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!

What Can Unitarity Tell Us About Amplitudes?

Let’s start by analysing the discontinuities in amplitudes, viewed as a function of external momenta. The basic Feynman rules tell us that $1$-loop processes yield amplitudes of the form

$\displaystyle \int d^4 l \frac{A}{l^2(p+q-l)^2}$

where $A$ is some term independent of $l$. This yields a complex logarithm term, which thus gives a branch cut as a function of a Mandelstam variable $(p+q)^2$.

It’s easy to get a formula for the discontinuity across such a cut. Observe first that amplitudes are real unless some internal propagator goes on shell. Indeed when an internal line goes on shell the $i\epsilon$ prescription yields an imaginary contribution.

Now suppose we are considering some process as a function of an external momentum invariant $s$, like a Mandelstam variable. Consider the internal line whose energy is encoded by $s$. If $s$ is lower than the threshold for producing a multiparticle state, then the internal line cannot go on shell. In that case the amplitude and $s$ are both real so we may write

$\displaystyle \mathcal{A}(s) = \mathcal{A}(s^*)^*$

Now we analytically continue $s$ to the whole complex plane. This equation must still hold, since each side is an analytic function of $s$. Fix $s$ at some real value greater than the threshold for multiparticle state production, so that the internal line can go on shell. In this situation of course we expect a branch cut.

Our formula above enforces the relations

$\displaystyle \textrm{Re}\mathcal{A}(s+i\epsilon) = \textrm{Re}\mathcal{A}(s-i\epsilon)$

$\displaystyle \textrm{Im}\mathcal{A}(s+i\epsilon) = -\textrm{Im}\mathcal{A}(s-i\epsilon)$

Thus we must indeed have a branch cut for $s$ in this region, with discontinuity given by

$\displaystyle \textrm{Disc}\mathcal{A}(s) = 2\textrm{Im}\mathcal{A}(s) \qquad (*)$

Now we’ve got a formula for the discontinuity across a general amplitude branch cut, we’re in a position to answer our original question. What can unitarity tell us about discontinuities?

When I say unitarity, I specifically mean the unitarity of the $S$-matrix. Remember that we compute amplitudes by sandwiching the $S$-matrix between incoming and outgoing states defined at a common reference time in the far past. In fact we usually discard non-interacting terms by considering instead the $T$-matrix defined by

$\displaystyle S = \mathbf{1}+iT$

The unitarity of the $S$-matrix, namely $S^\dagger S = \mathbf{1}$ yields for the $T$-matrix the relation

$\displaystyle 2\textrm{Im}(T) = T^\dagger T$

Okay, I haven’t quite been fair with that final line. In fact it should make little sense to you straight off! What on earth is the imaginary part of a matrix, after all? Before you think to deeply about any mathematical or philosophical issues, let me explain that the previous equation is simply a notation. We understand it to hold when evaluated between any incoming and outgoing states. In other words

$\displaystyle 2 \textrm{Im} \langle \mathbf{p}_1 \dots \mathbf{p}_n | T | \mathbf{k}_1 \dots \mathbf{k}_m\rangle = \langle \mathbf{p}_1 \dots \mathbf{p}_n | T^\dagger T | \mathbf{k}_1 \dots \mathbf{k}_m\rangle$

But there’s still a problem: how do you go about evaluating the $T^\dagger T$ term? Thinking back to the heady days of elementary quantum mechanics, perhaps you’re inspired to try inserting a completeness relation in the middle. That way you obtain a product of amplitudes, which are things we know how to compute. The final result looks like

$\displaystyle 2 \textrm{Im} \langle \mathbf{p}_1 \dots \mathbf{p}_n | T | \mathbf{k}_1 \dots \mathbf{k}_m\rangle = \sum_l \left(\prod_{i=1}^l \int\frac{d^3 \mathbf{q}_i}{(2\pi)^3 2E_i}\right) \langle \mathbf{p}_1 \dots \mathbf{p}_n | T^\dagger | \{\mathbf{q_i}\} \rangle \langle \{\mathbf{q_i}\} | T | \mathbf{k}_1 \dots \mathbf{k}_m\rangle$

Now we are in business. All the matrix elements in this formula correspond to amplitudes we can calculate. Using equation $(*)$ above we can then relate the left hand side to a discontinuity across a branch cut. Heuristically we have the equation

$\displaystyle \textrm{Disc}\mathcal{A}(1,\dots m \to 1,\dots n) = \sum_{\textrm{states}} \mathcal{A}(1,\dots m \to \textrm{state})\mathcal{A}(1,\dots n \to \textrm{state})^* \qquad (\dagger)$

Finally, after a fair amount of work, we can pull out some useful information! In particular we can make deductions based on a loop expansion in powers of $\hbar$ viz.

