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I'm given $\Psi(x,t)$ as a proposal for a wave function. $\Psi(x,t)=\int_{1}^{1+\Delta k} e^{i(kx-wt)} k^2 dk$

Now I try to compute $\Psi^*(x,t)\Psi(x,t)$ wich is the product

$(\int_{1}^{1+\Delta k} e^{-i(kx-wt)} k^2 dk) (\int_{1}^{1+\Delta k} e^{i(kx-wt)}k^2 dk)$

In wich way should I transform this to a double integral? Taking into account that $w=w(|k|)$

Thanks for your time.

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There is no comment about $\Delta k$ so I assumed is a real number because $k$ is. –  Jorge Sep 25 '11 at 12:48
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This is a product of Fourier integrals, so you can obtain its Fourier transform as the convolution of the two Fourier transforms. –  joriki Sep 25 '11 at 13:01
    
In Fourier Transforms, as far as I know, the range of integration is $\mathbb{R}$ not just $(1,1+\Delta k)$ –  Jorge Sep 25 '11 at 13:04
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1 Answer 1

It is easier to solve the one dimensional integral, and then to perform the multiplication:

$\int_1^{1+\Delta} e^{ikx-i\omega t} k^2 dk = - e^{-i\omega t} \frac{\partial}{\partial x^2}( \int_1^{1+\Delta} e^{ikx} dk) = - \Delta e^{-i\omega t} \frac{\partial}{\partial x^2}\left ( e^{[i(1+\Delta/2)x]} \frac{sin(\frac{\Delta x}{2})}{\frac{\Delta x}{2}}\right) $

All is left is to perform the differentiationwith respect to x.

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Thank you very much, nice idea! –  Jorge Sep 25 '11 at 15:13
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