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!

Completing the square for integration of e^( )

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm working on problem 2.22 from Griffith's Intro. to Quantum Mechanics (a free particle problem). I am stuck on the final integral from part b. Part a of the problem is normalizing:
    A*e-a*x2 which I did. Part b wants the general, time-dependent wave function.

    2. Relevant equations
    Griffith says that integrals of the form
    ∫e-(a*x2+b*x) can be solved by "completing the square." Griffith gives the example of defining y to be (a)(1/2)*(x+(b/(2a)) and using that definition to convert (a*x2+b*x) to y2 - (b2/(4*a)) I presume we are supposed to substitute in the converted expression into the integral and work from there.


    3. The attempt at a solution
    I have the following integral:
    ∫e-k2/(4*a)+i*(k*x-(h*k2)/(2*m)*t)
    I presume using the "completing the square" method is supposed to make this integral work out; but I don't see how this will help. If I do convert this into something like:
    ey + (something about a, h, and m),
    how do I integrate with y?

    Thanks,
    Vance
     
  2. jcsd
  3. Sep 23, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The exponential, [itex]e^{-k^2(y^2- b^2/4a})[/itex] can be separated as [itex]e^{-k^2y^2}e^{k^2b^2/4}[/itex] and then the constant, [itex]e^{k^2b^2/4a}[/itex] can be taken out of the integral, leaving [itex]\int e^{-k^2y^2}dy[/itex]. That integral cannot be done in terms of 'elementary' functions (polynomials, fractions, trig functions, exponential, logarithms). That integral is defined to be the "Error function" or "erf(x)".
     
  4. Sep 23, 2012 #3
    Thank-you so much. The missing part was the change of variables.
    Vance
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Completing the square for integration of e^( )
Loading...