For every operator ##A##, ##[A,A^n]=0##. And if operators commute they have complete eigen- spectrum the same. But if I look for ##p## and ##p^2## in one dimension ##sin kx## is eigen- function of ##p^2##, but it isn't eigen-function of ##p##.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]p^2 \sin kx=number \sin kx[/tex]

[tex]p\sin kx \neq number \sin kx[/tex]

where

[tex]p=-i\hbar\frac{d}{dx}[/tex]

[tex]p^2=-\hbar^2\frac{d^2}{dx^2}[/tex]

**Physics Forums - The Fusion of Science and Community**

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

# Commutator question

Loading...

Similar Threads - Commutator question | Date |
---|---|

I Question about commuting operators | Aug 7, 2016 |

Quantum circuit implementation of commuting measurements (a novice question) | Nov 20, 2014 |

Some questions about commutation relation | Aug 6, 2014 |

Questions about Commutators | Jun 28, 2014 |

Commutator question | Jan 12, 2013 |

**Physics Forums - The Fusion of Science and Community**