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Evaluating complex integral problem

  1. Nov 18, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm having some trouble evaluating the integral
    $$\int^\infty_{-\infty} \frac{\sqrt{2a}}{\sqrt{\pi}}e^{-2ax^2}e^{-ikx}dx$$
    Where a and k are positive constants

    2. Relevant equations
    I've been given the following integral results which may be of help
    $$\int^\infty_{-\infty} e^{-x^2} dx= \sqrt{\pi}$$$$\int^\infty_{-\infty} e^{-x^2}e^{-ikx} dx = \sqrt{\pi}e^{-\frac{k^2}{4}}$$
    3. The attempt at a solution
    The best I can come up with is
    $$\int^\infty_{-\infty} \frac{\sqrt{2a}}{\sqrt{\pi}}e^{-2ax^2}e^{-ikx}dx=\sqrt{2a}e^{-\frac{k^2}{8a}}$$
    But I'm not sure this is correct. Would someone be able to explain the steps to take to evaluate this integral so I can attempt to work it out myself. Thanks.
     
  2. jcsd
  3. Nov 18, 2016 #2

    PeroK

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    Are you talking about using those standard integrals you've posted? Or do you want to derive those standard integrals?

    In the first case, it's just a simple substitution, isn't it?

    In terms of the answer, I think that factor of ##\sqrt{2a}## should have cancelled out.
     
  4. Nov 18, 2016 #3
    Sorry, I'll give you the full story. Its an assignment question in which I have to use the overlap rule to find the probability that a measurement of the particles energy will give the ground state. The given wave functions are:
    $$Ψ(x,0)=\left(\frac{\sqrt{2a}}{\sqrt{\pi}}\right)^\frac{1}{4}e^{-ax^2}e^{-ikx}$$$$ψ(x)=\left(\frac{\sqrt{2a}}{\sqrt{\pi}}\right)^\frac{1}{4}e^{-ax^2}$$
    The other integrals ##\int^{\infty}_{-\infty}e^{-x^2}dx=\sqrt{\pi}## and ##\int^{\infty}_{-\infty}e^{-x^2}e^{-ikx}dx=\sqrt{\pi}e^{-\frac{k^2}{4}}## where given in the question as integral values which may be useful.

    The overlap integral I'm using is:
    $$p=\left| \int^{\infty}_{-\infty}ψ^*(x)Ψ(x,0)dx \right|^2 $$
    After substituting in the wave function values and simplifying a bit I've eventually ended up with:
    $$p=\left| \frac{\sqrt{2a}}{\sqrt{\pi}} \int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx \right|^2$$
    Which is where I'm getting stuck.

    I was thinking I need to do something to ##\int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx## and get it into the form ##\int^{\infty}_{-\infty}e^{-x^2}e^{-ikx}dx##. I've tried completing the square and using substitution which is how I got to the answer ##\sqrt{2a}e^{-\frac{k^2}{8a}}## but I'm not confident I've done it right. Really I just need some help evaluating ##\int^{\infty}_{-\infty}e^{-2ax^2}e^{-ikx}dx##. Thanks.
     
  5. Nov 18, 2016 #4

    PeroK

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    Just do the substitution ##u = \sqrt{2a}x##.

    I think you just made an algebraic error the first time.

    By the way, when you have a parameter in a standard integral, ##k## in this case, it's best to replace that with a parameter you are not going to use in your working. E.g. Use ##\lambda## here.

    That makes it easier than using ##k## for two different things.
     
  6. Nov 18, 2016 #5
    Thanks for the tip!
    I think this is where I'm getting confused. I thought the substitution ##u=\sqrt{2a}x## would mess the integral up and make it ##\frac{1}{\sqrt{2a}}\int^{\infty}_{-\infty}e^{-u^2}e^{-ikx} dx##?
     
  7. Nov 18, 2016 #6

    PeroK

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    If you can't do integration then QM will be hard indeed. You must substitute ##u## all the way thru the integral. Just replacing ##x## in one place is no good.
     
  8. Nov 18, 2016 #7
    Ok, thanks for your help with this.
     
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