Timeline for Why is the tachyon vertex operator $\int d^2z :e^{ik.X(z,\bar{z})}:$ integrated?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2020 at 9:33 | vote | accept | awsomeguy | ||
| Aug 30, 2020 at 15:25 | comment | added | Prof. Legolasov | @Rick in perturbative bosonic string theory, yes | |
| Aug 30, 2020 at 15:24 | comment | added | Rick | Are tachyons a "real" particle in string theory? | |
| Aug 30, 2020 at 11:44 | comment | added | awsomeguy | Let us continue this discussion in chat. | |
| Aug 30, 2020 at 11:41 | comment | added | Prof. Legolasov | @aweomeguy it corresponds to a type of local operators (for example, $\partial X(z)$ is an operator at point $z$, and $\partial X$ is a type of operators with no point specified), but that correspondence is not trivial. In fact, in correlation brackets, a stringy operator is a worldsheet integral of the CFT local vertex operator that it “corresponds” to. | |
| Aug 30, 2020 at 11:38 | comment | added | awsomeguy | To try and rephrase, my understanding was that although the stringy state does not have a worldsheet point associated with it, it corresponds to a local vertex operator. | |
| Aug 30, 2020 at 11:36 | comment | added | Prof. Legolasov | @awsomeguy I’m not exactly sure what you mean. Integrated is the opposite of local, right? This stringy state corresponds to an integral of the CFT state over the worldsheet, but not to a local CFT operator insertion acting on the CFT vacuum | |
| Aug 30, 2020 at 11:34 | comment | added | awsomeguy | I see that the state $|0p_1; 0p_2>$ does not have a point associated to it - but can it not still be written as a local operator on the vacuum state as per the state-operator map? i.e. the state corresponds to a vertex that is not already integrated? | |
| Aug 30, 2020 at 11:11 | comment | added | Prof. Legolasov | @awsomeguy hm... what you've written down looks like a legit state in string theory, yes. Note that it doesn't have a worldsheet point associated to it -- this state corresponds to an already integrated vertex operator in the CFT. To answer your first question -- string theory differs from the CFT by the presence of constraints. These can be derived from the Polyakov action, and the best way to quantize them is to use BRST (anystring theory textbook should contain details on BRST string quantization). In CFT, there are no such constraints. | |
| Aug 30, 2020 at 10:20 | comment | added | awsomeguy | And so what string theory states are we calculating the overlap between here, in the last expression? Is $|0p_1;0p_2>$ not a state in the string theory? | |
| Aug 30, 2020 at 10:05 | comment | added | awsomeguy | What exactly do you mean by a state/operator in string theory vs CFT? | |
| Aug 30, 2020 at 9:42 | history | answered | Prof. Legolasov | CC BY-SA 4.0 |