A Tale of Two Calculations

This post is mathematically advanced, but may be worth a skim if you’re a layman who’s curious how physicists do real calculations!

Recently I’ve been talking about the generalized unitarity method, extolling its virtues for $1$-loop calculations. Despite all this hoodoo, I have failed to provide a single example of a successful application. Now it’s time for that to change. I’m about to show you just how useful generalized unitarity can be, borrowing examples from $\mathcal{N}=4$ super-Yang-Mills (SYM) and $SU(3)$ Yang-Mills (YM).

We’ll begin by revising the general form of the generalized unitarity method. In picture form

What exactly does all the notation mean? On the left hand side, I’m referring to the residue of the integrand when all the loop momenta $l_i$ for $i = 1,2,3,4$ are taken on-shell. On the right hand side, I take a product of tree level diagrams with external lines as shown, and sum over the possible particle content of the $l_i$ lines. Implicit in each of the blobs in the equation is a sum over tree level diagrams.

We’d like to use this formula to calculate $1$-loop amplitudes. But hang on, doesn’t it only tell us about residues of integrands? Naively, it seems like that’s too little information to reconstruct the full result.

Fear not, however – help is at hand! Back in 1965, Don Melrose published his first paper. He presciently observed that loop diagrams in $D$ dimensions could be expressed as linear combinations of scalar loop diagrams with $\leq 4$ sides. Later Bern, Dixon and Kosower generalized this result to take account of regularization.

Let’s express those words mathematically. We have

$\displaystyle \mathcal{A}_n^{1\textrm{-loop}} = \sum_i D_i I_4(K^i) + \sum_j C_j I_3 (K^j) + \sum_m B_m I_2 (K^m) + R_n + O(\epsilon)\qquad (*)$

where $I_a$ are integrals corresponding to particular scalar theory diagrams, $K_a^i$ indicate distribution of momenta on external legs, $R_n$ is a rational function and $\epsilon$ a regulator.

The integrals $I_4$, $I_3$ and $I_2$ are referred to as box, triangle and bubble integrals respectively. This is an obvious homage to their structure as Feynman diagrams. For example a triangle diagram looks like

where $K_1$, $K_2$, $K_3$ label the sums of external momenta at each of the vertices. The Feynman rules give (in dimensional regularization)

$\displaystyle I_3(K_1, K_2, K_3) = \mu^{2 \epsilon}\int \frac{d^{4-2\epsilon}l}{(2\pi)^{4-2\epsilon}}\frac{1}{l^2 (l-K_1)^2 (l+K_3)^2}$

We call result $(*)$ above an integral basis expansion. It’s useful because the integrands of box, triangle and bubble diagrams have different pole structures. Thus we can reconstruct their coefficients by taking generalized unitarity cuts. Of course, the rational term cannot be determined this way. Theoretically we have reduced our problem to a simpler case, but not completely solved it.

Before we jump into a calculation, it’s worth taking a moment to consider the origin of the rational term. In Melrose’s original analysis, this term was absent. It appears in regularized versions, precisely because the act of regularization gives rise to extra rational terms at $O(\epsilon^0)$. Such terms will be familiar if you’ve studied anomalies.

We can therefore loosely say that rational terms are associated with theories requiring renormalization. (This is not quite true; see page 44 of this review). In particular we know that $\mathcal{N}=4$ SYM theory is UV finite, so no rational terms appear. In theory, all $1$-loop amplitudes are constructible from unitarity cuts alone!

Ignoring the subtleties of IR divergences, let’s press on and calculate an $\mathcal{N}=4$ SYM amplitude using unitarity. More precisely we’ll tackle the $4$-point $1$-loop superamplitude. It’s convenient to be conservative and cut only two propagators. To get the full result we need to sum over all channels in which we could make the cut, denoted $s = (12)$, $t = (13)$ and $u=(14)$.

To make our lives somewhat easier, we’ll work in the planar limit of $\mathcal{N}=4$ SYM. This means we can ignore any diagrams which would be impossible to draw in the plane, in particular the $u$-channel ones. We make this assumption since it simplifies our analysis of the color structure of the theory. In particular it’s possible to factor out all $SU(3)$ data as a single trace of generators in the planar limit.

Assuming this has been done, we’ll ignore color factors and calculate only the color-ordered amplitudes. We’ve got two channels to consider $s$ and $t$. But since the trace is cyclic we can cyclically permute the external lines to equate the $s$ and $t$ channel cuts. Draw a picture if you are skeptical.

