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What Is The Issue With Renormalisation

If you have an interest in physics you have likely come across renormalisation before, although what it really is would probably have not been explained.  Instead you would read Dirac never accepted it, Feynman called it a ‘dippy process’, and if you look under a quantum theorists rug you will find all these infinities.

I read it all and thought – surely it can’t be that bad, and decided to have a look at some papers on it.  Typical of what I found was the following:

http://arxiv.org/abs/hep-th/0212049

From the section on conventional renormalisation:
‘In other words, the bare mass has to diverge in such a way that its divergence cancels the divergent loop correction to yield a finite result. It amounts to shuffling the infinities to unobservable quantities like the bare mass. This is the part in renormalization theory which is very difficult to comprehend at the first sight.’

Shuffling infinities around?  I was despondent – Feynman was right – its dippy.  Having a background in math it’s also not legitimate math.

Ok – you do a bit more searching and you find a guy called Wilson sorted this out.  But exactly what did he do – how did he resolve it?  There doesn’t seem to be much detail at an elementary level explaining it.  This is a pity, because once you get past the mind boggling and mind numbing manipulations that is done in Quantum Field Theory, it really is pretty simple, as I hopefully will explain.

Perturbation Theory Order Notation

First we need to get out of the way a standard notation used in perturbation theory.  If you have a polynomial in x whose lowest term is of order n then that is written as O(x^n).  For example if you have a polynomial like x^3 + 3*x^4 + 2*x^10, it’s a polynomial of lowest order 3, and is written as O(x^3).

A Look At A Quantum Field Theory Infinity

Ok without further ado just what is the issue.  What is your typical QFT infinity.  Well normally you will find these horrid detailed manipulations involving Feynman diagrams with complex improper integrals in 4 dimensions that would make even the most enthusiastic mathematician wince.  Do I have to wade through this to get to the nut?
Fortunately you don’t – you simply need to see the final result.  The one I will look at is from meson-meson scattering and you will find the detail on page 145 of Zee – Quantum Field Theory In A Nutshell.  I will however write the equation in a simpler and I think even more transparent form than Zee uses.

M(K) = iλ + iλ^2*f(K) + limit Λ →∞ iCλ^2 log(Λ^2) + O(λ^3).

Here M is what’s called the scattering amplitude, K is a momentum, λ the coupling constant in meson theory, f a well behaved function, C a constant the exact value of which isn’t germane, and limit Λ →∞ iCλ^2 log(Λ^2) the result of evaluating an improper integral over momentum, where the formal definition of such as a limit has been included for reasons that will shortly be obvious.

We can see the problem immediately – the limit blows up.

This is exactly what caused Dirac to despair. If you take the limit you get nonsense. What if you don’t take the limit, but instead assign Λ some value. Such is called introducing a cut-off in the theory. Yes you get a finite answer – but as Dirac expressed it – the answer depends strongly on the cut-off. So you think about it a bit – what exactly is the cut-off – its a cut-off in momentum. Are we really sure integrating to infinity in an improper integral is correct. Large values of momentum mean large values of energy – are we really that sure our theory is correct for energies way beyond we can reach. It doesn’t seem likely. A cut-off looks the reasonable bet. OK – but what cut-off.

Before seeing how Wilson looks at renormalisation, and how it resolved this, I will for simplicity divide the equation through by i, call M(K)/i, λ’, and instead of taking the limit will have Λ fixed at some large, but unknown value.  This gives:

λ'(K) = λ + λ^2*f(K) + Cλ^2 log(Λ^2) + O(λ^3).   (1)

Now we are ready for the trick of renormalisation.  Wilson’s idea was to take the cut-off seriously.  We don’t know what cut-off to use – but we think there should be one that gives the correct value of λ’.  Suppose Λ is that value and we choose some K denoted by U to give:

λ'(U) = λ + λ^2*f(U) + Cλ^2 log(Λ^2) + O(λ^3).  (2)

λ'(U) is given a fancy name – λr the renormalised coupling constant.  A fancy name for what really is a simple concept.  Its something that can be measured so using it in formula is not a problem.

