Inverse Foourier Transform

1. Apr 4, 2008

Gatts

Inverse Fourier Transform

I have to calculate (don't take in account the units, obviously the're extrange)

$$\phi (r)=\frac{1}{(2\pi\hbar)^{3/2}}\int d^{3}p\hspace{7mm}{e^{i\frac{-p\cdot r}{\hbar}}\Psi(p)}$$

$$\Psi(p)=\frac{B}{(1+\frac{p^2}{m^2})^2}$$
I know that

$$d^{3}p=p^{2}Sin(\theta)dpd\theta d\phi$$

So i do

$$\phi (r)=\frac{1}{(2\pi\hbar)^{3/2}} \int p^{2}Sin(\theta)dpd\theta d\phi\hspace{7mm}{e^{i\frac{-p\cdot r Cos(\theta)}{\hbar}}\frac{B}{(1+\frac{p^2}{m^2})^2} }$$

but i use the change of variables

$$u=Cos(\theta);du=-Sin(\theta)d\theta$$

And the the integral take the form

$$\phi (r)=\frac{2\pi}{(2\pi\hbar)^{3/2}} \int_{1}^{-1}\int_{0}^{\infty}p^{2}dpdu \hspace{7mm}{e^{i\frac{-p\cdot r u}{\hbar}}\frac{B}{(1+\frac{p^2}{m^2})^2} }$$

Last edited: Apr 4, 2008
2. Apr 4, 2008

Gatts

Integrating

$$\phi (r)=\frac{2\pi} {(2\pi\hbar)^{3/2}} B \int_{0}^{\infty}pdp \hspace{7mm} \frac{\left(e^{i\frac{p r}{\hbar}}-e^{i\frac{p r}{\hbar}} \right){\frac{i p r}{hbar}\frac{1}{(1+\frac{p^2}{m^2})^2}$$

Multipliying for $$\frac{2}{2}$$ to form a cosine, so it is

$$\phi (r)=\frac{2\pi} {(2\pi\hbar)^{3/2}}\frac{2 B\hbar}{i r} \int_{0}^{\infty}pdp \hspace{7mm} Cos(\frac{p r}{\hbar})\frac{1}{(1+\frac{p^2}{m^2})^2} }$$

Last edited: Apr 4, 2008
3. Apr 4, 2008

Gatts

Bu the las integral, it is just a nightmare... the results froma Mathetmatica 6.1 is:

$$\frac{1}{2} m^2 \sqrt{\pi } MeigerG[{{0},{}},{{0,1},{\frac{1}{2}},\frac{m^2 r^2}{4\hbar^2}]$$

i as expecting the base level for the H atom $$\phi (r)$$, so if somebody could help me plz....

Last edited: Apr 4, 2008
4. Apr 4, 2008

Gatts

Na i found my error was a Sine not a cosine what I get afeter I do the first integral in du, I don't know why the page don't uptadete what i've write, but that was the error, mulltiply bi 2i/2i and you will get the base state for the H atom, THE ANSWER...
first part of the trip...