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Homework Help: Anti-commutation of parity operator

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data
    The parity operator is defined as [tex]P \psi (x) = \psi (-x)[/tex]. Show that [tex]P[/tex] and [tex]p_x[/tex] anti-commute, that is, [tex] \{ P,p_x \} = Pp_x + p_xP = 0 [/tex].

    2. Relevant equations
    [tex]P \psi (x) = \psi (-x)[/tex]
    [tex]p_x = - i \hbar \frac{\partial}{\partial x}[/tex]

    3. The attempt at a solution
    [tex] \{ P,p_x \} \psi(x) = ( Pp_x + p_xP ) \psi(x) = -i \hbar \left[ P \frac{\partial}{\partial x} \psi (x) + \frac{\partial}{\partial x} [ P \psi (x) ]\right] = -i \hbar \left[ \frac{\partial}{\partial (-x)} \psi (-x) + \frac{\partial}{\partial x} \psi (-x) \right] = -i \hbar \left[ - \frac{\partial}{\partial x} \psi (-x) + \frac{\partial}{\partial x} \psi (-x) \right] = 0 [/tex]

    Is it valid to say that [tex] P \frac{\partial}{\partial x} \psi (x) = \frac{\partial}{\partial (-x)} \psi (-x) [/tex] ?
  2. jcsd
  3. Apr 22, 2010 #2
    for the part where you found (pxP) this one is correct ..

    as for your question, I wont say that im 100% sure that it is valid .. but I would say that im 80% agree with what you suggested in order to get your answer ..

    what I know about the parity operator is that when you apply it on psi(x) you get psi(-x), so whenever you have x you simply change it to -x and vice versa (that would be applied to other things such as velocity and force) ..
  4. Apr 22, 2010 #3
    I'm under the impression that the parity operator transforms [tex]x \rightarrow -x[/tex].

    Let's say that [tex] f(x) = \frac{\partial \psi (x) }{\partial x}[/tex] .

    Then [tex]P f(x) = f(-x) = \frac{\partial \psi (-x) }{\partial (-x)} = - \frac{\partial \psi (-x) }{\partial x}[/tex] , right?

    I feel that something is wrong here.
  5. Apr 22, 2010 #4
    I told you im not 100% sure .. but i still think it is right the way you did ..

    hopefully there would be someone else to confirm what I think ..
  6. Apr 22, 2010 #5


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    Right. I don't see anything wrong. Use a trial function of x, like a few terms of a polynomial, that doesn't have definite parity and see how it works.
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