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Parity of wavefunctions

  1. Apr 11, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    A quantum harmonic oscillator is in a superposition of states(below):
    [itex]\Psi(x,t)[/itex] = 1/[itex]\sqrt{2}[/itex] ([itex]\Psi_{0}(x,t) + \Psi_{1}(x,t)[/itex]

    [itex]\Psi_{0}(x,t)[/itex] = [itex]\Phi(x) * e^{-iwt/2} [/itex] and [itex]\Psi_{1}(x,t)[/itex] = [itex]\Phi_{1}(x) * e^{-i3wt/2} [/itex]

    Show that <x> = C cos(wt) ...


    2. Relevant equations

    Negative parity: f(-x) = -f(x)
    Positive parity : f(x) = f(-x)

    3. The attempt at a solution
    On the mark schemes it shows that x|ψ(x)^2| = -(-x|ψ(-x)^2) are asymmetric functions,so when I expand the integral they should vanish,however, I thought that normalization condition allows the integral to be definite i.e = 1.


    I have a very basic understanding of parity, it would be great if someone could explain to me just what role it plays.


    P.S I have looked around for resources,and so far I have gathered that there are two distinct eigen-values when parity operator works(+-1).

    Thanks.
     

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  2. jcsd
  3. Apr 11, 2013 #2
    Would you be able to show at least some work? Write out the terms in [itex]\left\langle\psi \left|x\right| \psi \right>[/itex].

    if [itex]f(x) = - f(-x)[/itex], then what is
    [itex]\int_{-\infty}^\infty f(x) dx[/itex] ?

    if [itex]f(x) = - f(-x)[/itex] and [itex]g(x) = g(-x)[/itex], what is
    [itex]\int_{-\infty}^\infty f(x)g(x) dx[/itex] ?
     
  4. Apr 11, 2013 #3

    vela

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    You're comparing two different integrals. The normalization condition is
    $$\int_{-\infty}^\infty \lvert \psi(x) \rvert^2\,dx = 1$$ while the integral from the expectation value is
    $$\int_{-\infty}^\infty x\lvert \psi(x) \rvert^2\,dx.$$ Why would you think the first integral precludes the second one from vanishing? They have different integrands.
     
  5. Apr 12, 2013 #4
    I was in a rush hence my lack of input.

    Here is what I get after expansion:

    [itex]\Psi(x,t)[/itex] = 1/[itex]\sqrt{2}[/itex] ([itex]\Psi_{0}(x,t) + \Psi_{1}(x,t)[/itex]

    [itex]\Psi_{0}(x,t)[/itex] = [itex]\Phi(x) e^{-iwt/2} [/itex] and [itex]\Psi_{1}(x,t)[/itex] = [itex]\Phi_{1}(x) e^{-i3wt/2} [/itex]

    [itex]\left\langle\psi \left|x\right| \psi \right> = 1/\sqrt{2} \int_{-\infty}^\infty dx [ x (\Phi_{0}\Phi_{1} + \Phi_{0}\Phi_{1}e^{-i\omega t} + \Phi_{1}\Phi_{0}e^{i\omega t} + \Phi_{1}\Phi_{1} [/itex]]

    I understand that the integrand of a product of even-odd function is zero,but I still don't see why the two integrals above vanish,are they assumed to be asymmetric, is there any justification?
     
  6. Apr 12, 2013 #5
    Actually... I think I get it now...
    (below is what I think happens}

    Is it because [itex] \Phi(x)[/itex] = [itex] -\Phi(-x)[/itex](odd) and f(x) = f(-x) (even)...
    Integral of odd*even = integral of odd = zero .
    Also for the other two integrals(with exponential) they are defined as "symmetric limited function", something I am not familiar with.

    I appreciate the responses.

    EDIT: I was just restating the properties of odd/even functions,don't know why I had integral put in(fixed),sorry.
     
    Last edited: Apr 12, 2013
  7. Apr 12, 2013 #6

    vela

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    What is that notation even supposed to mean?
     
  8. Apr 12, 2013 #7
    Hi vela
    I have fixed few typos so it should be clear now.
     
    Last edited: Apr 12, 2013
  9. Jun 25, 2013 #8
    Hi
    Good thinking but not so much & clear definition.
     
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