Operation on Complex conjugate

Question

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?

In part it's conventional. – ZeroTheHero Dec 06 '17 at 21:11 — ZeroTheHero, Dec 06 '17 at 21:11

score 2 · Answer 1 · answered Dec 06 '17 at 15:43

2

$$ \int\psi^* \hat{F} \varphi = \int (\hat{F}^+ \psi)^* \varphi $$

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

answered Dec 06 '17 at 15:43

Ice-Nine

411

score 0 · Answer 2 · edited Dec 28 '17 at 11:45

0

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.

edited Dec 28 '17 at 11:45

eranreches

4,199

answered Dec 06 '17 at 16:18

physicopath

2,352

1

"...becomes an object for which no computation can be performed." What do you mean? Surely, you can multiply a matrix 1x2 times 2x2 matrix. – Ice-Nine Dec 06 '17 at 18:53
thx for pointing out my mistake, I have removed that part completely – physicopath Dec 06 '17 at 21:25

Operation on Complex conjugate

2 Answers2