- #1
Gatts
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Inverse Fourier Transform
I have to calculate (don't take in account the units, obviously the're extrange)
[tex]
\phi (r)=\frac{1}{(2\pi\hbar)^{3/2}}\int d^{3}p\hspace{7mm}{e^{i\frac{-p\cdot r}{\hbar}}\Psi(p)} [/tex]
[tex]
\Psi(p)=\frac{B}{(1+\frac{p^2}{m^2})^2}
[/tex]
I know that
[tex]
d^{3}p=p^{2}Sin(\theta)dpd\theta d\phi[/tex]
So i do
[tex]
\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}
} [/tex]
but i use the change of variables
[tex]
u=Cos(\theta);du=-Sin(\theta)d\theta [/tex]
And the the integral take the form
[tex]
\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}
} [/tex]
I have to calculate (don't take in account the units, obviously the're extrange)
[tex]
\phi (r)=\frac{1}{(2\pi\hbar)^{3/2}}\int d^{3}p\hspace{7mm}{e^{i\frac{-p\cdot r}{\hbar}}\Psi(p)} [/tex]
[tex]
\Psi(p)=\frac{B}{(1+\frac{p^2}{m^2})^2}
[/tex]
I know that
[tex]
d^{3}p=p^{2}Sin(\theta)dpd\theta d\phi[/tex]
So i do
[tex]
\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}
} [/tex]
but i use the change of variables
[tex]
u=Cos(\theta);du=-Sin(\theta)d\theta [/tex]
And the the integral take the form
[tex]
\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}
} [/tex]
Last edited: