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I'm studying Lattice QCD and got stuck in understanding the process of going from a Minkowski space-time to an Euclidean space-time. My procedure is the following:

I considered the Wick rotation in quantum mechanics $x_0 \to -i x_4$. From this, I thought it would be reasonable to assume that for the potential vector, the Wick rotation would be $A_0 \to -i A_4$, since $A_\mu$ is a four-vector like $x_\mu$. This implies $F_{0 i}F^{0 i} \to -F_{4 i}F_{4 i}$ and assuming a metric $g^{\mu \nu} = \;\mbox{diag}(1,-1,-1,-1)$, this results in $F_{\mu \nu}F^{\mu \nu} \to -F_{\mu \nu}F_{\mu \nu}$. Now, considering that $d^4x = dt\, d^3x \to -i d\tau\, d^3x$ the action should transform as

\begin{equation} i S = -\frac{i}{2}\int d^4x \;\mbox{Tr}(F_{\mu \nu}F^{\mu \nu}) \to \frac{1}{2}\int d^4x \;\mbox{Tr}(F_{\mu \nu}F_{\mu \nu}) = S_E\,, \end{equation} where $S_E$ is the Euclidean action, which is a positive number. So, $iS \to S_E$ instead of the expected $iS \to -S_E$. I am obviously doing something wrong. I suspect it could be in the transformation of $d^4x$, but I cannot see why it would be wrong. One thing that I noticed is that if I use the metric $g^{\mu \nu} = \;\mbox{diag}(-1,1,1,1)$, then I get the proper signal. But this is changing the metric in the middle of the calculation, which would be wrong without compensating with an appropriate minus signal and then the issue would persist.

I have issues with the Fermionic sector as well. I considered $\partial_0 \to -i\partial_4$ following the transformation of $x_0$. Also, I saw in the books (Gattringer, Rothe) that it was needed that $\gamma^0 \to \gamma_4$ and $\gamma^i \to i \gamma_i$ so the definition for the $\gamma$ matrices could change from $\{\gamma^\mu,\gamma^\nu\} = 2 g^{\mu \nu} \to \{\gamma_\mu, \gamma_\nu\} = 2 \delta_{\mu \nu}$. It seens reasonable. The problem is that the transformation in the action becomes

\begin{equation} iS = i\int d^4x \; \bar{\psi}(i\gamma^\mu \partial_\mu + g_0 \gamma^\mu A_\mu - m)\psi \to \int d^4x \;\bar{\psi}(\gamma_\mu \partial_\mu - i g_0 \gamma_\mu A_\mu - m)\,, \end{equation}

which is not the Euclidean action. I tried using $A_0 \to i A_4$ in the hope I could have made some mistake in the logic above, but with no luck. So what is the prescription to perform the Wick rotation? How to figure out which transformations I should perform in a wick rotation?

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I) Bosonic part:

  1. When we Wick-rotate, it is more natural to use signature convention $(-,+,+,+)$ for the Minkowski (M) metric $g^M_{\mu\nu}$, and $(+,+,+,+)$ for the Euclidean (E) metric $g^E_{\mu\nu}$, cf. e.g. my Phys.SE answer here. We will use Greek indices $\mu,\nu=0,1,2,3$, to denote curved spacetime indices; Roman indices $a,b=0,1,2,3$, for flat spacetime indices; and Roman indices $j,k=1,2,3$, for spatial indices. We shall focus on spacetime (as opposed to its Fourier transformed space). We shall not relabel $x^0=x^4$ to avoid messing with the orientation.

  2. Standard conventions for the Wick rotation in spacetime are [1] $$\begin{align} \exp\left(\color{red}{-}\frac{1}{\hbar}S_E\right)~=~&\exp\left(\frac{\color{red}{i}}{\hbar}S_M\right)\quad\text{Boltzmann factor}, \cr \color{red}{i}S_E~=~&S_M \quad\text{action }(w=-1), \cr t_E~=~&\color{green}{i}t_M \quad\text{time }(w=0), \cr \color{red}{i}\color{green}{i}\underbrace{{\cal L}^E}_{\equiv{\cal L}^E_{0123}}~=~&\underbrace{{\cal L}^M}_{\equiv{\cal L}^M_{0123}} \quad\text{Lagrangian density }(w=-1), \cr d^4x_E~=~&\color{green}{i}d^4x_M \quad\text{spacetime $4$-form }(w=0), \cr \color{red}{i}\mathbb{L}_E~=~&\mathbb{L}_M \quad\text{Lagrangian $4$-form }(w=-1), \cr V_E^0~=~&\color{green}{i}V_M^0 \quad\text{zero-comp. of contravariant vector }(w=0), \cr V^M_0~=~&\color{green}{i}V^E_0 \quad\text{zero-comp. of covariant vector }(w=0), \end{align}\tag{1}$$ and a straightforward generalization to a tensor field of Wick weight $w$ $$T_E^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s} ~=~\color{red}{i}^w \color{green}{i}^{\#\{\mu=0\} -\#\{\nu=0\}} T_M^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s}. \tag{2}$$ Note in particular that the action (which in other respects is a scalar object) is assigned the Wick weight $w=-1$ to ensure that the Boltzmann factor transforms correctly.

