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Tricky integral in QM variational method

  1. Sep 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the variational parameters [itex]\beta[/itex], [itex]\mu[/itex] for a particle in one in one dimension whose group-state wave function is given as:

    [itex]\varphi[/itex]([itex]\beta[/itex],[itex]\mu[/itex])=Asin(βx)exp(-[itex]\mu[/itex][itex]x^{2}[/itex]) for x≥0.

    The wavefunction is zero for x<0.

    2. Relevant equations
    The Hamiltonian is given as:

    H=-[itex]\frac{\\hbar^{2}}{2m}[/itex][itex]\frac{d^{2}}{dx^{2}}[/itex]+V(x)

    Where the potential field is defined as follows: V(x)=+∞ for x<0
    and V(x)=[itex]\frac{-f}{(x+a)^{2}}[/itex] for x≥0

    The terms f, a are positive constants.

    3. The attempt at a solution
    I am familiar with the general procedure. I know that

    E(β,μ)=<T> + <V>

    Where E are the eigen energies and T, V are the kinetic and potential energies, respectively.

    To minimize the Hamiltonian one takes partial derivatives of E with respect to β and μ, setting each term equal to zero to determine the variational parameters. This is all straightfoward.

    My confusion lies in the evaluation of the normalization coefficient A from <[itex]\varphi[/itex]|[itex]\varphi[/itex]>=1

    The closest tabulated integral has "x" in the exponential term, not "x[itex]^{2}[/itex]"

    Any help on solving this integral would be greatly appreciated.
     
  2. jcsd
  3. Sep 24, 2012 #2

    gabbagabbahey

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    Hi akehn, welcome to PF! :smile:

    Integrals of the form [itex]\int_0^\infty e^{-x^2}dx[/itex] can be easily computed by first calculating the square of the integral, transforming the problem from one of calculating a one-dimensional integral to one of calculating a two-dimensional integral over a region, since the 2D integral can be done easily by switching to polar coordinates

    [tex]\begin{align} \left( \int_0^{\infty} e^{-x^2}dx \right)^2 &= \left( \int_0^{\infty} e^{-x^2}dx \right) \left( \int_0^\infty e^{-x^2}dx \right) \\ &= \left( \int_0^{\infty} e^{-x^2}dx \right)\left( \int_0^\infty e^{-y^2}dy \right) \\ &= \int_0^{\infty} \int_0^{\infty} e^{-(x^2 + y^2)}dxdy \\ &= \int_0^{\infty} \int_0^{\frac{\pi}{2}} e^{-r^2}r dr d\theta \end{align}[/tex]

    You can utilize this technique for your integral, by using Euler's formula to write [itex]\sin^2 (\beta x)[/itex] in terms of complex exponentials, then "completing the square on the exponent" and making a substitution to get your integrand in the correct form.
     
    Last edited: Sep 24, 2012
  4. Sep 24, 2012 #3
    Ah ha! Thank you very much.
     
  5. Sep 24, 2012 #4

    gabbagabbahey

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    You're welcome! Note that I made a small edit to my post to correct typos, the only one of significance being that the upper limit on the polar angle in the last integral should be [itex]\frac{\pi}{2}[/itex] (90°), not [itex]\frac{\pi}{4}[/itex] (45°).
     
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