Now (as Zee expresses it his textbook) behold the wonders of renormalisation.  All it is, is a simple bit of algebra to replace the coupling constant with λr.

First, its not hard to see, O(λr^n) = O(λ^n) and λ^2 = λr^2 + O(λr^3) so if you subtract (2) from (1) you get

λ'(K) = λr + λr^2*(f(K) -f(U)) + O(λr^3).

The cut-off is gone.  Everything is expressed in terms of quantities we know.  That we don’t know what cut-off to use now doesn’t matter.

Conclusion

We now have a formula without infinities and the cut-off removed.  This is what renormalisation is about.  By assuming some cut-off and expressing our formulas in quantities that can be measured, in the above, the renormalised coupling constant, we can actually predict things.  There is nothing mathematically dubious happening.  It’s a very reasonable and straight forward process.

There is more, indeed a lot more, that can be said.  If anyone wants to go further I have found the following good:

http://arxiv.org/abs/hep-th/0212049

Hopefully this very basic introduction has demystified what is really a simple concept.

My favourite interest is exactly how can we view the world so what science tells us is intuitive.

24 replies
1. Drakkith says:

Wow, that's all renormalization is? Just saying, "Let's not use our theory at absurdly high energies"?

2. Jimster41 says:

Can you explain more why the improper integral over momentum blows up?

3. bhobba says:

Hi Guys

Thanks so much for your likes – its appreciated.

One thing I wanted to mention, but didn’t since I wanted to keep it short, is why, since the solution is actually quite straightforward, it took so long to sort out.

To see the issue, we defined λr in a certain way to be a function of λ and the cutoff. To second order, once we invert the function, λ = λr + aλr^2.

λ’ = λ + (f(K) + Clog(Λ^2))λ^2 = λr + aλr^2 + (f(K) + Clog(Λ^2))λr^2.

Now since λr = λ'(U), λr = λr + aλr^2 + (f(U) + Clog(Λ^2))λr^2 to second order, so

a = -(f(U) + Clog(Λ^2))

Since λ = λr + aλr^2 this means, to second order at least, λ secretly depends on the cutoff, and as the cutoff goes to infinity, so does the coupling constant. Normally when pertubation theory is used, you want what you perturb about to be a lot less than one. If not the higher orders do not get progressively smaller. If, for some reason, that’s not the case, it will likely fail, and you have made a lousy choice of perturbation variable. At the energies we normally access this corresponds to a cutoff that’s not large, and you get a λ that’s small, so you think its ok to perturb about. But, as we have seen, it secretly depends on the cutoff, in fact going to infinity with the cutoff. This makes it a downright lousy choice of what to perturb about. Not understanding this is what caused all these problems and took so long to sort out.

Thanks
Bill

4. bhobba says:

Wow, that’s all renormalization is? Just saying, “Let’s not use our theory at absurdly high energies”?

Yes.

Hopefully the post I did above helped understand why it remained a mystery for so long.

Stuff we use in our equations such as the electron mass and charge secretly depended on the cut-off. We did experiments at low energy from which we obtained small values, which when you look at the equations, means in effect its a low cut-off – otherwise you would get large values. This fooled people for yonks.

The way it works, in QED for example, is by means of what’s called counter-terms:
[URL]http://isites.harvard.edu/fs/docs/icb.topic1146665.files/III-5-RenormalizedPerturbationTheory.pdf[/URL]

You start out with a Lagrangian expressed in terms of electron mass and charge. But we know those values actually depend on the cut-off. We want our equations to be written in terms of what we actually measure which pretty much means measure at low energies. This is called the renormalised mass and charge. We write our Lagrangian in terms of those and you end up with equations in those quantities plus what are called counter terms that are cut-off dependant. You do your calculations and of course it still blows up. But now you have these counter terms that can be adjusted to cancel the divergence. Basically this is subtracting equation (2) from (1) in my paper. The cut-off dependant quantities cancel and you get finite answers.

Tricky – but neat.

Thanks
Bill

5. bhobba says:

Can you explain more why the improper integral over momentum blows up?

You have to read the reference I gave – its simply the result of calculating the equation.

