Renormalisation Group II: Loops, Infinities and Cutoffs from Scratch

Part 2 of eight. previous: Why scale matters · Course home · Glossary · next: Block spins

Part 1 pulled Feynman rules out of thin air and used words like loop, Euclidean, cutoff and regularise without explaining them. This part builds all of it slowly, starting from the path integral itself: first where the diagrams and their rules come from, then why loops diverge, what a cutoff is and what renormalisation actually does. If you have Peskin & Schroeder or Ashok Das's Field Theory: A Path Integral Approach, the matching sections are mentioned as we go, but you don't need them.

📝 Note

What's in this part and what you can skip first time. This is the longest page in the course, so here's the map:

1. Where Feynman diagrams come from. The rules of Part 1, derived.

2. Where loop integrals come from. Why a loop leaves one momentum free.

3. The Wick rotation. Why we work in "imaginary time" and why that links particles to magnets.

4. Why loops blow up and how to tell how badly, just by counting powers.

5. The cutoff and the other ways to make an integral finite.

6. One loop integral from start to finish. The standard recipe. Skippable on a first reading: it's technique and you can come back when you need it.

7. The same integral, three ways, showing what does and doesn't depend on the regulator. This is the heart of the page.

8. Renormalisation, plus what changes beyond one loop. The "beyond one loop" part is skippable first time.

9. An interactive widget where you move the cutoff and watch the bare parameters chase it.

If you only read two sections, read 1 and 7.

1. Where Feynman diagrams come from

Part 1 pulled three rules out of thin air: a line means \(1/(k^2+m^2)\), a vertex means \(\lambda_0/4!\) with a minus sign and a loop means "integrate over \(k\)". None of that was justified. Here is where all of it comes from, in one calculation.

Everything in this theory is an average over field shapes, weighted by \(e^{-S}\):

\[ \langle\,\cdots\rangle = \frac{\int\mathcal D\phi\;(\cdots)\;e^{-S_0[\phi]-S_{\rm int}[\phi]}}{\int\mathcal D\phi\;e^{-S_0[\phi]-S_{\rm int}[\phi]}},\qquad S_{\rm int}=\frac{\lambda_0}{4!}\int\mathrm d^dz\,\phi(z)^4. \]

Picture it

\(\int\mathcal D\phi\) looks frightening, but it's just "try every possible shape of the field and weight each one". Think of an exam where instead of one answer you average over every student's answer sheet, weighting each by how good it is. The weight \(e^{-S}\) is high for smooth, low-energy shapes and tiny for wild ones.

Two facts do all the work, both proved in Part 5:

1. Without the interaction, the average of two fields is the propagator: \(\langle\phi(x)\phi(y)\rangle_0=G(x-y)\), which in momentum space is \(1/(k^2+m^2)\). That's where the line comes from.

2. Wick's theorem: with the free weight, the average of any number of fields is the sum over all ways of pairing them up, each pair giving a propagator. Averages of an odd number vanish. To pair two fields, also called contracting them, means nothing more than taking those two out of the list and writing \(G\) for them. Part 5 derives the theorem in three lines from a source trick, so you never have to take it on faith.

Now expand the exponential, since \(\lambda_0\) is small:

\[ e^{-S_{\rm int}} = 1 - \frac{\lambda_0}{4!}\int\mathrm d^dz\,\phi(z)^4 + \dots \]

Every vertex comes with a minus sign, simply because the exponential expands with alternating signs. That's the minus sign Part 1 used without explaining.

Take the two-point function \(\langle\phi(x)\phi(y)\rangle\) and keep the first-order term:

\[ -\frac{\lambda_0}{4!}\int\mathrm d^dz\;\big\langle\phi(x)\,\phi(y)\,\phi(z)\phi(z)\phi(z)\phi(z)\big\rangle_0 . \]

Six fields, so Wick's theorem gives \(5\times3\times1=15\) pairings. Sort them into two kinds:

• 3 pairings join \(x\) straight to \(y\) and pair the four \(z\)'s among themselves. The diagram falls into two disconnected pieces and such pieces are exactly cancelled by the denominator of the average. Ignore them.

• 12 pairings connect everything: \(x\) pairs with one of the four \(\phi(z)\) (4 choices), \(y\) with one of the remaining three (3 choices) and the last two \(\phi(z)\) pair with each other (1 way). That's \(4\times3\times1=12\).

So the term is

\[ -\frac{\lambda_0}{4!}\cdot 12\int\mathrm d^dz\;G(x-z)\,G(z-y)\,G(0) = -\frac{\lambda_0}{2}\int\mathrm d^dz\;G(x-z)\,G(z-y)\,G(0). \]

Look at what each piece became. The two propagators \(G(x-z)\) and \(G(z-y)\) are the particle travelling in and out. The factor \(G(0)=G(z-z)\) is a propagator that starts and ends at the same point: the loop. In momentum space a propagator at zero separation is

\[ G(0)=\int\frac{\mathrm d^dk}{(2\pi)^d}\,\frac{1}{k^2+m^2}, \]

which is exactly the loop integral of Part 1. And the leftover number, \(12/4!=\tfrac12\), is Part 1's factor.

🧠 Defn

The symmetry factor is the number of Wick pairings that give a particular diagram, divided by the factorials put into the couplings. It exists only to correct over-counting: the \(4!\) in \(\lambda_0/4!\) was put there so that the commonest diagrams come out with tidy factors.

So the Feynman rules are not magic. A diagram is a bookkeeping picture for one group of Wick pairings: lines are propagators, dots are vertices from expanding \(e^{-S_{\rm int}}\) (each with a minus sign and a \(\lambda_0\)), closed loops are propagators that return to their own vertex and the symmetry factor counts how many pairings gave the same picture. Everything in Parts 1, 6 and 7 is this calculation, done for bigger diagrams.

Otaku corner

Wick's theorem is the Death Note of QFT: an enormous, tedious problem reduced to "write down every pairing and the rules do the rest". The catch, as any student who has counted the symmetry factor of a two-loop diagram at 2 am will tell you, is that the rules are simple but the bookkeeping is merciless.

2. Where loop integrals come from

Draw a Feynman diagram and at every dot (vertex) apply one rule: momentum in equals momentum out. For a diagram with no closed loops (a tree diagram), that rule fixes every internal momentum. Nothing is left to integrate.

Close a loop and something new happens. Momentum conservation at each dot still holds, but it no longer fixes everything:

Picture it

Picture a circle of friends passing money around. Each person gives away exactly what they receive, so nobody gains or loses. But the amount going round the circle can be anything: ten rupees, a thousand, a million. The rule at each person doesn't fix it. A loop momentum is exactly that: an amount circulating around the loop that the local rules leave free.

Quantum mechanics says: when something isn't fixed, add up all the possibilities. So every loop comes with an integral over its free momentum,

\[ \int\frac{\mathrm d^4k}{(2\pi)^4}\,(\text{the propagators that depend on }k). \]

The loop is a virtual particle that appears and disappears and \(k\) is its momentum. Because it's never observed, every value of \(k\) contributes, including huge ones. That word "every" is where the trouble in Part 1 came from.

One more useful fact: loops are the quantum part. If you keep track of \(\hbar\), a diagram with \(L\) loops carries an extra \(\hbar^{L}\) compared with the tree diagram with the same outside lines. Tree diagrams are classical physics and loops are the quantum jitter on top (Das §10.3).

3. Why we rotate to Euclidean space

In ordinary (Minkowski) spacetime, the propagator of a particle with mass \(m\) is

\[ \frac{i}{(k^0)^2-\mathbf k^2-m^2+i\varepsilon}. \]

The tiny \(+i\varepsilon\) is Feynman's rule for which way particles go in time. Think of the integral over the energy \(k^0\) as a path along the real line in the complex \(k^0\) plane.

