- #1
nhrock3
- 415
- 0
there is a function [tex]\Psi =\frac{c}{\sqrt{r}}e^{\frac{-r}{b}}[/tex]
find the probaility in [tex]\frac{b}{2}<r<\frac{3b}{2}\\[/tex] region
the rule states [tex]\int_{-\infty}^{+\infty}|\Psi|^2dv=1\\[/tex]
[tex]1=\int_{-\infty}^{+\infty}|\frac{c}{\sqrt{r}}e^{\frac{-r}{b}}|^2dv[/tex]
then they develop it as
[tex]c^2\int _{all space}\frac{1}{r}e^{\frac{-2r}{b}}2\pir^2dr=4\pic^2\int_{0}^{+\infty}re^{\frac{-2r}{b}}dr\\[/tex]
they as it because of spherical coordinates
but i can't see here the jacobian of spherical coordinates.
i can't see here the x,y,z transition to r ,theta,phi
i can't see it in the last equation
find the probaility in [tex]\frac{b}{2}<r<\frac{3b}{2}\\[/tex] region
the rule states [tex]\int_{-\infty}^{+\infty}|\Psi|^2dv=1\\[/tex]
[tex]1=\int_{-\infty}^{+\infty}|\frac{c}{\sqrt{r}}e^{\frac{-r}{b}}|^2dv[/tex]
then they develop it as
[tex]c^2\int _{all space}\frac{1}{r}e^{\frac{-2r}{b}}2\pir^2dr=4\pic^2\int_{0}^{+\infty}re^{\frac{-2r}{b}}dr\\[/tex]
they as it because of spherical coordinates
but i can't see here the jacobian of spherical coordinates.
i can't see here the x,y,z transition to r ,theta,phi
i can't see it in the last equation