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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?

Axion
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2 Answers2

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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.

Logan M
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    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 02 '13 at 23:47
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    @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 03 '13 at 00:02
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    @Axion I agree with Prahar's comment. Perhaps another way to say this is that in a CFT, 0 is literally a point, in that we can compute correlation functions between fields at 0 and other points. In an ordinary QFT, past infinity isn't such a thing. The idea of contracting past infinity to a single point seems inherently nonlocal, in that if I send two wave packets back in time in opposite directions in flat space, I expect them to be getting farther away from each other, not converging to the same point... – Logan M Dec 03 '13 at 00:39
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    I'm not saying that we never do things like this in physics, but that it isn't exactly what is meant by "local" (at least to me). For example, in the standard Penrose diagram for Minkowski space, past timelike infinity is mapped to a single point, but if you want to talk about local processes occurring in a neighborhood of that point you really have to blow it up to resolve that. There's also a technical issue as to in what sense the limits converge. On AdS this would be a whole different story of course, but in flat space it doesn't make much sense to regard past infinity as a single point. – Logan M Dec 03 '13 at 00:40
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    The space-time dimension is not an essential point, but the conformal invariance is quite necessary. Please check the impressive and clear Eqs.(2.8.17)-(2.8.28) of Pochinski I. – Yuan Yao Jul 11 '20 at 13:51
  • @SmartYao I think you may have misunderstood what I wrote. The question is clearly directed at 2D CFTs so my answer uses that language also. The generalization to higher dimensions is obvious; you merely use the corresponding radial quantization procedure. As you say, conformal invariance is crucial; part of the question is why it fails for non-CFTs, which the second paragraph of my answer explains. – Logan M Jul 12 '20 at 22:23
  • @LoganM Sorry for the confusion, and actually I just commented on Prahar's comment above, where the space-time dimension was emphasized. – Yuan Yao Jul 13 '20 at 14:36
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A small enhancement to the answer above, also meant as a answer to the comment of @LoganM. In general you can do the same steps, you can do the coordinate transformation to $z,\bar{z}$, and then look at the limit z to 0. This seems to give only one point, hence a local operator but if we look at the metric under this coordinate transformation we see that the metric diverges here. Even though the circle shrinks to a point when taking $z\bar{z} \rightarrow 0$, the metric diverges simultaneously.

This is analogous to the case in Penrose Diagramms. Even though diagrammatically all points in space seems to converge to one event at $t\rightarrow \infty $, the distance between each spatial point in the diagram diverges, while the points become infinitive narrow. What really happens to all points in space is, that their separation stays about the same also at the limit $t\rightarrow \infty $.

In general you can bring an event that was infinite far away (i.e. $t\rightarrow \infty $) to a finite distance (i.e. $z \rightarrow 0$), but you will encounter a singularity in the metric then (that is what we do in Penrose diagrams). This gives then an easy way to find the operator from the state, because we time evolved a finite distance in time which is obviously invertible. This operator sits only at z=0 but as argued this does not imply that the operator is local.

The relevant aspect of the CFT is therefore, that we can render the metric finite even though you are looking at $t\rightarrow \infty $ where it should explode otherwise.

The transformation that is allowed because we are dealing with a CFT brings all the points at $t\rightarrow \infty $ to the same event. Before this transformation the point z=0 only looks like a point because we hide the real extent of it in the singular metric. Now it is indeed a point and the operator at z=0. Hence, we can conclude that the operator is local from the fact that it sits only at z=0.

Qwe90909
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