Step 1: find the poles, rather than being told where they are

The propagator blows up where its denominator vanishes. Set it to zero with \(E_k=\sqrt{\mathbf k^2+m^2}\):

\[ (k^0)^2 - E_k^2 + i\varepsilon = 0 \qquad\Longrightarrow\qquad k^0 = \pm\sqrt{E_k^2 - i\varepsilon}. \]

Now expand the square root for small \(\varepsilon\), using \(\sqrt{a-x}\approx\sqrt a - \tfrac{x}{2\sqrt a}\):

\[ \sqrt{E_k^2 - i\varepsilon} \approx E_k - \frac{i\varepsilon}{2E_k}. \]

So the two poles sit at

\[ k^0 = +E_k - \frac{i\varepsilon}{2E_k} \quad\text{(just \textit{below} the real axis, on the right)}, \qquad k^0 = -E_k + \frac{i\varepsilon}{2E_k} \quad\text{(just \textit{above} it, on the left)}. \]

That is the whole content of the \(i\varepsilon\): it decides which side of the real axis each pole falls on. Nothing else in the calculation depends on it.

Step 2: notice which quarters of the plane are empty

Put those two poles on a picture. One is in the bottom-right quarter, the other in the top-left. The top-right with the bottom-left quarters contain nothing at all.

That is the fact the whole rotation rests on. If the \(i\varepsilon\) had the other sign, the poles would sit in the other two quarters, the free quarters would be different, so the rotation would have to go the other way.

Step 3: why you are allowed to rotate (the step usually skipped)

Cauchy's theorem says that the integral around a closed loop containing no poles is zero. Our path is not closed, so we have to build a closed one out of four pieces:

1. the real axis from \(-R\) to \(+R\), which is the integral we want,

2. a quarter-circle arc of radius \(R\) sweeping up through the empty top-right quarter,

3. the imaginary axis coming back down from \(iR\) to \(-iR\),

4. another quarter-circle through the empty bottom-left quarter.

No poles are enclosed, so those four pieces add to zero. That gives "real axis integral \(=\) imaginary axis integral" only if the two arcs contribute nothing. So check it, rather than hoping.

On the arc, \(|k^0| = R\). The integrand falls off like \(1/R^2\), since the denominator is dominated by \((k^0)^2\) when \(R\) is large. The length of a quarter-circle grows like \(R\). So the arc contributes at most about

\[ \frac{\pi R/2}{R^2 - E_k^2} \;\xrightarrow[R\to\infty]{}\; 0 . \]

The arcs vanish. Now the rotation is justified.

📝 Note

This is the condition, not a formality. Had the integrand fallen off only as \(1/R\), the numerator \(R\) from the arc length would have cancelled it, so this estimate would no longer show the arcs vanishing. Note carefully what that does and does not mean: it means the bound has failed, not that the rotation is impossible. When the arcs cannot be shown to vanish, you either keep them as extra terms or introduce a regulator that makes the integral well enough behaved to try again. Every Wick rotation quietly rests on this step, so it is the one to look at when a rotation misbehaves.

Picture it

Imagine a rubber band stretched along the real line, with two nails (the poles) hammered into the board. You can swing the band round as much as you like, as long as it doesn't catch on a nail. The two empty quarters let it swing a quarter turn onto the imaginary line without catching anything and the answer stays the same. Checking the arcs is checking that the far ends of the band are not nailed down too.

Step 4: do the rotation, keeping every \(i\)

Set \(k^0 = ik^4\) with \(k^4\) real. Two things change, both of which must be tracked.

The measure. \(k^0 = ik^4\) gives \(\mathrm d k^0 = i\,\mathrm d k^4\). That factor of \(i\) is not optional, it travels with the integral for the rest of the calculation.

The invariant. \((k^0)^2 = (ik^4)^2 = -(k^4)^2\), so

\[ (k^0)^2 - \mathbf k^2 - m^2 \;=\; -(k^4)^2 - \mathbf k^2 - m^2 \;=\; -\big(k_E^2 + m^2\big), \qquad k_E^2 \equiv (k^4)^2 + \mathbf k^2 . \]

This is the Wick rotation. Note what \(k_E^2\) is: a sum of four squares, so it is positive, while it treats the time direction exactly like the three space directions. Track the factors and the propagator becomes

\[ \frac{i}{k^2-m^2+i\varepsilon} \;\longrightarrow\; \frac{-i}{k_E^2+m^2}, \]

whose \(-i\) then meets the \(i\) from the measure, so that the combination actually appearing under the integral, \(\mathrm d k^0\,i/(k^2-m^2+i\varepsilon)\), becomes the clean \(\mathrm d k^4/(k_E^2+m^2)\). Either way the useful part is the same: no poles anywhere on the real axis, no \(i\varepsilon\) left to worry about, because \(k_E^2+m^2\) never vanishes for real \(k_E\).

A worked example, both ways

Nothing above is convincing until you watch the two routes land on the same number. Take the simplest case, dropping the \(\mathbf k\) integral entirely and writing \(E\) for \(E_k\):

\[ I = \int_{-\infty}^{\infty}\frac{\mathrm d k^0}{(k^0)^2 - E^2 + i\varepsilon}. \]

Route A, staying in Minkowski space and using residues. Keep \(\varepsilon\) finite for now, writing \(a = \sqrt{E^2 - i\varepsilon}\), so the denominator factorises cleanly as \((k^0-a)(k^0+a)\) with neither pole sitting on the path. This matters: a pole on the contour cannot simply be ignored, which is exactly what the \(i\varepsilon\) is there to prevent.

Close the contour with a big semicircle in the upper half plane. It contributes nothing, by the same \(1/R\) argument as in Step 3. The only pole inside is the left one, at \(k^0 = -a\), whose residue is

\[ \left.\frac{1}{k^0-a}\right|_{k^0=-a} = -\frac{1}{2a}. \]

Cauchy's residue theorem gives \(2\pi i\) times that:

\[ I_\varepsilon = 2\pi i\cdot\left(-\frac{1}{2a}\right) = -\frac{i\pi}{a}. \]

Only now let \(\varepsilon\to0^+\), so that \(a\to E\):

\[ I = -\frac{i\pi}{E}. \]

Route B, rotating first. Substitute \(k^0 = ik^4\), so \(\mathrm d k^0 = i\,\mathrm d k^4\) and \((k^0)^2 - E^2 = -(k^4)^2 - E^2\):

\[ I = \int_{-\infty}^{\infty}\frac{i\,\mathrm d k^4}{-(k^4)^2 - E^2} = -i\int_{-\infty}^{\infty}\frac{\mathrm d k^4}{(k^4)^2 + E^2}. \]

That last integral is one you already know, \(\int\mathrm d x/(x^2+a^2) = \pi/a\). So

\[ I = -i\cdot\frac{\pi}{E} = -\frac{i\pi}{E}. \]

The same answer. Once the deformation has been justified, Route B needs no residues, no pole bookkeeping, no \(i\varepsilon\): just a substitution followed by a first-year integral. The pole bookkeeping did not disappear, it was spent once, in Steps 1 to 3, to earn the right to rotate. Afterwards every integral is an ordinary one, which is the entire reason anybody rotates.

🤔 Problem

Do one yourself. Evaluate \(\displaystyle\int_{-\infty}^{\infty}\frac{\mathrm d k^0}{\big[(k^0)^2-E^2+i\varepsilon\big]^2}\) by rotating, then check the power of \(E\) by dimensions.

