Phy5645/probabilitycurrentdensity

Lets consider that our wavefunction in spherical coordinates is$\psi ={\frac {e^{\pm i{\overrightarrow {k}}.{\overrightarrow {r}}}}{r}}{\text{ }}$ where ${\overrightarrow {k}}={\frac {k{\overrightarrow {r}}}{r}}$

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if we want to calculate the probability current density ${\text{j}}$ for this wavefunction;

we can define probability current density of a wave function as ${\text{j=}}{\frac {\hslash }{2im}}(\psi ^{*}\triangledown \psi -\psi \triangledown \psi ^{*})$ and plug the wavefunction into this equation. Then as you see we need to calculate the

gradient. The gradient is $\triangledown \psi =\triangledown {\frac {e^{\pm i{\overrightarrow {k}}.{\overrightarrow {r}}}}{r}}={\frac {1}{r}}\triangledown e^{\pm i{\overrightarrow {k}}.{\overrightarrow {r}}}+e^{\pm i{\overrightarrow {k}}.{\overrightarrow {r}}}\triangledown {\frac {1}{r}}=(\pm i{\overrightarrow {k}}-{\frac {\overrightarrow {r}}{r^{2}}}){\frac {e^{\pm i{\overrightarrow {k}}.{\overrightarrow {r}}}}{r}}$.

Thus, since ${\overrightarrow {p}}=\hslash {\overrightarrow {k}}$, then here the probability current density comes out;

${\overrightarrow {j}}{\text{=}}\pm {\frac {\hslash }{m}}{\overrightarrow {k}}{\frac {1}{r^{2}}}=\pm {\overrightarrow {\upsilon }}{\frac {1}{r^{2}}}$.

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On the other hand, if we want to writen down the expression of number of particles which move per second through a sphere;

${\text{N=}}{\overrightarrow {j.}}{\overrightarrow {n}}(1sn)={\overrightarrow {j}}{\frac {\overrightarrow {r}}{r}}(1sn)=\pm \upsilon {\frac {1}{r^{2}}}$ So, as a result, through a unit surface, the number of particles can be defined by

${\text{N}}_{surface}{\text{=N}}_{s}=\pm (4\pi r^{2}){\frac {\upsilon }{r^{2}}}=\pm 4\pi \upsilon$

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