It is said (in e.g. Hawking, 1979, Euclidean quantum gravity) that the integral: $$ \int \mathcal{D}\phi \exp(iS[\phi])\tag{1} $$ for real fields in Minkowski space does not converge, but the Wick rotated version:

$$ \int \mathcal{D}\phi \exp(-S_E[\phi])\tag{2} $$ does. This confuses me since (1) and (2) are simply related by rotating a contour through complex space and (assuming no poles) should therefore take the same value. Thus how can (1) not converge and (2) converge. Further more from what I can tell Feynman rules in Minkowski space are derived using (1) - how is this allowed if the integral itself does not converge?

  • $\begingroup$ I think the point concerning Feynman rules is dealt with by the prescription $m^2\rightarrow m^2-i\varepsilon$ (Ketov, 2000; pg535) $\endgroup$ – Quantum spaghettification Dec 17 '17 at 12:44

The two integrals are not related by a change of contour, or at least not in any obvious way. These are functional integrals over some space of fields $V_{\mathbb{R}}=\{\phi\ |\ \phi: \mathbb{R}^d\rightarrow \mathbb{R}\}=:\Gamma_1$ namely a (flat) infinite-dimensional integration contour inside the complexification $V_{\mathbb{C}}=\{\phi\ |\ \phi: \mathbb{R}^d\rightarrow \mathbb{C}\}$. There is no (possibly curved) second contour $\Gamma_2$ in $V_{\mathbb{C}}$ which would give you the other integral. Wick rotation is more subtle because it involves analytic continuation in the arguments of the function $\phi$ which itself is the variable of integration. Namely, one does something like turning $\phi(x_0,x_1,\ldots,x_{d-1})$ into $\phi(\pm i x_0,x_1,\ldots,x_{d-1})$.

Moreover, integrals like $\int \mathcal{D}\phi\ \exp(iS[\phi])$ or $\int \mathcal{D}\phi\ \exp(-S_E[\phi])$ don't make sense at all by themselves. Even in the better behaved Euclidean case, and in the absence of UV and IR cutoff, what would an equation like $$ \int \mathcal{D}\phi\ \exp(-S[\phi])\ =\ \frac{3}{4} $$ mean? Why not $\frac{\pi}{2}$ or $10^{100}$ while we're at it?

What may make sense are ratios like $$ \frac{\int \mathcal{D}\phi\ F[\phi]\ \exp(iS[\phi])}{\int \mathcal{D}\phi\ \exp(iS[\phi])} $$ or $$ \frac{\int \mathcal{D}\phi\ F[\phi]\ \exp(-S_E[\phi])}{\int \mathcal{D}\phi\ \exp(-S_E[\phi])} $$ for $suitable$ functionals $F[\phi]$. Luckily this is what physics needs, e.g., correlation functions. I think that what Hawking is trying to say, in a very mathematically careless fashion, is the following fact. In some cases one can make sense of the Euclidean ratio as an honest integral with respect to a ($\sigma$-additive) probability measure namely $$ \int_{S'(\mathbb{R}^d)}\ F[\phi]\ d\mathbb{P}(\phi)\ . $$ On the other hand, for the Minkowski ratio one cannot do that with a $\sigma$-additive complex measure, even for a free theory. This was first noted by Cameron see, e.g., this article.


Notice that even though the Euclidean momentum and the 4-momentum in Minkowski space are related by a rotation of the 0-component, the integrals are different. In more detail, under the substitution $p^0=\mathrm{i}\,p^0_\mathrm{E}$, the integral becomes $$ \int\limits_{-\infty}^\infty\!\mathrm{d}p^0=\mathrm{i}\,\int\limits_{\mathrm{i}\,\infty}^{-\mathrm{i}\,\infty}\!\mathrm{d}p^0_\mathrm{E}\ne \mathrm{i}\,\int\limits_{-\infty}^{\infty}\!\mathrm{d}p^0_\mathrm{E} \;.$$ Suppose you were to compute the integral over a Gaussian, then you would do $$ \int\limits_{-\infty}^\infty\!\mathrm{d}x\,e^{x^2}\to \mathrm{i}\,\int\limits_{-\infty}^\infty\!\mathrm{d}x_\mathrm{E}\,e^{-x^2_\mathrm{E}}\;,$$ where the left-hand side is convergent while the right-hand side is not. (And you're right in saying that the $\varepsilon$ term makes the integral converge.)

  • $\begingroup$ May I ask why wick rotating such that t --> it imply that p --> ip0 ? $\endgroup$ – user148792 Aug 31 '18 at 12:25

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