Show solution

Rotate: \(k^0 = ik^4\), so \(\mathrm d k^0 = i\,\mathrm d k^4\) and the bracket becomes \(-\big[(k^4)^2+E^2\big]\). Squaring that bracket kills the minus sign, since \((-1)^2 = 1\):

\[ \int\frac{i\,\mathrm d k^4}{\big[(k^4)^2+E^2\big]^2} = i\int\frac{\mathrm d k^4}{\big[(k^4)^2+E^2\big]^2}. \]

The standard integral is \(\int_{-\infty}^{\infty}\mathrm d x/(x^2+a^2)^2 = \pi/(2a^3)\), so the answer is \(i\pi/(2E^3)\).

Dimensions: the original integrand goes like \(1/(\text{energy})^4\) with one power of energy from \(\mathrm d k^0\), leaving \(1/(\text{energy})^3\). The answer has \(E^3\) downstairs, as required. Notice also that squaring flipped the overall sign compared with the first example, because the minus from the rotation was raised to an even power.

Two checks worth doing. Differentiating the first result with respect to \(E^2\) gives \(\partial_{E^2}(-i\pi/E) = i\pi/(2E^3)\), the same answer. And note that what was squared here is the denominator as written; squaring the full Feynman propagator \(i/(k^2-m^2+i\varepsilon)\) brings an extra \(i^2=-1\) along with it.

What the rotation means, beyond the algebra

There is a reason this trick keeps appearing in a course about magnets.

The quantum weight is \(e^{iS}\), an oscillating phase, which makes every integral a delicate business of cancellation. Under \(t \to -i\tau\), that phase turns into

\[ e^{iS} \;\longrightarrow\; e^{-S_E}, \]

a real, decaying weight. But \(e^{-S_E}\) is a Boltzmann factor, exactly the object from Part 3 that gives a magnet its probability of sitting in some arrangement. So the rotation does not just make integrals easier. It is the precise statement of why a quantum field theory in \(d\) spacetime dimensions is the same mathematics as a statistical system in \(d\) space dimensions, which is the bridge this whole course walks across.

Doing the rotation by computer

You don't have to take Cauchy's word for it. Wolfram can do the complex integral directly, \(i\varepsilon\) and all:

(* 1. along the real k0 line, with the i eps kept *)
Integrate[1/(k0^2 - e^2 + I eps), {k0, -Infinity, Infinity}, Assumptions -> {e > 0, eps > 0}]
(* Pi/Sqrt[-e^2 + I eps]   ->   -I Pi/e  as eps -> 0 *)

(* 2. along the imaginary line, k0 = I k4, dk0 = I dk4 *)
Integrate[I/(-k4^2 - e^2), {k4, -Infinity, Infinity}, Assumptions -> e > 0]
(* -I Pi/e : the same *)

(* 3. where are the poles?  (e = 1, eps = 1/10) *)
k0 /. Solve[k0^2 - 1 + I/10 == 0, k0] // N
(* {-1.001 + 0.050 I, 1.001 - 0.050 I} : top-left and bottom-right, as in the picture *)

(* 4. a full 4D integral, done both ways (Peskin eq. 6.49) *)
inner = Integrate[1/(k0^2 - w^2 + I eps)^3, {k0, -Infinity, Infinity}, Assumptions -> {w > 0, eps > 0}];
inner0 = Limit[inner, eps -> 0, Direction -> "FromAbove"];                    (* -3 I Pi/(8 w^5) *)
Integrate[4 Pi p^2 (inner0 /. w -> Sqrt[p^2 + Dl]), {p, 0, Infinity}, Assumptions -> Dl > 0]/(2 Pi)^4
(* -I/(32 Pi^2 Dl) *)
(-1)^3 I 2 Pi^2 Integrate[k^3/(k^2 + Dl)^3, {k, 0, Infinity}, Assumptions -> Dl > 0]/(2 Pi)^4
(* -I/(32 Pi^2 Dl) : the Wick-rotated answer agrees *)

In the last check, the first calculation does the energy integral as a genuine contour integral in Minkowski space, then the three momentum integrals. The second does the whole thing as one rotated, four-dimensional Euclidean integral. They agree exactly. That's the Wick rotation, checked.

After the rotation we can use four-dimensional spherical coordinates, just as you use \(\mathrm d^3k = 4\pi k^2\,\mathrm d k\) in three dimensions:

\[ \int\mathrm d^4k_E = 2\pi^2\int_0^\infty k^3\,\mathrm d k, \]

because the "surface area" of a unit sphere in four dimensions is \(2\pi^2\). In \(d\) dimensions it's \(S_d = 2\pi^{d/2}/\Gamma(d/2)\).

📝 Note

The \(\Gamma\) function, if you haven't met it. It's the factorial, extended so that it works for any number, not just whole ones. For positive integers, \(\Gamma(n)=(n-1)!\), so \(\Gamma(1)=1\), \(\Gamma(2)=1\), \(\Gamma(3)=2\). It also satisfies \(\Gamma(z+1)=z\,\Gamma(z)\), which is the factorial's own rule.

That one relation is all we need later. Rewrite it as \(\Gamma(z)=\Gamma(z+1)/z\) and let \(z\) be small: the top tends to \(\Gamma(1)=1\), so

\[ \Gamma(z)\approx\frac1z \quad\text{for small } z. \]

So \(\Gamma\) blows up at zero like \(1/z\). That single pole is where every infinity in dimensional regularisation comes from: the loop integrals produce \(\Gamma(2-\tfrac d2)\), which at \(d=4\) is \(\Gamma(0)\). (A sharper version, \(\Gamma(z)=\tfrac1z-\gamma_E+\dots\) with \(\gamma_E=0.5772\), is used in section 6.)

Why this also joins QFT to statistical physics

Rotating time the same way, \(\tau = it\), turns the quantum weight \(e^{iS}\) into \(e^{-S_E}\): a real, positive weight, just like the Boltzmann factor \(e^{-E/k_BT}\) of statistical mechanics.

Picture it

Tong says that after the rotation the only difference between a quantum field theory and a hot magnet is the words we use. In a magnet, the field jitters because of heat and the temperature sets how much. In a quantum field, it jitters because of quantum uncertainty and \(\hbar\) sets how much. Same maths, different story.

This is why this course can move between magnets and particles so freely (Peskin §9.3, Das §14.1, Tong §2.3). From now on every momentum is Euclidean and I drop the \(E\).

4. Why loop integrals blow up

Count the powers of \(k\). The tadpole from Part 1 has one loop and one propagator:

\[ \int\mathrm d^4k\,\frac{1}{k^2+m^2}\sim\int^{\infty}\frac{k^3\,\mathrm d k}{k^2} = \int^{\infty}k\,\mathrm d k \to\infty. \]

At large \(k\) the measure gives four powers (\(k^3\,\mathrm d k\)) and the propagator takes away only two, so the integral grows like \(k^2\). We call that a quadratic divergence. The bubble in the picture above has two propagators, so it goes like \(\int k^3\mathrm d k/k^4=\int\mathrm d k/k\). That grows like \(\ln k\): a logarithmic divergence.

Picture it

A logarithmic divergence is sneaky. Peskin (§12.1) points out that such an integral gets the same amount from every doubling of \(k\): from 1 to 2, from 2 to 4, from 1000 to 2000. It's like a savings plan that pays you one rupee every time your age doubles. You never get rich quickly, but the total keeps growing forever.

