Let's say we have a symmetric potential, in position representation [tex]V(x)=V(-x)[/tex] and let [tex]P[/tex] be the parity operator.(adsbygoogle = window.adsbygoogle || []).push({});

Then quite clearly [tex]PV=VP[/tex] but I was told the stronger statement [tex]V=PV[/tex] is not true, but I thought

[tex]V=\int_{-\infty}^{\infty} V\left|x\right\rangle\left\langle x \right| dx[/tex]

(where I have used completeness and linearity of the integral... though I'm having second thoughts about linearity - can I just move the integral through V?)

[tex]V=\int_{-\infty}^{\infty} V(x)\left|x\right\rangle\left\langle x \right| dx[/tex]

[tex]V=\int_{-\infty}^{\infty} V(-x)\left|x\right\rangle\left\langle x \right| dx[/tex]

[tex]V=\int_{-\infty}^{\infty} (PV(x))\left|x\right\rangle\left\langle x \right| dx[/tex]

[tex]V=\int_{-\infty}^{\infty} P(V(x)\left|x\right\rangle)\left\langle x \right| dx[/tex]

[tex]V=\int_{-\infty}^{\infty} P(V\left|x\right\rangle)\left\langle x \right| dx[/tex]

[tex]V=\int_{-\infty}^{\infty} (PV)\left|x\right\rangle\left\langle x \right| dx[/tex]

[tex]V=(PV)[/tex]

If V is not the same as PV, why not?

Cheers

**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!

# Operators - Parity

Loading...

Similar Threads for Operators Parity |
---|

I Why do we need the position operator? |

I How to pick an operator onto which the wavefunction collapses |

I The Ehrenfest Theorem |

A Inverse momentum operator |

A Field quantization and photon number operator |

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