07/02/2009

Due to the relevance of the argument, after a nice discussion with a contribution of Carl Brannen, I decided to pursue this matter further. Indeed, the only way to have a covariant formulation of a gradient expansion is adding a time variable and taking the true time variable Wick rotated. In this way, for d=1+1 wave equation you will use d=2+1 wave equation and so on. In d=3+1 you will use d=4+1 wave equation. Let me explain with some equations what I mean. I consider again d=1+1 case as

$\partial^2_{tt}u-\partial^2_{xx}u=0$

but, instead to apply a gradient expansion to it, I apply this to the equation

$\partial^2_{tt}u-\Delta_2u=0$

being $\Delta_2 = \partial_{xx}+\partial_{yy}$. As usual, I rescale time variable as $t\rightarrow\sqrt{\lambda}t$ and I take a solution series

$u=u_0+\frac{1}{\lambda}u_1+\frac{1}{\lambda^2}u_2+\ldots.$

Now I will get the set of equations

$\partial^2_{tt}u_0=0$

$\partial^2_{tt}u_1=\Delta_2u_0$

$\partial^2_{tt}u_2=\Delta_2u_1$

and so on. Let us note that, in this case, we can introduce two new spatial variables as $z=x+iy$ and $\bar z=x-iy$. These are conjugate variables as you know. So, already at the leading order I have solved my equation. Indeed, I note that

$\Delta_2=\partial_z\partial_{\bar z}$

and so the Laplacian has the solution $f(z)+g(\bar z)$ being f and g arbitrary functions. In this case the gradient expansion gives immediately the exact result making its application trivial as should be. Indeed, I take $t=0$ in the perturbation series and put $iy=t$ and I get

$u=f(x+t)+g(x-t)$

that is the exact solution. Nice, it works! This means that a quantum field theory using gradient expansion exists and it is a strong coupling expansion. This result is surely less trivial than the one obtained above.