Peskin (§10.1) turns this counting into a rule. For a diagram with \(L\) loops and \(P\) internal lines, the superficial degree of divergence is

\[ D = (\text{powers of }k\text{ on top}) - (\text{powers on the bottom}) = 4L-2P. \]

If \(D>0\) the integral diverges like \(\Lambda^D\), if \(D=0\) like \(\ln\Lambda\) and if \(D<0\) it usually converges. For \(\phi^4\) theory, use two facts. First, \(L=P-V+1\): each internal line brings a momentum, each of the \(V\) vertices imposes one conservation law and one of those laws is just overall conservation. Second, \(4V=N+2P\): each vertex has four legs, each internal line uses two and each of the \(N\) outside lines uses one. Eliminating \(L\) and \(P\) gives

\[ \boxed{\;D = 4-N.\;} \]

This is a remarkable result. How badly a diagram diverges depends only on how many lines stick out of it, not on how complicated it is inside.

outside lines \(N\)	\(D\)	what it corrects	how it diverges
2	2	the mass	like \(\Lambda^2\)
4	0	the coupling \(\lambda\)	like \(\ln\Lambda\)
6, 8, ...	negative	nothing new	finite

So in the whole theory, only two kinds of number ever blow up: the mass and the coupling. (The overall size of the field also needs a small fix, starting at two loops.) That short list is what "renormalisable" means and in Part 5 you'll see it's the same list as the couplings that survive coarse-graining.

📝 Note

What an infinity is telling you. A loop momentum \(k\) describes a fluctuation of size about \(1/k\). Integrating to \(k=\infty\) means including fluctuations down to zero size. So the theory is claiming to know nature at arbitrarily short distances and it doesn't. The same thing happens in classical physics: the electric field energy of a point charge, \(\int_r^\infty\frac{\varepsilon_0}{2}E^2\,\mathrm d V=\frac{e^2}{8\pi\varepsilon_0 r}\), blows up as \(r\to0\). Nobody concluded the electron has infinite mass. They concluded that "point charge" stops being a good description at some small distance.

5. The cutoff and other ways to regularise

To get a finite answer we must change the theory at very short distances, in a way controlled by one adjustable number. That step is called regularisation. Then we ask which predictions depend on the adjustable number and which don't.

Picture it

How long is the coastline of the Sundarbans? It depends on your ruler and with all those creeks and islands the answer is wildly ruler-dependent. Measure with a 100 km ruler and you skip every bay. Use a 1 m ruler and you trace every rock and the length grows. As the ruler shrinks, the length keeps growing, just like a divergent integral as \(\Lambda\) grows. The ruler length is the regulator. The coastline's length is not a sensible question on its own. But "is Britain's coast more wiggly than Norway's, measured with the same ruler?" has a perfectly good answer.

Here are four common ways to do it.

A hard momentum cutoff. Only integrate up to \(|k|=\Lambda\). This is the easiest one to picture and it's the one Wilson's picture uses. Its weakness is that in gauge theories it breaks gauge symmetry, which is why particle physicists often prefer the next two.

A lattice. Put the field on a grid with spacing \(a\). Then no wave can be shorter than the grid allows, so \(\Lambda\sim\pi/a\). For a magnet this isn't a trick at all, because the atoms really do sit on a lattice. It's also how QCD is simulated on computers.

Pauli-Villars. Subtract the same diagram with an imaginary, very heavy particle of mass \(M\):

\[ \frac1{k^2+m^2}\to\frac1{k^2+m^2}-\frac1{k^2+M^2}=\frac{M^2-m^2}{(k^2+m^2)(k^2+M^2)}. \]

For \(k\ll M\) nothing changes. For \(k\gg M\) the integrand falls off two powers faster, which is enough to make the bubble converge. Here \(M\) plays the role of \(\Lambda\).

Dimensional regularisation. The bubble diverges in four dimensions but converges in fewer, because there is less room (\(k^{d-1}\,\mathrm d k\)) at large \(k\). So do the integral in a general dimension \(d\), where the answer is a finite formula in \(d\) and only at the end let \(d\) approach 4. The infinity then shows up as a term like \(1/\epsilon\), with \(\epsilon=4-d\). For the tadpole:

\[ \int\frac{\mathrm d^dk}{(2\pi)^d}\,\frac1{k^2+m^2} = \frac{\Gamma(1-\tfrac d2)}{(4\pi)^{d/2}}\,(m^2)^{d/2-1}\ \xrightarrow{\ d=4-\epsilon\ }\ -\frac{m^2}{8\pi^2\epsilon}+\text{finite}. \]

This method respects every symmetry and Part 7 uses it. The price is that it needs an arbitrary mass scale \(\mu\) to keep the units right in \(d\neq4\) and that \(\mu\) turns out to be where the "running" of couplings comes from.

6. Doing a loop integral from start to finish

Every one-loop integral in this course and almost every one in Peskin, is done with the same five-step recipe (Peskin §6.3 lists it right after computing the electron's magnetic moment):

1. Draw the diagram and write down the integral.

2. Use Feynman parameters to squeeze all the denominators into one.

3. Complete the square in that denominator by shifting the loop momentum.

4. Tidy up the numerator: throw away odd powers of the loop momentum and simplify even ones.

5. Wick rotate, use four-dimensional spherical coordinates and apply one master formula.

Let's do every step, with nothing skipped, on the bubble from the picture in section 2. We're already in Euclidean space, so the Wick rotation of step 5 is done.

Step 1: write the integral

With incoming momentum \(p=p_1+p_2\), the two lines of the loop carry \(k\) and \(k+p\), so the bubble is

\[ B(p) = \int\frac{\mathrm d^dk}{(2\pi)^d}\,\frac{1}{(k^2+m^2)\big((k+p)^2+m^2\big)}. \]

We write \(d\) dimensions rather than 4 so we can use dimensional regularisation at the end.

Step 2: Feynman parameters

The trouble is that there are two different denominators, one centred on \(k=0\) and one on \(k=-p\), so the integrand has no simple symmetry. Feynman's trick squeezes them into one. For any two positive numbers \(A\) and \(B\),

\[ \frac{1}{AB} = \int_0^1\frac{\mathrm d x}{\big[xA+(1-x)B\big]^2}. \]

Let's check it. The derivative of \(\dfrac{-1}{(A-B)\,[xA+(1-x)B]}\) with respect to \(x\) is \(\dfrac{1}{[xA+(1-x)B]^2}\), because the inside of the bracket has derivative \(A-B\). So the integral is that antiderivative at \(x=1\) minus its value at \(x=0\):

\[ \int_0^1\frac{\mathrm d x}{[xA+(1-x)B]^2} = \frac{-1}{(A-B)A}-\frac{-1}{(A-B)B} = \frac{-B+A}{(A-B)AB} = \frac{1}{AB}. \quad\checkmark \]

Picture it

It's like mixing two paints. Instead of dealing with a red tin and a blue tin separately, you consider every possible mixture, from all red (\(x=1\)) to all blue (\(x=0\)) and add them all up with the right weights. Each single mixture is simple to handle, even though the two separate tins weren't. The number \(x\) is called a Feynman parameter and it just labels the mixture.

For three or more denominators the same idea works with more parameters (Peskin eq. 6.41):

\[ \frac{1}{A_1A_2\cdots A_n} = (n-1)!\int_0^1\mathrm d x_1\cdots\mathrm d x_n\,\delta\Big(\sum x_i-1\Big)\,\frac{1}{\big[x_1A_1+\dots+x_nA_n\big]^n}. \]

The delta function just says the mixture fractions add up to one. (Checked in Wolfram for \(n=3\).)

