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I am trying to reconcile two perspectives on the ground state defined through the path integral. In Tom Hartman's gravity lectures (http://www.hartmanhep.net/topics2015/gravity-lectures.pdf) he says that any state $| Y \rangle$ evolved for a long enough euclidean time is projected onto the ground state, since $$| Y \rangle = \sum_n y_n |n \rangle \rightarrow \mathrm{e}^{-\tau E_n}y_0 | 0\rangle$$ in the limit $\tau \rightarrow \infty$. Hence, whatever boundary data $\phi_1, \, \phi_2$ you take, we have $$\lim_{\tau \rightarrow \infty}\langle \phi_2 | \mathrm{e}^{- \tau H} | \phi_1 \rangle = \lim_{\tau \rightarrow \infty} \int_{\phi(-\tau/2)=\phi_1}^{\phi(\tau/2)=\phi_2}\mathcal{D}\phi\,\mathrm{e}^{-S_E[\phi]} = \langle0 | 0\rangle$$ up to some normalization.

Now consider radial quantization: euclidean time is taken to be in the radial direction in the plane. Suppose also scale invariance. Then the point at the origin corresponds to the far past while the point at infinity corresponds to the far future (e.g., consider the map from the infinite cylinder to the plane where one end of the cylinder is mapped to the origin of the plane and the other end of the cylinder is mapped to the point at infinity). In this case, the ground state is given only by path integrals over the ball containing the origin with no operator insertions. The state-operator correspondence associates $|\phi\rangle \leftrightarrow \phi(0)$. But since inserting $\phi(0)$ at the origin is essentially inserting it into the very-far-past, shouldn't the previous argument come into effect? If it did, we would be forced to conclude that $\phi(0) | 0 \rangle = | 0 \rangle$ for any $\phi(0)$. This is wrong, but why? Please help clarify my thinking.

Maybe another way of stating the question: How do we prove in radial quantization that the ground state is given only by a path integral over the ball containing no operator insertions (equivalently, only inserting the identity)?

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    $\begingroup$ Because $\phi(0)$ contains not only $a$, but $a^+$ too, no? $\endgroup$ Commented Apr 2, 2022 at 22:07

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This is a good question, and I think that the answer is a bit subtle. The crucial point is how fast the state approaches the vacuum in comparison with the behavior of other states at large distance.

Think of the limit $$ \phi(\tau)|0\rangle = \sum_n |n\rangle \langle n | \phi(\tau)|0\rangle \sim \sum_n e^{-\tau E_n} |n\rangle $$ inside a correlation function: $$ \langle 0 | \mathcal{O}_1(x_1) \cdots \mathcal{O}_n(x_n) \phi(\tau)|0\rangle \sim \sum_n e^{-\tau E_n} \langle 0 | \mathcal{O}_1(x_1) \cdots \mathcal{O}_n(x_n) |n\rangle $$ The important thing to know is that in QFT correlation functions are tempered distribution: they grow at most like a power of the distance when the points are taken away from each other. In comparison, the damping $e^{-\tau E_n}$ at large euclidean time is exponential for all states with positive energy $E_n > 0$. Since exponential decay always dominate over any power, the limit $$ \langle 0 | \mathcal{O}_1(x_1) \cdots \mathcal{O}_n(x_n) \phi(\tau)|0\rangle \xrightarrow{\tau \to \infty} \langle 0 | \mathcal{O}_1(x_1) \cdots \mathcal{O}_n(x_n) |0\rangle $$ is true in ordinary quantization. I believe this is the reasoning behind Hartman's statement.

In radial quantization, however, the "time" $\tau$ is not a physical distance, but the logarithm of the distance: $$\tau = -\log(r)$$ with $\tau \to \infty$ corresponding to an operator inserted at the origin ($r = 0$). This means that we have the asymptotic behavior $$ \langle 0 | \mathcal{O}_1(x_1) \cdots \mathcal{O}_n(x_n) \phi(r)|0\rangle \sim \sum_n r^{E_n} \langle 0 | \mathcal{O}_1(x_1) \cdots \mathcal{O}_n(x_n) |n\rangle $$ You could be tempted to conclude that $\phi(0)|0\rangle = |0\rangle$ in the limit $r \to 0$, because all $E_n > 0$. But the correlation functions in the sum on the right-hand side might diverge like powers of $r$ in the limit $r \to 0$ (and they do, think of a conformal 2-point function proportional to $r^{-2\Delta}$). So the limit is not valid in the case of "radial quantization" time.

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  • $\begingroup$ This is perfect, thank you especially for wading through my unclear explanation. It is a big help. $\endgroup$
    – Diffycue
    Commented Apr 4, 2022 at 18:29

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