$\displaystyle \mathcal{A}(m,n) = \sum_{L=0}^\infty \hbar^L \mathcal{A}^{(L)}(m,n)$

where $\mathcal{A}^{(L)}(m,n)$ is the $L$-loop amplitude with $m$ incoming and $n$ outgoing particles. Expanding equation $(\dagger)$ order by order in $\hbar$ we obtain

$\displaystyle \textrm{Disc}\mathcal{A}^{(0)}(m,n) = 0$

$\displaystyle \textrm{Disc}\mathcal{A}^{(1)}(m,n) = \sum_{\textrm{states}} \mathcal{A}^{(0)}(m,\textrm{state})\mathcal{A}^{(0)}(n,\textrm{state})^*$

and so forth. The first equation says that tree amplitudes have no branch cuts, which is immediately obvious from the Feynman rules. The second equation is more interesting. It tells us that the discontinuities of $1$-loop amplitudes are given by products of tree level amplitudes! We can write this pictorially as

Here we have specialized to $m=2$, $n=3$ and have left implicit a sum over the possible intermediate states. This result is certainly curious, but it’s hard to see how it can be useful in its current form. In particular, the sum we left implicit involves an arbitrary number of states. We’d really like a simpler relation which involves a well-defined, finite number of Feynman diagrams.

It turns out that this can be done, provided we consider particular channels in which the loop discontinuities occur. For each channel, the associated discontinuity is computed as a product of tree level diagrams obtained by cutting two of the loop propagators. By momentum conservation, each channel is uniquely determined by a subset of external momenta. Thus we label channels by their external particle content.

How exactly does this simplification come about mathematically? To see this we must take a more detailed look at Feynman diagrams, and particularly at the on-shell poles of loop integrands. This approach yields a pragmatic method, at the expense of obscuring the overarching role of unitarity. The results we’ve seen here will serve as both motivation and inspiration for the pedestrian perturbative approach.

We leave those treats in store for a future post. Until then, take care, and please don’t violate unitarity.

The Calculus of Particle Scattering

Quantum field theory allows us to calculate “amplitudes” for particle scattering processes. These are mathematical functions that encode the probability of particles scattering through various angles. Although the theory is quite complicated, miraculously the rules for calculating these amplitudes are pretty easy!

The key idea came from physicist Richard Feynman. To calculate a scattering amplitude, you draw a series of diagrams. The vertices and edges of the diagram come with particular factors relevant to the theory. In particular vertices usually carry coupling constants, external edges carry polarization vectors, and internal edges carry functions of momenta.

From the diagrams you can write down a mathematical expression for the scattering amplitude. All seems to be pretty simple. But what exactly are the diagrams you have to draw?

Well there are simple rules governing that too. Say you want to compute a scattering amplitude with $2$ incoming and $2$ outgoing particles. Then you draw four external lines, labelled with appropriate polarizations and momenta. Now you need to connect these lines up, so that they all become part of one diagram.

This involves adding internal lines, which connect to the external ones at vertices. The types and numbers of lines allowed to connect to a vertex is prescribed by the theory. For example in pure QCD the only particles are gluons. You are allowed to connect either three or four different lines to each vertex.

Here’s a few different diagrams you are allowed to draw – they each give different contributions to the overall scattering amplitude. Try to draw some more yourself if you’re feeling curious!

Now it’s immediately obvious that there are infinitely many possible diagrams you could draw. Sounds like this is a problem, because adding up infinitely many things is hard! Thankfully, we can ignore a lot of the diagrams.

So why’s that? Well it transpires that each loop in the diagram contributes an extra factor of Planck’s constant $\hbar$. This is a very small number, so the effect on the amplitude from diagrams with many loops is negligable. There are situations in which this analysis breaks down, but we won’t consider them here.