For our bubble, take \(A=(k+p)^2+m^2\) and \(B=k^2+m^2\):

\[ B(p) = \int_0^1\mathrm d x\int\frac{\mathrm d^dk}{(2\pi)^d}\,\frac{1}{\big[x\big((k+p)^2+m^2\big)+(1-x)\big(k^2+m^2\big)\big]^2}. \]

Step 3: complete the square

Multiply out the square bracket. The \(k^2\) and \(m^2\) terms appear with total weight \(x+(1-x)=1\) and \((k+p)^2=k^2+2k\cdot p+p^2\), so

\[ x\big((k+p)^2+m^2\big)+(1-x)\big(k^2+m^2\big) = k^2 + 2x\,k\cdot p + x\,p^2 + m^2. \]

Now complete the square, exactly as you would for \(k^2+2bk\): \(\;k^2+2x\,k\cdot p = (k+xp)^2 - x^2p^2\). So

\[ k^2 + 2x\,k\cdot p + x\,p^2 + m^2 = (k+xp)^2 + x(1-x)\,p^2 + m^2. \]

Define the shifted loop momentum \(\ell = k+xp\) and the constant

\[ \Delta = m^2 + x(1-x)\,p^2. \]

The denominator is now just \((\ell^2+\Delta)^2\). Shifting the integration variable doesn't change the integral, since \(\mathrm d^d k=\mathrm d^d\ell\) and both run over all of space:

\[ B(p) = \int_0^1\mathrm d x\int\frac{\mathrm d^d\ell}{(2\pi)^d}\,\frac{1}{(\ell^2+\Delta)^2}. \]

Picture it

Shifting \(k\) to \(\ell\) is like moving the origin of a map to the centre of a city. The city is the same, but now it looks round and symmetric around the origin, which makes it far easier to describe.

Step 4: tidy up the numerator

Here the numerator is just \(1\), so there's nothing to do. In QED (Part 7) the numerator contains \(\ell^\mu\) and \(\ell^\mu\ell^\nu\), so two rules are needed:

\[ \int\frac{\mathrm d^d\ell}{(2\pi)^d}\,\frac{\ell^\mu}{D^3} = 0, \qquad \int\frac{\mathrm d^d\ell}{(2\pi)^d}\,\frac{\ell^\mu\ell^\nu}{D^3} = \frac{g^{\mu\nu}}{d}\int\frac{\mathrm d^d\ell}{(2\pi)^d}\,\frac{\ell^2}{D^3}. \]

These are Peskin's equations 6.45 and 6.46. They get quoted as "by symmetry" far too quickly, so here is where they come from. Everything rests on one fact established in Step 3: after the shift, \(D\) depends only on \(\ell^2\), never on the direction of \(\ell\).

Two pieces of small print before we start. First, the argument below is cleanest after the Wick rotation of section 3, where all \(d\) directions really are alike and the natural object is the Euclidean \(\delta^{ab}\); the covariant statement with \(g^{\mu\nu}\) follows from it. Second, it assumes the integral is defined, which for the rank-two case at \(d=4\) means dimensionally regularised, since that integral is logarithmically divergent. A regulator that respects the symmetry is what lets you shuffle the integrand this way at all.

The first rule. Send \(\ell \to -\ell\). The measure \(\mathrm d^d\ell\) is unchanged, since flipping every axis is just a relabelling. \(D\) is unchanged, because it only knows \(\ell^2\). The numerator \(\ell^\mu\) flips sign. So the whole integral equals minus itself:

\[ I^\mu = -I^\mu \qquad\Longrightarrow\qquad I^\mu = 0. \]

Nothing deeper than that. For every \(\ell\) contributing \(+3\), the point \(-\ell\) contributes \(-3\).

The second rule, the one that looks like magic. Take it in two pieces.

Off the diagonal, say \(\mu = 1\) with \(\nu = 2\). Now flip only the second axis, \(\ell^2 \to -\ell^2\), leaving the rest alone. Again the measure and \(D\) do not notice, because \(D\) depends on the length. The numerator \(\ell^1\ell^2\) flips sign. Same argument as before, so every off-diagonal component vanishes.

On the diagonal, the components cannot be calculated one at a time, but they do not need to be, because in Euclidean space they are all equal. Nothing in the integral distinguishes one axis from another: the only thing \(D\) knows is the length of \(\ell\), which treats every direction alike. So whatever number \(\int \ell^1\ell^1/D^3\) comes to, \(\int\ell^2\ell^2/D^3\) comes to the same. That holds for all \(d\) of them.

Put those together. The answer is zero off the diagonal, with \(d\) equal entries on it, which in Euclidean space is precisely the statement that it is proportional to \(\delta^{ab}\). Rotating back, the covariant version reads

\[ \int\frac{\mathrm d^d\ell}{(2\pi)^d}\frac{\ell^\mu\ell^\nu}{D^3} = A\,g^{\mu\nu}\quad\text{for some number }A. \]

(Note that the Minkowski \(g^{\mu\nu}\) does not have \(d\) equal diagonal entries, since its time entry has the opposite sign. The "all directions alike" step is the Euclidean one. What survives the rotation is the statement that no vector is available to build the answer from, leaving the metric as the only rank-two object in sight.)

Finding \(A\) takes one line. Contract both sides with \(g_{\mu\nu}\). On the left, \(g_{\mu\nu}\ell^\mu\ell^\nu = \ell^2\). On the right, \(g_{\mu\nu}g^{\mu\nu} = g^\mu{}_\mu = d\), the trace of the identity in \(d\) dimensions rather than of the metric. So

\[ \int\frac{\mathrm d^d\ell}{(2\pi)^d}\frac{\ell^2}{D^3} = A\,d \qquad\Longrightarrow\qquad A = \frac1d\int\frac{\mathrm d^d\ell}{(2\pi)^d}\frac{\ell^2}{D^3}, \]

which is the rule. The \(1/d\) is not a convention or a fudge. It is there because the trace of the metric is \(d\), so the one number \(A\) has to be shared out over \(d\) equal diagonal entries.

Picture it

Strip the integral away and it is an averaging fact about a sphere. Scatter points evenly on a sphere in 3D. The average of \(xy\) is zero, because for every point with \(xy>0\) there is a mirror point with \(xy<0\). The averages of \(x^2\), \(y^2\) and \(z^2\) must be equal, since the sphere looks the same along each axis, while together they add up to the average of \(r^2\). So each one is \(\langle r^2\rangle/3\). That \(3\) is the number of dimensions. In \(d\) dimensions it is \(d\), which is the whole content of the \(1/d\) above.

🤔 Problem

Check it yourself. Take the 3D Gaussian \(f(\boldsymbol\ell) = e^{-\ell^2}\) in place of \(1/D^3\), where the integrals can all be done in closed form. Verify that \(\int\mathrm d^3\ell\;\ell_x\ell_y\,f = 0\) while \(\int\mathrm d^3\ell\;\ell_x^2\,f = \tfrac13\int\mathrm d^3\ell\;\ell^2 f\).

Show solution

The Gaussian factorises, \(e^{-\ell^2} = e^{-\ell_x^2}e^{-\ell_y^2}e^{-\ell_z^2}\), so every integral is a product of one-dimensional ones. Writing \(G_n=\int_{-\infty}^{\infty}x^n e^{-x^2}\mathrm d x\), we need \(G_0=\sqrt\pi\), \(G_1=0\) and \(G_2=\tfrac{\sqrt\pi}{2}\).

For the off-diagonal piece, \(\int \ell_x\ell_y f = G_1\,G_1\,G_0 = 0\), since each odd integral vanishes on its own.

