Hellp with fourier integral via completing the square

  • Thread starter mmwave
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  • #1
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Main Question or Discussion Point

This comes from quantum mechanics but it's basically a Fourier integral I can't quite do...

F(k) = 1/sqrt(2a*[pi]) * [inte] exp( -(ax^2+ikx) dx over infinite limits. i is sqrt(-1)

to do this, I complete the square getting

exp( -(sqrt(a)*(x +ik/(2a))^2 * exp(k^2 / (4a))

sticking this in the integral and integrating over x I get

F(k) = 1/sqrt(2a*[pi]) * exp(k^2 / 4a )

I like the k^2 / a part but the factor of 4 seems wrong as well as the sign of the exponent possibly.

exp(-a x^2) should transform to exp(k^2/a)?

Help with using completing the square to do this integral would be greatly appreciated.

I used [inte] exp(-y^2) = sqrt([pi]) from a table.
 

Answers and Replies

  • #2
Hurkyl
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Your completion of the square looks wrong. Try completing this square (and testing your answer) as a seperate problem... working subproblems seperately tend to make it easier to see what's wrong.

(Incidentally, Mathematica agrees with the factor of 4)

Oh, I also presume that a is supposed to be positive? (It makes a big difference)
 
  • #3
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Ok,

x^2 + ikx/a = (x + ik/2a)^2 + k^2/4a^2

so my exponential becomes

exp(-a * (k/2a)^2) * exp(-a * (x + ik/2a))

the integral then is exp(-a * (k/2a)^2) * sqrt([pi]/a)

(i'm ignoring the constant in front of the integral for now)

and a is positive constant

the final answer becomes 1/sqrt(2*[pi]) * sqrt([pi]/a) * exp(-k^2/4a)

I like that better, exp(k^2) is not absolutely integralable.
 

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