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  What uniquely defines a CFT?

+ 6 like - 0 dislike

So, I am quite new to CFT (and a as descriptive answer as possible would be appreciated). I want to know what uniquely defines a CFT in 2D and otherwise.

  1. Firstly in 2D, What defines a CFT? So I know we start with a central charge, and a set of primary fields with given conformal dimensions. What else is required?

  2. I have heard that your spectrum of primary fields, and three point correlators uniquely defines a CFT. Can someone elaborate on this? How can you find OPEs of arbitrary string of fields from three point correlators? Is this true in all dimensions?

  3. How does this fit in with the method of conformal bootstrap? AFAIK, conformal bootstrap is imposing some adhoc dynamical relations on the operator algebra coefficients. I have just seen this word thrown around, can someone explain properly how does this help in defining a CFT?

This post imported from StackExchange Physics at 2014-06-21 08:59 (UCT), posted by SE-user ramanujan_dirac
asked Jun 19, 2014 in Theoretical Physics by ramanujan_dirac (235 points) [ no revision ]

2 Answers

+ 3 like - 0 dislike

I am no expert, but here is what I understand about classifying CFTs in two dimensions (so I will only try to answer your first question, I do not know enough to tackle the others):

The defining property of a CFT is that the primary fields are invariant under the transformations generated by the Virasoro algebra $\mathrm{Vir}_c$ (or rather the sum of the algebra with is conjugate, $\mathrm{Vir}_c \oplus \bar{\mathrm{Vir}}_c$) spanned by $L_n, n \in \mathbb{Z}$ and a central charge operator $C$ with the commutation relations $[L_n,L_m] = (m-n)L_{m+n} + \frac{C}{12}(m^3 - m) \delta_{m+n,0}$ and $[C,L_n]=0$. Since $C$ is central, it is a Casimir operator, hence it has only one eigenvalue, whith is denoted $c \in \mathbb{R}$ , the central charge. The classification of CFTs is now achieved by understanding the representation theory of the Virasoro algebra, since, for the algebra to act upon the fields, they must transform as elements of some representation of $\mathrm{Vir}_c$.

A crucial property of useful QFTs is that they must be unitary, i.e. the space of states (which, by the state-field correspondence of CFTs, is in bijection to the space of fields) must possess a positive-definite inner product and $L_n^\dagger = L_{-n}$ must hold on them. So, we seek unitary representations of the Virasoro algebra. In addition, it is enough (and desirable) to know all irreducible representations, i.e those that do not possess a true subspace that is also a representation - meaning that this subspace is closed with respect to the action of the $L_n$ on it.

One then begins to build representations out of the so-called Verma modules $V(h,c)$. To get these, first define a vector $|h\rangle$ to be a highest weight vector, i.e. $L_0|h\rangle = h|h\rangle$ and $L_n|h\rangle = 0 \forall n > 0$, and then let these modules be the span of $\{L_{n_1}\dots L_{n_j}|h\rangle | n_1 \geq n_2 \geq \dots \geq n_j > 0\}$. This is a representation of $\mathrm{Vir}_c$ by construction, but in general neither unitary nor irreducible.

We can define an inner product on these modules by setting $\langle h | h \rangle = 0$ and extending this to all other states through the commutation relations and the unitarity condition $L_n^\dagger = L_{-n}$ - for any given Verma element $|v\rangle = L_{n_1}\dots L_{n_j}|h\rangle$, we will eventually arrive at some number if we stubbornly evaluate $\langle v | v \rangle$. For example, with $|v\rangle = L_{-1}|h\rangle$, one finds $\langle v | v \rangle = 2h$ (try it!). This is obviously negative for $h < 0$, thus this will not be a unitary representation. For $h = 0$, it is zero, and we can (in the sense that vectors with zero norm are unphysical, nicely corresponding to quantization of Yang-Mills theories) simply factor these out - $|v\rangle$ generates the subrepresentation $V(-1,c) \subset V(0,c)$. Indeed $V(0,c) / V(-1,c)$ will, for suitable values of $c$ be an irreducible representation.

