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Let $\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$. I am trying to find the probability current. http://en.wikipedia.org/wiki/Probability_current

I checked on wikipedia, and they give 2 formulas.

$j = \frac{\bar h}{2mi}(\psi^*\frac{\partial\psi}{\partial x} - \psi \frac{\partial\psi^*}{\partial x})$ $j = \frac{\bar h}{2mi}(\psi^*\nabla\psi - \psi\nabla\psi^*)$ and on another thread, the person uses $j= \frac{i}{2}(\psi\partial_x\psi^*-\psi^*\partial_x\psi) \; ,$ Probability flux

Which formula should I use?

If I want to find $\nabla\psi$, does it equals to $\nabla \psi = (\partial_x \psi, \partial_t \psi)$?

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    These are all basically the same. The first form is explicitly one-dimensional; the second form would apply equally well to $x \in \mathbb{R}^n$ (in which case $j$ would also be a vector); the third form uses dimensionless units in which $\hbar=m=1$. The gradient doesn't include the time dimension... $\nabla\psi = \partial_{x}\psi$ in one dimension.2012-01-03

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These three formulas are all equivalent, in your case. $ \nabla \psi = \dfrac{ \partial \psi}{ \partial x } = \partial_x \psi. $ If you are working in more than one dimension (which you are not since $\psi$ depends only on $x$) you would have that $ \nabla \psi = ( \partial_x \psi, \partial_y \psi ). $ The third equation is also using the normalization $\hbar / 2m = 1$.