  3. The Levi-Civita symbol $$\begin{align} \epsilon_E^{\mu_0\mu_1\mu_2\mu_3}~=~&\color{red}{i}^{-1}\color{green}{i}\epsilon_M^{\mu_0\mu_1\mu_2\mu_3}\qquad (w=-1),\cr \epsilon^E_{\mu_0\mu_1\mu_2\mu_3}~=~&\color{red}{i}\color{green}{i}^{-1}\epsilon^M_{\mu_0\mu_1\mu_2\mu_3}\qquad (w=1), \end{align} \tag{3}$$ is not Wick-rotated, as the values of the symbol only consist of $\pm 1$ and $0$. Both curved and flat indices are Wick-rotated. For this reason the determinant of the vielbein $$e~=~\det(e^a_{\mu})\quad\text{and}\quad\sqrt{|g|}~=~\sqrt{\det(g_{\mu\nu})\det(\eta^{ab})}\tag{4}$$ are invariant under Wick-rotation. The Hodge star operator $$\begin{align} \star(dx^{\mu_1}\wedge\ldots\wedge dx^{\mu_r}) ~=~&\frac{\sqrt{|g|}}{(4-r)!}g^{\mu_1\nu_1}\ldots g^{\mu_r\nu_r}\epsilon_{\nu_1\ldots\nu_4}dx^{\nu_{r+1}}\wedge\ldots\wedge dx^{\nu_4}, \cr (\star\Omega)_{\nu_{r+1}\ldots\nu_4} ~=~&\frac{\sqrt{|g|}}{r!}\Omega_{\mu_1\ldots\mu_r} g^{\mu_1\nu_1}\ldots g^{\mu_r\nu_r}\epsilon_{\nu_1\ldots\nu_4}, \end{align} \tag{5}$$ transforms as $$ \star_E~=~\color{red}{i}\star_M \qquad (w=1). \tag{6}$$

  4. Example: Topological/theta/Chern-Simons term. Consider a Lagrangian 4-form term of the form of a $4$-form $$\begin{align} \color{red}{i}\mathbb{L}_E~\stackrel{(1)}{=}~&\mathbb{L}_M~=~\Omega_M~=~d^4x_M ~\Omega^M_{0123}\cr ~\stackrel{(1)}{=}~&d^4x_E ~\Omega^E_{0123}~=~\Omega_E. \end{align}\tag{7} $$ The corresponding Lagrangian density term reads $$\begin{align} \color{red}{i}\color{green}{i}{\cal L}_E ~\stackrel{(1)}{=}~& {\cal L}_M ~=~ \frac{1}{4!}\epsilon^{\mu_0\mu_1\mu_2\mu_3} \Omega^M_{\mu_0\mu_1\mu_2\mu_3}\epsilon_{0123} ~=~ \Omega^M_{0123}\cr ~\stackrel{(1)}{=}~&\color{green}{i}\Omega^E_{0123} ~=~\frac{\color{green}{i}}{4!}\epsilon^{\mu_0\mu_1\mu_2\mu_3} \Omega^E_{\mu_0\mu_1\mu_2\mu_3} \epsilon_{0123}.\end{align}\tag{8} $$

  5. Example: Kinetic $F\wedge \star F$ term. $${\cal L}_E~\stackrel{(1)}{=}~-{\cal L}_M~=~\frac{e}{4}F^M_{\mu\nu}F_M^{\mu\nu}~\stackrel{(1)}{=}~\frac{e}{4}F^E_{\mu\nu}F_E^{\mu\nu}.\tag{9}$$

  6. Example: QED in flat space. Let us here only consider QED (abelian gauge theory), and leave it to the reader to generalize to QCD (non-abelian gauge theory). The zero-component of the gauge variables (with indices down) is a co-vector/one-form and should transform like a time derivative $$ \frac{\partial}{\partial t_M}~\stackrel{(1)}{=}~\color{green}{i} \frac{\partial}{\partial t_E}\tag{10}$$ under Wick rotation. This implies $$ \color{green}{-}A^0_M~=~A^M_0~\stackrel{(1)}{=}~\color{green}{i}A^E_0~=~\color{green}{i}A^0_E, \qquad F^M_{0j}~\stackrel{(1)}{=}~\color{green}{i}F^E_{0j},\tag{11}$$ Therefore the Maxwell Lagrangian density transforms as $$ {\cal L}_M~=~-\frac{1}{4}F^M_{\mu\nu}F_M^{\mu\nu}~=~\frac{1}{2}F^M_{0j}F^M_{0j}-\frac{1}{4}F_{jk}F_{jk}, \tag{12}$$ $$ \begin{align} {\cal L}_M~=~&{\cal T}_M-{\cal V},\cr {\cal T}_M~=~&\frac{1}{2}F^M_{0j}F^M_{0j}, \cr {\cal V}~=~&\frac{1}{4}F_{jk}F_{jk};\end{align}\tag{13}$$ and $$ {\cal L}_E~=~\frac{1}{4}F^E_{\mu\nu}F_E^{\mu\nu}~=~\frac{1}{2}F^E_{0j}F^E_{0j}+\frac{1}{4}F_{jk}F_{jk},\tag{14}$$ $$ \begin{align} {\cal L}_E~=~&{\cal T}_E+{\cal V},\cr {\cal T}_E~=~&\frac{1}{2}F^E_{0j}F^E_{0j}, \cr {\cal V}~=~&\frac{1}{4}F_{jk}F_{jk}.\end{align}\tag{15}$$ In particular, an Euclidean Lagrangian density ${\cal L}_E$ looks like a standard Lagrangian density (i.e. kinetic term minus potential term), with an apparent potential equal to minus ${\cal V}$.

