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 $$Z[0]:=\langle\text{VAC, out}|\text{VAC, in}\rangle=\int D\varphi\;\exp\left(i\int d^4x \,\mathcal{L}(\varphi)\right)$$ 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.

  • $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Feb 2 '15 at 23:08
  • $\begingroup$ @ACuriousMind: How does one recognize $\operatorname{det}\mathcal{D}$ as the Faddeev-Popov determinant? $\endgroup$
    – Ryan Unger
    Feb 2 '15 at 23:20
  • $\begingroup$ 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) $\endgroup$
    – ACuriousMind
    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).

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

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