Very briefly if you have a look at the resulting equation you can simplify it to M(K) = iλ + iλ^2*f(K) + 1/2*λ^2 ∫d^4k 1/k^4 (the integral is from -∞ to ∞) where k is the 4 momentum (its done by breaking the integral into two parts – one for very large k and the other for k less than that). That’s a tricky integral to do but after a lot of mucking about, is i π^2 ∫ 1/k^2 dk^2 where the integral is from 0 to ∞. This is limit Λ→∞ i π^2 log (Λ^2).

But as far as understanding renormalisation is concerned its not germane to it. You can slog through the detail but it wont illuminate anything.

Thanks
Bill

6. stevendaryl says:

There are a few things that are a little mysterious about it, still. So if you assume that you understand the low-momentum behavior of a system, but that its high-momentum behavior is unknown, it makes sense not to integrate over all momenta. But why is imposing a cut-off the right way to take into account the unknown high-energy behavior?

The second thing that’s a little mysterious is the relationship between renormalization and the use of “dressed” propagators. It’s been a while since I studied this stuff (a LONG while), but as I remember it, it goes something like this:

You start describing some process (such as pair production, or whatever) using Feynman diagrams drawn using “bare” masses and coupling constants. You get loops that would produce infinities if you integrated over all momenta. Then you renormalize, expressing things in terms of the renormalized (measured) masses and coupling constants. This somehow corresponds to a similar set of Feynman diagrams, except that

[LIST=1]
[*]The propagator lines are interpreted as “dressed” or renormalized propagators
[*]You leave out the loops that have no external legs (the ones that would give infinite answers when integrating over all momenta)
[/LIST]
The first part is sort of by definition: The renormalization program is all about rewriting amplitudes in terms of observed masses and coupling constants. But the second is a little mysterious. In general, if we have two power-series:

$A = sum_n A_n lambda_0^n$
$lambda = sum_n L_n lambda_0^n$

we can rewrite $A$ in terms of $lambda$ instead of $lambda_0$ to get something like:

$A = sum_n B_n lambda^n$

For a general power series, you wouldn’t expect the series in terms of $B_n$ to be anything like the series in terms of $A_n$. But for QFT, it seems that they are basically the same, except that the $B_n$ series skips over the divergent terms.

I’m guessing that the fact that the renormalized Feynman diagrams look so much like the unrenormalized ones is a special feature of propagators, rather than power series in general.

7. bhobba says:

There are a few things that are a little mysterious about it, still. So if you assume that you understand the low-momentum behavior of a system, but that its high-momentum behavior is unknown, it makes sense not to integrate over all momenta. But why is imposing a cut-off the right way to take into account the unknown high-energy behavior?

I think the answer lies in the condensed matter physics these ideas sprung from. I don’t know the detail but the claim I have read is pretty much any theory in the low energy limit will look like that.

Certainly it’s the case here – to second order we have:

λ’ = λr + λr^2 f(K) – λr^2 f(U) = λr + λr^2 f(K) + λr^2 C*Log (Λ^2) for some cut-off Λ.

Its the same form as the un-renormalised equation. This is the self similarity you hear about, that is said to be what low energy equations must be like.

The second thing that’s a little mysterious is the relationship between renormalization and the use of “dressed” propagators. It’s been a while since I studied this stuff (a LONG while), but as I remember it, it goes something like this:

Wilson’s view where the cut-off is taken seriously looks at it differently.

If you look at the bare Lagrangian its simply taken as an equation valid at some cut-off – we just don’t know the cut-off. You write it in terms of some renormalised parameters by means of counter-terms where the cut-off is explicit ie the counter terms are cut-off dependant. You then shuffle these counter terms around to try to get rid of the cut-off – similar to what I did.

I am not that conversant with the detailed calculations of this method but I think its along the lines of the following based on the example before.

I will be looking at second order equations. First let λr be some function of λ so λ = λr + aλr^2 for some a ie λ = (1 + aλr)λr where aλr is the first order of the counter term in that approach since we are looking only at second order.

You substitute it into your equation to get the renormalised equation with the counter term:

λ’ = λr + (f(K) + C*log Λ^2 + a) λr^2.

