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For a wavefunction:

$$\Psi(\textbf{x}) = e^{ikz} + \dfrac{f(\theta)}{r}e^{ikr}$$

Where $z = r\cos(\theta)$.

The probability current $J$ is then given by:

$$J(\textbf{x}) = J_1(\textbf{x}) + J_2(\textbf{x}) + J_{12}(\textbf{x})$$

Where $J_1$ is the current due to the first term(plane wave) and the second term is the current due to the second term (spherical wave) and the third one is due to the interference of the 2 waves.

I calculated $J_1$ like this:

$$J_1(\textbf{x}) = \dfrac{\hbar}{m}\Im((e^{ikr\cos(\theta)})^*(\nabla \cdot e^{ikr\cos(\theta)}))$$ $$J_1(\textbf{x}) = \dfrac{\hbar}{m}\Im((e^{-ikr\cos(\theta)}) (\dfrac{\partial}{\partial r}+\dfrac{\partial}{\partial \theta})e^{ikr\cos(\theta)})$$ $$J_1(\textbf{x}) = \dfrac{\hbar}{m}\Im(ik\cos(\theta) + ik\sin(\theta))$$ $$J_1(\textbf{x}) = \dfrac{\hbar k}{m}(\sin(\theta) + \cos(\theta))$$

Now to calculate the total flux through a sphere of radius $R$:

$$\Phi = \int J_1\cdot dA$$ $$\Phi = J_1 A = J_1\pi R^2$$ $$\Phi = \dfrac{\pi\hbar kR^2}{m}(\sin(\theta) + \cos(\theta))$$

My question is, is this the correct way to do it and is it correct?

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1 Answer 1

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You are on the right track, the problem so far is that the gradient in spherical coordinates is not $\partial/{\partial r} + \partial/{\partial \theta}$, so your second step becomes

$$ J_1(\textbf{x}) = \dfrac{\hbar}{m}\Im\left\{(e^{-ikr\cos(\theta)}) \left(\dfrac{\partial}{\partial r} \hat r +\color{red}{\frac{1}{r}}\dfrac{\partial}{\partial \theta} \hat θ \right)e^{ikr\cos(\theta)}\right\} $$

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  • $\begingroup$ Did you mean it is with $1/r$ or is it not? $\endgroup$ Dec 16, 2018 at 11:09
  • $\begingroup$ @MohammadAreebSiddiqui It is with $1/r$ $\endgroup$
    – caverac
    Dec 16, 2018 at 11:36

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