4
$\begingroup$

These questions are linked, so I've asked them in a single post:

Why is a critical one-dimensional many-body system a two-dimensional conformal field theory?- Why the switch from 1D to 2D?

What does 2+1 dimensional mean? Two dimensions of space and one of time? Or is it a strange way of saying three dimensional?

Can a critical many-body system in thermal equilibrium be a 2+1 dimensional system? - Assuming the "+1" is time, considering a Wick's rotation, does the +1 become a measure of temperature?

Improve this question
$\endgroup$
10
  • $\begingroup$ Are you familiar with path integral approach to thermal/statistical field theory? If not, that would be why this is puzzling to you --- and an answer can focus on that. $\endgroup$
    – genneth
    Commented Dec 12, 2011 at 11:51
  • $\begingroup$ @genneth I am familiar with path integrals - I thought that was the answer to the 2nd and third questions, but nowhere seems to say it explicitly. But it's the first question I'm really stuck on, and I don't know enough about CFT to know if it's somehow related to path integrals too. $\endgroup$
    – Calvin
    Commented Dec 12, 2011 at 13:05
  • $\begingroup$ a CFT is just a field theory with some special symmetries. As Ron alludes to in his answer, the equilibrium properties of a quantum field theory in $d$ dimensions at finite temperature is given by a $d+1$-dimensional classical field theory (if this is not clear, please say so). $\endgroup$
    – genneth
    Commented Dec 12, 2011 at 17:00
  • $\begingroup$ @genneth So a finite temperature d-dimensional many-body system is not necessarily a field theory, and this is why a (d+1)-dimensional classical field theory describes it in path integral form? $\endgroup$
    – Calvin
    Commented Dec 13, 2011 at 21:12
  • $\begingroup$ I'm not sure I understood what you said; but assuming I did: yes --- a quantum field theory in $d$-dimensions has the same partition function as a classical field theory in $d+1$-dimensions. $\endgroup$
    – genneth
    Commented Dec 13, 2011 at 23:57

2 Answers 2

Reset to default
2
$\begingroup$

At (only) the critical point of a many body system, there is no longer a preferred length scale, hence the system develops an additional scaling symmetry. These scaling symmetries distinguish (in 2D) the conformal field theories among all other field theories.

This is the reason why at the critical point, a many-body theory with 2 space dimensions becomes equivalent to a conformal field theory in 1+1D. (More specifically, they are related by a Wick rotation.)

See http://en.wikipedia.org/wiki/Scale_invariance for some more details.

Improve this answer
$\endgroup$
1
$\begingroup$

The field theory doesn't have to be conformal--- a quantum theory is always defined by a path integral in one dimension more, which is the time variable. The reason is that the partition function is $\mathrm{tr}(e^{-\beta H})$, which is the imaginary time periodic boundary conditions for a path integral in one dimension more, namely the time dimension. The notation "2+1" means 2 space and 1 time dimensions.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks @Ron, I understand that not all field theories are conformal. I've just read a few papers that say they consider a one dimensional many-body system and then say it's a 2D conformal field theory - do you know why? $\endgroup$
    – Calvin
    Commented Dec 12, 2011 at 11:38
  • $\begingroup$ @Calvin: You have to post the paper--- not all one dimensional many-body systems are conformal field theories, and these authors might be making a mistake. I can't say without reading the paper. $\endgroup$
    – Ron Maimon
    Commented Jan 11, 2012 at 15:13
  • $\begingroup$ For example, the paper arXiv:hep-th/0603001v2 which says "For one-dimensional (1D) quantum many-body systems at criticality (i.e. 2D CFT), it is known..." $\endgroup$
    – Calvin
    Commented Jan 11, 2012 at 21:55
  • $\begingroup$ Any ideas? I'm thinking their 2D CFT means 1+1 dimensions. $\endgroup$
    – Calvin
    Commented Jan 29, 2012 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.