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?


6 Answers 6


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.

  • $\begingroup$ What is meant by 'singular' in this context? Ps. I much appreciated your paper, I feel like it helped my understanding a lot. $\endgroup$ Jun 22, 2016 at 0:33
  • 1
    $\begingroup$ @RobinEkman: Singular means that the formal expression for the Hamiltonian resulting from the interaction specified is not a self-adjoint operator in a dense domain. Thus the dynamics is formally ill-defined and must be renormalized by taking care to represent everything properly as a limit that makes sense. $\endgroup$ Jun 22, 2016 at 5:38

A lot of difficulties in renormalization get clarified, when one uses the point of view of "Reparametrization of the theory".

Let me illustrate what I mean:

If one naively obtains a quantum field theory from a classical field theory by a quantization procedure, one gets unrealistic results for measurable quantities. This shouldn't be surprising: Why should classical parameters of our theory be the correct ones for our quantum field theory?

So we want to reparametrize our theory, such that finite new parameters give realistic results.

Consider a Lagrangian Density $\mathcal{L} = \frac{1}{2} A \partial_\mu \phi \partial^\mu \phi + \frac{1}{2} B^2 \phi^2 + \frac{1}{4!} C \phi^4$.

Now measurable quantities of the corresponding quantum field theory will depend on those parameters, e.g.

  • "physical mass" = pole of propagator = P(A,B,C)

  • "Wave function renormalization" = Residuum of propagator = Z(A,B,C)

  • "4-Point Vertex at certain mandelstams" = G(A,B,C)

Now we want to invert those functions to get a new parametrization A(P,Z,G), B(P,Z,G), C(P,Z,G). Since P, Z, G are measurable in the experiment, they should be finite and a good parametrization of our Quantum Field Theory.

Now at first consider a nonperturbative context: Here we should be able to invert P, Z, G exactly to obtain a reparametrization in which our measurable quantities are hopefully finite.

In a perturbative context we can calculate P,Z,G up to order, let's say, N. Then the "Renormalization conditions" (which are more "Reparametrization conditions") define a different reparametrization for each order N. If we renormalize up to order N, then we've choosen a reparametrization which makes everything finite till order N. Maybe this reparametrization will also make stuff finite in higher order, but maybe not.

Nota Bene: Stuff like MS or MS-bar just defines different renormalization/reparametrization conditions, namely implicitely via counterterms, since every substraction scheme determines the structure of the counterterms and the structure of the counterterms determines the reparametrization.

P.S.: In above nomenclature normally one chooses the Reparametrization condition Z(A,B,C) = 1.

P.P.S.: This post is more an illustration than a real rigorous argumentation. But everything I said can be made more rigorous.


Before I try to answer your question, one thing:

Does Ryder really calculate the $\mathcal{O}(\lambda^2)$ to the propagator as the first contribution in perturbation theory, because there is actually a $\mathcal{O}(\lambda^1)$ to the propagator and the $\mathcal{O}(\lambda^2)$-loop is as far as I am concerned a two-loop diagram, i.e. having two loop momenta, whhich is actually hard to work out as a first example to introduce the procedure of renormalization. But to be honest, this $\mathcal{O}(\lambda^1)$-contribution does not lead to any renormalization of the mass, so you won't learn as much from this, but I am not sure if Ryder mention this.

But let me try to answer your question with a slightly different point of view:

The thing with renormalization is, that you see that your loop integrals diverge, so you impose a momentum-cut-off $\Lambda$, with the motivation in mind, that the theory has only a limited range of validity. And as you already mentioned, the result will then be dependent of this cut-off, so you "hide" the dependence on the cut-off in the redefinition of the parameters (masses and couplings) of your theory. The argument is that the parameters you write in the Lagrangian (say, $\lambda_0$ and $m_0$) have no physical meaning, but the parameters you measure by experiment ($\lambda$ and $m$) have and they are shifted. So up to this point it is rigth to say that the parameters are dependend on the cut-off, but the problem you mention does not arise.

Let me point this out at the following example: For the $\phi$-propagator (the Green's-function) you get some modification due to the loop integral $\Pi(p^2;\Lambda)$: $$\frac{i}{p^2-m^2+i\epsilon} \quad\to\quad \frac{i}{p^2-m^2-\Pi(p^2;\Lambda)+i\epsilon} $$ where you can interprete the term $m^2 + \Pi(p^2;\Lambda) $ as the effective $m_\mathrm{eff}^2$ of the propagator. (To the form of the propagator including the loops, it is more or less a geometric series of free propagator and lopp insertions) To go on, it is now necessary to declare, how you define your parameter (in this case the mass $m$) by imposing a certain renormalization condition. So for example a particular choice is that you say that for $p^2 = m^2$ $m_\mathrm{eff}^2 = m^2$ or $\Pi(p^2=m^2;\Lambda) = 0$. But even in this case you will see a $\Lambda$-dependendent term in $\Pi(p^2;\Lambda)$ To get rid of this you, redefine the loop integral as $$ \Pi(p^2;\Lambda) \quad\to\quad \overline{\Pi}(p^2) = \lim_{\Lambda\to\infty} \left( \Pi(p^2;\Lambda) - \Pi(p^2=m^2;\Lambda) \right) $$

The limit $\Lambda\to\infty$ can be done, because you subtract more or less the divergent part of the loop integral. And that is the important point you not only say that you redefine your parameters a dependent on the cut-off, but you also impose a renormalization condition which in the end leads to a $\Lambda$-independent quantity.

