# In path-integral, when do we have to insert fact $i$ in front of the action $S$ in the exponent?

I have got stuck in these concepts for a fews days: Wick rotation, Euclidean spacetime and QED in gravity.

Generally, in Minkowski space time, there is a factor $$i$$ in front of the action $$S$$, e.g., the path integral looks like $$$$\int \mathcal{D}{(\mbox{fields})} \exp\{iS_{Mink}\}$$$$

Now we perform a Wick rotation $$t=-ix^4$$, the metric shall go from $$(-,+,+,+)$$ to $$(+,+,+,+)$$ which is positive-definite and is known as "Euclidean spacetime". Doing some algebra, the path-inetgral will look like $$$$\int \mathcal{D}{(\mbox{fields})} \exp\{-S_{Euc}\}.$$$$ where $$-S_{Euc}=iS_{Mink}$$ and $$Euc=$$ Euclidean spacetime.

My confusion is: Does the name "Euclidean spacetime" depend on the positivity of metric? Suppose I have a Dirac theory in gravitational field of positive-definite metric $$g^{\mu\nu}$$ $$$$S_{Dirac}= \int d^4x \sqrt{g}~~i\bar{\psi}\gamma^{\mu}(\nabla_{\mu}-ieA_{\mu})\psi,~~~\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu \nu}$$$$ Which one should I choose for path-integral, $$\exp\{{iS_{Dirac}}\}$$ or $$\exp\{-S_{Dirac}\}$$?

1. The Minkowski Boltzmann factor is always $$\exp\left(\frac{i}{\hbar}S_M\right)~=~\exp\left(\frac{i}{\hbar}\int\! dt_M~L_M\right) \tag{1}$$ while the Euclidean Boltzmann factor is always $$\exp\left(-\frac{1}{\hbar}S_E\right)~=~\exp\left(-\frac{1}{\hbar}\int\! dt_E~L_E\right). \tag{2}$$ Concerning the Wick-rotation \begin{align} -S_E~=~&iS_M, \cr t_E~=~&it_M, \cr L_E~=~&-L_M,\end{align} \tag{3} see also e.g. this Phys.SE post.
2. In a nutshell, the imaginary unit $$i=\sqrt{-1}$$ in the Minkowski Boltzmann factor (1) is a remnant of the $$i$$ in the unitary evolution operator $$\hat{U}~=~\exp\left(-\frac{i}{\hbar}\hat{H}\Delta t_M\right),\tag{4}$$ cf. the standard derivation of the path integral formalism from the operator formalism.
3. Concerning the $$i$$'s in the Minkowski Dirac action $$S_{{\rm Dir},M}$$:
• Sorry, I am still confused. 1. By the "Boltzmann factor", are you talking about statistical mechanics? I have less knowledge about that. 2. But to me, it seems the appearence of factor $i$ depends on which spacetime we are working in? 3. In an asymptotically euclidean gravitational field, there would be no $i$, since my action should reduce to $S_E$ on the boundary in stead of $S_M$, right? Nov 13, 2020 at 11:49
• I think OP’s question was — what if we defined QFT to live on Euclidean space rather than Minkowski (as opposed to analytically continuing Minkowski QFT to Euclidean signature)? I don’t know if this is possible, in fact, some of the axioms of QFT depend on Minkowski signature and seem to not have a Euclidean counterpart. However, quantum gravity models that live on Euclidean space exist, eg Ponzano-Regge. These are defined on a differential manifold and signature is a constraint on their dynamics. And in the path integral of Ponzano-Regge you still have a factor of $i$, not $-1$. Nov 14, 2020 at 23:28