In quantum mechanics, the Hamilltonian operator is constructed as the infinitesimal generator of the time translation group, which is a 1-parameter group. Yet it can still depend on time. So you have a situation where the generator of a 1-parameter group can depend on the parameter. Yet the momentum operator, the generator of infinitesimal spatial translations, cannot depend on the parameter(s) of the spatial translation group, namely position. And similarly the angular momentum operator cannot depend on angle (or direction), the parameter(s) of the rotation group. Is there a fundamental reason for this, or is it simply that we happen to already know the properties of the dynamical variables from classical mechanics, so we don't bother with it? I'd be disappointed if it was the latter, because that might undermine the elegance of treatments like Sakurai or Townsend (at the graduate and undergrad levels respectively) which try to derive QM from minimal first principles.(adsbygoogle = window.adsbygoogle || []).push({});

Any help would be greatly appreciated.

Thank You in Advance.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Why isn't momentum a function of position?

Loading...

Similar Threads for isn't momentum function |
---|

I Spin Angular Momentum Dirac Equation |

A Inverse momentum operator |

I Does the Schrödinger equation link position and momentum? |

I Feynman Diagram-Momentum conservation in primitive vertex |

I Why isn't the photon energy h nu over two? |

**Physics Forums | Science Articles, Homework Help, Discussion**