In the defination of vertex algebra,we call the vertex operator state-field correspondence,does that mean that it is an injective map?? Are there some physical intepretations about state-field correspondence ?.Or why we need state-field correspondence in physical viewpoint?? Does it have some relations to highest weight representations?

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This question is too broad. I think that you should pick up a book in conformal field theory, such as this one: books.google.com/books?id=keUrdME5rhIC where you will find the answers to your questions. Perhaps also these lectures by Paul Ginsparg arxiv.org/abs/hep-th/9108028 might be useful. –  José Figueroa-O'Farrill Feb 27 '10 at 13:13

Yes, the state-field map $v\mapsto Y(v,z)$ is an injective map, since by the axioms of VOA, $Y(v,z)1|_{z=0}=v$.

The state-field correspondence appears in 2-dimensional field theory because such a field theory attaches an amplitude to a "pair of pants" (a 2-sphere with 3 holes). Namely, if you regard two of the holes as "incoming" and one as "outgoing" (i.e., you regard the surface as a worldsheet of an interaction of two strings in which they unite into one), then to this surface corresponds an operator $A: H\otimes H\to H$, where $H$ is the Hilbert space of the theory (attached to the circle). This can be viewed as a map $A: H\to {\rm End}(H)$, which is the state-field correspondence. In conformal field theory, the field map $Y$ above is obtained in the limit when the holes become very small (so the surface becomes the Riemann sphere without 3 points, say, $0,z,\infty$). For instance, in the minimal models $H=\oplus_{i=0}^n V_i\otimes V_i^\ast$, where $V_i$ are all the irreducible unitary highest weight representations of the Virasoro algebra of central charge $c=1-6/(m+2)(m+3)$, where $m\ge 1$ is an integer. The restriction of $A$ to the summand $V_0\otimes V_0^\ast$ at all three holes turns out to be a tensor product $Y\otimes Y^\vee$, where $Y=Y(z): V_0\to {\rm End}(V_0)[[z,z^{-1}]]$, which equips $V_0$ with a structure of a vertex operator algebra.

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I want to elaborate a little on Pavel's excellent answer.

We can think (very schematically) of local operators in an n-dimensional field theory the following way. We have an n-1 manifold M with some additional structures (topological, conformal, metric etc), to which our field theory assigns a vector space Z(M) of states. Given x in M and a time t in the interval, we can ask for local operators on Z(M) at the point x and time t. This can be visualized (following field theory axiomatics) as follows: we cross M with an interval, and cut out a tiny ball around the point (x,t) in this cylinder. We obtain a cobordism (with additional structure) between M times $S^{n-1}$ and M. We can then use the field theory axioms to turn states $Z(S^{n-1})$ into operators on $Z(M)$. Physically we think of inserting measurements on fields on spacetime M times interval that only ask about the value of fields in a small (punctured) neighborhood of (x,t).

In general this is a very complicated structure. But if we're in a topological field theory, then this picture is independent of lots of things - such as most importantly the size and shape of the ball we removed (as well as its position). In a 2d CFT at least we know this structure is independent of size and shape of the disc we've cut out, and depends holomorphically on z=(x,t) a point in a Riemann surface.

For simplicity though let's stick to TFT, since this picture works equally well in any dimension. If we apply this idea to the case $M=S^{n-1}$ itself, we find that $Z(S^{n-1})$ has an algebra structure --- in fact an algebra structure parametrized by cutting a ball out of a cylinder (if you look at this carefully you find the topologist's notion of $E_n$ algebra -- for $n=1$ it's simply associative, for $n=2$ it's "braided" (commutative in a coarse sense) and it gets more and more commutative as n increases). Moreover Z(M) for ANY M is now a module over this algebra.

This is how I think of state-field correspondence: states on the n-1 sphere are equivalent to local operators in the field theory acting on any space (these are the fields). In chiral CFT we find the notion of vertex algebra immediately from this -- it's a conformal refinement of the abstract notion of $E_2$ (or "braided") algebra we derived above from TFT, where now things depend holomorphically rather than locally constantly on parameters..

EDIT: One more piece of data here is the unit - there's a canonical state on the (n-1) sphere, given by considering it as the boundary of the ball we cut out (ie doing the path integral on the ball with boundary conditions on the sphere..) This is the vacuum state. It's easy to see it corresponds to the identity operator on $Z(M)$ for any $M$, and is the unit for the algebra structure on $Z(S^{n-1})$. We now recover the injectivity of the state-field correspondence: we consider the pair of pants (punctured cylinder) picture above for $M=S^{n-1}$ itself, and apply a given operator $v\in Z(S^{n-1})$ to the vacuum incoming state, obtaining an outgoing state which is again v. Saying this more carefully in the 2d CFT case recovers the vacuum axiom of a vertex algebra, which Pavel explains gives injectivity of the state-field correspondence.

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Very minor correction: the source of the cobordism in the second paragraph is a disjoint union. –  S. Carnahan Mar 14 '10 at 19:44