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Wave Function - Normalisation & Calculation of Expectation Values

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data

    i. Confirming the wavefunction is normalised

    ii. Calculating the expectation values: [tex]<\hat{x}> , <\hat{x^{2}}> , <\hat{p}> , <\hat{p^{2}}>[/tex] as a function of [tex]\sigma[/tex]

    iii. Interpreting the results in regards to Heisenberg's uncertainty relation.

    2. Relevant equations

    wave function [tex]\psi(x) = (2\pi\sigma)^{\frac{-1}{4}}exp[\frac{-x^{2}}{4\sigma}][/tex]

    3. The attempt at a solution

    i. I know that a wave function [tex]\psi(x)[/tex] is normalised if:

    [tex]|\psi(x)|^{2} = 1[/tex]

    So I have tried to modulus [tex]\psi(x)[/tex] and then square it, but this doesn't equal to 1. Or at least I'm just doing it wrong.. I'm thinking maybe have to take the real part of the exponential term in the equation.. or something like that, but I don't know.

    ii. I'm not sure how to calculate these for this wave function.
     
    Last edited: Feb 21, 2010
  2. jcsd
  3. Feb 21, 2010 #2

    gabbagabbahey

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    No, the wavefunction is normalized if [itex]\langle\psi|\psi\rangle=1[/itex]. You can expand this in the position basis by using the identity,

    [tex]\hat{1}=\int_{-\infty}^{\infty}|x\rangle\langle x|dx[/tex]

    (Where [tex]\hat{1}[/itex] is the identity operator) Using this, along with the fact that [tex]\hat{1}|\psi\rangle=|\psi\rangle[/itex], you have

    [tex]1=\langle\psi|\hat{1}|\psi\rangle=\int_{-\infty}^{\infty}\langle\psi|x\rangle\langle x|\psi\rangle dx=\int_{-\infty}^{\infty}|\psi(x)|^2dx[/tex]

    This derivation is done in every introductory QM text I can recall, so the fact that you don't know it by heart yet suggests you need to study your text and notes.
     
  4. Feb 21, 2010 #3
    I actually just missed off the indefinate integral from what I posted by mistake.

    I know it shouldn't be that hard to show, but I just can't figure out what steps to take at the moment!
     
    Last edited: Feb 21, 2010
  5. Feb 21, 2010 #4

    gabbagabbahey

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    Okay, so did you try evaluating the integral? If so, what did you get?
     
  6. Feb 21, 2010 #5
    I think I've figured it out now..

    [tex]|\psi(x)| = (2\pi\|\sigma|)^{-\frac{1}{4}}exp^{(-\frac{1}{4})Re(\frac{-x^{2}}{\sigma})} = (2\pi\sigma)^{-\frac{1}{4}}exp^{(-\frac{1}{4})(\frac{-x^{2}}{\sigma})}[/tex]

    Then can square this:

    [tex]((2\pi\sigma)^{-(\frac{1}{4})})^{2} = (2\pi\sigma)^{-(\frac{1}{2})}[/tex]

    Also:

    [tex](exp^{(-\frac{1}{4})(\frac{-x^{2}}{\sigma})})^{2} = exp^{(\frac{x^{2}}{4\sigma})}[/tex]

    So:

    [tex]\integral exp^{(\frac{x^{2}}{4\sigma})} = \sqrt{2\sigma\pi}[/tex]

    And:

    [tex](2\pi\sigma)^{(\frac{-1}{2})} = \frac{1}{\sqrt(2\pi\sigma)}[/tex]

    Therefore:

    [tex]\integral |\psi(x)|^{2} = \frac{(\sqrt{2\sigma\pi})}{(\sqrt{2\sigma\pi})} = 1[/tex]

    How's that?
     
  7. Feb 21, 2010 #6

    gabbagabbahey

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    No,

    [tex]\left(e^{-\frac{x^2}{4\sigma}\right)^2=e^{-\frac{x^2}{2\sigma}[/tex]

    This makes absolutely no sense. On the RHS of this equation you have a function with a numerical value that depends on the value of [itex]x[/itex]. On the LHS you have a constant. How can these two quantities possibly be equal to each other for all [itex]x[/itex]?
     