So we can get a good approximation to a scattering amplitude by evaluating diagrams with only a small number of loops. The simplest have $0$-loops, and are known as tree level diagrams because they look like trees. Here’s a QCD example from earlier

Next up you have $1$-loop diagrams. These are also known as quantum corrections because they give the QFT correction to scattering processes from quantum mechanics, which were traditionally evaluated using classical fields. Here’s a nice QCD $1$-loop diagram from earlier

If you’ve been reading some of my recent posts, you’ll notice I’ve been talking about how to calculate tree level amplitudes. This is sensible because they give the most important contribution to the overall result. But the real reason for focussing on them is because the maths is quite easy.

Things get more complicated at $1$-loop level because Feynman’s rules tell us to integrate over the momentum in the loop. This introduces another curve-ball for us to deal with. In particular our arguments for the simple tree level recursion relations now fail. It seems that all the nice tricks I’ve been learning are dead in the water when it comes to quantum corrections.

But thankfully, all is not lost! There’s a new set of tools that exploits the structure of $1$-loop diagrams. Back in the 1950s Richard Cutkosky noticed that the $1$-loop diagrams can be split into tree level ones under certain circumstances. This means we can build up information about loop processes from simpler tree results. The underlying principle which made this possible is called unitarity.

So what on earth is unitarity? To understand this we must return to the principles of quantum mechanics. In quantum theories we can’t say for definite what will happen. The best we can do is assign probabilities to different outcomes. Weird as this might sound, it’s how the universe seems to work at very small scales!

Probabilities measure the chances of different things happening. Obviously if you add up the chances of all possible outcomes you should get $1$. Let’s take an example. Suppose you’re planning a night out, deciding whether to go out or stay in. Thinking back over the past few weeks you can estimate the probability of each outcome. Perhaps you stay in $8$ times out of $10$ and go out $2$ times out of ten. $8/10 + 2/10 = 10/10 = 1$ just as we’d expect for probability!

Now unitarity is just a mathsy way of saying that probabilities add up to $1$. It probably sounds a bit stupid to make up a word for such a simple concept, but it’s a useful shorthand! It turns out that unitarity is exactly what we need to derive Cutkosky’s useful result. The method of splitting loop diagrams into tree level ones has become known as the unitarity method.

The nicest feature of this method is that it’s easy to picture in terms of Feynman diagrams. Let’s plunge straight in and see what that looks like.

At first glance it’s not at all clear what this picture means. But it’s easy to explain step by step. Firstly observe that it’s an equation, just in image form. On the left hand side you see a loop diagram, accompanied by the word $\textrm{Disc}$. This indicates a certain technical property of a loop diagram that it’s useful to calculate. On the right you see two tree diagrams multiplied together.

Mathematically these diagrams represent formulae for scattering amplitudes. So all this diagram says is that some property of $1$-loop amplitudes is produced by multiplying together two tree level ones. This is extremely useful if you know about tree-level results but not about loops! Practically, people usually use this kind of equation to constrain the mathematical form of a $1$-loop amplitude.

If you’re particularly sharp-eyed you might notice something about the diagrams on the left and right sides of the equation. The two diagrams on the right come from cutting through the loop on the left in two places. This cutting rule enables us to define the unitarity method for all loop diagrams. This gives us the full result that Cutkosky originally found. He’s perhaps the most aptly named scientist of all time!

We’re approaching the end of our whirlwind tour of particle scattering. We’ve seen how Feynman diagrams give simple rules but difficult maths. We’ve mentioned the tree level tricks that keep calculations easy. And now we’ve observed that unitarity comes to our rescue at loop-level. But most of these ideas are actually quite old. There’s just time for a brief glimpse of a hot contemporary technique.

In our pictorial representation of the unitarity method, we gained information by cutting the loop in two places. It’s natural to ask whether you could make further such cuts, giving more constraints on the form of the scattering amplitude. It turns out that the answer is yes, so long as you allow the momentum in the loop to be a complex number!

You’d be forgiven for thinking that this is all a bit unphysical, but in the previous post we saw that using the complex numbers is actually a very natural and powerful mathematical trick. The results we get in the end are still real, but the quickest route there is via the complex domain.

So why do the complex numbers afford us the extra freedom to cut more lines? Well, the act of cutting a line is mathematically equivalent to taking the corresponding momentum $(l-K)$ to be on-shell; that is to say $(l-K)^2 =0$. We live in a four-dimensional world, so $l$ has four components. That means we can solve a maximum of four equations $(l-K)^2 =0$ simultaneously. So generically we should be allow to cut four lines!