One now considers these unitarity condition level by level - so we look at the norm of $L_{-2}|h\rangle$, $L^2_{-1}|h\rangle$, $L_{-3}|h\rangle$ and so on. This can actually be done in closed form (by something called the Kac determinant formula), and with a quite complicated analysis one finally concludes that:

  • There are no unitary CFTs for a highest weight $h < 0$ or for a central charge $c < 0$
  • For $c > 1, h \geq 0$, all Verma modules are unitary and irreducible
  • For $c = 1, h \geq 0$, all Verma modules are unitary, but reducible if $h = \frac{n^2}{4}, n \in \mathbb{N}$
  • For $c \in (0,1)$, only those with $c = 1 - \frac{6}{m(m+1)}, m \in \mathbb{N}_{\geq2}$ and $h = \frac{((m+1)r - ms)^2-1}{4m(m+1)}, r \in {1,2,\dots,m-1}, s\in {1,2,\dots,r}$ are unitary and irreducible. If $m = \frac{p}{p-q}$, then $r < p$ and $s < q$ produces unitary, but in general reducible representations.

So, this contains the answer to your first question: Specifying the central charge (i.e. setting a definite Virasoro algebra as symmetry) alone is, in general, not enough, one also needs to specify which conformal weights will occur. But, as there is no natural reason to prefer one weight over another, all possible weights will occur. Therefore, the representations with $c\in(0,1)$ are particularly interesting, since the allowed weights are already heavily constrained by the Virasoro symmetry. They are called minimal models.

For example, $c = 1/2$ has three distinct representations with $ h = 0,\frac{1}{2},\frac{1}{16} $, and their direct sum is where the theory of a Majorana fermion CFT takes place. This is also the theory of the continuum limit of the Ising model.

Hopefully, this is a somewhat satisfying answer to your question what lies at the heart of defining a 2D CFT.

This post imported from StackExchange Physics at 2014-06-21 08:59 (UCT), posted by SE-user ACuriousMind
answered Jun 20, 2014 by ACuriousMind (910 points) [ no revision ]
+ 3 like - 0 dislike

To give a short clarification on the latter two questions (what I describe holds for CFTs in d>=2 dimensions:

2) In a unitary theory the dilatation operator can be diagonalized and all two point functions are determined by the dimensions of the operators. For nonunitary theories the dilatation operator can have Jordan cells and this leads to logarithimic conformal field theories. Similarly, the three point functions have their position dependence determined up to an overall factor by the global conformal invariance. An overall scale factor is left undetermined, thats the OPE coefficient or coupling constant for the three fields in the correlator. If you consider a correlator like <$\phi(x_1)\phi(x_2)\phi(x_3)$> you can use the operator product expansion on two of the operators (the expansion is valid if you can separate the two fields by a sphere) and you get $\phi(x_1)\phi(x_2)$~$C_{\phi\phi\phi}\phi(x)$+... and use the fact that two point functions are only nonvanishing for identical primary fields (or their descendants) to see the OPE coefficients should show up in the three point functions.

Once you know the three point functions you therefore know the OPE expansion and can use it to reduce a string of fields. Knowing all three point functions and the spectrum therefore determines all the local data of the theory, not necessarily all the data.

3) The conformal bootstrap equation comes from requiring associativity of the operator product expansion. If I have a four point function I have three possible channels for the OPE expansion and it should not matter which one I choose. Requiring that the decomposition in all channels gives the same answer leads to the bootstrap equations. Associativity of the OPE is equivalent to locality, this I believe was first proven by Mack in http://www.sciencedirect.com/science/article/pii/0550321377902383#. Of course the OPE expansion is not always valid and the region of convergence of different channels may not overlap, but you can analytically continue the correlator so that the regions overlap.

For more details and equations I'd look at Slava's notes at https://sites.google.com/site/slavarychkov/

This post imported from StackExchange Physics at 2014-06-21 08:59 (UCT), posted by SE-user David Meltzer
answered Jun 20, 2014 by David Meltzer (40 points) [ no revision ]

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