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Outcomes of measuring ##p_x## and their probabilities

  1. Oct 9, 2016 #1
    1. The problem statement, all variables and given/known data
    The operator for the momentum of a particle in 1D is ##\hat{p}_x = -i\hbar\frac{d}{dx}## has the eigenfunction ##\psi(x) = e^{ikx}##.

    The Schrodinger equation for a free particle in 1D is
    ##\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi##
    Which has the solution ##\psi(x) = cos(kx)##.

    a) Find the corresponding energy eigenvalue E.
    b)What are possible outcomes of measuring ##\hat{p}_x## and the relative probabilities of these outcomes?
    c)Is it true that the radial probability density for the electron in the lowest-energy atomic state of hydrogen is
    ##P(r) = \frac{1}{\pi a_0^3}e^{-2r/a_0}##, if the normalised wavefunction is

    ##\psi(r) = \sqrt{\frac{1}{\pi a_0^3}}e^{-r/a_0}##

    And what's the most probable value of ##r##?

    2. Relevant equations


    3. The attempt at a solution
    For a) I subbed ##\psi(x)## into the LHS and set it equal to the RHS, which gave

    ##\frac{\hbar^2k^2}{2m} \psi(x) = E \psi(x)##

    So the eigenvalue is ##\frac{\hbar^2k^2}{2m}##.

    b) The possible outcomes are the eigenvalues of the momentum operator which are ##\hbar k##. For their relative probabilities I think you have to calculate coefficients of something, can't remember what, which are given by

    ##c_n = \int_{-\infty}^{\infty} e^{ikx}\cos{(kx)} dx##

    Which is a tricky integral and in any case I don't know why those are the probabilities, I'm just basing it on an example we were given. I'd have to repeat the integral again with ##e^{-ikx}## afterwards. So is this the right method and if so how do I integrate that? And why are those the probabilities?

    c) I think the probability density function is correct, calculated using ##P(r) = \psi^{*}\psi##. Is the most probable value of ##r## given by finding the maximum of the probability density function? I'm not sure differentiating and setting = 0 really gets me anywhere because it's an exponential and only has a turning point not a maximum, right?

    Thanks for any help!
     
  2. jcsd
  3. Oct 9, 2016 #2

    TSny

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    For (b), can you write cos(kx) in terms of eigenfunctions of ##\hat{p}_x##?

    For (c), note that they are talking about the "radial probability density" rather than just the "probability density".
     
  4. Oct 9, 2016 #3
    ##cos(kx) = \frac{1}{2} (e^{ikx}+e^{-ikx})##, those are both eigenfunctions aren't they? The eigenfunctions added together are still an eigenfunction. But what would I do with it then, integrate it?

    Radial probability density as opposed to probability density... is the answer the Bohr radius then? By definition, I suppose. So it's not a calculation question really? Or can that be shown from the probability density?
     
    Last edited: Oct 9, 2016
  5. Oct 9, 2016 #4

    TSny

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    Yes.
    Check to see if that is true.
    Review the basic postulates of quantum mechanics. http://sydney.edu.au/science/chemistry/~mjtj/CHEM3117/Resources/postulates.pdf
    In particular, postulate 3 in this link is relevant. There is a kink in applying this postulate to your wavefunction due to the fact that your wavefunction is not normalizable. But, you can still deduce the answer for part (b) of the question.

    A brief review of the radial probability density (distribution) is given here
    https://www.dartmouth.edu/~genchem/0405/spring/6belbruno/radial.pdf
     
  6. Oct 10, 2016 #5
    So I need a complete set of eigenvectors. I have a set of eigenfunctions though, don't I? And based on the postulate I would square the coefficients, i.e. ##\frac{1}{2}##, giving me a probability of ##\frac{1}{4}## for each outcome. If it were normalisable the probability of each outcome would be ##\frac{1}{2}##. So I'm ##\frac{1}{4}## is wrong and there's something you have to do to account for non-normalisability? Why is it even possible for the wavefunction to be non-normalisable for this single particle in 1D? I thought those sorts of wavefunctions tended to describe beams of particles.

    Oh, that's not the correct radial probability density then... It should be multiplied by ##4\pi r^2dr## giving radial probability density of
    ##\frac{4}{a_0^3}r^2e^{-\frac{2r}{a_0}}dr##
    Which if I differentiate w.r.t. r and set equal to zero gives the Bohr radius ##a_0## as the maximum!
     
    Last edited: Oct 10, 2016
  7. Oct 10, 2016 #6

    TSny

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    What's important is that you can express your wavefunction as a linear superposition of eigenfunctions of ##\hat{p}_x##.
    The squares of the coefficients give you the probabilities only if the wavefunction has been normalized.
    Yes
    When the wavefunction is not normalized, then the squares of the coefficients are proportional to the probabilities. In your case, that's all you need in order to answer the question since the probabilities must add to 1.
    It's a peculiarity of dealing with a free particle in infinite space. As you can see, there is no way to normalize the wavefunction cos(kx).
    Not necessarily. I guess your wavefunction could be thought of as a superposition of two beams traveling in opposite directions.

    The ##dr## would not be included in the radial density function. The density function is the coefficient of ##dr## in the expression for the probability of finding the particle between ##r## and ##r+dr##.
    Yes, that's right.
     
  8. Oct 10, 2016 #7
    The only thing I can think, then, is that the probability of each outcome is still ##\frac{1}{2}##, which is clearly proportional to ##\frac{1}{4}## and results in addition to 1. But the proportionality thing isn't stated in the postulates, is it?
     
  9. Oct 10, 2016 #8
    Actually, if this is sort of two beams moving in opposite directions, then each would have an amplitude given by their coefficients. So another way to arrive at ##\frac{1}{2}## would be to say the probability of moving in one direction is
    ##\frac{|A|^2}{|A|^2+|B|^2}##
    If A and B are the amplitudes of the left and right moving beams respectively. Which, subbing in ##A = B = \frac{1}{2}## gives ##\frac{1}{2}##. Does that work? In the sense of being correct physics?

    Regardless, thank you very much for your help and patience!
     
  10. Oct 10, 2016 #9

    TSny

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    Yes, that's a good way of thinking about it.
     
  11. Oct 10, 2016 #10
    Great! Thanks again for your help! :)
     
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