So we’re down to considering the $s$-channel unitarity cut. Explicitly the relevant formula is

where $\mathcal{A}_4$ is the tree level $4$-particle superamplitude. Now observe that by necessity $\mathcal{A}_4$ must be an MHV amplitude. Indeed it is only nonvanishing if exactly two external particles have $+$ve helicity. Leaving the momentum conservation delta function implicit we quote the standard result

$\displaystyle \mathcal{A}_4(-l_1, 1, 2, l_2) = \frac{\delta^{(8)}(L)}{\langle l_1 1\rangle\langle 1 2\rangle\langle 2l_2\rangle\langle l_2 l_1 \rangle}$

where $\delta^{(8)}(L)$ is a supermomentum conservation delta function. We get a similar result for the other tree level amplitude, involving a delta function $\delta^{(8)}(R)$. Now by definition of the superamplitude, the sum over states can be effected as an integral over the Grassman variables $\eta_{l_1}$ and $\eta_{l_2}$. Under the integral signs we may write

$\displaystyle \delta^{(8)}(L) \delta^{(8)}(R) = \delta^{(8)}(L+R)\delta^{(8)}(R) = \delta^{(8)}(\tilde{Q})\delta^{(8)}(R)$

where $\delta^{(8)}(\tilde{Q})$ is the overall supermomentum conservation delta function, which one can always factor out of a superamplitude in a supersymmetric theory. The remaining delta function gives a nonzero contribution in the integral. To evaluate this recall that the Grassman delta function for a process with $n$ external particles has the form

$\displaystyle \delta^{(8)}(R) = \prod_{A=1}^4 \sum_{i

We know that Grassman integration is the same as differentiation, so

$\displaystyle \int d^4 \eta_{l_1} d^4 \eta_{l_2} \delta^{(8)}(R) = \langle l_1 l_2 \rangle ^4$

Now plugging this in to the pictured formula we find the $s$-channel residue to be

$\displaystyle \textrm{Res}_s = \frac{\delta^{(8)}(\tilde{Q})\langle l_1 l_2 \rangle^2}{\langle 12 \rangle\langle 34 \rangle \langle l_1 1 \rangle \langle 2 l_2 \rangle \langle l_2 4 \rangle \langle 3 l_1 \rangle} \qquad (\dagger)$

Now for the second half of our strategy. We must compare this to the residues from scalar box, triangle and bubble integrands. We aim to pull out a kinematic factor depending on the external momenta, letting the basis integrand residue absorb all factors of loop momenta $l_1$ and $l_2$. But which basis integrands contribute to the residue from our unitarity cut?

This is quite easy to spot. Suppose we consider the residue of a loop integrand after a generic unitarity cut. Any remaining dependence on loop momentum $l$ appears as factors of $(l-K)^{-2}$. These may be immediately matched with uncut loop propagators in the basis diagrams. Simple counting then establishes which basis diagram we want. As an example

$\displaystyle \textrm{factor of }(l-K_1)^{-2}(l-K_2)^{-2}\Rightarrow 2 \textrm{ uncut propagators} \Rightarrow \textrm{box diagram}$

We’ll momentarily see that this example is exactly the case for our calculation of $\mathcal{A}_4^{1\textrm{-loop}}$. To accomplish this, we must express the residue $(\dagger)$ in more familiar momentum space variables. Our tools are the trusty identities

$\displaystyle \langle ij \rangle [ij] =(p_i + p_j)^2$

$\displaystyle \sum_i \langle ri \rangle [ik] = 0$

The first follows from the definition of the spinor-helicity formalism. Think of it as a consequence of the Weyl equation if you like. The second encodes momentum conservation. We’ve in fact got three set of momentum conservation to play with. There’s one each for the left and right hand tree diagrams, plus the overall $(1234)$ relation.

To start with we can deal with that pesky supermomentum conservation delta function by extracting a factor of the tree level amplitude $\mathcal{A}_4^{\textrm{tree}}$. This leaves us with

$\displaystyle \textrm{Res}_s = \mathcal{A}_4^{\textrm{tree}} \frac{\langle 23 \rangle \langle 41 \rangle \langle l_1 l_2 \rangle^2}{ \langle l_1 1 \rangle \langle 2 l_2 \rangle \langle l_2 4 \rangle \langle 3 l_1 \rangle}$

Those factors of loop momenta in the numerator are annoying, because we know there shouldn’t be any in the momentum space result. We can start to get rid of them by multiplying top and bottom by $[l_2 2]$. A quick round of momentum conservation leaves us with

$\displaystyle \textrm{Res}_s = \mathcal{A}_4^{\textrm{tree}} \frac{\langle 23 \rangle \langle 41 \rangle [12] \langle l_1 l_2 \rangle}{(l_2 + p_2)^2\langle l_2 4 \rangle \langle 3 l_1 \rangle}$

That seemed to be a success, so let’s try it again! This time the natural choice is $[3l_1]$. Again momentum conservation leaves us with

$\displaystyle \textrm{Res}_s = \mathcal{A}_4^{\textrm{tree}} \frac{\langle 23 \rangle \langle 41 \rangle [12] [34]}{(l_2 + p_2)^2 (l_1+p_3)^2}$

Overall momentum conservation in the numerator finally leaves us with

$\displaystyle \textrm{Res}_s = -\mathcal{A}_4^{\textrm{tree}} \frac{\langle 12 \rangle [12] \langle 23 \rangle [23]}{(l_2 + p_2)^2 (l_1+p_3)^2} = -\mathcal{A}_4^{\textrm{tree}} \frac{st}{(l_2 + p_2)^2 (l_1+p_3)^2}$

where $s$ and $t$ are the standard Mandelstam variables. Phew! That was a bit messy. Unfortunately it’s the price you pay for the beauty of spinor-helicity notation. And it’s a piece of cake compared with the Feynman diagram approach.

