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The Parity Operator: Find the average value of the parity.

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves in the potential energy V(x)= [itex]\frac{1}{2}[/itex] mω2x2
    . The ground-state wave function is
    [itex]\psi[/itex]0(x)=([itex]\frac{a}{π}[/itex])1/4e-ax2/2
    and the first excited-state wave function is
    [itex]\psi[/itex]1(x)=([itex]\frac{4a^3}{π}[/itex])1/4e-ax2/2
    where a = mω/[itex]\hbar[/itex]

    What is the average value of the parity for the state

    ψ(x)=[itex]\frac{\sqrt{3}}{2}[/itex][itex]\psi[/itex]0(x)+ [itex]\frac{1-i}{2\sqrt{2}}[/itex][itex]\psi[/itex]1(x)

    2. Relevant equations

    ∏[itex]\psi[/itex](x)=[itex]\psi[/itex](-x)
    ∏[itex]\psi[/itex]λ(x) = [itex]\psi[/itex]λ(x)

    3. The attempt at a solution

    First off I'm extremely confused on how to approach this to the point of where I don't know what I'm solving for so I'm someone can help me understand the problem and what it's asking me to do.

    I just finished reading the parity operator section and all I understand was that ∏ inverts the coordinates of the wave function. I also got that ∏^2 is the identity vector which means that the eigenvalues have to be ±1 and that an even eigenfunction corresponds to 1 while an odd function corresponds to -1 and The eigenfunction of the parity operator form a complete set. That's it from the book but it shows no examples or anything remotely close to what this questions asks.

    Am I trying to solve for ∏? Why are the ground state and first state included in this problem? Please help. Anything will be helpful.
     
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 23, 2012 #2

    vela

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    Your expression for ##\psi_1(x)## is incorrect. There should be a factor of x in there.

    You're being asked to find ##\langle \psi | \hat{\Pi} | \psi \rangle ##.

    Think about how you'd calculate the average energy ##\langle \psi | \hat{H} | \psi \rangle ## of the state. You're being asked to do the same sort of calculation except this time with the parity operator instead of the Hamiltonian.

    Hint: The energy eigenstates of the harmonic oscillator are also parity eigenstates.
     
  4. Oct 31, 2013 #3
    Use Expansion Coefficients

    First off, ##\psi##0 and ##\psi##1 are eigenfunctions of [itex]\Psi[/itex]. You can see their expansion coefficients given in the equation for [itex]\Psi[/itex], I will refer to these as cn

    The average value of the parity, or <[itex]\Pi[/itex]> will be Ʃ|cn|2an, where the a's are the eigenvalues of the given eigenfunctions.

    To find the eigenvalues use [itex]\prod\psi = a\psi[/itex]

    This should simplify the problem quite a bit -- it's actually very straightforward if you approach it this way, and you don't have to deal with any messy integrals.
     
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