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If I have a Hamiltonian $H$, the corresponding time-evolution operator is $e^{-iHt}$. If one defines the evolution operator in imaginary time, one uses $e^{-H\tau}$, where $\tau = it$.

It is commonly said that $e^{-H\tau}$ is non-unitary (see, for example, paragraph one of this arXiv post, or this StackExchange post). But, if $\tau = it$, shouldn't it follow that $e^{-H\tau} = e^{-iHt}$? The only explanation for why $e^{-H\tau}$ is non-unitary that I can think of is that really we are putting $\text{Im}(\tau)$ into the exponent instead of $\tau$. But if that is the case, then why do we do that, and why isn't it written explicitly as $e^{-H\text{Im}(\tau)}$?

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    $\begingroup$ Writing $\tau=it$ is not a good idea and leads to precisely this sort of confusion. The point is that $\tau$ is real. $\endgroup$
    – jacob1729
    Sep 2 at 20:14
  • $\begingroup$ @jacob1729 I think $t$ is real, and so then $\tau$ is imaginary. So I'm confused since won't the real part of $\tau$ always be $0$? $\endgroup$ Sep 2 at 20:16
  • $\begingroup$ @BioPhysicist That was a typo in my question. I meant to write $\text{Im}$ instead of $\text{Re}$. $\endgroup$ Sep 2 at 20:20
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If $H$ is hermitian then $U=e^{-itH}$ is unitary if and only if $t$ is real. Making a change of variables $t=i\tau$ won't change that. The point is that when you do a Wick rotation to imaginary time you are not making a simple change of variables - a change of variables after all can't actually affect the physics.

The basic place where an imaginary time quantity arises is the thermal density matrix $$ \rho = \frac{e^{-\beta H}}{Z}$$ with $\beta=1/(k_BT)$ the inverse temperature, which to have physical meaning must be real. This is the same thing as $U$ for an imaginary time $t=-i\beta$. This should be enough to convince you that in the vast majority of cases when talking about imaginary time one really does consider the time to be imaginary, and not purely real as needed for $U$ to be unitary.

In the context often encountered in QFT courses one is interested in time-dependent quantities, here the Wick rotation is less physical and more of a mathematical trick - you decide that the observables $O(t)$ asked for are hard to compute along the real line and instead compute them along the imaginary axis $O(it)$ and hope that the resulting formulas are analytically continuable to the entire complex plane.

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  • $\begingroup$ It seems to me you write that the wick rotation is "physical" in one context and non-physical in another. Is there another more clear way of expressing the difference in your examples? :) $\endgroup$
    – BjornW
    Oct 8 at 8:26
  • $\begingroup$ @BjornW I updated the answer a little - the point is that its physical when you are really computing an equilibrium statistical quantity that depends upon the physical temperature $\beta$ which must be real. It's mathematica when you are computing time-dependent quantities where now $t$ is real but you treat it as imaginary. $\endgroup$
    – jacob1729
    Oct 8 at 12:41
  • $\begingroup$ Thanks! BTW observables like scattering amplitudes in QFT aren't solvable by this method, right? Why? (Maybe another question for that :) $\endgroup$
    – BjornW
    Oct 8 at 20:58
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If $\hat H$ is hermitian, then $U(t)=e^{i \hat H t}$ is unitary for $t\in \mathbb{R}$ because \begin{align} U^\dagger(t)=\left(U^*(t)\right)^\top=e^{-i \hat H^\dagger t}= e^{-i \hat H t}=U(-t)=U^{-1}\, . \end{align}

If $\tau\in \mathbb{R}$ then $e^{\hat H\tau}$ is not unitary because $\left(e^{\hat H \tau}\right)^\dagger = e^{\hat H \tau} \ne \left(e^{\hat H \tau}\right)^{-1}$.

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$e^{-iHt}$ is Unitary for real $t$. Saying that $\tau=it$ doesn't change this. This is because if $t$ is real then $\tau$ is purely imaginary.

$e^{-iHt}$ is not Unitary for purely imaginary $t$. Saying that $\tau=it$ doesn't change this. This is because if $t$ is imaginary then $\tau$ is real.

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