# Can interacting QFT be formulated in terms of the path integral or hamiltonian of a relativistic particle? [duplicate]

I read that free QFT need not be formulated in terms of fields. One can derive the same propagator as the path integral of the single free particle action $$\int d{\tau} \eta _{\mu \nu} x^{•\mu}x^{•\nu}$$, as long as one also allows paths that go backward in time.

More specifically, the propagator of free QFT is:

$$\int d^4p \frac{e^{-ipx}}{p^2-m^2+i\epsilon}$$

This is also equal to:

$$\int _0^{\infty} d\tau \langle x_2,t_2|e^{-iH\tau}|x_1,t_1\rangle$$

$$H$$ is $$\partial_{\mu}\partial^{\mu} + m^2 -i\epsilon$$. This operator generates translations in proper time $$\tau$$. Then we integrate over $$\tau$$ to sum over all paths connecting the two spacetime points. Time is treated as just another co-ordinate, so backward-in-time paths are allowed. This calculation is also equal to the path integral of the free particle action.

This calculation also happens to equal $$\langle 0|T(\phi(x),\phi(y))|0\rangle$$, for a quantum field $$\phi$$. This "co-incidence" forms the basis of the quantised field forumation of QFT.

The above makes the "quantised field" formulation of QFT seem like a mathematical trick, a co-incidence. Can interacting QFT also be thought of this way, or do we need quantised fields for that?

• Jul 24, 2022 at 7:19
• Are you asking whether it is possible to describe interacting QFT in terms of a first-quantized language? Without referring to fields, but rather treating particles as the fundamental entities? If this is it, I suggest editing your question to make these aspects more clear, so to keep it distinct of the duplicate Jul 30, 2022 at 1:50
• @NíckolasAlves Thanks. I changed the title. I think the body of the post is very different from the body of the other post. Jul 30, 2022 at 3:30