Quantum field theory and gradient expansion

21/02/2009

In a preceding post (see here) I showed as a covariant gradient expansion can be accomplished maintaining Lorentz invariance during computation. Now I discuss here how to manage the corresponding generating functional

Z[j]=\int[d\phi]e^{i\int d^4x\frac{1}{2}[(\partial\phi)^2-m^2\phi^2]+i\int d^4xj\phi}.

This integral can be computed exactly, the theory being free and the integral is a Gaussian one, to give

Z[j]=e^{\frac{i}{2}\int d^4xd^4yj(x)\Delta(x-y)j(y)}

where we have introduced the Feynman propagator \Delta(x-y). This is well-knwon matter. But now we rewrite down the above integral introducing another spatial coordinate and write down

Z[j]=\int[d\phi]e^{i\int d\tau d^4x\frac{1}{2}[(\partial_\tau\phi)^2-(\partial\phi)^2-m^2\phi^2]+i\int d\tau d^4xj\phi}.

Feynman propagator solving this integral is given by

\Delta(p)=\frac{1}{p_\tau^2-p^2-m^2+i\epsilon}

and a gradient expansion just means a series into p^2 of this propagator. From this we learn immeadiately two things:

  • When one takes p=0 we get the right spectrum of the theory: a pole at p_\tau^2=m^2.
  • When one takes p_\tau=0 and Wick-rotates one of the four spatial coordinates we recover the right Feynman propagator.

All works fine and we have kept Lorentz invariance everywhere hidden into the Euclidean part of a five-dimensional theory. Neglecting the Euclidean part gives us back the spectrum of the theory. This is the leading order of a gradient expansion.

So, the next step is to see what happens with an interaction term. I have already solved this problem here and was published by Physical Review D (see here). In this paper I did not care about Lorentz invariance as I expected it would be recovered in the end of computations as indeed happens. But here we can recover the main result of the paper keeping Lorentz invariance. One has

Z[j]=\int[d\phi]e^{i\int d\tau d^4x\frac{1}{2}[(\partial_\tau\phi)^2-(\partial\phi)^2-m^2\phi^2-\frac{\lambda}{2}\phi^4]+i\int d\tau d^4xj\phi}

and if we want something not trivial we have to keep the interaction term into the leading order of our gradient expansion. So we will break the exponent as

Z[j]=\int[d\phi]e^{i\int d\tau d^4x\frac{1}{2}[(\partial_\tau\phi)^2-\frac{\lambda}{2}\phi^4]-i\int d\tau d^4x\frac{1}{2}[(\partial\phi)^2+m^2\phi^2]+i\int d\tau d^4xj\phi}

and our leading order functional is now

Z_0[j]=\int[d\phi]e^{i\int d\tau d^4x\frac{1}{2}[(\partial_\tau\phi)^2-\frac{\lambda}{2}\phi^4]+i\int d\tau d^4xj\phi}.

This can be cast into a Gaussian form as, in the infrared limit, the one of our interest, one can use the following small time approximation

\phi(x,\tau)\approx\int d\tau' d^4y \delta^4(x-y)\Delta(\tau-\tau')j(y,\tau')

being now

\partial_\tau^2\Delta(\tau)+\lambda\Delta(\tau)^3=\delta(\tau)

that can be exactly solved giving back all the results of my paper. When the Gaussian form of the theory is obtained one can easily show that, in the infrared limit, the quartic scalar field theory is trivial as we obtain again a generating functional in the form

Z[j]=e^{\frac{i}{2}\int d^4xd^4yj(x)\Delta(x-y)j(y)}

being now

\Delta(p)=\sum_n\frac{A_n}{p^2-m^2_n+i\epsilon}

after Wick-rotated a spatial variable and having set p_\tau=0. The spectrum is proper to a trivial theory being that of an harmonic oscillator.

I think that all this machinery does work very well and is quite robust opening up a lot of possibilities to have a look at the other side of the world.


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