For the diagonal, \(\int\ell_x^2 f = G_2\,G_0\,G_0 = \tfrac{\sqrt\pi}{2}\pi = \tfrac{\pi^{3/2}}{2}\). By the same computation \(\int\ell_y^2f\) and \(\int\ell_z^2f\) give the identical number, so

\[ \int \ell^2 f = \int(\ell_x^2+\ell_y^2+\ell_z^2)f = 3\times\frac{\pi^{3/2}}{2}, \]

and therefore \(\int\ell_x^2 f = \tfrac13\int\ell^2f\) exactly. The \(3\) appeared for the reason given above: three equal contributions sharing one total.

Step 5: spherical coordinates and the master formula

In \(d\) dimensions, \(\mathrm d^d\ell = S_d\,\ell^{d-1}\mathrm d\ell\), where \(S_d=2\pi^{d/2}/\Gamma(d/2)\) is the area of the unit sphere (this is \(2\pi^2\) for \(d=4\)). So

\[ \int\frac{\mathrm d^d\ell}{(2\pi)^d}\,\frac{1}{(\ell^2+\Delta)^n} = \frac{S_d}{(2\pi)^d}\int_0^\infty\frac{\ell^{d-1}\,\mathrm d\ell}{(\ell^2+\Delta)^n}. \]

Substitute \(t=\ell^2/\Delta\), so \(\ell=\sqrt{\Delta t}\) and \(\mathrm d\ell = \frac{\sqrt\Delta}{2\sqrt t}\,\mathrm d t\). The radial integral becomes

\[ \int_0^\infty\frac{\ell^{d-1}\,\mathrm d\ell}{(\ell^2+\Delta)^n} = \frac12\,\Delta^{d/2-n}\int_0^\infty\frac{t^{d/2-1}\,\mathrm d t}{(1+t)^n} = \frac12\,\Delta^{d/2-n}\,\frac{\Gamma(\tfrac d2)\,\Gamma(n-\tfrac d2)}{\Gamma(n)}. \]

The last step uses a standard integral, \(\int_0^\infty t^{a-1}(1+t)^{-a-b}\,\mathrm d t=\Gamma(a)\Gamma(b)/\Gamma(a+b)\) (the "beta function"). Multiply by \(S_d/(2\pi)^d = 2/\big((4\pi)^{d/2}\Gamma(d/2)\big)\) and the \(\Gamma(d/2)\) and the \(2\) cancel:

\[ \boxed{\;\int\frac{\mathrm d^d\ell}{(2\pi)^d}\,\frac{1}{(\ell^2+\Delta)^n} = \frac{\Gamma(n-\tfrac d2)}{(4\pi)^{d/2}\,\Gamma(n)}\;\Delta^{d/2-n}.\;} \]

This is the master formula. Every one-loop integral in the course is a special case. With \(n=1\) it's the tadpole formula of section 5. With \(n=2\) it gives our bubble:

\[ B(p) = \frac{\Gamma(2-\tfrac d2)}{(4\pi)^{d/2}}\int_0^1\mathrm d x\;\Delta^{d/2-2}. \]

Step 6: let \(d\) approach 4

Put \(d=4-\epsilon\), so \(2-\tfrac d2 = \tfrac\epsilon2\). There are three factors to expand, keeping terms up to \(\epsilon^0\):

• \(\Gamma(\tfrac\epsilon2) = \tfrac2\epsilon-\gamma_E+O(\epsilon)\), where \(\gamma_E=0.5772\dots\) is Euler's constant. The Gamma function has a pole at zero and this is where the infinity lives.

• \((4\pi)^{-d/2} = (4\pi)^{-2}(4\pi)^{\epsilon/2} = \frac1{16\pi^2}\big(1+\tfrac\epsilon2\ln4\pi+\dots\big)\), using \(a^{\epsilon}=e^{\epsilon\ln a}\approx1+\epsilon\ln a\).

• \(\Delta^{-\epsilon/2} = 1-\tfrac\epsilon2\ln\Delta+\dots\), the same trick.

Multiply the three. The \(\tfrac2\epsilon\) from the first factor times the \(\tfrac\epsilon2\) terms of the others gives finite pieces, so

\[ \boxed{\;B(p) = \frac{1}{16\pi^2}\left[\frac2\epsilon - \gamma_E + \ln4\pi - \int_0^1\mathrm d x\,\ln\big(m^2+x(1-x)p^2\big)\right]+O(\epsilon).\;} \]

(Strictly, the logarithm is \(\ln(\Delta/\mu^2)\), with the scale \(\mu\) from section 5 keeping the units right.) The infinity is the \(2/\epsilon\). Everything that depends on the physics, meaning the masses and momenta, sits in the finite logarithm. Every step above was checked in Wolfram:

Integrate[1/(x A + (1 - x) B)^2, {x, 0, 1}, Assumptions -> {A > 0, B > 0}]     (* 1/(A B) *)
Expand[x ((k + p)^2 + m^2) + (1 - x) (k^2 + m^2) /. k -> l - x p]            (* l^2 + m^2 + x(1-x)p^2 *)
Sd = 2 Pi^(d/2)/Gamma[d/2];
Sd/(2 Pi)^d Integrate[l^(d - 1)/(l^2 + D0)^n, {l, 0, Infinity},
    Assumptions -> {D0 > 0, n > d/2, d > 0}] // FullSimplify
(* Gamma[n - d/2] D0^(d/2 - n)/((4 Pi)^(d/2) Gamma[n]) : the master formula *)
Series[Gamma[eps/2] (4 Pi)^(-(4 - eps)/2) D0^(-eps/2), {eps, 0, 0}]
(* (2/eps - EulerGamma + Log[4 Pi] - Log[D0])/(16 Pi^2) *)

Peskin's §6.3 uses exactly these steps on a harder integral, the correction to the electron's magnetic moment and ends with Schwinger's famous result \(\frac{g-2}{2}=\frac{\alpha}{2\pi}\approx0.0011614\) (exercise 2.6). Part 7 uses them on the QED vacuum polarisation.

7. Same integral, three regulators

This is the key lesson of the page, done with real numbers. Take the log-divergent bubble with no momentum flowing in,

\[ I(m) = \int\frac{\mathrm d^4k}{(2\pi)^4}\,\frac1{(k^2+m^2)^2}, \]

and regularise it three different ways. (This is the bubble \(B(p)\) of section 6 at \(p=0\), so the dimensional-regularisation line is just section 5's answer with \(\Delta=m^2\).) Each was done symbolically in Wolfram:

\[ \begin{aligned} \text{cutoff:}&\quad I = \frac1{16\pi^2}\left[\ln\!\Big(1+\frac{\Lambda^2}{m^2}\Big)-\frac{\Lambda^2}{\Lambda^2+m^2}\right]\ \xrightarrow{\Lambda\gg m}\ \frac1{16\pi^2}\left[\ln\frac{\Lambda^2}{m^2}-1\right]\\ \text{Pauli-Villars:}&\quad I = \frac1{16\pi^2}\ln\frac{M^2}{m^2}\\ \text{dimensional:}&\quad I = \frac1{16\pi^2}\left[\frac2\epsilon-\gamma_E+\ln4\pi-\ln\frac{m^2}{\mu^2}\right] \end{aligned} \]

Each one is infinite when its regulator is removed. But look at how each depends on \(m\). Every one contains \(-\frac1{16\pi^2}\ln m^2\), plus a constant that depends on the regulator but not on \(m\). So a difference such as \(I(m_1)-I(m_2)=\frac1{16\pi^2}\ln\frac{m_2^2}{m_1^2}\) comes out finite and it's the same in all three.

