Timeline for Free field rigorous quantization - possibly a misunderstanding?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 9, 2021 at 0:30 | answer | added | JustWannaKnow | timeline score: 2 | |
| Mar 8, 2021 at 18:13 | comment | added | JustWannaKnow | @PedroLauridsenRibeiro I editted the post and corrected what you pointed out: $E$ is defined for $f \in \mathscr{S}(\mathbb{R}^{4})$ instead of $\mathscr{H}$. | |
| Mar 8, 2021 at 18:11 | history | edited | JustWannaKnow | CC BY-SA 4.0 |
added 3 characters in body
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| Mar 8, 2021 at 12:54 | comment | added | JustWannaKnow | $a(E\mbox{Re}(f)-iE\mbox{Im}(f)) = a(C\{E\mbox{Re}(f)+iE \mbox{Im}(f)\}) = a (CEf)$, as proposed by Reed & Simon. But I had to use the fact that $Ef \in \mathscr{H}_{C}$. If not, it is not clear to me how to justify this factor. | |
| Mar 8, 2021 at 12:52 | comment | added | JustWannaKnow | Moreover, take the annihilation operator $a$ for example. iF $f\in \mathscr{S}(\mathbb{R}^{4})$ is arbitrary, we have (I'm gonna omit the $\sqrt{2}^{-1}$ for simplicity): $a^{\dagger}(Ef) + a(Ef) =a^{\dagger}(E\mbox{Re}(f)) + i a^{\dagger}(E\mbox{Im}(f))+ a(E\mbox{Re}(f)) + i a(E\mbox{Im}(f))$. Now, focusing on $a$, we can put these two factors together $a(E\mbox{Re}(f)-i\mbox{Im}(f))$ since $a$ is antilinear. But then, why the factor $C$ arises? This seems to arise only if $E\mbox{Re}(f) = CE\mbox{Re}(f)$ and $E\mbox{Im}(f) = C\mbox{Im}(f)$, since, in this case (cont) | |
| Mar 8, 2021 at 12:46 | comment | added | JustWannaKnow | @PedroLauridsenRibeiro thanks for the comment. The point is that when Reed & Simon introduce $\varphi$ and $\pi$ in terms of the Segal quantization operator, they introduce for elements of $f \in \mathscr{H}_{C}$. That is my problem with the definition of $\varphi$ and $\pi$. Because $Ef$ is not necessarily $\mathscr{H}_{C}$, then these maps seems ill-defined. | |
| Mar 8, 2021 at 4:02 | comment | added | Pedro Lauridsen Ribeiro | First of all, notice that the domain of $E$ is $\mathscr{S}(\mathbb{R}^4)$, not $\mathscr{H}=$ codomain of $E$, and the conjugation $C$ acts on $\mathscr{H}$, not on $\mathscr{S}(\mathbb{R}^4)$, so that the composites $CE$ and $C\mu E$ are well defined. Moreover, the formulae for $\varphi_m(f)$ and $\pi_m(f)$ in terms of creation and annihilation operators are valid for all $f\in\mathscr{S}(\mathbb{R}^4)$, not only for those such that $Ef\in\mathscr{H}_C$ (if that's the case, then $\varphi_m(f)$ and $\pi_m(f)$ are self-adjoint). Thanks to $C$, $\varphi_m$ and $\pi_m$ are complex linear. | |
| Mar 7, 2021 at 17:34 | history | asked | JustWannaKnow | CC BY-SA 4.0 |