# Path integral in Euclidean field theory

I'm very unexperienced in QFT, but I'm reading Salmhofer's book on renormalization and, at the very begining of the book, he discusses Feynamn path integral formulation on quantum mechanics to motivate the use of functional integrals. In QM, Feynman's path integral is a way of characterizing the kernel $$K(t;x,y)$$ of the propagator operator $$U(t) = e^{\frac{it}{\hbar}\hat{H}}$$. This kernel is interpreted as the transition amplitude of a particle to go from point $$x$$ at $$t_{0}=0$$ to a point $$y$$ at time $$t>0$$. Feynman's path integral as I describe here is not well-defined mathematically; however, if one take $$t \to it$$ then the associate path integral is well-defined in terms of Wiener measures. Thus, sometimes it is preferable to work on Euclidean spaces.

Now, Salmhofer states:

"In quantum field theory, one is not dealing with a single particle, but with infinitely many particles, because one has to account for the creation and annihilation of particles. One can formaly write down a Hamiltonian, but it becomes very difficult to give a mathematical definition of it. We shall simply define the theory by the functional integral."

I'd like to understand this last sentence in bold type. In ordinary QM, the action (for, e.g. a free particle) is given by: $$S(t,\phi) = \int_{0}^{t}[\frac{1}{2}m|\phi'(s)|^{2}-V(\phi(s))]ds$$ So, I'm assuming that the action for field is some integral with respect to $$x$$ of a field $$\phi=\phi(x)$$, where $$x \in \mathbb{R}^{d}$$. But what it is not clear is:

(1) What does it mean for a funcional integral to define the theory and

(2) What's the new interpretation of the functional integral? Is it also a transition amplitude, but now for fields?

• I think what they mean is: You can approach a quantum field theory via a canonical quantisation approach through the Hamiltonian, or a path integral approach through the action. Instead of starting with canonical quantisation, and then defining the path integral and demonstrating that this formulation is equivalent to canonical quantisation, they will start with the path integral approach, and more importantly use the path integral to define the quantum field theory itself. Hence, they're avoiding dealing with the lack of mathematical foundation of a quantum field theory and path integral.
– g.s
Jan 7 at 9:03
• I think it's important to note that in most quantum field theories a path integral is not even a mathematically well defined object. In fact it's not even an integral. So the authors are basically saying "we don't care about any of that mathematics stuff, we'll just define the whole quantum field theory as given by this object we call a path integral" and then they just go from there.
– g.s
Jan 7 at 9:07
• Could you expand on "not even an integral" and give some references where this is discussed? Jan 7 at 9:47

(1) The key object of a quantum field theory is the correlation function. For instance, in the simplest case it the object $$\langle \phi(x)\phi(y)\rangle,$$ where brackets means the averaging over all possible configurations of fields. Writing down the path integral in the first line, you propose a way to calculate all of these correlators (explicit computation of them is the much more subtle question). Roughly speaking, all the information of a theory is in its action $$S$$. And the path integral approach allows you to write down the expressions for any correlator in your theory (in general, there are many hidden details and subtle questions)
(2) Yes, the path integral is the transition amplitude. For instance, we can consider the process in scalar field theory of $$\phi\phi\rightarrow\phi\phi$$, which means the scattering of two scalars. In order to find the amplitude, you should (roughly speaking, I omit many details!) compute $$\langle\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle,$$ which means that initial coordinates of particles are $$x_1$$ & $$x_2$$, whereas final coordinates are $$x_3$$ & $$x_4$$.