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!

Homework Help: 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:


    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


    User Avatar
    Homework Helper
    Gold Member

    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


    User Avatar
    Homework Helper
    Gold Member

    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°).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook