Convergence of Euclidean path integral

Question

Can you explain why they say, that the Euclidean path integral converges?

Naively, it diverges because in the Lagrangian, when calculating the kinetic term, we take the difference between neighbouring points $x$ and square it.

The neighbouring points $x_1$ and $x_0$ can be anything and because we're integrating over all possible $x$'s, there will always be a combination of $x$'s that will give low action $S$, and hence $\exp(-S)$ close to unity, as there is an infinite number of such contributions, to the sum (integral) it would diverge.

Answer 1

The claim "The Euclidean path integral converges but the Minkowski path integral does not" is in general unproven. The physicists' handwaving argument for this is usually little more than that the Minkowskian $\mathrm{e}^{\mathrm{i}S[q(t)]}$ oscillates around the unit circle in the complex plane, while the Euclidean $\mathrm{e}^{-S[q(t)]}$ is a real function that goes to zero as the value of the action goes towards $+\infty$, so the "only" challenge is to show that as $q(t)\to\infty$ (whatever that means), $S[q(t)]\to +\infty$ fast enough to ensure convergence, which is supposed to be "easier" than controlling the oscillation of the Minkowskian version in the complex plane.

In some sense this is true, as all rigorous constructions of the path integral I'm aware of construct a Euclidean path integral and then use the Osterwalder-Schrader reconstruction theorem to compute the Minkowskian n-point functions as analytic continuations of the Euclidean n-point functions, i.e. "Wick rotation" $t\mapsto \mathrm{i}t$ of the Euclidean n-point functions yields the Minkowksian n-point functions, but they never construct a Minkowskian path integral.

The canonical reference for the rigorous construction of Euclidean path integrals is "Quantum Physics - A functional integral point of view" by Glimm and Jaffe. They rigorously construct the path integral for 2-dimensional QFTs with polynomial interaction terms and for specific 3-dimensional theories. This involves more subtle constructions than the usual arguments in physics texts already at the level of free theories, see e.g. this answer of mine.

No general proof of the existence of the path integral for higher-dimensional theories is known in either the Euclidean nor the Minkowskian version, particularly not for theories 4-dimensional QFTs similar to the Standard Model. Producing such a proof would constitute a solution to the Yang-Mills millenium problem.

Answer 2

The convergence of the Euclidean path integral is almost obvious in the operator formalism. Consider a theory with Euclidean action $L_E[x,\dot{x}]$ and hamiltonian $\hat{H}$. The Euclidean path integral with periodic boundary conditions can be written as $$\int_{x(0)=x(T)}[dx(\tau)]\ e^{-\int_0^T L_E[x(\tau),\hat{x}(\tau)]}=\mathrm Tr[e^{-T\hat{H}}]=\sum_n g_n e^{-TE_n} $$ where the $E_n$ are the eigenvalues of the hamiltonian and $g_n$ their degeneracy. When both $g_n$ and $E_n$ grow polinomially, the sum in the rhs converge, and so does the lhs.

To show understand why the existence of an infinite number of low-action field configurations does not cause the path integral to diverge, we can look at chapter 7, appendix 4 of Sidney Coleman's "Aspects of Symmetry". There, Coleman shows that in the Euclidean path integral of a harmonic oscillator the field configurations with finite action form a subset of zero measure. Coleman considers a harmonic oscillator with Euclidean lagrangian $$ L_E=\frac{1}{2}\dot{x}^2+\frac{1}{2}\omega^2x^2 $$ To compute its path integral, we define the orthonormal eigenfunctions $x_n(\tau)$ satisfying $$\left(-\frac{d^2}{d\tau^2}+\omega^2\right)x_n(\tau)=\lambda_nx_n(\tau) $$ where all $\lambda_n >0 $. Any path $x(\tau)$ can be expressed as a linear combination of these eigenfunctions $$x(\tau)=\sum_{n\in\mathbb{Z}}a_nx_n(\tau) $$ By orthonormality, the Euclidean action $S_E$ evaluated along a generic path is then $$S_E=\frac{1}{2}\sum_{n\in\mathbb{Z}}\lambda_na_n^2 $$ A necessary condition for a field configuration to have finite action is that $$\lambda_na_n^2 < a^2 $$ where $a$ is some upper bound. We define the path integral measure $$\int[dx(\tau)]\equiv \prod_{n\in\mathbb{Z}}\int_{-\infty}^{+\infty}\frac{da_n}{\sqrt{2\pi}} $$ and compute the ratio $$\frac{\displaystyle\int_{\text{finite action}}[dx(\tau)]e^{-S_E[x,\dot{x}]}}{\displaystyle\int [dx(\tau)]e^{-S_E[x,\dot{x}]}} \leq \prod_{n\in\mathbb{Z}}\frac{\displaystyle\int_{-\frac{a}{\sqrt{\lambda_n}}}^{+\frac{a}{\sqrt{\lambda_n}}}da_n\ e^{-\frac{1}{2}\lambda_na_n^2}}{\displaystyle\int_{-\infty}^{+\infty}da_n\ e^{-\frac{1}{2}\lambda_na_n^2}} $$ where i canceled the $1/\sqrt{2\pi}$ factors in the rhs. Changing the variables to $b_n=\frac{a_n}{\sqrt{\lambda_n}}$ in both the numerator and the denominator one finds $$\frac{\displaystyle\int_{\text{finite action}}[dx(\tau)]e^{-S_E[x,\dot{x}]}}{\displaystyle\int [dx(\tau)] e^{-S_E[x,\dot{x}]}} \leq \prod_{n\in\mathbb{Z}}\frac{\displaystyle\int_{-a}^{+a}db_n\ e^{-\frac{1}{2}b_n^2}}{\displaystyle\int_{-\infty}^{+\infty}db_n\ e^{-\frac{1}{2}b_n^2}}=0 $$ where I canceled the $1/\sqrt{\lambda_n}$ factors in the rhs. The rhs vanishes because it is an infinite product of factors $<1$. This is not a rigorous proof because it applies only to free field theories on compact manifolds, but it has the right spirit.