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Probability of finding a particle in a certain state, using projection

  1. Dec 7, 2009 #1
    I was reading about a certain methood that uses projection to calculate the probability of finding a particle in a certain state. The explanation is not detailed enough for me to get my head around how to use it, but maybe some of you people are familiar with the methood? The methood goes like this:

    We have that [tex]\epsilon \subset H[/tex] is a subspace with an orthonormal basis [tex]\{\phi_j\}[/tex] (here H is Hilbert space). We define an operator [tex]\Pi:H\rightarrow \epsilon[/tex] like this:


    Let's say that [tex]A[/tex] is a measurable property of a particle, and [tex]\hat A[/tex] is the corresponding operator. Let's make [tex]a[/tex] to be one of the eigenvalues of [tex]\hat A[/tex] and [tex]\epsilon_a[/tex] is the corrisponding eigenspace. That is: [tex]\epsilon_a [/tex] is the subspace in H that is spanned by all eigenfunctions of [tex]\hat A[/tex] with eigenvalue [tex]a[/tex]. If the particle has the wavefunction [tex]\psi_t[/tex] then the probability to measurement of [tex]A[/tex] will give the result [tex]a[/tex] is:


    Now there is given an example to explain this. We are given wave function


    And the parity operator [tex]\hat p[/tex] which functions like this:

    [tex]\hat p \Psi(x) = \Psi(-x)[/tex]

    Now it's easy to show that the eigenvalues of the parity operator are [tex]\pm 1[/tex] and the corresponding eigenfunctions are:
    all even functions for eigenvalue +1
    all odd functions for eigenvalue -1

    Now the probability of getting [tex]p = +1[/tex] is given by:

    [tex]P(p=+1) = \parallel \hat\Pi_{+1}\Psi\parallel^2 = \parallel Ae^{-x^2/a^2}\parallel^2[/tex]

    Now what I don't get is how [tex]\hat \Pi_{+1}\Psi[/tex] was calculated. Anyone care to shed a light on this for me?
  2. jcsd
  3. Dec 7, 2009 #2


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    The projection operator is defined to give zero when applied to an odd function and to give the function back when it is applied to an even function. So

    [tex] \Pi_{+1} ( xe^{-x^2/a^2} )= 0 [/tex]


    [tex] \Pi_{+1} ( e^{-x^2/a^2} )= e^{-x^2/a^2} [/tex]
  4. Dec 7, 2009 #3
    But what about the sum:


    I'm not sure how I would use this in general with other problems.
  5. Dec 7, 2009 #4
    I think I got it now, but I still find the definition of [tex]\Pi[/tex] a little bit odd.
  6. Dec 7, 2009 #5
    I doubt you are the only one.

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