0
$\begingroup$

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?

$\endgroup$
0
0
$\begingroup$

To develop some intuition I recommend to read the first several chapters of A. Zee "Quantum field theory in a nutshell" and, if you prefer more rigorous narration, Peskin & Shroeder.

(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 $x_3$ & $x_4$.

It is not a rigorous and complete answer, just a sketch with recommendations. Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.