Picture it

Measure the heights of two people standing in two different buildings, each from its own ground floor. The numbers depend on which building's floor you measure from. But the difference in their heights doesn't, as long as you measure both from the same floor. The regulator is the choice of floor. Physics only ever asks about the difference.

ang = 2 Pi^2/(2 Pi)^4;                                   (* Vol(S^3)/(2Pi)^4 *)
Icut = ang Integrate[k^3/(k^2 + m^2)^2, {k, 0, L}, Assumptions -> {m > 0, L > 0}];
Ipv  = ang Integrate[k^3 (1/(k^2 + m^2)^2 - 1/(k^2 + M^2)^2), {k, 0, Infinity},
                     Assumptions -> {m > 0, M > m}];     (* Log[M^2/m^2]/(16 Pi^2) *)
Idr  = Gamma[2 - d/2]/(4 Pi)^(d/2) (m^2)^(d/2 - 2) mu^(4 - d);
Series[Idr /. d -> 4 - eps, {eps, 0, 0}]                 (* 2/eps - EulerGamma + ... *)
(* all three give dI/d(m^2) = -1/(16 Pi^2 m^2) once the regulator is large *)

That's the whole trick of renormalisation. The part that depends on the regulator is a constant and a constant can be soaked up into a number we never measure directly.

8. Renormalisation: trade the made-up numbers for measured ones

The numbers \(m_0\) and \(\lambda_0\) written in the Lagrangian are called bare parameters. Nobody measures them. What you measure is, for example, the position of the pole in the propagator (\(m_{\rm phys}\)) and a collision rate at some energy (which defines \(\lambda_{\rm phys}\)). Loops connect the two:

\[ m^2_{\rm phys} = m_0^2 + \delta m^2(\Lambda), \qquad \lambda_{\rm phys} = \lambda_0 + \delta\lambda(\Lambda). \]

Picture it

Think of a rosogolla. The one you buy in Kolkata comes soaked in syrup and it's always sold, served and eaten that way. If you weigh it, you weigh the sponge plus the syrup it has soaked up. The weight of the "dry" sponge alone is a number nobody ever measures, because the rosogolla never leaves the syrup. An electron is the same: it can never be separated from its cloud of virtual particles, so its "bare" mass is like the dry sponge, a bookkeeping number and nothing more. (Physicists without a sweet tooth use a ball pushed through water, which drags water along and feels heavier by half the mass of the water it displaces.)

Ling-Fong Li's lecture notes (arXiv:1208.4700) point out that renormalisation isn't special to particle physics. An electron moving through a crystal responds to forces as if it had an effective mass \(m^*\), different from its mass \(m\) in empty space, because it's constantly tugging on the lattice. In a crystal, both \(m\) and \(m^*\) can be measured, since you can take the electron out. Quantum field theory differs in two ways: the shift from bare to physical is infinite (it comes from the high-momentum modes) and you can never switch the interactions off, so the bare value is never measurable.

The recipe has four steps:

1. Regularise, so every quantity is finite and depends on \(\Lambda\).

2. Work out the measured quantities in terms of the bare ones.

3. Turn that around: write the bare parameters in terms of the measured ones.

4. Rewrite every other prediction using the measured parameters.

After step 4, the \(\Lambda\) drops out of every prediction, at least in a renormalisable theory like \(\phi^4\) (Peskin §10.2 does this in full). The infinity hasn't been hidden under the carpet. It has been moved into the link between two sets of numbers and only one of those sets is ever measured.

Dmitry Shirkov, one of the founders of the renormalisation group, put it simply in his fifty-year retrospective in the CERN Courier: physical quantities come out as finite functions of the new "renormalised" couplings, with "all infinities being swallowed by the Z factors". The Z factors are the numbers that connect bare and renormalised quantities, such as \(m_0^2=Z_m m^2\).

Why only a few numbers ever need fixing

Here's a lovely fact from Li's notes that explains why renormalisation only needs a few counterterms. Take the bubble from section 6 and differentiate it with respect to the outside momentum. Each derivative puts one more power of the loop momentum in the denominator, so the integral becomes more convergent. Using section 5's formula,

\[ \frac{\partial B}{\partial p^2} = \frac{\Gamma(2-\tfrac d2)\,(\tfrac d2-2)}{(4\pi)^{d/2}}\int_0^1\mathrm d x\;x(1-x)\,\Delta^{d/2-3}. \]

Now \(\Gamma(2-\tfrac d2)(\tfrac d2-2) = -\Gamma(3-\tfrac d2)\) and \(\Gamma(3-\tfrac d2)\to\Gamma(1)=1\) as \(d\to4\). So there's no pole left:

\[ \frac{\partial B}{\partial p^2} \xrightarrow{\ d\to4\ } -\frac{1}{16\pi^2}\int_0^1\frac{x(1-x)\,\mathrm d x}{m^2+x(1-x)p^2}\qquad\text{(finite).} \]

So the infinity in \(B(p)\) lives entirely in its value at one point, \(B(0)\) and the difference \(B(p)-B(0)\) is finite. (Checked in Wolfram.) In general, an integral that diverges like \(\Lambda^D\) only has infinities in the first few terms of its Taylor series in the outside momenta, up to power \(D\). That's why a finite number of counterterms, one for each of those first few terms, is enough.

Picture it

It's like the fare meter in a Kolkata taxi. Whatever the ride, the fare is a fixed flag-down charge plus something that depends on the distance. If someone tampered with the meter's starting number, every fare would be off by the same amount and you could fix every bill by correcting that one starting number. The divergence is the tampered flag-down charge and the counterterm is the correction. The distance-dependent part, the real physics, was never broken.

Counterterms as extra Feynman rules

There's a tidy way to organise all this, called renormalised perturbation theory (Li §2.2, Peskin §10.2). Write the Lagrangian in terms of the measured mass \(m\) and coupling \(\lambda\) and move everything that absorbs infinities into separate counterterms:

\[ \mathcal L = \underbrace{\tfrac12(\partial\phi)^2+\tfrac12m^2\phi^2+\frac{\lambda}{4!}\phi^4}_{\text{same form, measured numbers}} \;+\; \underbrace{\tfrac12\delta_Z(\partial\phi)^2+\tfrac12\delta_m\phi^2+\frac{\delta_\lambda}{4!}\phi^4}_{\text{counterterms}}. \]

The counterterms are treated just like new interactions, with their own Feynman rules. They are drawn as a circle with a cross:

Beyond one loop

At two loops something new happens: a diagram can contain a smaller divergent diagram inside it. Take two bubbles in a row, a four-point diagram proportional to \(\lambda^3\Gamma(p)^2\), where \(\Gamma(p)\) is one bubble (Li's Fig. 9). Its overall degree of divergence is \(D=4-4=0\), which suggests one derivative should make it finite. But it doesn't, because each bubble diverges on its own and a derivative only helps one of them at a time.

The rescue comes from the one-loop counterterm. In the two-loop calculation, the counterterm vertex can replace either of the two bubbles:

Add the three together and complete the square:

\[ \lambda^3\,\Gamma(p)^2 - 2\lambda^3\,\Gamma(0)\,\Gamma(p) = \lambda^3\big[\Gamma(p)-\Gamma(0)\big]^2 - \lambda^3\,\Gamma(0)^2. \]

The bracket is finite (that's the difference \(B(p)-B(0)\) from above) and the last term doesn't depend on \(p\) at all, so a new two-loop counterterm cancels it. So with the lower-order counterterms included, the leftover infinities are again just constants, the first term of a Taylor series and the same trick works at every order. This is the BPH recipe (Bogoliubov, Parasiuk and Hepp, later completed by Zimmermann):

1. Build the diagrams from the renormalised Lagrangian.

2. At one loop, find the infinite first Taylor terms and add counterterms to cancel them.

3. At two loops, include diagrams with the one-loop counterterms inside, then add new counterterms for what's left. Repeat at every order.

