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Dirac notation - expectation value of kinetic energy

  1. Jan 18, 2016 #1
    It's my first post so big thanks in advance :)

    1. The problem statement, all variables and given/known data

    So the question states "By interpreting <pxΨ|pxΨ> in terms of an integral over x, express <Ekin> in terms of an integral involving |∂Ψ/∂x|. Confirm explicitly that your answer cannot be negative in value." ##The 'px's should have hats to indicate an operator##
    If it helps, it's part of a question where Ψ is the wavefunction of a particle in a 1D pot energy well and the question is testing on a chapter about introducing and working with Dirac notation.

    2. Relevant equations
    None that I can think of

    3. The attempt at a solution
    The maths of my solution so far is
    CJx2b66.png

    So, the part I'm stuck with is proving explicitly that it can't be negative in value. Obviously the factor to the left is always positive but it's the integral that's the problem. I know the integrand is always real and positive as it's a modulus squared, I just can't see any way to prove that the integral of that, over infinity, is never negative.

    It's due in for next Monday so any help would be greatly appreciated!


     
  2. jcsd
  3. Jan 18, 2016 #2

    TSny

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    Looks good.

    You are integrating from ##- \infty## to ##+ \infty##. So, what is the sign of ##dx## in the integral?
     
  4. Jan 19, 2016 #3
    I'm not sure what you mean - I'd presume ##\text{d}x## is always positive, as it's effectively a small interval ##\delta x## centered on ##x##.

    So are you saying if I take the integral to be the infinite sum of small sections ##| \partial \Psi / \partial x|^2 ~ \delta x##, then each section must be positive, as both ##| \partial \Psi / \partial x|^2## and ##\delta x## are always positive?

    Thanks
     
  5. Jan 19, 2016 #4

    TSny

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    Yes.

    Another way to think about it is that ##|\partial \Psi / \partial x|^2## is a non-negative function of x. So, the integral is the "area under" the graph of a function that never goes below the x axis. So, the area is positive according to the sign conventions of area-under-a-graph as used in calculus.
     
  6. Jan 19, 2016 #5
    Ah right, I think I get you.

    So, to check I understand it - if the limits of integration were reversed ##\int^{-\infty}_{\infty}## (so you're integrating across the graph in the negative x direction), would the integral be negative?
     
  7. Jan 19, 2016 #6

    TSny

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    Yes, that's right.
     
  8. Jan 19, 2016 #7
    Awesome. Thanks for all the help :smile:
     
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