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Operators - Parity

  1. Oct 19, 2007 #1
    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.
    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]

    If V is not the same as PV, why not?
  2. jcsd
  3. Oct 19, 2007 #2
    If P acts like PV(x)=V(-x), then obviously PV=V for symmetric potentials, and your calculation was unnecessarily complicated. But I don't know how precisely your P is defined. Could it be, that it was defined so that it works like this PV(x)P^-1 = V(-x)?

    hmh.. in fact this comes down to the question about what you want V to be. If you think that it is operator, that maps Psi into V*Psi, then PV=VP seems to be the only way.

    In position representation V is an operator that maps the Psi like this

    \Psi(x) \mapsto V(x)\Psi(x)

    Now you want to know what PV is. P maps the state V*Psi like this

    V(x)\Psi(x)\mapsto V(-x)\Psi(-x)

    so, in other words

    PV\Psi(x) = V(-x)\Psi(-x) = V(-x)P P^{-1}\Psi(-x) = V(-x)P\Psi(x)

    \implies PV(x)=V(-x)P \quad\Leftrightarrow\quad PV(x)P^{-1}=V(-x)

    Here the notation is bad, because it would be better to not have the (x) after the V, but its easier to write the operator V(-x) with this notation.

    If you wanted to have PV=V, for symmetric potentials, this would imply


    which is wrong, because for symmetric potentials we have

    Last edited: Oct 19, 2007
  4. Oct 19, 2007 #3
    Yeah, I realise my calculation was overly complicated, but I felt like I was trying to justify the obvious and didn't know how to do it.
    That actually makes a lot of sense - I was thinking completely wrong about how the operators work (dropping the vector it operates on). It makes sense to me now: if you trasform a state psi by [tex]\psi\rightarrow P\psi[/tex], the states must transform as [tex]A \rightarrow P^{-1}AP[/tex] for consistency (which is the same as your expression for parity since the parity transformation is its own inverse)
    Thanks for clearing it up.
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