Divergences inside divergences can also be nested, one completely inside the other:

The rule that says when a multi-loop integral really converges is Weinberg's theorem: the whole diagram and every sub-diagram inside it must have negative degree of divergence. Proving that BPH always works is famously hard and Li's notes and Peskin §10.4 go further if you're curious. The takeaway for this course is simple: in a renormalisable theory, a finite list of counterterms, fixed order by order, removes every infinity.

Otaku corner

Death Note fans will recognise the counterterm's attitude. The infinity shows up, grinning and the counterterm calmly cancels it, exactly as planned. The trick is that the plan was written before the calculation, one order at a time.

9. Hold the physics fixed and move the cutoff

Keep \(m_{\rm phys}\) and \(\lambda_{\rm phys}\) fixed, since those are what experiments tell you and ask what the bare parameters must be for different cutoffs. From Part 1,

\[ m_0^2(\Lambda) = m^2_{\rm phys}-\frac{\lambda}{32\pi^2}\left[\Lambda^2-m^2\ln\!\Big(1+\frac{\Lambda^2}{m^2}\Big)\right]. \]

The bare coupling has to change with \(\Lambda\) too, only much more slowly, like a logarithm (Part 7 works out exactly how). The bare parameters are functions of the cutoff, arranged so that low-energy physics never notices the cutoff. That sentence is the renormalisation group. Wilson's big step, in Parts 5 and 6, was to take it literally and ask what happens when you lower \(\Lambda\) bit by bit.

Picture it

It's like a camera's exposure setting. Change the shutter speed and you must change the aperture to keep the photo equally bright. The shutter speed is \(\Lambda\), the aperture is the bare parameter and the brightness of the photo is the measured physics. The rule linking shutter and aperture is the RG.

Slide the cutoff up to the Planck scale (\(\log_{10}\Lambda\approx19\)) with a mass of 125 GeV, like the Higgs. The bare mass has to cancel the loop correction to about thirty digits. That's the "hierarchy problem" from exercise 1.4 and Part 8 explains why mass terms and only mass terms, are this sensitive.

Exercises

🤔 Problem

2.1. Use \(D=4L-2P\), \(L=P-V+1\) and \(4V=N+2P\) to derive \(D=4-N\) for \(\phi^4\) theory. Then redo it in \(d\) dimensions, where \(D=dL-2P\) and show \(D=d-\frac{d-2}{2}N+(d-4)V\). What changes when \(d<4\) and when \(d>4\)?

Show solution

From \(4V=N+2P\), \(P=(4V-N)/2\). Then \(L=P-V+1=V-N/2+1\) and \(D=4L-2P=4V-2N+4-4V+N=4-N\). In \(d\) dimensions, \(D=d(V-N/2+1)-(4V-N)=d-\frac{d-2}2N+(d-4)V\), which is Peskin's general formula. For \(d<4\) the \(V\) term is negative, so more vertices make diagrams more convergent and only a few diagrams diverge at all. Such a theory is called super-renormalisable. For \(d>4\) the \(V\) term is positive, so every extra vertex makes things worse and you would need infinitely many corrections: non-renormalisable. The dividing line is where \([\lambda]=4-d\) changes sign, which is the relevant/irrelevant boundary of Part 5.

🤔 Problem

2.2. Apply one Pauli-Villars subtraction to the tadpole, \(\int\frac{\mathrm d^4k}{(2\pi)^4}\left[\frac1{k^2+m^2}-\frac1{k^2+M^2}\right]\). Count powers of \(k\): is it finite now? How many subtractions does a quadratic divergence need?

Show solution

One subtraction makes the integrand \(\frac{M^2-m^2}{(k^2+m^2)(k^2+M^2)}\sim\frac{M^2}{k^4}\) at large \(k\) and \(\int k^3\,\mathrm d k/k^4\) still diverges, now only logarithmically. One subtraction removes two powers of \(k\). That was enough for the bubble (\(D=0\)) but not for the tadpole (\(D=2\)). You need a second heavy particle, with two masses \(M_1,M_2\) and coefficients chosen so that both the \(1/k^2\) and the \(1/k^4\) tails cancel. The habit to learn is to count powers before you trust a regulated integral.

🤔 Problem

2.3. Using the dimensional-regularisation formula for \(I(m)\), show that \(\mu\frac{\mathrm d I}{\mathrm d\mu}=\frac{1}{8\pi^2}\). Do the same with the cutoff version, using \(\Lambda\frac{\mathrm d I}{\mathrm d\Lambda}\). What do you notice?

Show solution

\(I\) depends on \(\mu\) only through \(+\frac1{16\pi^2}\ln\mu^2\), so \(\mu\,\mathrm d I/\mathrm d\mu=\frac{2}{16\pi^2}=\frac1{8\pi^2}\). The cutoff version at large \(\Lambda\) is \(\frac1{16\pi^2}(\ln\Lambda^2-\ln m^2-1)\), so \(\Lambda\,\mathrm d I/\mathrm d\Lambda=\frac1{8\pi^2}\) as well. The number in front of the logarithm is the same in every method. It's the one part of a log-divergent integral with real physical meaning and the beta functions of Part 7 are built from it.

🤔 Problem

2.4. Change the contour in the Wolfram code: put the \(+i\varepsilon\) with the opposite sign, \(1/(k_0^2-E^2-i\varepsilon)\). Where do the poles go now and which way would you have to rotate?

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The poles move to \(+E+i\varepsilon\) (top-right) and \(-E-i\varepsilon\) (bottom-left). Now the top-left and bottom-right quarters are empty, so the real line must be rotated clockwise onto the imaginary axis, \(k^0=-ik^4\). The Euclidean integral you get is the complex conjugate of before. The Feynman sign of \(i\varepsilon\) is what picks the anticlockwise rotation and it's the same choice that makes positive-energy particles move forward in time.

🤔 Problem

2.5. (In words.) A friend says: "Renormalisation just means subtracting infinity from infinity, so you can't trust it." Reply in three sentences, using the coastline or the two-buildings picture.

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One possible answer. Nothing infinite is ever subtracted: with a regulator in place every number is finite, just as a coastline measured with a fixed ruler has a finite length. We then only compare measurable things, like the difference \(I(m_1)-I(m_2)\) above and those come out the same whatever regulator you used. What's left over, the way numbers depend on the scale, is a real prediction (the running of couplings) and it has been measured (Part 7).

🤔 Problem

2.6. Schwinger's number. Peskin §6.3 carries the recipe through for the electron's magnetic moment and finds, at zero momentum transfer, \(F_2(0)=\frac{\alpha}{2\pi}\int\mathrm d x\,\mathrm d y\,\mathrm d z\,\delta(x+y+z-1)\,\frac{2m^2z(1-z)}{m^2(1-z)^2}\). Do the integrals and show \(F_2(0)=\alpha/2\pi\). With \(\alpha=1/137.036\), what is \(\frac{g-2}{2}\)?

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The integrand doesn't depend on \(x\) or \(y\) separately, so first use the delta function to do the \(x\) integral: it sets \(x=1-y-z\), which must be at least 0, so \(y\) runs from \(0\) to \(1-z\). The integrand simplifies to \(\frac{2z(1-z)}{(1-z)^2}=\frac{2z}{1-z}\). The \(y\) integral just multiplies by the length of its range, \(1-z\), giving \(2z\). Then \(\int_0^1 2z\,\mathrm d z=1\), so \(F_2(0)=\frac{\alpha}{2\pi}\). Numerically \(\frac{1}{2\pi\times137.036}=0.0011614\), against the measured \(0.0011597\). The small difference comes from two-loop and higher terms, which have now been computed up to five loops and agree with experiment to about one part in a billion. (Checked in Wolfram.)