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$\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$

If I want to find $\nabla\psi$, am I correct just to take the partial derivative of $\psi$ with respect to $x$, $\partial_x\psi$?

Does $\nabla\psi = Axe^{i(kx-\omega t)}+Bxe^{-i(kx+\omega t)}$?

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1 Answer 1

$\nabla \psi$ is a vector: $\nabla \psi = (\partial_x \psi, \partial_t \psi)$.

In general if $f : \mathbb{R}^n \to \mathbb{R}$ is a function mapping $(x_1, \dots, x_n) \to f(x_1, \dots, x_n)$, then $$ \nabla f = ( \partial_1 f, \partial_2 f, \dots, \partial_n f)$$

For more, see here.

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