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In a QFT with Euclidean signature, the correlation functions can only be well-defined in a time-ordered manner (This is Claim 1 on Page 2 of Simmons-Duffin's lecture note). For example, a scalar 2pt function $\langle O(x_1)O(x_2)\rangle$ $(x\equiv(\tau,\vec x))$should be regarded as $$ \ \langle O_1( x_1)O_2( x_2)\rangle\equiv\langle\Omega|T_EO_1( x_1)O_2( x_2)|\Omega\rangle= \begin{cases} \langle\Omega|O_1(x_1)O_2(x_2)|\Omega\rangle,&\tau_1>\tau_2,\\ \langle\Omega|O_2(x_2)O_1(x_1)|\Omega\rangle,&\tau_1<\tau_2. \end{cases} $$
where $T_E$ is the Euclidean time-ordering, $|\Omega\rangle$ the vacuum.

On the other hand, correlation functions should be constrained by the global symmetries of the theory. For example, given a $d$-dimensional CFT and a conformal transformation $f(x)$, the correlation functions of primary operators should satisfy $$ \langle O_1(x_1)...O_n(x_n)\rangle=\Big|\frac{\partial f^{\mu}(x_1)}{\partial x_1^\nu}\Big|^{\Delta_1/d}...\Big|\frac{\partial f^{\mu}(x_n)}{\partial x_n^\nu}\Big|^{\Delta_n/d}\langle O_1(f(x_1))...O_n(f(x_n))\rangle.\tag I $$ where $\Delta_i$ is the scaling dimension of $O_i$.

My problem occurred when I was trying to derive (I) using the operator equation for primaries $$ U_fO(x)U_f^{-1}=\Big|\frac{\partial f^{\mu}}{\partial x^\nu}\Big|^{\Delta/d}O(f(x)),\tag{II} $$ where $U_f$ is the representation of the conformal transformation $f$ in Hilbert space. Here is my derivation (for simplicity, take $n=2$). Assuming $\tau_1>\tau_2$, one can remove the time-ordering $T_E$ and insert $U_f$ into the correlator $$ \begin{align} \langle O_1(x_1)O_2(x_2)\rangle=&\langle\Omega| O_1(x_1)O_2(x_2)|\Omega\rangle\\ =&\langle\Omega| U_fO_1(x_1)U_f^{-1} U_fO_2(x_2) U_f^{-1}|\Omega\rangle\\ =&\Big|\frac{\partial f^{\mu}(x_1)}{\partial x_1^\nu}\Big|^{\Delta_1/d}\Big|\frac{\partial f^{\mu}(x_2)}{\partial x_2^\nu}\Big|^{\Delta_2/d}\langle\Omega| O_1(f(x_1))O_2(f(x_2))|\Omega\rangle. \tag{III} \end{align} $$ From the second line to the third line, I use Eq. (II) and the fact that $U_f|\Omega\rangle=|\Omega\rangle$ (since $f$ is a global symmetry). Obviously, the time ordering is missing in the final result, and it should be wrong because even if we have $\tau_1>\tau_2$, it doesn't ensure $\tau_{f(x_1)}>\tau_{f(x_2)}$. So $\langle\Omega| O_1(f(x_1))O_2(f(x_2))|\Omega\rangle$ could be ill-defined for some choices of $f$ and $x_i$.

Where did I go wrong?

EDIT: In many textbooks (e.g., the Yellow Book (YB) by Di Francesco), Eq. (I) is derived by the Euclidean Path integral method. Assuming the conformal invariance of the action and the functional integration measure, Eq. (I) is merely a consequence of the conformal transformation law for primaries (II). (see the argument for (eq. 4.48) in YB). My main concern is to find an operator method to derive Eq. (I). However, what I derived was Eq. (III) when I attempted to do this.

