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Complete the square

  1. Oct 27, 2007 #1
    this isn't really a homework problem

    it's just me trying to understand a part of an example problem from a modern physics book

    it's an integral of a wave packet, blah blah blah

    but they go on to say "...to evaluate the integral, we first complete the square in the exponent as..."

    [tex]ikx - a^2k^2 = - (ak - \frac{ix}{2a} )^2 - \frac{x^2}{4a^2} [/tex]

    how in the world does one arrive at that???
     
    Last edited: Oct 27, 2007
  2. jcsd
  3. Oct 27, 2007 #2

    cristo

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  4. Oct 27, 2007 #3
    i appreciate the link

    but it doesn't help

    edit: nm, i see what's going on....but can someone explain wtf
     
    Last edited: Oct 27, 2007
  5. Oct 27, 2007 #4

    Kurdt

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    If you can elaborate on what exactly it is you're struggling with then perhaps someone can help you.
     
  6. Oct 27, 2007 #5
    getting started lol

    how would you work with the [tex]a^2k^2[/tex]
     
  7. Oct 27, 2007 #6

    arildno

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    We can start with:
    [tex]ikx-a^{2}k^{2}=-((ak)^{2}-ikx))=-((ak)^{2}-2(ak)\frac{ix}{2a})[/tex]
    What must you add&subtract in order to generate an expression in which k is hidden away linearly within a square?
     
  8. Oct 27, 2007 #7
    where did you get the [tex]\frac{ix}{2a}[/tex]
     
  9. Oct 28, 2007 #8

    arildno

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    [tex]ikx=1*ikx=\frac{2a}{2a}*ikx=2(ak)\frac{ix}{2a}[/tex]
     
  10. Oct 28, 2007 #9

    Kurdt

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    If you take a look at a general example it may help clarify. We use the fact that:

    [tex] (x+b)^2 = x^2+2bx +b^2 [/tex]

    which can be written as,

    [tex] (x+b)^2 -b^2 = x^2 +2bx [/tex]

    Now if you look at what arildno has done in post number 6, hes written your equation in the form [itex] x^2+2bx[/itex] where [itex]x=-(ak)[/itex] and [itex]b=\frac{ix}{2a}[/itex]. Now I'm sure you can confirm for yourself from here where they obtained the equation in your original post.
     
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