  • $\begingroup$ Oh I meant he calculates only the one loop and absorbs everything else into $O(\lambda^2)\,.$ However I'm still confused about why the approximation should work: To me it looks like we have a perturbation series of $f(m,\lambda)\sim\Sigma_n \lambda^n a_n(m)$ and we are sending $\lambda$ to infinity. Is sending $m\to\infty, \lambda\to\infty$ in the way that is done equivalent to writing it as a perturbation series in the physical variables $m_{\text{phys}},\lambda_{\text{phys}},$ ie. is it like $\lim_{m,\lambda\to\infty} f(m,\lambda)\sim\sum_n \lambda_{\text{phys}}^n a_n(m_{\text{phys}})$? $\endgroup$
    – JLA
    Jun 14, 2016 at 0:27
  • $\begingroup$ Yes renormalization is a very confusing topic. I for myself won't call myself an expert in this and I understand that this procedure of regulating the integrals and renormalize the measurable quantities, such that the dependence on the regulator vanishes seems like swiping many divergent terms under the carpet and hoping that no one will know. But let me put it this way, suppose that the loop integrals have form: regular term for p + divergent term (which is regulated). what you then do is to subtract the divergent term as I pointed out above with the redefinition of the loop-integral. $\endgroup$
    – P.Stoecker
    Jun 15, 2016 at 20:35
  • $\begingroup$ I don't know how much you read about renormalized perturbation theory, but usually one introduces the so called counter-terms, i.e. you rephrase the Lagrangian by 'shifting' from the bare to the true parameters and these extra terms can than be interpreted as interactions and additional Feynman-rules. And when you write down a particular process you have to include the counter-terms as well. But the "coupling" of this counter-terms is than fixed by your renormalization condition. This is for example done in 10.2 of Peskin & Schroeder. Maybe in this framework you don't have this confussion. $\endgroup$
    – P.Stoecker
    Jun 15, 2016 at 20:49

I would like to stress the difference between

1) Perturbative Renormalization

2) Non-perturbative Renormalization

By Perturbative Renormalization I mean removing infinities from the computation of a correlator/amplitude, order by order. This is done by introducing counterterms, i.e. re-writing the bare parameters of the lagrangian as $\lambda_{Bare} = \lambda_{obs} + \delta \lambda$, choosing a regularization scheme with its "cut-off" scale $\Lambda$, and imposing the value of a (possible finite number of) correlators/amplitudes. You have to actually measure these values in order to establish them - you cannot compute them from the theory. In term of these parameters, any other correlator/amplitude can be computed obtaining a finite result, no matter how high is the order you get to.

Why this works can be interpreted either with a complicate diagrammatical analysis or with a floating-cutoff approach, a la Wilson.

Now let's turn to the second point.

Once you solved the first problem (making finite the quantities order by order), and you start computing correlators/amplitudes, you discover that the (now finite) results involves "large logaritms" like $log(m/m_0)$ where $m_0$ is the renormalization scale, i.e. the tipical scale of the quantities you measure and impose which could be, for example, correlators/amplitudes involving fields/particles with a similar common energy scale, and $m$ is the typical scale of the correlator/amplitude you want to compute. This is a problem because, even tough the quantities you are computing are finite at every order, you cannot neglect higher orders which involve more and more of these "large logaritms".

The solution to this problem is to smoothing the passage from the renormalization scale to the scale of interest, that is to pass from $m_0$ to a scale $m'_0 = m_0 + \delta m$ (now the logaritms are small), and from here to $m'_0 + \delta_m$ and on and on until you get to $m$. This "smoothing" procedure is the Renormalization Group Flow, and can be thought as an expansion of the correlators/amplitudes not in terms of $\lambda_{Bare}$ or $\lambda_{obs}$ but rather of coupling costants which depend (or run) on the scale $m$ of the observables you want to compute.

Now, if this running is such that the coupling costants remains small then the perturbative expansion will remain valid and so "perturbative renormalization" holds, otherwise other technique have to be devised in order to produce quantitative previsions from your theory.


The trick is in the introduction of a renormalization scale.

Once the perturbation theory has been regularized, one obtains a momentum (and cut-off) dependent interaction of the (schematic) form in 4D $$\lambda(p)=\lambda_0+\alpha\lambda_0^2 \ln(\Lambda^2/p^2), $$ where $\lambda_0$ is the bare interaction, and $\alpha$ some numerical factors.

What one does is to fix the value of the interaction at a given energy $p^2=\mu^2$, such that $\lambda(\mu)\ll 1$. Expressing $\lambda(p)$ in terms of $\lambda (\mu)$ one obtains $$\lambda(p)=\lambda(\mu)+\alpha \lambda(\mu)^2\ln(\mu^2/p^2),$$ which is indeed a good perturbation theory.

Note that this perturbation theory works fine only for energies close enough to $\mu$, and that the reference point $\lambda(\mu)$ should be given by the experiments, or a more fundamental theory.

  • $\begingroup$ Ok I'll continue reading about that. I guess maybe I am confused about how you know, when making a prediction to what order you should calculate the perturbation series, and if adding more terms makes the value of the truncated series significantly different. $\endgroup$
    – JLA
    Jun 14, 2016 at 3:34
  • $\begingroup$ @JLA: since the perturbation expansion in terms of $\lambda(\mu)$ is fine (for $p$ close to $\mu$), adding more terms won't change much. However, one can resums the logs, to improve the series, using RG like equations. $\endgroup$
    – Adam
    Jun 14, 2016 at 6:14

To understand why renormalization may work you might first consider simpler situations in Classical/Quantum Mechanics. In this case there are explicitly solvable toy models where one can see exactly what happens and why. See my paper "A Toy Model of Renormalization and Reformulation" on arXiv.

About how to cope with growing terms, see my short note here.


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