The contradiction between Eq. (III) and Eq. (I) can somewhat be more clearly demonstrated using the Heaviside step function $\theta(x)$. Again, take $n=2$ for simplicity. The Euclidean 2pt functions can be written as follows $$ \begin{align} \langle O_1(x_1)O_2(x_2)\rangle=&\langle\Omega|T_EO_1(x_1)O_2(x_2)|\Omega\rangle\\=&\theta(\tau_1-\tau_2)\langle\Omega|O(x_1)O(x_2)|\Omega\rangle+\theta(\tau_2-\tau_1)\langle\Omega|O(x_2)O(x_1)|\Omega\rangle.\tag{IV} \end{align} $$ Then using the same trick (inserting $U_f^{-1}U_f$ into the correlator), one obtains $$ \begin{align} &\langle O_1(x_1)O_2(x_2)\rangle\\ =&\theta(\tau_1-\tau_2)\langle\Omega|U_f^{-1}U_fO_1(x_1)U_f^{-1}U_fO_2(x_2)U_f^{-1}U_f|\Omega\rangle\\ +&\theta(\tau_2-\tau_1)\langle\Omega|U_f^{-1}U_f O_2(x_2)U_f^{-1}U_f O_1(x_1)U_f^{-1}U_f|\Omega\rangle\\ =&\Big|\frac{\partial f^{\mu}(x_1)}{\partial x_1^\nu}\Big|^{\Delta_1/d}\Big|\frac{\partial f^{\mu}(x_2)}{\partial x_2^\nu}\Big|^{\Delta_2/d}\Big(\theta(\tau_1-\tau_2)\langle\Omega|O_1(f(x_1))O_2(f(x_2))|\Omega\rangle\\ +&\theta(\tau_2-\tau_1)\langle\Omega|O_2(f(x_2))O_1(f(x_1))|\Omega\rangle\Big).\tag{V} \end{align} $$ However, according to the definition (IV), we have $$ \begin{align} &\langle O_1(f(x_1))O_2(f(x_2))\rangle\\ =&\theta(\tau_{f(x_1)}-\tau_{f(x_2)})\langle\Omega|O_1(f(x_1))O_2(f(x_2))|\Omega\rangle\\ +&\theta(\tau_{f(x_2)}-\tau_{f(x_1)})\langle\Omega|O_2(f(x_2))O_1(f(x_1))|\Omega\rangle.\tag{VI} \end{align} $$

Comparing (V) and (VI), we may naively claim that in general $$ \langle O_1(x_1)O_2(x_2)\rangle\neq \Big|\frac{\partial f^{\mu}(x_1)}{\partial x_1^\nu}\Big|^{\Delta_1/d}\Big|\frac{\partial f^{\mu}(x_2)}{\partial x_2^\nu}\Big|^{\Delta_2/d}\langle O_1(f(x_1))O_2(f(x_2))\rangle, $$ since $f(x)$ in general does not need to be a chronological spacetime transformation, i.e, $\theta(\tau_{f(x_1)}-\tau_{f(x_2)})\neq \theta(\tau_{1}-\tau_{2})$.

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  • $\begingroup$ Related/possible duplicate: physics.stackexchange.com/q/653189/50583 $\endgroup$ Commented Nov 15 at 10:25
  • $\begingroup$ @ACuriousMind, hi, thanks for the link. However, I think I do understand why out-of-time-ordered correlators are ill-defined in the Euclidean signature. Here, the main problem might be how to obtain a time-ordered correlator when considering symmetry operations. $\endgroup$ Commented Nov 15 at 11:21
  • $\begingroup$ You can't really use this operator logic to derive this equatino. This is because the operator $U_f$ is in general unbounded and its domain include many states that you are interested in. For instance, if $f$ breaks time ordering, then right-hand side of your (III) is not even well-defined. $\endgroup$ Commented Nov 20 at 17:26
  • $\begingroup$ Hi, Peter @PeterKravchuk, thank you very very much for the comments. This is exactly where I am confused. If $f$ breaks the time ordering, I will obtain an ill-defined correlator even if I started from a well-defined one. I've been trying to find a problem with the derivation of eq. (III), but I can't. As you can see, the only inputs to derive (III) are the insertion of the identity $U_f^{-1}U_f$ into the correlator and the fact that $U_f$(and its inverse) preserves the vacuum. $\endgroup$ Commented Nov 21 at 4:56
  • $\begingroup$ Also, you are absolutely right, $U_f$ is in general unbounded. But it's not clear to me how this property breaks the derivation of eq. (III) I posted. $\endgroup$ Commented Nov 21 at 4:59

2 Answers 2

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TL;DR: OP has a valid point:

  1. OP's eq. (I) gets modified in case of achronological spacetime transformations $f$ if the correlation function has an Euclidean time ordering $T_E$ inside a la eq. (5) in Ref. 1, as displayed in OP's eq. (VI).