However, the equations $(l-K)^2 =0$ are quadratic. This means we are only guaranteed a solution if the momentum $l$ can be complex. So to use a four line cut, we must allow the loop momentum to be complex. With our simple $2$ line cuts there was enough freedom left to keep $l$ real.

The procedure of using several loop cuts is known as the generalized unitarity method. It’s been around since the late 90s, but is still actively used to determine scattering amplitudes. Much of our current knowledge about QCD loop corrections is down to the power of generalized unitarity!

That’s all for now folks. I’ll be covering the mathematical detail in a series of posts over the next few days.

My thanks to Binosi et al. for their excellent program JaxoDraw which eased the drawing of Feynman diagrams.

Complex Scattering For Beginners

Quantum field theory is a description of interacting particles. These are the fundamental constituents of our universe. They are real, in the sense that their properties are described by real numbers. For example, a typical particle has a momentum through spacetime described by a vector with four real components.

In my line of work we’re generally interested in finding out what happens when particles scatter. There are various rules that enable you to determine a numerical amplitude from diagrams of the process. These so-called “Feynman rules” combine the real quantities in different ways depending on the structure of the theory.

Trouble is, it can be quite tricky to do the exact calculations from these rules. It’s a bit like trying to put together a complicated piece of Ikea furniture with no idea what the end product is meant to look like! In a sense the task is possible, but you’d be hard pressed not to go wrong. Plus it would take you ages to finish the job.

What we really need is some extra pointers that tell us what we’re trying to build. Turns out that we can get that kind of information by performing a little trick. Instead of keeping all of our particle properties real, we bring in the complex numbers.

The complex numbers are like a souped up version of the real numbers. The extra ingredient is a new quantity $i$ which squares to $-1$. This might all sound rather contrived at the moment, but in fact mathematically the complex numbers are a lot nicer behaved. By bringing them into play you can extract more information about your original scattering process for free!

Let’s go back to our Ikea analogy. Suppose that you get a mate in to help with the job. Your task is still the same as ever, but now as you construct it you can share tips. This makes everything easier. Moreover you can pool your guesses about what piece of furniture you’re building. The end result is still the same (hopefully!) but the extra input helped you to get there.

The Kallen-Lehmann Representation

Enough waffle, let’s get into some maths. Warning: you might find this hard going if you’re a layman! Consider the propagator of a generic (interacting) quantum field theory

$\displaystyle \mathcal{A}(x,y) = \langle 0 | T \phi (x)\phi (y) | 0 \rangle$

We’d like to look at it’s analytic properties as a function of the momentum $p$ it carries. The first step is to use the standard completeness relation for the quantum states of the theory

$\displaystyle \mathbf{1} = | 0 \rangle \langle 0 | + \sum_{\lambda} \int\frac{d^3p}{2(2\pi)^3E_{\mathbf{p}}(\lambda)} |\lambda_{\mathbf p}\rangle \langle \lambda_{\mathbf p} |$

where $|\lambda_{\mathbf{p}}\rangle$ is a general (possibly multiparticle) eigenstate of the Hamiltonian with momentum $\mathbf{p}$. Inserting this in the middle of the propagator we get

$\displaystyle \mathcal{A}(x,y) = \sum_{\lambda} \int\frac{d^3p}{2(2\pi)^3E_{\mathbf{p}}(\lambda)} \langle 0 | \phi(x) |\lambda_{\mathbf p}\rangle \langle \lambda_{\mathbf p} | \phi(y) | 0 \rangle$

where we have assumed that the VEV $\langle \phi(x) \rangle$ vanishes, which is equivalent to no interactions at $\infty$. This is a very reasonable assumption, and in fact is a key assumption for scattering processes. (It’s particularly important in the analysis of spontaneous symmetry breaking, for example).

Now a little bit of manipulation (exercise: use the transformation of the quantities under the full Poincare group) gives us that

$\displaystyle \langle 0 | \phi(x) |\lambda_{\mathbf p}\rangle = \langle 0 | \phi(0) | \lambda_{0}\rangle e^{-ip.x}|_{p_0 = E_{\mathbf{p}}}$

Now substituting and introducing an integration over $p_0$ we get

$\displaystyle \mathcal{A}(x,y) = \sum_{\lambda} D(x-y, m_{\lambda}^2)Z$

where $D(x-y,m_{\lambda}^2)$ is a Feynman propagator and $Z = |\langle0|\phi(0)|\lambda_0\rangle|^2$ is a renormalization factor. We’ll safely ignore $Z$ for the rest of this post, since it doesn’t contribute to the analytic behaviour we’re interested in.

