1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

QM expectation values of x

  1. Apr 9, 2009 #1
    1. The problem statement, all variables and given/known data
    calculate <x>, when [tex]\Psi[/tex](x,t)=A*exp(-([tex]\sqrt{}Cm[/tex]/2h)x[tex]^{}2[/tex]

    2. Relevant equations
    <x>=[tex]\int[/tex][tex]\Psi^{}*[/tex]x[tex]\Psi[/tex]dx over all space..

    3. The attempt at a solution
    ok know how to do this but how do i do the intergral... my maths isnt so good, and the book does it very very vague... i know <x>=0 but dont know how to prove it.
  2. jcsd
  3. Apr 9, 2009 #2
    so far ive got down to

    A2[tex]\int[/tex]exp(-([tex]\sqrt{}Cm[/tex]/2h)x2)x dx, fomr -a/2 to a/2
  4. Apr 9, 2009 #3
    Hi fredrick.
    Just want to make sure, you aren't going to integrate the whole space, right?
    Well, there are several ways of solving this problem.
    So the easiest one is, well, do the math.
    So it seems like you are stuck and don't know how to do the integration.

    Just a very small hint, what is the anti derivative of exp(ax^2)*x?
    not sure? Well, obviously, it is going to relate to exp(ax^2), right? There is no other way to get this term, right?
    So it is probably something with exp(ax^2).
    Know, try to differentiate your "guess" function. And remember the chain rule. And see if it is the same as exp(a*x^2)x
    Once you get your guessed function right, I think you can find the anti derivative of your function.
    And the rest is pretty much plug in

    I'll tell the other way, which doesn't even involve doing all the dirty job after you finish your calculation :D
    Good luck!
  5. Apr 10, 2009 #4


    User Avatar
    Science Advisor
    Gold Member

    Actually, if you have an x as well as an exp(x^2) as the integrand, there is no need to even worry about integrating exp(x^2).

    Make a substitution u=ax^2, du=2ax and simply do it that way.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook