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Why do we sandwich operators in quantum mechanics in such a way that the operator acts on the wavefunction and not on its complex conjugate?

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  • $\begingroup$ In part it's conventional. $\endgroup$ – ZeroTheHero Dec 6 '17 at 21:11
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$$ \int\psi^* \hat{F} \varphi = \int (\hat{F}^+ \psi)^* \varphi $$

also, if F is an Hermitian operator $$ \int \Psi \hat{F} \Psi^* = \int \Psi^* \hat{F}\Psi $$

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Take the kinetic energy operator and ground state wave function for a particle in a box with width of $L$ as an example (in 1D) $$\hat{T}=-\frac{\hbar}{2m}\frac{d}{dx^2}$$ and $$\psi=\sqrt{\frac{2}{L}}\sin\frac{\pi}{L}x$$ Obviously, $\hat{T}\psi$ means something and $\psi^{*}\hat{T}$ means something else.

Please look at my previous answer for a more complete explanation of the notation.

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    $\begingroup$ "...becomes an object for which no computation can be performed." What do you mean? Surely, you can multiply a matrix 1x2 times 2x2 matrix. $\endgroup$ – Ice-Nine Dec 6 '17 at 18:53
  • $\begingroup$ thx for pointing out my mistake, I have removed that part completely $\endgroup$ – physicopath Dec 6 '17 at 21:25

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