So why is this useful? One natural way to extract information from this formula might be to distinguish one-particle states. Let’s see how that helps. Recall that our states $|\lambda_{\mathbf{p}}\rangle$ are eigenvalues of the energy-momentum operator $(H, \mathbf{P})$. Generically we get one-particle states of mass $m$ arranged along a hyperboloid in energy-momentum space, due to special relativity. We also have multiparticle states of mass at least $2m$ forming a continuum at higher energy and momenta. (This is obvious if you consider possible vector addition of one-particle states).

Now we can use this newfound knowledge to rewrite the sum over $\lambda$ as an isolated one-particle term, plus an integral over the multiparticle continuum as follows

$\displaystyle \mathcal{A}(x,y) = D(x-y,m^2)Z + \int_{4m^2}^\infty dM^2 D(x-y, M^2)$

This is starting to look promising. Transforming to momentum space is the last step we need to extract something useful. We find

$\displaystyle \mathcal{A}(p^2) = \frac{iZ}{p^2-m^2} + \int_{4m^2}^{\infty}dM^2 \frac{iZ}{p^2 - M^2}$

Considering the amplitude as an analytic function of the “Mandelstam variable” $p^2$ we find an isolated simple pole from an on-shell single particle state, plus a branch cut from multiparticle states.

It’s easy to generalize this to all Feynman diagrams. The key point is that all the analytic structure of an amplitude is encoded by the propagators. Indeed, the vertices and external legs merely contribute polarization vectors, internal symmetry factors and possibly positive factors of momentum. Singularities and branch cuts can only arise from propagators.

So what’s the big deal?

We’ve done a lot of work to extract some seemingly abstract information. But now it’s time for a substantial payoff! The analytic structure of Feynman diagrams can help us to determine their values. I won’t go into details here, but I will briefly mention one important application.

Remember that the scattering matrix in any sensible theory must conserve probabilities, and so be unitary. This requirement, coupled with our observations about Feynman diagrams tells us a lot about perturbative results. The result is usually known as the optical theorem and allows you to extract information about the discontinuities of higher loop diagrams from those at lower loops.

Still this seems rather esoteric, until you turn the whole procedure on it’s head. Suppose you are trying to guess a $1$-loop amplitude. You know it’s general form perhaps, but need to fix some constants. Well from the $S$-matrix unitarity we know it has a branch cut and that the discontinuity is encoded by some tree level diagrams. These diagrams are essentially given by “cutting” the loop diagram.

So go ahead and compare the discontinuity you have with the product of the relevant tree diagrams. This will give you constraints on the constants you need to fix. Do this enough times, for different “cuts” and you will have fixed your $1$-loop amplitude. Simple!

This method is known as generalized unitarity. It’s a vital tool in the modern amplitudes box, and has been used successfully to attack many difficult loop calculations. I’ll return to it more rigorously later, and promise to show you a genuine calculation too.

Why I Like Supersymmetry

Supersymmetry can be variously described as beautiful, convenient, unphysical and contrived. The truth is that nobody really knows whether we’re likely to find it in our universe. Like most theoretical physicists I hope we do, but even if we don’t it can still be a useful mathematical tool.

There are tons of reasons to like supersymmetry, as well as a good many arguments against it. I can’t cover all of these in a brief post, so I’m just going to talk about one tiny yet pretty application I glanced at today.

Let’s talk about scattering processes again, my favourite topic of (physics) conversation. These are described by quantum field theory, which is itself based on very general principles of symmetry. In the standard formulation (imaginatively called the Standard Model) these symmetries involve physical motions in spacetime, as well as more abstract transformations internal to the theory. The spacetime symmetries are responsible for giving particles mass, spin and momentum, while the internal ones endow particles with various charges.

At the quantum level these symmetries actually provide some bonus information, in the form of certain identities that scattering processes have to satisfy. These go by the name of Ward identities. For example QED has a both a gauge and a global $U(1)$ symmetry. The Ward identity for the global symmetry tells you that charge must be conserved. The Ward identity for the gauge symmetry tells you that longitudinally polarized photons are unphysical.

If you’re a layman and got lost above then don’t worry. All you need to know is that Ward identities are cool because they tell you extra things about a theory. The more information you have, the more constrained the answer must be, so the less work you have to do yourself! And this is where supersymmetry comes into the picture.

