See the formulas (95) and (96) of this notes https://arxiv.org/abs/1602.07982. When one try to perform the Wick rotation $t=-i\tau$ to the field in Minkowski/Lorentzian spacetime $$\mathcal{O}_L(t, \mathbf{x})=e^{i H t} \mathcal{O}_L(0,\mathbf{x}) e^{-i H t}=(e^{-i H t})^\dagger \mathcal{O}_L(0,\mathbf{x}) e^{-i H t}.$$ If we do it for the first equality we have $$\mathcal{O}_L(-i\tau, \mathbf{x})=e^{i H (-i\tau)} \mathcal{O}_L(0,\mathbf{x}) e^{-i H (-i\tau)}=e^{H \tau} \mathcal{O}_L(0,\mathbf{x}) e^{-H \tau}.$$ But for the second one we have $$\mathcal{O}_L(-i\tau, \mathbf{x})=(e^{-i H (-i\tau)})^\dagger \mathcal{O}_L(0,\mathbf{x}) e^{-i H (-i\tau)}=(e^{-H \tau})^\dagger \mathcal{O}_L(0,\mathbf{x}) e^{-H \tau}=e^{-H \tau} \mathcal{O}_L(0,\mathbf{x}) e^{-H \tau},$$ then the results are different. Which one is correct?
1 Answer
In theory this should be the second one, that is, after applying $\dagger$ in the first expression, the exponent should really say $e^{iHt^*}$ with an implicit complex conjugate that gets erased by the assumption that the time coordinate is real. Obviously in a Wick rotation it is not real.
This also roughly matches the stat mech intuition that you should have something looking like $$ \langle Q \rangle=\frac1Z\langle\psi|e^{-\beta H/2} \hat Qe^{-\beta H/2} |\psi\rangle,$$ getting Boltzmann factors in both sides because the wave function picks them up for you.
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$\begingroup$ But if we assume in the beginning $\Phi$ field is a constant, then only the first one makes sense since we expect the rotated field is also a constant. Is this right? $\endgroup$– InuyashaCommented Nov 26, 2022 at 19:49