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$\mathbf{P}_{ab} = \int_{a}^{b} |\psi(p)|^2\, dp$
$[br]\left[ - \frac{\hbar^2}{2m} \nabla^2 + V\left(\mathbf{r}\right) \right] \psi\left(\mathbf{r}\right) = E \psi \left(\mathbf{r}\right).[br]$
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