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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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  • $\begingroup$ Mapping are usually written in the form $x_0\to f(x_0)$. You write $x_0\to -i x_4$. Do you mean $x_0\to i x_0=x_4$ ? $\endgroup$ – Antonio Ragagnin Apr 2 '14 at 7:50
  • $\begingroup$ Also, in the $F_{0 i}$ formula I get: $F_{0 i} F^{0 i} = -F_{0 i} F_{0 i}$ which becomes $-\:(-i)F_{4 i} (-i)F_{4 i}=F_{4 i} F_{4 i}$ (no minus sign). Am I wrong? $\endgroup$ – Antonio Ragagnin Apr 2 '14 at 8:08
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I) Bosonic part: When we Wick-rotate, it is more natural to use sign convention $$\tag{1} \eta_{\mu\nu}~=~{\rm diag}(-1,+1,+1,+1)$$

for the Minkowski (M) metric, and

$$\tag{2} \delta_{\mu\nu}~=~{\rm diag}(+1,+1,+1,+1)$$

for the Euclidean (E) metric. Here we will use Greek indices $\mu,\nu=0,1,2,3$, to denote spacetime indices, and Roman indices $j,k=1,2,3$, for spatial indices. Standard conventions for the Wick rotation are

$$\tag{3} -S_E~=~iS_M, \qquad t_E~=~it_M, \qquad {\cal L}_E~=~-{\cal L}_M. $$

Let us here only consider QED (abelian gauge theory), and leave it to the reader to generalize to QCD (nonabelian 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

$$\tag{4} \frac{\partial}{\partial t_M}~=~i \frac{\partial}{\partial t_E}$$

under Wick rotation. This implies

$$\tag{5} -A^0_M~=~A^M_0~=~iA^E_0~=~iA^0_E, \qquad F^M_{0j}~=~iF^E_{0j},$$

Therefore the Maxwell Lagrangian density transforms as

$$\tag{6} {\cal L}_M~=~-\frac{1}{4}F^M_{\mu\nu}F_M^{\mu\nu}~=~\frac{1}{2}F^M_{0j}F^M_{0j}-\frac{1}{4}F_{jk}F_{jk}, $$

$$\tag{7} \qquad {\cal L}_M~=~{\cal T}_M-{\cal V},\qquad {\cal T}_M~=~\frac{1}{2}F^M_{0j}F^M_{0j}, \qquad {\cal V}~=~\frac{1}{4}F_{jk}F_{jk};$$

and

$$\tag{8} {\cal L}_E~=~\frac{1}{4}F^E_{\mu\nu}F_E^{\mu\nu}~=~\frac{1}{2}F^E_{0j}F^E_{0j}+\frac{1}{4}F_{jk}F_{jk},$$

$$\tag{9} \qquad {\cal L}_E~=~{\cal T}_E+{\cal V},\qquad {\cal T}_E~=~\frac{1}{2}F^E_{0j}F^E_{0j}, \qquad {\cal V}~=~\frac{1}{4}F_{jk}F_{jk},$$

which is consistent with the last equality of eq. (3). 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. Ref. 1.

References:

  1. P. van Nieuwenhuizen and A. Waldron, A continuous Wick rotation for spinor fields and supersymmetry in Euclidean space, arXiv:hep-th/9611043.
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  • $\begingroup$ Thanks for the answer. It helped me a lot to understand better the procedure. Based in it, I managed to solve most of the problem using the metric $g^{\mu \nu} = \;\mbox{diag}(-1,1,1,1)$. I managed to get into the final results using the transformations for the $\gamma$ matrices in the question. However, with this metric, they obey the relation $\{\gamma^\mu,\gamma^\nu\} = -2 g^{\mu \nu}$ and after the transformation I obtain $\{\gamma^\mu_E,\gamma^\nu_E\} = 2 \delta^{\mu \nu}$ instead of $\{\gamma^\mu_E,\gamma^\nu_E\} = -2 \delta^{\mu \nu}$. How can this be justified? $\endgroup$ – WilhelmM Apr 2 '14 at 18:55
  • $\begingroup$ After some time, I had time to go back to the problem with a fresh mind and figured out the problem with the gamma matrices. Thank you for the help. $\endgroup$ – WilhelmM Oct 6 '14 at 16:54
  • $\begingroup$ "The zero-component of the gauge variables (with indices down) is a co-vector/one-form and should transform like a time derivative." Can you elaborate on this? It makes perfect sense physically but I find it difficult to justify mathematically. Now that Wick rotation is a change of variable in integration, why part of the integrand should change accordingly? $\endgroup$ – Chong Wang Mar 13 '18 at 6:02
  • $\begingroup$ Great answer! Does (5) imply that the electric field in euclidean spacetime changes sign under time reversal? $\endgroup$ – Faser Mar 8 at 7:07

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