# Operator-state correspondence in QFT

The operator-state correspondence in CFT gives a 1-1 mapping between operators $\phi(z,\bar{z})$ and states $|\phi\rangle$, $$|\phi\rangle=\lim_{z,\bar{z}\mapsto 0} \phi(z,\bar{z}) |0\rangle$$ where $|0\rangle$ is the $SL(2,\mathbb{Z})$ invariant vacuum.

Why can't we have a similar operator-state correspondence in non-CFT QFTs? Can't we just map operators to states by acting with the operator on the vacuum state?

The operator-state correspondence says that all states in the theory can be created by operators which act locally in a small neighborhood of the origin. That is to say that the entire Hilbert space of a CFT can be thought of as living at a single point. The key here is that for CFTs we have radial quantization, and states evolve radially outwards unitarily from the origin. This corresponds to the limit $z, \bar z \rightarrow 0$.
If you wanted to do the same for an ordinary QFT, the analagous thing would be associating a Heisenberg picture operator $\Phi$ with the state $\displaystyle \lim_{t \rightarrow -\infty} \Phi(t) | 0 \rangle$. The biggest problem here is that now one can't think of these as local operators acting at a single point if you want to get the full Hilbert space of the theory. Obviously one always has a map from operators to states just by acting the operators on the vacuum as above, but only for CFTs does the map go the other way that every state corresponds uniquely to a single local operator.
• Thanks Logan. But why can't we think of the point $-\infty$ as the single point from which we can create the states from the vacuum and then evolve them unitarily in time? – Axion Dec 2 '13 at 23:47
• @Axion - The $t\to-\infty$ limit is not a point in Minkowski space-time since we still have a whole 3-dimensional space-time at $t\to-\infty$ given by coordinates $(x,y,z)$ – Prahar Dec 3 '13 at 0:02