# How to evaluate path integrals numerically?

• A
Since we only know Gaussian integration, could one get Green's function numerically with interacting action. Usual perturbation theory is tedious and limited, could one get high accurate result with PC beyond perturbation?

## Answers and Replies

Orodruin
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https://en.m.wikipedia.org/wiki/Lattice_QCD

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.

• Demystifier and atyy
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It seems that lattice path integrals are not used in condensed matter. Does someone know why is that?

atyy
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It seems that lattice path integrals are not used in condensed matter. Does someone know why is that?

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

king vitamin
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It seems that lattice path integrals are not used in condensed matter. Does someone know why is that?

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.

• atyy
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Why does the sign problem not appear in QCD?

king vitamin
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The sign problem appears in QCD at finite density, $\mu \neq 0$. Recall that the chemical potential is introduced by taking
$$H \rightarrow H - \mu Q$$
where $Q$ is the particle density (the U(1) Noether current). So for the Dirac equation, $Q = \int d^3x \, \bar{\psi} \gamma^0 \psi$, so the Euclidean Dirac action becomes
$$\mathcal{S} = -\int d^{4} x \bar{\psi} \left( \gamma^{\nu} D_{\nu} + \gamma^0 \mu + m \right) \psi$$
($D_{\nu}$ is the covariant derivative, so this analysis includes coupling to gauge fields). Then in the path integral,
$$Z = \int\mathcal{D}A \, \mathcal{D}\bar{\psi} \, \mathcal{D}\psi \, e^{-\mathcal{S}} = \int \mathcal{D}A \, \mathrm{det}\left( \gamma^{\nu} D_{\nu} - \gamma^0 \mu + m \right)$$

Now the issue is that the Dirac operator satisfies
$$\left( \gamma^{\nu} D_{\nu} + \gamma^0 \mu + m \right)^{\dagger} = \left( -\gamma^{\nu} D_{\nu} + \gamma^0 \mu + m \right) = \gamma^5 \left( \gamma^{\nu} D_{\nu} - \gamma^0 \mu + m \right) \gamma^5$$
so
$$\mathrm{det}\left( \gamma^{\nu} D_{\nu} - \gamma^0 \mu + m \right)^{\dagger} = \mathrm{det}\left( \gamma^{\nu} D_{\nu} + \gamma^0 \mu + m \right)$$
So the determinant for Dirac fermions is only real at $\mu = 0$. 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.

• Demystifier and Orodruin
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@king vitamin I have some technical questions about the last two lines of your derivation. What is the point of introducing ##\gamma^5## matrices and how do they dissappear in the last line? Does your last line imply that ##{\rm det}(-\gamma^{0\dagger})={\rm det}(\gamma^0)##? If so, how is that compatible with ##\gamma^{0\dagger}=\gamma^0##?

king vitamin
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In the last line I am using $(\gamma^5)^{-1} = \gamma^5$ and the fact that determinants are invariant under similarity transforms,
$$\mathrm{det}(M) = \mathrm{det}(S^{-1}M S) \qquad \forall S \in \mathrm{GL}(n).$$
for $M$ an $n\times n$ matrix.

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

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! $\mathrm{det}(aM) = a^n \mathrm{det}(M)$ for an $n\times n$ matrix.

• Demystifier
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antihermiticity of $\gamma^{\nu}D_{\nu}$.
Is it a consequence of Euclidean metric? With Minkowski (+---) metric I don't think it's true, because then
$$\gamma^{0\dagger}=\gamma^0 ,\;\; \gamma^{i\dagger}=-\gamma^i, \;\; D_{\nu}^{\dagger}=-D_{\nu}$$
so ##\gamma^{\nu}D_{\nu}## is neither hermitian nor anti-hermitian for Minkowski gamma matrices.

king vitamin
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Indeed, the hermiticity properties of the Minkowski-signature gamma matrices are not even invariant under Lorentz transformations. Recall that boosts generators are not Hermitian, so the spinorial representation matrices $S$ corresponding to boosts are non-unitary, so even if $\gamma^{\mu}$ is (anti)hermitian, the matrix $S^{-1}\gamma^{\mu}S$ will not be in general. But in contrast to Spin(1,3), representations of the group Spin(4) can be chosen so that all elements are unitary so that (anti)hermiticity is preserved. (I believe you can choose all of the Euclidean gamma matrices to be antihermitian, and the arguments in my previous posts go through identically.) The fact that Spin(4) has this nice property which Spin(3,1) does not have is related to the compactness of the former.

Of course, Euclidean signature is important before even mentioning fermions because we need to get rid of that factor of $i$ 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.

• Demystifier
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Thanks, that was very illuminating!