  2. For above reason, one might for simplicity limit the study to chronological spacetime transformations $f$.

Concerning Euclidean $n$-point correlation functions, let us mention the following:

  1. Assuming that the self-adjoint Hamiltonian $H$ is bounded from below but unbounded from above, we need the Euclidean $n$-point correlation function $$\begin{align} &\langle\Omega|A^H_n(\tau_n)A^H_{n-1}(\tau_{n-1})\cdots A^H_2(\tau_2)A^H_1(\tau_1)|\Omega\rangle\cr ~=~&\langle\Omega|A^H_n(0)U(\tau_n\!-\!\tau_{n-1})A^H_{n-1}(0)\cdots A^H_2(0)U(\tau_2\!-\!\tau_1)A^H_1(0)|\Omega\rangle\end{align}\tag{A}$$ (in the Heisenberg picture) to be time-ordered $$ {\rm Re}(\tau_n)~>~{\rm Re}(\tau_{n-1})~>~\ldots~>~{\rm Re}(\tau_2)~>~{\rm Re}(\tau_1), \tag{B}$$ in order for the Euclidean time evolution operator$^1$ $$U(\tau_2\!-\!\tau_1) ~=~e^{-\frac{1}{\hbar}(\tau_2-\tau_1)H}\tag{C}$$ to be bounded/exponentially decaying/mathematically well-defined, cf. Claim 1 and eq. (2) in Ref. 1.

  2. Out-of-Euclidean-time-ordered $n$-point correlation function can be defined via analytic continuation, cf. Claim 2 and eq. (6) in Ref. 1.

  3. If we additionally assume causality, this means that local operators at different real Euclidean spacetime points commute, cf. eq. (3.5) in Ref. 2. This would at least resolve OP's paradox in that case.

  4. See also e.g. my related Phys.SE answer here.

References:

  1. D. Simmons-Duffin, 2019 TASI Lectures on CFT in Lorentzian Signature; Chapter 2.

  2. T. Hartman, S. Jain & S. Kundu, Causality Constraints in CFT, JHEP 05 (2016) 099, arXiv:1509.00014; Chapter 3 (hat tip:Connor Behan).

$^1$ In eq. (A) we have assumed that the vacuum $|\Omega\rangle$ is a zero-energy eigenstate. In eq. (C) we have for simplicity assumed that the Hamiltonian $H$ has no explicit time dependence.

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  • $\begingroup$ Thanks for the ref! I'm reading Hartman, Jain, and Kundu (HJK)'s paper right now. However, I have some questions about your answer: 1. You mentioned at the beginning that "OP is right". Do you mean Eq. (III) is correct? However, from my viewpoint, Eq. (III) clearly contradicts Eq. (I), then how could (III) be correct? 2. You mentioned that "Out-of-Euclidean-time-ordered $n$pt correlation function can be defined via analytic continuation, cf. Claim 2 and eq. (6) in Ref. 1." However, would it be more appropriate to call it an out-of-Lorentzian-time-ordered correlation function? $\endgroup$ Commented Nov 17 at 15:11
  • $\begingroup$ (Connecting to the previous text)-Since the analytic continuation is done in the region $Re(\tau_1)>Re(\tau_2)...>Re(\tau_n)$. Therefore, the Euclidean parts of $\tau_i$ are always "Euclidean-time-ordered". 3. Does Eq.(3.5) in Ref.2 contradict Claim 1 in Ref.1? $\endgroup$ Commented Nov 17 at 15:19
  • $\begingroup$ 1. Eq. (III) is in general not correct. 2. No, the Q&A are here in Euclidean signature. 3. Not in the underlying substance. $\endgroup$ Commented Nov 17 at 15:22
  • $\begingroup$ I'm sorry, then, regarding to 1: What do you mean by saying that OP is right? Regarding to 2: In Claim 2 in Ref. 1, the author said "$F(\tau_1,...,\tau_n)$ can be analytically continued to a holomorphic function of its arguments in the region $Re(\tau_1)>Re(\tau_2)>...>Re(\tau_n)$". It seems that the Euclidean parts of $\tau_{1,2,...n}$ should be "Euclidean-time-ordered" even after the continuation. $\endgroup$ Commented Nov 17 at 15:38
  • $\begingroup$ 1. I updated the answer. 2. No, that's essentially an overinterpretation of what Ref. 1 is trying to say here. $\endgroup$ Commented Nov 17 at 17:51
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The answer of @Qmechanic from this S.E. question inspires this answer:

