Timeline for Problems with common mode expansion for 2D plane CFT for non integer weight?
Current License: CC BY-SA 4.0
23 events
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| Apr 24, 2021 at 1:50 | history | edited | user1379857 | CC BY-SA 4.0 |
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| Apr 24, 2021 at 1:48 | vote | accept | user1379857 | ||
| Apr 24, 2021 at 1:48 | answer | added | user1379857 | timeline score: 0 | |
| Feb 18, 2021 at 17:29 | history | edited | Urb | CC BY-SA 4.0 |
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| Feb 18, 2021 at 17:21 | history | edited | user1379857 | CC BY-SA 4.0 |
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| Feb 18, 2021 at 16:33 | comment | added | Wakabaloola | i think the relevant case might be when $\phi(z,\bar{z})$ are 'components' of a tensor in a given chart, in the sense that the conformally-invariant quantity in that chart is: $\phi(z,\bar{z})(dz)^h(d\bar{z})^{\tilde{h}}$. With the Laurent expansion of $\phi(z,\bar{z})$ as in (1), the modes scale in a convenient way under $z\rightarrow \lambda z$. | |
| Feb 18, 2021 at 0:58 | comment | added | Prahar | @user1379857 - yes, "it's a convention" was a naive answer (though correct to some extent since the multivaluedness could potentially have been absorbed into a crazy transformation of $\phi_{m,n}$). The better answer is given by the multi-valuedness argument AND my second comment regarding the action of $\phi_{m,n}$ on the vacuum state. | |
| Feb 18, 2021 at 0:56 | comment | added | user1379857 | I see what you are saying now. But doesn't that contradict the claim that whether you use (1) or (2) is a matter of convention? | |
| Feb 18, 2021 at 0:50 | comment | added | Prahar | @user1379857 - I am NOT saying that. I am simply saying that when $h \in {\mathbb R}$, the field is not single valued as $z \to e^{2\pi i}z$, ${\bar z} \to {\bar z}$ and this multi-valuedness is reflected in the expansion (1), but not in (2). Therefore (1) is correct and (2) is not. | |
| Feb 18, 2021 at 0:49 | comment | added | user1379857 | I'm confused. Are you saying that you can't have $h, \bar{h} \in \mathbb{R}$? Because I'm sure you can (like in free boson cft) | |
| Feb 18, 2021 at 0:45 | comment | added | Prahar | @user1379857 - I don't see why it would matter that the transformation isn't real? In 2D CFT, we analytically continue $z$ and ${\bar z}$ to be INDEPENDENT complex coordinates. Such analytic continuations are quite necessary to define CFTs and to perform calculations. At the end of the computation, we then reduce to the real slice $z^* = {\bar z}$. It is perfectly acceptable (and important when one studies CFTs) to consider transformations like $z \to e^{2\pi i} z$ and ${\bar z} \to {\bar z}$. | |
| Feb 18, 2021 at 0:42 | comment | added | user1379857 | $z \mapsto e^{i \theta} z$ and $\bar {z} \mapsto z$ isn't real transformation so I don't see how you need to stipulate you return to your starting point for $\theta = 2 \pi$. Let me note that $h \in \mathbb{R}$ isn't such a strange condition, for instance in string theory with the constant mode you can have $h = \bar{h} = \alpha k^2 / 4$. The condition $h - \bar{h} \in \mathbb{Z}$ is a common one | |
| Feb 18, 2021 at 0:38 | comment | added | user1379857 | Because the terms corresponding to $z^{-N}$ for $N > 0$ are attatched to operators which annihilate the vacuum, and the terms $z^{N}$ for $N > 0$ go away do to $ z \to 0$. | |
| Feb 18, 2021 at 0:37 | comment | added | d_b | I am puzzled by your point 2. Say $h=1$, then the terms proportional to $z^{-1}$, $z^{-2}$, $\ldots$ do not annihilate the vacuum. I don't see why you say only the $z^{0}$ term remains | |
| Feb 18, 2021 at 0:36 | comment | added | Prahar | I am taking $z \to e^{2\pi i} z$ and ${\bar z} \to {\bar z}$. | |
| Feb 18, 2021 at 0:34 | comment | added | user1379857 | If we use (1), then assuming it's not multivalued under $z \mapsto e^{2 \pi i } z$ we get $h - \bar{h} \in \mathbb{Z}$. However individually both $h, \bar{h}$ can be real | |
| Feb 18, 2021 at 0:30 | comment | added | Prahar | @user1379857 If you are interested in $h \in {\mathbb R}$ you need to understand what that means. It means that your field has particular properties as you rotate $z \to e^{2\pi i} z$. In particular, for generic $h \in {\mathbb R}$, the field is multi-valued. All of these properties are captured by the expansion (1), but not (2). | |
| Feb 18, 2021 at 0:17 | comment | added | user1379857 | I'm skeptical that it is even a matter of convention. The analyticity properties seem different to me depending on whether one uses (1) or (2). I suspect only one can be correct. I am also interested in $h \in \mathbb{R}$, not necessarily a half integer. $h$ doesn't have to be rational. | |
| Feb 18, 2021 at 0:05 | comment | added | Prahar | If $h$ (or ${\bar h}$) is half-integer, then the field $\phi(z,{\bar z})$ is not single-valued on the plane and this is crucial (rotation by $2\pi$ on the cylinder is not the identity for fermionic states). This multi-valuedness is clearly absent in your (2), but is there in (1). | |
| Feb 18, 2021 at 0:01 | comment | added | Gold | @user1379857 why you don't want an answer that references the cylinder? | |
| Feb 17, 2021 at 23:47 | comment | added | Prahar | Another possible answer to your question is that we would like to choose notation so that $\phi_{m,n} \ket{0}$ for all $m,n>0$. | |
| Feb 17, 2021 at 23:45 | comment | added | Prahar | If you don't want an answer to reference the cylinder, then the only correct answer to your question is "it's a convention". | |
| Feb 17, 2021 at 23:40 | history | asked | user1379857 | CC BY-SA 4.0 |