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?


1 Answer 1


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.

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    $\begingroup$ 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? $\endgroup$
    – Axion
    Dec 2, 2013 at 23:47
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    $\begingroup$ @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 $(x,y,z)$ $\endgroup$
    – Prahar
    Dec 3, 2013 at 0:02
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    $\begingroup$ @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... $\endgroup$
    – Logan M
    Dec 3, 2013 at 0:39
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    $\begingroup$ 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. $\endgroup$
    – Logan M
    Dec 3, 2013 at 0:40
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    $\begingroup$ 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. $\endgroup$
    – Yuan Yao
    Jul 11, 2020 at 13:51

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