I've read about renormalization of $\phi^4$ theory, ie. $\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-m^2\phi^2-\frac{\lambda}{4!}\phi^4\,,$ particularly from Ryder's book. But I am confused about something:
Ryder begins by calculating the two point Green's function $G(x,y)$ to order $O(\lambda^2)$ (ie. the first order correction to the free propagator). Now if we take $\lambda$ to be small then this should be a good approximation, but $G(x,y)$ diverges, so he regularizes it by imposing a momentum cut-off $\Lambda$, and then makes $m$ a function of $\Lambda\,,$ ie. $m=m(\Lambda)\,.$ Then he goes on to to the same for the four point Green's function, and finds that $\lambda$ is also a function of the cut-off, ie. $\lambda=\lambda(\Lambda)\,.$ But at this point $\lambda(\Lambda)$ is no longer small when $\Lambda$ is large (in particular $\lambda\to\infty$ as $\Lambda\to\infty$), so what makes the perturbation series valid? How can we ignore the $O(\lambda^2)$ terms? I've read things like "renormalized up to 1 loop", but what about all the other loops, are they small? Or am I misunderstanding what's going on?
Perhaps it is like this: When we calculate the two point Green function G(x,y) after making a momentum cut-off at some large $\Lambda>\Lambda_0\,,$ where $\Lambda_0$ is larger than the momentum we are conducting the experiment at, we find that the mass has shifted to the physical mass $m_P=m+\lambda m^{(1)}(\Lambda_0)+O(\lambda^2)\,,$ where $m^{(1)}(\Lambda)$ is a first order correction term. Now we have a second equation for $\lambda$ given by some function $\lambda(\Lambda)$ that goes to infinity as $\Lambda\to\infty\,,$ but $\lambda(\Lambda_0)<1\,.$ Then we can ignore the $O(\lambda^2)$ term and just say $m_P=m+\lambda m^{(1)}(\Lambda_0)\,.$ Now since low energy physics should be independent of energy scale and $\Lambda_0$ is already large, we assume $m_P$ has the same form at all energy scales $\Lambda$ and this defines $m$ a functions of $\Lambda\,,$ so $m_P=m(\Lambda)+\lambda(\Lambda)m^{(1)}(\Lambda)\,,$ and then for every calculation we make to order one in $\lambda$ we substitute in this formula for $m_P$ take $\Lambda\to\infty$ and calculate the results. Now technically a better approximation to $m_P$ at the cut-off point is $m_P=m+\lambda m^{(1)}(\Lambda_0)+\lambda^2 m^{(2)}(\Lambda_0)\,,$ and so if we want a better result we should set $m_P=m(\Lambda)+\lambda(\Lambda) m^{(1)}(\Lambda)+\lambda^2(\Lambda) m^{(2)}(\Lambda)$ and do the same thing. Is it something like this?
Best Answer
Renormalization is always needed when the Hamiltonian is singular. Singular means that the formal expression for the Hamiltonian resulting from the interaction specified is not a self-adjoint operator in a dense domain. Then the dynamics is formally ill-defined and must be renormalized by taking care to represent everything properly as a limit that makes sense.
In particular, this is always the case in interacting relativistic quantum field theories in 3 or 4 space-time dimensions.
To understand why and how renormalization works one can first consider simpler situations in quantum mechanics. In this case there are explicitly solvable toy models (low rank singular perturbations of simple solvable systems) where one can see exactly what happens and why. See my paper Renormalization without infinities - a tutorial, which discusses this in detail.
About how (or whether) one can tell whether the terms in an asymptotic series will be small see the discussions in Chapter B5: Divergences and renormalization of my theoretical physics FAQ.