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The mathematical representation of a spherical wave travelling outwards from a point is given by $$ ψ(π‘Ÿ) = \frac{𝐴}{π‘Ÿ} . 𝑒^{π‘–π‘˜π‘Ÿ}$$

Find out the probability current density 𝑗(π‘Ÿ) and plot the result

I was taught that $$j(x,t) =\frac{i\hbar}{2m} [ { ψ\frac{dψ^{*}}{dx}}+{ ψ^{*}\frac{dψ}{dx}}]$$ plugging $ ψ(π‘Ÿ) = \frac{𝐴}{π‘Ÿ} . 𝑒^{π‘–π‘˜π‘Ÿ}$ in the above equation and solving gave me $$j(r)= \frac{-i{\hbar}A^{2}}{r^{3}m}$$ but I had a doubt that can probability current density can be imaginary , if it can be imaginary I was asked to plot the j(r) how am I supposed to do it.

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    $\begingroup$ your original expression for j is clearly imaginary. If you change the plus sign to a minus sign you will end up with something real rather than imaginary. j = ihbar/2m (psi psi*' - psi* psi') $\endgroup$
    – hft
    Feb 4 at 8:37

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The mistake here is applying a one-dimensional formula for current to a three-dimensional case. The full expression for (non-relativistic) probability current is $$ \mathbf{j}(\mathbf{r},t)=\frac{\hbar}{2mi}\left[\psi^*(\mathbf{r},t)\nabla\psi(\mathbf{r},t) - \psi(\mathbf{r},t)\nabla\psi^*(\mathbf{r},t)\right] $$

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