Now we want the cut-off term to go away – so we define a = a’ – C*log Λ^2 and get

λ’ = λr + (f(K) + a’)) λr^2.

We then apply what we say λr is to determine a’ – namely λr = λ'(U) so a’ = -f(U).

My reading of the modern way of looking at renormalisation using counter-terms is its along the lines of the above.

Thanks
Bill

8. atyy says:

There are a few things that are a little mysterious about it, still. So if you assume that you understand the low-momentum behavior of a system, but that its high-momentum behavior is unknown, it makes sense not to integrate over all momenta. But why is imposing a cut-off the right way to take into account the unknown high-energy behavior?

The idea is that below a certain energy, there are “emergent” degrees of freedom that are enough for describing the very low energy behaviour we are interested in. For example, even though the standard model has quarks, for condensed matter physics, we just need electrons, protons and neutrons. The cut-off represents the energy where we will be obliged to consider new degrees of freedom like supersymmetry or strings. If we knew the true degrees of freedom and the Hamiltonian at high energy, we could integrate over the high energies and by an appropriate change of “coordinates” obtain the emergent degrees of freedom and Hamiltonian at the cut-off. However, in practice we do not know the high energy details, so we make a guess about the low energy degrees of freedom and the low energy symmetries (here low means much lower than the high energy, but still much higher than the energy at which we do experiments). So we put in all possible terms into the Hamiltonian with the low energy degrees of freedom that are consistent with any known symmetry. In other words, we must do the integral over the unknown high energy degrees of freedom (as required in the path integral picture), and we attempt to do it by guessing the low energy degrees of freedom and symmetry.

It turns out that even if we use a simple Hamiltonian that is lacking many possible terms, if we have a cut-off that is low enough that we know there are not yet new degrees of freedom, but still much higher than the energy scale we are interested in, then we can show that the low energy effective Hamiltonian will contain all possible terms consistent with the symmetry – these turn out to be the counterterms.

The usual explanation of the counterterms is physically senseless. It is better to think of them as automatically generated by a high cutoff, and a flow to low energy. However, it is incredibly inefficient to start calculations by writing down all possible terms. The counterterm technology is a magically efficient way to get the right answer (like multiplication tables :oldtongue:).

Naturally, our guess about the low energy degrees of freedom may be wrong, and our theory will be falsified by experiment. However, a feature of this way of thinking is that a non-renormalizable theory like QED or gravity, by requiring a cut-off, shows the scale at which new physics must appear. In other words, although experiment can show us new physics way below the cut-off, the theory itself indicates new physics in the absence of experimental disproof.

9. wabbit says:

You have to read the reference I gave – its simply the result of calculating the equation.

Very briefly if you have a look at the resulting equation you can simplify it to M(K) = iλ + iλ^2*f(K) + 1/2*λ^2 ∫d^4k 1/k^4 (the integral is from -∞ to ∞) where k is the 4 momentum (its done by breaking the integral into two parts – one for very large k and the other for k less than that). That’s a tricky integral to do but after a lot of mucking about, is i π^2 ∫ 1/k^2 dk^2 where the integral is from 0 to ∞. This is limit Λ→∞ i π^2 log (Λ^2).

But as far as understanding renormalisation is concerned its not germane to it. You can slog through the detail but it wont illuminate anything.

Thanks
Bill

Isn’t there another way to look at it ? I am not familiar with renormalization but one argument I recall seeing went (very) roughly along the following lines :

Assume the integral is actually finite, because the “true” integrand is not 1/k^2 but some unknown function phi(k) such that phi(k)~1/k^2 for k not too large (or small, wherever it diverges), and phi(k) is summable (i.e. assume we don’t really know the high energy behavior, but whatever it is must give a finite answer, perhaps because spacetime is quantized or whatever reason).

Then the calculation still holds, there are no infinities but the unknown parts cancel off at low energy. The result is the same but the mysterious infinities have been replaced by unknown finite quantities.

My recollection of that argument is very hazy so I am unsure about it, but does this work (or rather, something similar, presumably after some renormalization of the argument :wink: )?