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?


1 Answer 1


Feynman originally though that this was possible, but it turns out that if you want to correctly reproduce Bose or Fermi statistics then closed loop diagrams are necessary.


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