Recent content by 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...

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 please...

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}...

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...

you made a error integrating, i do it by my self and the result is: \phi(p) = \frac{4\pi}{(2\pi\hbar)^{3/2}}\frac{1}{\pi a^3}\frac{2a^3 \hbar ^4}{\left(\hbar ^2 + a^2 p^2 \right)^2} \phi(k) = \frac{4\pi}{(2\pi\hbar)^{3/2}}\frac{1}{\pi a^3}\frac{2a^3 \hbar ^4}{\left(\hbar ^2 +\hbar ^2...