Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  State operator correpondence and $S^1 \times S^1$ to $R^2$

+ 2 like - 0 dislike
1138 views

I want to know the "state operator correspondence" in detail.

What i want to know is the correspondence of SS1 to R2 How the states and the operator match to each other?

I know that for SS1 via radial quantization one circle plays a role for time, which produce the states.

And for R2, we have operators.

But i don't know how these two (states in SS1 and operators in R2) match each other. Can you explain this in detail? (Recommendation of any references will be helpful to me!)

This post is imported from Physics StackExchange at 2016-06-02 04:57 (UTC) posted by SE User phy_math, SE users associated with first three comments below are ACuriousMind, phy_math and ACuriousMind respectively.

asked Jun 2, 2016 in Theoretical Physics by phy_math (185 points) [ revision history ]
edited Jun 2, 2016 by dimension10

I'm not sure what you are asking. The state-operator correspondence of CFTs means that there is a bijection between states of the theory and operators. What do you mean with the "corresponding of SS1 to R2", and how is this supposed to relate to the state-operator correspondence?

@ACuriousMind, What i want to know how states and operator corresponds. i.e, how we know there is a bijection map between them?

The proof of the state-operator correspondence should be in every good CFT resource. Please indicate what you don't understand about it. Also, you still have not explained what role SS1 or R2 are supposed to play here.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysi$\varnothing$sOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...