  8. Feb 21, 2010 #7
    [tex]\left(e^{-\frac{x^2}{4\sigma}\right)^2=e^{-\frac{2x^2}{4\sigma}} = e^{-\frac{x^2}{2\sigma}[/tex]

    My lecturer gave the hint:

    [tex]\integral exp(-ax^{2}) = \sqrt{\frac{\pi}{a}}[/tex]

    Hence from this I took [tex]a = \frac{1}{2\sigma}[/tex]

    Therefore:

    [tex]\integral exp(\frac{-x^{2}}{2\sigma}) = \sqrt{2\pi\sigma}[/tex]
     
  9. Feb 21, 2010 #8

    gabbagabbahey

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    Do you mean

    [tex]\int_{-\infty}^{\infty} e^{-ax^2}dx= \sqrt{\frac{\pi}{a}}[/tex]

    ???

    Try using \int_{-\infty}^{\infty} instead of \integral so it doesn't look like you are claiming [tex]e^{-ax^2}= \sqrt{\frac{\pi}{a}}[/tex].

    If this fixed post is what you meant, then yes, that's correct.
     
  10. Feb 21, 2010 #9
    Yes that is what I meant! =D

    Just got to find the expectation values now.. I'll give that a go..
     
  11. Feb 21, 2010 #10
    [tex]

    <\hat{x}> = \int_{-\infty}^{\infty} \psi^{*}x\psi dx

    [/tex]

    Therefore:

    [tex]\psi^{*}\psi= ((2\pi\sigma)^{\frac{-1}{4}})^{2}(exp(\frac{-x^{2}}{4\sigma}))^{2} =(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma})[/tex]

    [tex]x\psi^{*}\psi= x(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma})[/tex]

    [tex] \int x \psi^{*}\psi= \int x(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma}) = (2\pi\sigma)^{\frac{-1}{2}} \int x exp(\frac{-x^{2}}{2\sigma}) [/tex]

    [tex](2\pi\sigma)^{\frac{-1}{2}} \int x exp(\frac{-x^{2}}{2\sigma})


    = (2\pi\sigma)^{\frac{-1}{2}} [\frac{-2x^{3}}{2\sigma} exp^{(\frac{-x^{2}}{2\sigma})} - \frac{-2x^{2}}{2\sigma} exp^{(\frac{-x^{2}}{2\sigma})}]

    = (2\pi\sigma)^{\frac{-1}{2}} (x)

    = <\hat{x}>


    [/tex]

    .. hopefully this is correct.. or at least along the right lines?
     
    Last edited: Feb 21, 2010
  12. Feb 27, 2010 #11

    gabbagabbahey

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    If you are integrating with respect to [itex]x[/itex], then shouldn't there be some [itex]dx[/itex]'s in these integrals? And you are also neglecting to put the limits of integration in.

    This may seem nitpicky, but what good is performing a calculation if you can't communicate it clearly to other physicists? You could (theoretically) come up with some brilliant unified field theory that revolutionizes physics and fail to get it published, simply because the reviewers don't look past the basic errors in notation. For this reason, you will likely find that your professors will deduct marks for these somewhat inconsequential errors when they are present on your assignments. You should get in the habit of proofreading everything you write/ type before submitting it, whether it is to your professor, a scientific journal, or even just an internet forum.

    There are numerous calculation errors here. When calculating an integral, it is always a good idea to differentiate the result (before substituting limits) to make sure you get the original integrand back. If you don't, then you've done something wrong.

    So, do you get [itex]xe^{-x^2/2\sigma}[/itex] back when you differentiate [tex] \frac{-2x^{3}}{2\sigma} e^{\frac{-x^{2}}{2\sigma}} - \frac{-2x^{2}}{2\sigma} e^{\frac{-x^{2}}{2\sigma}}[/itex]?
     
  13. Feb 27, 2010 #12
    No..

    [tex] \left< x \right> = \int_{-\infty}^{\infty} x\left(2\pi\sigma\right)^{-\frac{1}{2}}e^{-\left(\frac{x^{2}}{2\sigma}\right) }dx[/tex]

    So:

    [tex] \left< x \right> = 0 [/tex]
     
  14. Feb 27, 2010 #13

    gabbagabbahey

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    That's much better!:approve:

    Now, how about the rest of the expectation values?
     
  15. Feb 27, 2010 #14
    Good! Yeah, basically I looked at my notes again and realised I was going about them completely the wrong way! Will get on with having a go at the other expectation values..
     
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