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Parity and Copehagen

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle is of mass m has the wave function [itex]\psi[/itex](x)=Ne-k|x-a| with (a>0):

    The particle is in the state [itex]\psi[/itex], and its parity is measured. What is the probability to find a positive parity?

    2. Relevant equations

    None that I can think of...

    3. The attempt at a solution

    My attempt at a solution is making the leap that this is a qualitative, rather than quantitative question...which leads me to want to say that there is a 50% chance that one would find a positive parity.

    Is my assumption that this question wouldn't have a quantitative solution correct? And of course, if it is, did I get the correct reasoning?
     
  2. jcsd
  3. Sep 18, 2011 #2

    vela

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    I don't think so. As you let a approach 0, the wave function becomes closer and closer to being an even function, so the probability should vary with a.

    You might want to think about the fact that you can decompose any function into even and odd parts using fe=[f(x)+f(-x)]/2 and fo=[f(x)-f(-x)]/2. This is just a suggestion. I'm not sure how the problem works out, actually.
     
  4. Sep 18, 2011 #3
    In which case my solution would need to be something along the lines of

    [itex]\int[/itex] (fe)(fo)dx (taken from 0 to a)?
     
  5. Sep 18, 2011 #4
    Wait, since I just want the probability of even parity it would be:

    [itex]\int[/itex]fe(x)2dx

    again from 0 to a.
     
  6. Sep 18, 2011 #5
    Functioning under the bold assumption that my previous thought was at least in the right ball game I arrive at the probability of positive parity being

    =(2N2/4)[itex]\int[/itex][e-2k|x-a|+e-2k|-x-a|]dx

    The integral appears to be a bit of a pain at first glance, but does this at least seem like the right direction to be going in?
     
  7. Sep 18, 2011 #6

    diazona

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    As vela said, this sounds like a fully quantitative question, so you're going to have to do some math.

    Here's something to think about: you know (I hope) that when you make a measurement corresponding to an operator, the result is one of the eigenvalues of that operator, and the probability of each result is the squared absolute value of the projection of the actual state on to the corresponding eigenstate. In other words, if an operator [itex]A[/itex] has eigenvalues [itex]a_i[/itex] with corresponding eigenstates [itex]|a_i\rangle[/itex], then the probability of getting a particular result is
    [tex]P(a_i) = |\langle\psi|a_i\rangle|^2[/tex]
    Hopefully you're familiar with all that?

    If the operator [itex]A[/itex] is degenerate, so that it has multiple eigenstates [itex]|a_i^k\rangle[/itex] with the same eigenvalue, then the probability is obtained by projecting on to the entire space of the desired eigenvalue. First you project the state [itex]|\psi\rangle[/itex] on to the space,
    [tex]|\psi_{a_i}\rangle = \sum_k \langle a_i^k|\psi\rangle|a_i^k\rangle[/tex]
    and then you can compute the probability as before:
    [tex]P(a_i) = |\langle\psi|\psi_{a_i}\rangle|^2[/tex]

    Now, how does this apply to your problem? Parity is represented by an operator which inverts a single spatial coordinate, e.g. [itex]x\to -x[/itex]. Mathematically you can represent it as
    [tex]P = \int|-x\rangle\langle x|\mathrm{d}x[/tex]
    but that's not necessary for this particular problem. What you do need to know is that the parity operator has two eigenvalues, +1 (even) and -1 (odd), which means this operator is highly degenerate. So according to what I wrote in the previous paragraph, in order to find the probability of even parity, you first need to determine the projection of your state on to the even-parity subspace. (Make sure you understand why this is what you need to do)

    Fortunately, for the parity operator, this projection is easy. You just use the formula vela gave you to find the even part of the wavefunction. Let's call this [itex]\psi_e(x)[/itex]. Then you can determine the inner product of this function with your original wavefunction,
    [tex]\langle\psi|\psi_e\rangle = \int(\cdots)\mathrm{d}x[/tex]
    (you get to fill in where the dots are), and finally plug it in to
    [tex]P(\text{even}) = |\langle\psi|\psi_e\rangle|^2[/tex]
    to determine the probability.
     
  8. Sep 18, 2011 #7
    I'm sold until understanding why I need to determine the projection of my state on to the even-parity subspace. Although it certainly explains why I was asked earlier in the problem set to prove that the parity operator projects onto even and odd subspaces.

    To the more mathematical-mechanical aspect

    <[itex]\psi[/itex]|[itex]\psi[/itex]e>=[itex]\int[/itex]dx[itex]\psi[/itex](x)[itex]\psi[/itex]e(x)

    then

    P(even)=[itex]\int[/itex]dx|[itex]\psi[/itex][itex]\psi[/itex]e|2

    with appropriate substitutions for [itex]\psi[/itex] and [itex]\psi[/itex]e?
     
    Last edited: Sep 18, 2011
  9. Sep 18, 2011 #8

    diazona

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    Well, in general, to determine the probability that you get an eigenvalue [itex]a_i[/itex], you need to project the state on to the subspace corresponding to that eigenvalue. Do you at least believe that, in general? The thing about projecting on to an even-parity subspace is just the application of that to this specific problem: the eigenvalue in question is +1.

    As for the math, [itex]\langle\psi|\psi_e\rangle = \int\psi(x)\psi_e(x)\mathrm{d}x[/itex] is almost correct. You missed one little thing (which is not going to make a difference in this case, but it's best not to get into the habit of missing little details :wink:). However, it does not follow from there that [itex]P(\text{even}) = \int|\psi(x)\psi_e(x)|^2\mathrm{d}x[/itex]. You did the substitution wrong. Check that step carefully.
     
  10. Sep 18, 2011 #9
    Ah ha, ok now I definitely will believe that explanation; in general and in this case. This makes more sense now than it did earlier!

    As for missing the small things; maybe that's why I like nuclear physics, nuclei are just so darn big! :)
     
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