Negative Exponentials

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Aug 6, 2018 at 13:36	comment	added	user267839		@knzhou: One point stays unclear. Indeed, using the expresion for $a^\dagger$ in terms of $\phi$ and $\pi$ as MRT explained below one get the relation $$\frac{d}{dt} a^\dagger = \frac{i}{\hbar} [H, a^\dagger] = \frac{i \omega}{\hbar} a^\dagger$$ and so we get indeed $$ \quad a^\dagger(t,p) = e^{i \omega t / \hbar} a^\dagger(0,p)$$ But the argument that by considering the standard $(+---)$ (so $x^{\mu} =(ct, -\vec{x})$) signature, this dependence "goes with $e^{ipx}$" isnt clear to me. Why it cant be also $e^{-ipx}$. Must the exponential fulfil this compatibility with four vectors?
Aug 6, 2018 at 11:04	comment	added	knzhou		@MRT Yes, it looks totally fine to me! I think it’s fundamentally the same proof too: everything boils down to the commutation relations between creation and annihilation operators.
Aug 6, 2018 at 10:23	comment	added	MRT		(+1), of course this is the most clear way to explain this. Is my expanation below correct or not in your opinion?
Aug 6, 2018 at 10:04	history	answered	knzhou	CC BY-SA 4.0