Perturbative expansion and self-contractions in functional integral

Question

Consider a one-dimensional integral $$I(g)=\int dx\, e^{-x^2-gx^4}$$ One can formally expand it perturbatively order by order in $g$ so that $$I(g)=\left<1\right>-g\left<x^4\right>+\frac{g^2}{2}\left<x^8\right>+O(g^3)$$ where $\left<f(x)\right>:=\int dx\, e^{-x^2}f(x)$. This expansion is asymptotic but that's fine. I thought that basically the same procedure applies to QFT. Consider $$Z(g)=\int \mathcal{D}\phi\,\, e^{-\int (\nabla \phi)^2+g\phi^4}$$ and expand it formally in $g$ $$Z(g)= \left<1\right>-g\left<\phi^4\right>+\frac{g^2}{2} \left<\phi^8\right>+O(g^3)$$ where $\left<f(\phi)\right>:=\int \mathcal{D}\phi\,\, e^{-\int (\nabla \phi)^2}f(\phi)$. My problem with that expression is that it involves correlation functions at coincident points (which then need to be integrated over space). But correlators at coincident points are infinite.

Are these superficial infinities related to the renormalization necessary in QFT? Or this is a different issue (I believe it is)? How the formulas should be corrected then?

Answer 1

You can absorb these divergences from self contractions (‘search for cephalopod Feynman diagrams and ‘complete normal ordering’ in google) into your bare couplings and wavefunction renormalisation (the required bare couplings need to exist in your theory if it is renormalisable). In the g=0 example the same applies, but now the bare coupling is a “cosmological constant” that you can include or ignore depending on the question and background spacetime of interest.

So yes, these are related to renormalisation of qft, but sometimes these diagrams do not affect the beta functions (i.e. do not affect the RG flow) in which case they might be dropped. They are usually identified with normal ordering (more generally ` complete normal ordering’) issues, which is a type of renormalisation but often more mild. These diagrams can affect and shift the vacuum around which you are doing perturbation theory if it so happens that you chose the wrong vacuum in your perturbation expansion.

In general, the procedure that removes all these self contraction diagrams (which automatically also ensures you are doing perturbation theory around the exact vacuum) is `complete normal ordering’.

Ellis, J., Mavromatos, N. & Skliros, D., Complete Normal Ordering 1: Foundations, Nucl.Phys. B909 (2016) 840-879

Answer 2

So this is a type of infinity that physicists don't usually worry about because they stem from the fact that we have to consider 2 (or more) particles at the same spacetime point.

Now why is this a problem you may ask? Well, in the current (ill-defined) construction of QFT, localising particles at the same point involves the high energy behaviour of the theory. You can see this by simply looking at Heisenberg uncertainty principle, or think in terms of wavelength and momentum.
Anyway, we do not have a UV-complete description of QFT, therefore these high energy fluctuations must be integrated out, or if you prefer, the theory that we use is already an integrated version of the full theory.

And to answer your question, the renormalisation group allows one to flow from a theory at one energy to another but it doesn't really make sense to extrapolate all the way up the energy scale because we have no idea what is out there so we must stop as some arbitrary scale basically.

Finally, since this singularities arise from our wanting to put 2 particle at the same spacetime point, one way around this problem is to consider strings, which can happily sit on top of one another. So string theory is UV complete (or thought to be, I'm a little unclear on that).

Answer 3

As suggested by Wakabaloola this seems to be a normal-ordering issue. I will review here a toy example that was helpful for me. Consider a propagator in a free scalar field theory $$G^{m^2}(x,y)=\left<\phi(x)\phi(y)\right>=\frac{\int\mathcal{D}\phi e^{-S_{m^2}[\phi]}\phi(x)\phi(y)}{\int\mathcal{D}\phi e^{-S_{m^2}[\phi]}},\qquad S_{m^2}=\frac12\int \nabla\phi^2+m^2\phi^2$$ For any $m$ this propagator is Green's function of the massive Laplacian $$(-\Delta+m^2)G^{m^2}(x,y)=\delta(x-y)$$ so we know what it is. Now let us replace $m^2\to m^2+\mu^2$ and treat $\mu^2$ as a small perturbation. Then we expect $$G^{m^2+\mu^2}(x,y)=G^{m^2}(x,y)+\mu^2\frac{\partial}{\partial m^2}G^{m^2}(x,y)+O(\mu^4)=\frac{\int\mathcal{D}\phi e^{-S_{m^2}[\phi]}\phi(x)\phi(y)\Big(1-\frac{\mu^2}2\int\phi^2+O(\mu^4)\Big)}{\int\mathcal{D}\phi e^{-S_{m^2}[\phi]}\Big(1-\frac{\mu^2}2\int\phi^2+O(\mu^4)\Big)}=\frac{\left<\phi(x)\phi(y)\right>-\frac{\mu^2}2\int_z\left<\phi(x)\phi(y)\phi^2(z)\right>+O(\mu^4)}{\left<1\right>-\frac{\mu^2}2\int_z\left<\phi^2(z)\right>+O(\mu^4)}$$ Although here the interaction term is just quadratic it still produces the problematic self-contractions. This example also illustrates that maybe computing just the partition function (the denominator) is no good, it's better to consider some observable.

One way to deal with this formal problem is by assuming that fields coming from the action are normal-ordered. This excludes their self-contractions. In particular it implies that denominator just $\left<1\right>$ so that the first non-trivial correction is $$-\frac{\mu^2}2\int_z\left<\phi(x)\phi(y)\phi^2(z)\right>\to-\mu^2\int_z\left<\phi(x)\phi(z)\right>\left<\phi(y)\phi(z)\right>=-\mu^2 \int_z G(x,z)G(z,y)$$ which is finite. Interestingly, comparing with the direct expansion of $G^{m^2+\mu^2}$ this implies a relation for the propagator $$\partial_{m^2}G^{m^2}(x,y)=-\int_z G^{m^2}(x,z)G^{m^2}(z,y)$$ which can indeed be derived by differentiating the Laplace equation w.r.t. $m^2$.

So at least in this case the naive perturbation theory with the normal-ordering prescription seems to be a valid method.