II) Fermionic part: Wick rotation of spinor fields is a well-known non-trivial problem, cf. e.g. Refs. 2-4.

References:

  1. W. Siegel, Fields; p. 329.

  2. P. van Nieuwenhuizen and A. Waldron, A continuous Wick rotation for spinor fields and supersymmetry in Euclidean space, arXiv:hep-th/9611043.

  3. A.J. Mountain, Wick rotation and supersymmetry, 1999; End of section 6.

  4. A. Bilal & S. Metzger, arXiv:hep-th/0307152; sections 2.2 + 2.3.

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  • $\begingroup$ Notes: $\quad-(S_E-\int\! dt_E~J_k\phi^k)$ $=-S_E+\int\! dt_E~J_k\phi^k$ $=i(S_M+\int\! dt_M~J_k\phi^k)$; External sources $J_k^E=J_k^M$ do not change! Correlation fcts $\langle\phi^{k_1}\dots\phi^{k_n}\rangle_J$ do not change. $\quad-W_c^E[J]=iW_c^M[J]$ [minus (plus) inspired by ERGE (linked cluster theorem)]; $\quad-(W_c^E[J]+\int\! dt_E~J_k\phi_{\rm cl}^k)$ $=-\Gamma_E[\phi_{\rm cl}]$ $=i\Gamma_M[\phi_{\rm cl}]$ $=i(W_c^M[J]-\int\! dt_M~J_k\phi_{\rm cl}^k)$. Note extra minus in Legendre transformation! 2 implicit time integrations: $\quad S^E_2=iS^M_2$. $\quad(S^E_2)^{-1}=i(S^M_2)^{-1}$. $\endgroup$ Commented Oct 8, 2021 at 7:23
  • $\begingroup$ Notes for later: $\quad G^E_c=W^E_{c,2}=iW^M_{c,2}=iG^M_c$. $\quad\Gamma^E_2=i\Gamma^M_2$. $\quad\Gamma_2=-W_{2,c}^{-1}=-G_c^{-1}=-G_0^{-1}+\Sigma$ and $S_2=-G_0^{-1}$ for both $M/E$. $\endgroup$ Commented Apr 12, 2022 at 13:36
  • $\begingroup$ Notes for later: $\quad\langle\phi^k\phi^{\ell}\rangle^{(0)}_{J=0}=i\hbar((S^M_2)^{-1})^{k\ell}=\hbar((S^E_2)^{-1})^{k\ell}$. $\quad\langle\phi^k\phi^{\ell}\rangle^c_{J=0}=\frac{\hbar}{i}(W^M_{c,2})^{k\ell}=-\hbar(W^E_{c,2})^{k\ell}$. physics.stackexchange.com/a/649644/2451 physics.stackexchange.com/a/111381/2451 physics.stackexchange.com/a/348518/2451 $\endgroup$ Commented Nov 7, 2022 at 12:09
  • $\begingroup$ Notes for later: Kin-pot terms: Anti-action $\quad s_M := -S_M$; $\quad S_E=:s_E=is_M:=-iS_M$; Anti-Lagrangian density $\quad {\cal L}_E=\ell:=-{\cal L}_M$; $\quad\ell={\cal V}+g^{00}{\cal T}$; $\endgroup$ Commented Jan 4 at 14:40
  • $\begingroup$ Notes for later: We usually implicitly assume that $t^M_2\geq t^M_1$ in the temporal integration region (corresponding to time-ordering $T$). If $t^M_2< t^M_1$ (corresponding to anti-time-ordering $AT$), we need opposite $i\epsilon$ prescription and opposite Wick rotation $iS_M=-{\rm sgn}(t^E_2\!-\!t^E_1)S_E<0$. So $t_M=\pm e^{i(\epsilon-\frac{\pi}{2})}t_E$ or $t_E=\pm e^{i(\frac{\pi}{2}-\epsilon)}t_M$ for $\begin{Bmatrix}T \cr AT\end{Bmatrix}$. $\endgroup$ Commented Jan 5 at 12:50

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