For simplicity, let's take a scalar primary two-point function in a 2D Euclidean CFT$^{1}$ as an example (we assume $\tau_{x_1}\neq\tau_{x_2}$):

$$ \begin{align} \langle\Omega|T_E\{ O_1(x_1)O_2(x_2)\}|\Omega\rangle=&\theta(\tau_{x_1}-\tau_{x_2})\langle\Omega|O_1(x_1)O_2(x_2)|\Omega\rangle\\ &+\theta(\tau_{x_2}-\tau_{x_1})\langle\Omega| O_2(x_2)O_1(x_1)|\Omega\rangle{\tag{H1}}. \end{align} $$ Inserting the identity $U_f^{-1} U_f$ into the above expression, and using the fact that the vacuum is invariant under global conformal transformations ($U_f|\Omega\rangle=|\Omega\rangle$), we get $$ \begin{align} \langle\Omega|T_E\{ O_1(x_1)O_2(x_2)\}|\Omega\rangle=&\theta(\tau_{x_1}-\tau_{x_2})\langle\Omega|U_fO_1(x_1)U_f^{-1}U_fO_2(x_2)U_f^{-1}|\Omega\rangle\\ &+\theta(\tau_{x_2}-\tau_{x_1})\langle\Omega| U_fO_2(x_2)U_f^{-1}U_fO_1(x_1)U_f^{-1}|\Omega\rangle{\tag{H2}}. \end{align} $$ To avoid getting eq. (V) in the problem description, we recast the r.h.s. of eq. ($\text{H2}$) with the help of the time-ordering operator $T_E$ $$ \begin{align} &\theta(\tau_{x_1}-\tau_{x_2})\langle\Omega|U_fO_1(x_1)U_f^{-1}U_fO_2(x_2)U_f^{-1}|\Omega\rangle+1\leftrightarrow 2\nonumber\\ =&\langle\Omega|T_{E}\{ U_f(\tau^+_{x_1})O_1(x_1)U_f^{-1}(\tau^-_{x_1})U_f(\tau^+_{x_2})O_2(x_2)U_f^{-1}(\tau^-_{x_2})\} |\Omega\rangle.\tag{H3} \end{align} $$ In the above, $\tau^{+(-)}_{x_i}$ denotes the time which is infinitesimally greater (less) than $\tau_{x_i}$. $U_f(\tau)$ denotes a symmetry operator (in the Heisenberg picture) defined at the time slice $\tau$. For example, if $f=f_\theta$ is a rotational mapping by rotating the $\theta$ angle counterclockwise, then$^2$ $$ U_{f_{\theta}}(\tau)=e^{\theta\cdot L(\tau)},\quad L(\tau):=-\int_{\Sigma_{\tau}}d x\Big(T^{\tau}_{~\tau}(\tau, x) x-T^{\tau}_{~x}(\tau,x)\tau\Big). $$
Now, we can apply the transformation law eq. (II) in the problem description to eq. ($\text{H3}$) here to get \begin{align} \Big|\frac{\partial f^\mu(x_1)}{\partial x_1^\nu}\Big|^{\Delta_1/2}\Big|\frac{\partial f^\mu(x_2)}{\partial x_2^\nu}\Big|^{\Delta_2/2}\big\langle\Omega\big|T_{E}\big\{O_1\big(f(x_1)\big)O_2\big(f(x_2)\big)\big\} \big|\Omega\big\rangle. \end{align}

$^1$ We replace $T_E\{\cdot\}$ with the radial ordering $R\{\cdot\}$ if the radial quantization is considered.

$^{2}$ We emphasize that $L(\tau)$ is actually time-independent (its Schrodinger counterpart $L^S(\tau):=e^{-H\tau} L(\tau) e^{H\tau}$ is time-dependent); however, to put $U_f$ into the time-ordering operation, we have to pick a time for computing it.

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