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Homework Help: Statistical interpretation (intro to quantum)

  1. May 2, 2012 #1
    When we calculate the average of anything: we add up (or integrate) the sum{ [all the things were taking the average of] * [the probability of getting that thing ]}.

    The thing about the average i'm that curious about is for example: the average height of 5 people turns out to be say 6 feet. suppose here it turns out that NONE of the people have a height of exactly 6 feet.

    So the "average" (expectation value) doesn't correspond to an actual physical value that any one person has.

    So why is it that when we find the expectation value (of say position) on a particle it corresponds to a physical value that we can measure; when this value doesn't (necessarily) correspond to any one actual value that the particle has?

    I think the root of my problem is; how is it that the square of our wave function corresponds to the probability of finding that particle (at a given time and place) ?

    Perhaps i'm confusing a single measurement on a particle compared to multiple measurements on a particle in multiple equally prepared states??

    Anyways any clarification would be appreciated, Thanks
    1. The problem statement, all variables and given/known data

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    3. The attempt at a solution
  2. jcsd
  3. May 3, 2012 #2
    The term 'expectation value' is an unfortunate one and really should be something more like 'ensemble average'. Imagine you know all of the accessible quantum states of a given system, and there are g of them. And you were to construct an ensemble of g systems one in each of the accessible states. If you were to measure the value of some quantity X for each of the systems and then add up the values and divide by g, this would be the ensemble average or 'expectation value' and there is no reason to expect that the number should be one that is possible to actually measure.

    In regards to your question about why the square of a position wave function gives a probability distribution of position, well it is a result of the generalized statistical interpretation. Here is how quantum theory works:

    Given an operator, if one can find all of its eigenvectors and eigenvalues, then one can express the state of a system as a linear combination (or integral if the spectrum of eigenvalues is continuous) of the eigenvectors of the operator. If fn is an eigenfunction with eigenvalue n, and the eigenvalue is non degenerate, then the probability to measure the value n is given by the square of the constant on fn in the linear combination. For example, if the operator is the position in one dimension: x, then the eigenfunctions are delta functions, because:

    [tex] x \delta(x-y) = y \delta(x-y) [/tex]

    and the eigenvalues are any real number y, so that the spectrum is continuous. Now we can write the wave function as an integral over eigenfunctions:

    [tex] \Psi(x)=\int c(y) \delta(x-y) dy [/tex]

    and as usual the coefficients c(y) can be found by fouriers trick (exploiting the "dirac" orthonormality of the eigenfunctions):

    [tex] c(y) = \int \Psi(x) \delta(x-y) dx = \Psi(y) [/tex]

    now by the statistical interpretation the probability that we measure a position between y and y+dy is:

    [tex] P(y+dy) = | c(y) |^2 dy = |\Psi(y)|^2 dy [/tex]

    which is the well known result. Hope that helps
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