If path integrals aren't well-defined, how can they have any physical meaning?

I am confused about a particular point about the nature of path integration. According to what I've read, what we really mean when we say functional integration is

\begin{equation} \int\mathcal{D}\phi= \int_{-\infty}^\infty\prod_x d\phi(x) \end{equation} in the sense that

\begin{equation} \int\prod_{i=1}^n dx_i=\underbrace{\int\cdots\int}_n dx_1\cdots dx_n \end{equation}

As I understand it, we integrate at each point over all field configurations at that point.

But supposedly one never actually does a path integral because this is not well-defined. $\prod_x$ makes no sense.

My Question:

Why doesn't $\prod_x$ make any sense? How can the integral have any meaning if its not well-defined?

When I said we don't actually do path integrals, what I meant to say is that we can do some very specific path integrals and the way we do them is rather ad-hoc. In other owrds, it's highly nontrivial and not straightforward. To show this, I'll do a very general path integral for you. (I had most of this typed up already for another reason.)

The vacuum transition amplitude for a set of quantum fields collectively denoted by $\varphi(x)$ is given by the path integral $$Z:=\langle\text{VAC, out}|\text{VAC, in}\rangle=\int D\varphi\;\exp\left(i\int d^4x \,\mathcal{L}(\varphi)\right)$$ where $D\varphi$ is the path measure $$D\varphi:=\prod_{x,\ell}d\varphi_\ell(x)$$ and $\ell$ runs over components and species. Let $A$ be a real symmetric $N\times N$ matrix and $x$ and $J$ be vectors with $N$ components. Then we have the following integral formula: $$\iint\dotsb\int dx_1dx_2\dotsb dx_N\,\exp\big((i/2)x^\text{T}Ax+iJ\cdot x\big)=\sqrt{\frac{(2\pi i)^N}{\text{det} \,A}}\exp\big(-(i/2)J^\text{T}A^{-1}J\big)$$ Now let us add to the integral of the Lagrangian in the exponent a current term that we will write as the integral of $J\varphi$. For example, in electromagnetism this would be $A\cdot J$ with $A_\mu$ the photon field and $J_\mu$ the conserved 4-current. This defines the partition function, which is a functional of the current: $$Z[J]:=\int D\varphi\;\exp\left(i\int d^4x \,\big[\mathcal{L}(\varphi)+J\varphi\big]\right)$$ Now we separate the Lagrangian into a free quadratic part $\mathcal{L}_0$ and a (possibly) more complicated interaction part $\mathcal{L}_1$: $$\mathcal{L}(\varphi)=\mathcal{L}_0(\varphi)+\mathcal{L}_1(\varphi)$$ For our purposes we set the interaction to zero: we only want single-particle states$^1$. We write the integral of the quadratic part very generally as $$\int d^4x\,\mathcal{L}_0=-\frac{1}{2}\int d^4x\,d^4y \sum_{\ell,m}\mathcal{D}_{\ell x,my}\varphi_\ell(x)\varphi_{m}(y)$$ and the source term as $$\int d^4x \,J\varphi=\int d^4x\, \sum_\ell J^\ell(x)\varphi_\ell(x)$$ Plugging this into the partition function $$Z[J]=\int D\varphi\;\exp\left(-\frac{i}{2}\int d^4x\,d^4y \sum_{\ell,m}\mathcal{D}_{\ell x,my}\varphi_\ell(x)\varphi_{m}(y)+i\int d^4x\, \sum_\ell J^\ell(x)\varphi_\ell(x)\right)$$ we note that if we think of the integrals as sums, we get a infinite multiple integral in the above form$^2$. Denote the inverse of $\mathcal{D}_{\ell x,my}$ by $\Delta_{\ell m}(x,y)$. They are inverses in the sense that $$\sum_{m}\int d^4y \,\mathcal{D}_{\ell x,my}\Delta_{mn}(y,z)=\delta^4(x-z)\delta_{\ell n}$$ The matrix $\Delta$ is called the propagator. In the absence of external fields, translation invariance will make $\mathcal{D}$ necessarily only a function of $x-y$, which can be written as a Fourier integral $$\mathcal{D}_{mx,ny}:=(2\pi)^{-4}\int d^4p\, e^{ip\cdot(x-y)}\mathcal{D}_{mn}(p)$$ and the inverse relation is then $$\Delta_{mn}(x,y)=(2\pi)^{-4}\int d^4p\, e^{ip\cdot(x-y)}\mathcal{D}_{mn}^{-1}(p)$$ where $\mathcal{D}^{-1}(p)$ is the ordinary matrix inverse of $\mathcal{D}(p)$. We can now evaluate the $Z[J]$ path integral. The factor in the finite dimensional case generalizes as $$\sqrt{\frac{(2\pi i)^N}{\text{det} \,A}}\longrightarrow \lim_{N\rightarrow\infty}\sqrt{\frac{(2\pi i)^N}{\text{det} \,\mathcal{D}}}$$ The numerator part diverges and obviously has no meaning$^3$. The denominator might appear more interesting, but it does not depend on the fields or the currents. We call this overall factor $\mathcal{C}$. The term in the exponent generalizes as $$-(i/2)J^\text{T}A^{-1}J\longrightarrow \frac{i}{2}\sum_{mn}\int d^4x\,d^4y\, J^m(x)\Delta_{mn}(x,y)J^n(y)$$ (note the sign change because $\mathcal{D}$ is defined with an extra minus sign). Putting this all together, we have $$Z[J]=\mathcal{C}\exp\left(\frac{i}{2}\sum_{mn}\int d^4x\,d^4y\, J^m(x)\Delta_{mn}(x,y)J^n(y)\right)=\mathcal{C}\exp\big(iW[J]\big)$$ where $W[J]$ is the quadratic functional $$W[J]:=\frac{1}{2}\sum_{mn}\int d^4x\,d^4y\, J^m(x)\Delta_{mn}(x,y)J^n(y)$$

1. The quantization of the interaction terms leads to the vertices of QFT.

2. Here $\mathcal{D}_{mx,ny}$ is treated as a matrix very loosely. What we should really do is introduce a lattice on spacetime such that the labels $x$ and $y$ are discrete.

3. I think this term has meaning in certain situations, but I cannot recall any at the moment.

• While a good answer to the general case, it might be worthwhile to mention that, in some cases, the path integral is rigorously defined and can be computed non-perturbatively. An example are two-dimensional pure Yang-Mills theories, and any topological field theory that has an action. The meaning of $\det{\mathcal{D}}$ depends on the context, but it can be the Faddeev-Popov determinant, for example, which again is relevant in the context of gauge theories. Feb 2 '15 at 23:08
• @ACuriousMind: How does one recognize $\operatorname{det}\mathcal{D}$ as the Faddeev-Popov determinant? Feb 2 '15 at 23:20
• Ah, sorry, I misspoke, the FP determinant is something else. This determinant appearing in the path integral measure is relevant for deriving the chiral anomaly through the Fujikawa method, though. (This time it's right) Feb 2 '15 at 23:29

According to Faddeev and Slavnov (Gauge fields: Introduction to Quantum Theory), "...all those properties of the Feynman integral that are used in practice in the perturbation theory are derived directly from the definition of the quasi-Gaussian integral and can be rigorously established independent on the issue of existence of Feynman integral measure. Therefore, in the framework of perturbation theory, the formalism of functional integration is a quite rigorous method, and results obtained using this method do not require additional proof." (I quote by a Russian edition).

• Is this because our Lagrangians are usually at most quadratic? Feb 2 '15 at 22:44
• @Ocelo7: I don't think so. It is because we only use perturbation theory. Feb 2 '15 at 23:43