Now we can immediately read off the dependence of the residue on loop momenta. We have two factors of the form $(l-K)^{-2}$ so our result matches only the box integral. Therefore the $4$-point $1$-loop amplitude in $\mathcal{N}=4$ SYM takes the form

$\displaystyle \mathcal{A}_4^{1\textrm{-loop}} = DI_4(p_1,p_2,p_3,p_4)$

We determine the kinematic constant $D$ by explicitly computing the $I_4$ integrand residue on our unitarity cut. This computation quickly yields

$\displaystyle \mathcal{A}_4^{1\textrm{-loop}} = st \mathcal{A}_4^{\textrm{tree}}I_4(p_1,p_2,p_3,p_4)$

Hooray – we are finally done. Although this looks like a fair amount of work, each step was mathematically elementary. The entire calculation fits on much less paper than the equivalent Feynman diagram approach. Naively you’d need to draw $1$-loop diagrams for all the different particle scattering processes in $\mathcal{N}=4$ SYM, including possible ghost states in the loops. This itself would take a long time, and that’s before you’ve evaluated a single integral! In fact the first computation of this result didn’t come from classical Feynman diagrams, but rather as a limit of string theory.

A quick caveat is in order here. The eagle-eyed amongst you may have spotted that my final answer is wrong by a minus sign. Indeed, we’ve been very casual with our factors of $i$ throughout this post. Recall that Feynman rules usually assign a factor of $i$ to each propagator in a diagram. But we’ve completely ignored this prescription!

Sign errors and theorists are best of enemies. So we’d better confront our nemesis and find that missing minus sign. In fact it’s not hard to see where it comes from. The only place in our calculation where extra factors of $i$ wouldn’t simply cancel comes from the cut propagators. Look back at the very first figure and observe that the left hand side has four more factors of $i$ than the right.

Of course we’ve only cut two propagators to obtain the amplitude. This means that we should pick up an extra factor of $(1/i)^2 = -1$. This precisely corrects the sign error than pedants (or experimentalists) would find irritating!

I promised an $SU(3)$ YM calculation, and I won’t disappoint. This will also provide a chance to show off generalized unitarity in all it’s glory. Explicitly we’re going to show that the NMHV gluon four-mass box coefficients vanish.

To start with, let’s disentangle some of that jargon. Remember that an $n$-particle NMHV gluon amplitude has $3$ negative helicity external gluons and $n-3$ positive helicity ones. The four-mass condition means that each corner of the box has more than two external legs, so that the outgoing momentum is a massive $4$-vector.

The coefficient of the box diagram will be given by a generalized unitarity cut of four loop propagators. Indeed triangle and bubble diagrams don’t even have four propagators available to cut, which mathematically translates into a zero contribution to the residue. The usual rules to compute residues tell us that we’ll always have a zero numerator factor left over in for bubble and triangle integrands.

Now the generalized unitarity method tells us to compute the product of four tree diagrams. By our four-mass assumption, each of these has at least $4$ external gluons. We must have exactly $4$ negative helicity and $4$ positive helicity gluons from the cut propagators since all lines are assumed outgoing. We have exactly $3$ further negative helicity particles by our NMHV assumption, so $7$ negative helicity gluons to go round.

But tree level diagrams with $\geq 4$ legs must have at least $2$ negative helicity gluons to be non-vanishing. This is not possible with our setup, since $7 < 8$. We conclude that the NMHV gluon four-mass box coefficients vanish.

Our result here is probably a little disappointing compared with the $\mathcal{N}=4$ SYM example above. There we were able to completely compute a $4$ point function at $1$-loop. But for ordinary YM there are many more subcases to consider. Heuristically we lack enough symmetry to constrain the amplitude fully, so we have to do more work ourselves! A full analysis would consider all box cases, then move on to nonzero contributions from triangle and bubble integrals. Finally we’d need to determine the rational part of the amplitude, perhaps using BCFW recursion at loop level.

Don’t worry – I don’t propose to go into any further detail now. Hopefully I’ve sketched the mathematical landscape of amplitudes clearly enough already. I leave you with the thought-provoking claim that the simplest QFTs are those with the most symmetry. As Arkani-Hamed, Cachazo and Kaplan explain, this is at odds with our childhood desire for simple Lagrangians!