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Is it true that $\frac{\partial\Psi_1^*}{\partial x}\Psi_2|^{\infty}_{-\infty}=\frac{\partial\Psi_2^*}{\partial x}\Psi_1|^{\infty}_{-\infty}$ for wavefunctions obeying the Schrodinger equation? For me it seems immediately obvious that not necessarily, but this seems to be necessary to show $\frac{d}{dt}\int_{-\infty}^{\infty}\Psi_1^*\Psi_2dx=0$

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When solving the Schrödinger equation, one must also impose boundary conditions. It is usually taken as an implicit assumption that the wavefunctions vanish at spatial infinity. In other words, $\Psi_i(\pm\infty) = 0$ for $i = 1,2$, which leads to the expression you wrote down.

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  • $\begingroup$ Of course! Thanks a lot. $\endgroup$
    – HakemHa
    Commented Nov 20, 2021 at 6:43

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