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 \begin{equation} \int \mathcal{D}{(\mbox{fields})} \exp\{iS_{Mink}\} \end{equation}

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 \begin{equation} \int \mathcal{D}{(\mbox{fields})} \exp\{-S_{Euc}\}. \end{equation} 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 $g^{\mu\nu}$ \begin{equation} S_{Dirac}= \int d^4x \sqrt{g}~~i\bar{\psi}\gamma^{\mu}(\nabla_{\mu}-ieA_{\mu})\psi,~~~\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu \nu} \end{equation} Which one should I choose for path-integral, $\exp\{{iS_{Dirac}}\}$ or $\exp\{-S_{Dirac}\}$?


1 Answer 1

  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 $S_{{\rm Dir},M}$:

    • It's EL equations should be the Dirac equation.
    • It should be real up to boundary terms.
  • 1
    $\begingroup$ 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? $\endgroup$
    – Sven2009
    Nov 13, 2020 at 11:49
  • 1
    $\begingroup$ 1. It is just a name here. 2. It depends on the spacetime signature, which I assume is either Minkowski or Euclidean. 3. Right. $\endgroup$
    – Qmechanic
    Nov 13, 2020 at 13:58
  • $\begingroup$ 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 $-1$. $\endgroup$ Nov 14, 2020 at 23:28

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