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In summary: Yes, the antihermiticity is a consequence of the Euclidean metric. In Minkowski space, the relevant relation is \gamma^{\nu} D_{\nu}^{\dagger} = -\gamma^{0} D_{0}.

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However, if you find perturbation theory tedious, lattice computations may not be your best go-to unless you really really like programming, running your code for weeks on a supercluster, and then resubmit because you found a small bug. I would not go near lattice computations, but I am very grateful that others do because they are essential for understanding QCD.

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This guy did:Demystifier said:

https://www.researchgate.net/publication/1943569_Lattice_path_integral_approach_to_the_one-dimensional_Kondo_model

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Demystifier said:

I'm not sure, but it may be because of the "sign problem": https://arxiv.org/abs/1105.1374.

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Demystifier said:

The path integral over a Euclidean lattice is used in quantum Monte Carlo simulations in condensed matter physics, but the sign problem can be a big issue.

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Why does the sign problem not appear in QCD?

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[tex]

H \rightarrow H - \mu Q

[/tex]

where [itex]Q[/itex] is the particle density (the U(1) Noether current). So for the Dirac equation, [itex]Q = \int d^3x \, \bar{\psi} \gamma^0 \psi[/itex], so the Euclidean Dirac action becomes

[tex]

\mathcal{S} = -\int d^{4} x \bar{\psi} \left( \gamma^{\nu} D_{\nu} + \gamma^0 \mu + m \right) \psi

[/tex]

([itex]D_{\nu}[/itex] is the covariant derivative, so this analysis includes coupling to gauge fields). Then in the path integral,

[tex]

Z = \int\mathcal{D}A \, \mathcal{D}\bar{\psi} \, \mathcal{D}\psi \, e^{-\mathcal{S}} = \int \mathcal{D}A \, \mathrm{det}\left( \gamma^{\nu} D_{\nu} - \gamma^0 \mu + m \right)

[/tex]

Now the issue is that the Dirac operator satisfies

[tex]

\left( \gamma^{\nu} D_{\nu} + \gamma^0 \mu + m \right)^{\dagger} = \left( -\gamma^{\nu} D_{\nu} + \gamma^0 \mu + m \right) = \gamma^5 \left( \gamma^{\nu} D_{\nu} - \gamma^0 \mu + m \right) \gamma^5

[/tex]

so

[tex]

\mathrm{det}\left( \gamma^{\nu} D_{\nu} - \gamma^0 \mu + m \right)^{\dagger} = \mathrm{det}\left( \gamma^{\nu} D_{\nu} + \gamma^0 \mu + m \right)

[/tex]

So the determinant for Dirac fermions is only real at [itex]\mu = 0[/itex]. For many calculations in lattice QCD you work at zero density and this is ok, but it seems that QCD at finite density is a major subject of interest with very rich many-body physics at play. I don't know much about the field of finite-density QCD, but some searching found an interesting discussion in Section IV of https://arxiv.org/abs/1101.0109 which mentions experimental conditions where this physics should emerge.

Obviously, in context of condensed matter the above manipulations only hold for systems which are Dirac-like at low energies, but more generally one can relate the sign problem to systems whose Euclidean path integrals have non-positive-definite Boltzmann weights.

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[tex]

\mathrm{det}(M) = \mathrm{det}(S^{-1}M S) \qquad \forall S \in \mathrm{GL}(n).

[/tex]

for [itex]M[/itex] an [itex]n\times n[/itex] matrix.

The point of doing this is to show what I intended to show, but with [itex]\mu=0[/itex] this constitutes a proof that the determinant of the Euclidean Dirac operator is real in spite of the antihermiticity of [itex]\gamma^{\nu}D_{\nu}[/itex].

Demystifier said:Does your last line implies that det(−γ0†)=det(γ0)det(−γ0†)=det(γ0){\rm det}(-\gamma^{0\dagger})={\rm det}(\gamma^0)? If so, how is that compatible with γ0†=γ0γ0†=γ0\gamma^{0\dagger}=\gamma^0?

It does imply that, which is perfectly compatible with the relation you gave. Don't forget that determinants are not linear! [itex]\mathrm{det}(aM) = a^n \mathrm{det}(M)[/itex] for an [itex]n\times n[/itex] matrix.

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Is it a consequence of Euclidean metric? With Minkowski (+---) metric I don't think it's true, because thenking vitamin said:antihermiticity of [itex]\gamma^{\nu}D_{\nu}[/itex].

$$\gamma^{0\dagger}=\gamma^0 ,\;\; \gamma^{i\dagger}=-\gamma^i, \;\; D_{\nu}^{\dagger}=-D_{\nu}$$

so ##\gamma^{\nu}D_{\nu}## is neither hermitian nor anti-hermitian for Minkowski gamma matrices.

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Of course, Euclidean signature is important before even mentioning fermions because we need to get rid of that factor of [itex]i[/itex] sitting in front of the action in the path integral. Oscillating Boltzmann weights need to be avoided for computational efficiency. Of course, while Euclidean-time is ok if you just want static observables, you need to analytically continue if you want access to dynamics, and analytically continuing numerical results is ill-defined in general. So this is another major stumbling block for these methods.

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Thanks, that was very illuminating!

The choice of numerical method depends on the type of path integral and the complexity of the integrand. Some common methods include Monte Carlo integration, Gaussian quadrature, and Simpson's rule. It is important to carefully consider the properties of the integral and the desired level of accuracy before selecting a method.

Discretizing the path involves dividing the integral into smaller segments and approximating the integral over each segment. This is necessary in numerical evaluation of path integrals as it allows for the use of numerical methods that are only applicable to finite intervals. The accuracy of the evaluation depends on the size of the segments chosen.

Singularities in the integrand can lead to inaccuracies in the numerical evaluation of path integrals. One approach is to avoid the singularity by choosing a different path or by transforming the integral. Another approach is to use specialized numerical methods that can handle singularities, such as adaptive quadrature.

No, different types of path integrals require different numerical methods for accurate evaluation. For example, Monte Carlo integration is more suitable for high-dimensional integrals, while Gaussian quadrature may be more efficient for low-dimensional integrals. It is important to choose a method that is appropriate for the specific integral being evaluated.

To improve the accuracy, one can decrease the size of the discretized segments, use a more precise numerical method, or increase the number of samples in Monte Carlo integration. It is also important to check for errors in the implementation of the chosen method and to verify the results using other methods or analytical solutions if available.

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