    My question is why doesn't a similar argument hold for the orbital angular momentum operator L? The difference is that L generates rotations that rotate only the object but not its orientation around an axis. However surely the expectation value of a scalar operator should still be invariant under this type of rotation, meaning L and S commute. However this is not the case, because I know that any component of L does not commute with J2.
  2. jcsd
  3. Apr 10, 2015 #2
    Scalar operators that involve the orientation of the object may not be invariant under the rotations generated by ##\vec L##. For example consider the angle between an electron's spin axis and its momentum. This angle is a scalar quantity (related to the scalar operator ##\vec S \cdot \vec P##) which is invariant under the full rotations generated by ##\vec J##. But it is not invariant under the rotations generated by ##\vec L##, which will rotate the momentum but not the spin. You can confirm that ##[\vec J, \vec S \cdot \vec P] = 0## while ##[\vec L, \vec S \cdot \vec P] \neq 0##.
  4. Apr 10, 2015 #3
    Beautiful, thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook