(Hopefully easy) integration question

  • Thread starter lotm
  • Start date
  • #1
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Heyo,

I'm having difficulty seeing how these two lines follow. I'm fairly sure I'm being an eejit and the answer's straightforward, but would appreciate a quick explanation of what's going on.

[tex]\frac{1}{2\pi}\int d^3p e^{-i(\emph{p}^2/2m)t} \\ \times e^{i\emph{p.(x-x_0)}} \\
= (\frac{m}{2 \pi it})^{3/2}e^{\frac{im(\emph{x-x_0}^2}{2t}}[/tex]

Thanks in advance.
 
Last edited:

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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Which two lines do you mean exactly?
I only see a single expression.
 
  • #3
7
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Yeah, I'm struggling a bit to get the tex code right. Screw it, I'll come back once I've figured out how to ask the question properly.
 
  • #4
306
1
Do you mean:

[tex]
\frac{1}{2\pi}\int d^3p e^{-i(\vec{p}^2/2m)t}
e^{i\vec{p}\cdot(\vec{x}-\vec{x_0})}
= \left (\frac{m}{2 \pi it}\right )^{3/2}e^{\frac{im(\vec{x-x_0})^2}{2t}}
[/tex]
 
  • #5
306
1
If that's the case then you can click on the equation to see the markup.

What you've got here, is a Fourier Transform of a Gaussian returning a Gaussian. It is easy though it might look daunting. Complete the square and use a little substitution and you'll be all set.
 

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