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

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# Operators - Parity

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