Timeline for Free Boson Correlator $ \langle X(z)X(w) \rangle =- \ln |z - w| $
Current License: CC BY-SA 3.0
13 events
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| Aug 12, 2014 at 14:31 | history | edited | Abdelmalek Abdesselam |
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| Jun 30, 2014 at 17:29 | answer | added | Abdelmalek Abdesselam | timeline score: 10 | |
| Jun 30, 2014 at 13:52 | comment | added | Marcel Bischoff | $\partial X$ gives you a Hilbert space, namely the bosonic Fock space. By extending the Hilbert space one can also make sense of Vertex operators, i.e. certain exponentials of the field $X$. | |
| Jun 30, 2014 at 11:40 | comment | added | john mangual | @MarcelBischoff $\langle x(z)x(w) \rangle = - \ln (z-w)$ is not always positive right? In fact $$\langle x(z)x(z+1) \rangle = - \ln 1 = 0$$ And it is hard to define the norm $||x(z)|| = \langle x(z)x(z) \rangle$ which would be divergent. So we can't put $x(z)$ into our Hilbert space. | |
| Jun 30, 2014 at 8:33 | comment | added | Marcel Bischoff | X is not a quantum field in the sense that the two-point function is not a positive measure and thus does not yield a positive inner product. | |
| Jun 30, 2014 at 2:15 | answer | added | user1504 | timeline score: 9 | |
| Jun 29, 2014 at 20:58 | history | edited | john mangual | CC BY-SA 3.0 |
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| Jun 29, 2014 at 20:37 | comment | added | André Henriques | Let's see... the free boson CFT being a non-compact CFT, the state-field correspondence is no longer surjective: every state gives rise to a field, but not every field comes from a state (the vacuum field being the prototypical example: there's no vacuum state in the free boson CFT). So you're saying that $X$ is a field? But again, you can only take correlators of the form $\langle$state$|$product of fields$|$state$\rangle$, so that doesn't justify writing $\mathbb{E}[X(z,\overline{z})X(w, \overline{w})]$... | |
| Jun 29, 2014 at 16:50 | comment | added | john mangual | @AndréHenriques p22: the holomorphic part $x(z)$ is not a conformal field. Under a conformal map the metric transforms like $$ ds^2 \mapsto \frac{\partial f}{\partial z} \frac{\partial \overline{f}}{\partial \overline{z}} ds^2 $$ conformal "fields" $\phi(z, \overline{z})$ transform like differential forms, so that $ \phi(z, \overline{z}) dz^h d\overline{z}^{h'}$ is invariant: $$ \phi \mapsto \big(\frac{\partial f}{\partial z}\big)^h \big( \frac{\partial \overline{f}}{\partial \overline{z}}\big)^{h'} \phi $$ Maybe $x(z)$ can have a logarithmic singularity or something. | |
| Jun 29, 2014 at 14:24 | history | edited | john mangual | CC BY-SA 3.0 |
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| Jun 29, 2014 at 14:02 | comment | added | john mangual | @AndréHenriques the equations of motion are the Cauchy Riemann equations, and the fields are sums of anti-holomorphic and one anti-holomorphic function $X(z, \overline{z}) = x(z)+\overline{x}(\overline{z})$. These have correlators $\langle x(z)x(w)\rangle = \log (z-w)$. I think you can derive this using an explicit power series $x(z) = \sum a_n z^n$ where $a_n$ are independent Gaussian random variables. This hides that $\vec{a} \in \ell^2$ is also a Gaussian random vector. One of my questions is whether $\vec{a}$ exists. | |
| Jun 29, 2014 at 13:33 | comment | added | André Henriques | My understanding is that $X$ is a not a quantum field of the free boson CFT [see middle of p.22 of Ginsparg's lectures], only $\partial X$ is. The first question to answer is then: what kind of mathematical object is $X$, and what does it mean (when is it legal) to insert it into a correlator? | |
| Jun 29, 2014 at 12:19 | history | asked | john mangual | CC BY-SA 3.0 |