Timeline for Creation and annihilation operators as operator-valued distributions
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2020 at 17:39 | comment | added | MathMath | Yes! You are right! You made me realize that my last expression should be $X(h^{*}) = a(h)$ instead of $X(h)=a(h^{*})$. Moreover, this makes sense in the sense that, as $X(h^{*})$ is a distribution, it is usually written (formally) as the right hand side of this last expression! Thanks so much for helping me! | |
| Oct 16, 2020 at 17:25 | comment | added | Igor Khavkine | I think you get the point. But just for the record, for your last displayed equation to be consistent, it should have $h(x)$ not $\overline{h(x)}$ on the right-hand side. If you really follow enough sources, you'll find that some people make $a(h)$ linear and others anti-linear. Pick your favorite convention, and mentally adjust when reading other presentations. | |
| Oct 16, 2020 at 17:01 | comment | added | MathMath | @IgorKhavkine now that I edited the post, I think I got your point. If $X(h) = a(h^{*})$ then I write $X(h^{*})= a(h)$ as (\ref{2}) and everything seems to be consistent again! In fact, it seems just a matter of notation! | |
| Oct 16, 2020 at 16:54 | history | edited | MathMath | CC BY-SA 4.0 |
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| Oct 16, 2020 at 16:31 | comment | added | MathMath | @IgorKhavkine thank you for your comment! First, I think you meant $a$ instead of $a^{*}$ right? Second, I don't cite any reference in my post because I was following a lot of them, but all of them seem to agree on this notations. About the definition of $X(h)$, I agree to you that this is possible, but it seems to have some troublesome consequences. I'll edit my post to elaborate a little more on this! | |
| Oct 16, 2020 at 16:08 | comment | added | Igor Khavkine | You confusion seems primarily about the notation. But many authors use slight variations on these notations, so it's impossible to tell if there's any contradiction, since you don't cite the source that seems to have confused you. In any case, even if the author you are reading has introduced an anti-linear mapping for $a^*$, you are absolutely free to mentally introduce a new symbol $X$, obeying $X(h) := (a(h^*))^*$, so that $X$ is also linear when $a$ is. Then, while reading, you can mentally re-express $a^*$ (however it was defined) in terms of $X$ wherever necessary. | |
| Oct 16, 2020 at 15:47 | history | asked | MathMath | CC BY-SA 4.0 |