Supersymmetry is another (very special) type of symmetry that pairs up fermions (matter) and bosons (forces). Because it’s a symmetry it has associated Ward identities. These relate different scattering amplitudes. The upshot is that once you compute one amplitude you get more for free. The more supersymmetry you have, the more relations there are, so the easier your job becomes.

So what’s the use if supersymmetry isn’t true then? Well, in general terms it’s still useful to look at these simplified situations because it might help us discover tools that would be hard to uncover otherwise. Take the analogy of learning a language, for example. One way to do it is just to plunge headlong in and try to pick things up as you go along. This way you tend to get lots of everyday phrases quickly, but won’t necessary understand the structure of the language.

Alternatively you can go to classes that break things down into simpler building blocks. Okay spending one hour studying the subjunctive alone might not seem very useful at first, but when you go back to having a real conversation you’ll pick up new subtleties you never noticed before.

If you’re still unconvinced here’s a (somewhat trivial) concrete example. Recall that you can show that purely positive helicity gluon amplitudes must vanish at tree level in QCD. The proof is easy, but requires some algebraic fiddling. The SUSY Ward identity tells us immediately than in a Super-Yang-Mills (SYM) theory this amplitude must vanish to all orders in the loop expansion. So how do we connect back to QCD?

Well the gluon superpartners (gluinos) have quadratic coupling to the gluon, so an all gluon scattering amplitude in SYM can’t include any gluinos at tree level. (Just think about trying to draw the diagram if you’re confused)! In other words, at tree level the SYM amplitude is exactly the QCD amplitude, which proves our result.

Not sure what will be on the menu tomorrow – I’m guessing that either color-ordering or unitarity methods will feature. Drop me a comment if you have a preference.

Spinor Helicity Formalism – Twistors to the Rescue

Dang! Didn’t get my teeth into enough supersymmetry today. I’m standing at the gateway though, so I’ll be able to tell you much more tomorrow. For now, let’s backtrack a bit and take a look at spinor helicity formalism.

First things first, I need to remind you that on-shell matter particles (specifically spin $\frac{1}{2}$ fermions) are represented as spinors in quantum field theory. For a massless fermion, the spinor encodes the momentum and helicity of the particle. We introduce the so called spinor helicity notation

$\displaystyle [p| = +\textrm{ve helicity particle with momentum }p \\ \langle p| = -\textrm{ve helicity particle with momentum }p$

Their hermitian conjugate spinors give the corresponding antiparticles, as you’d expect if you’re familiar with QFT. One can thus naturally define the inner product of a particle and antiparticle state by contracting their corresponding spinors. We see these contractions a lot in the Parke-Taylor formula, for example.

Now it turns out that every null future pointing vector can be represented in terms of it’s corresponding spinor helicity as according to the identification

$\displaystyle p \leftrightarrow |p]\langle p|\qquad (A)$

This can be made formal easily using the Weyl equation that the spinor states must satisfy. But what exactly is the use of this?

Well we saw that writing down scattering amplitudes in the spinor helicity formalism was particularly easy, since we could keep the amplitudes manifestly “on-shell” throughout the process. However I did sweep under the carpet a little algebraic manipulation. I can be more explicit about that now. The only difficult steps in the simplification I omitted are due to momentum conservation.

Usually momentum conservation for an $n$ particle process takes the form

$\displaystyle \sum_{i=1}^n p_i = 0$

This is easy to implement in the usual Feynman diagram formalism, because it is linear! But in the spinor helicity world, we see that this formula becomes quadratic on account of the identification (A) above. This quadratic relation is somewhat troublesome to deal with, and requires annoying identity manipulation to impose.

But what if we want to have our cake and eat it? Imagine a world where we could have the spinor helicity simplifications and yet keep the simplicity of linear momentum conservation. Fortunately for us, such a world exists. We can get there by means of an abstract tool called the dual momentum twistor. There’s not enough time to tell you about that now, but watch out for its appearance in a later post.

So tomorrow will be some supersymmetry then, and maybe a short aside on calculating graviton scattering amplitudes easily. Easily, you say? Well it’s a doddle… at tree level… with a small number of particles…

My thanks to Andi Brandhuber for an enlightening discussion on this point.

Recursion Relations for Amplitudes

So yesterday’s discussion was all about how simple some gluon scattering processes (or amplitudes) look. In particular the maximal helicity violating (MHV) ones are special, because hundreds of terms cancel down to give a single neat result. Today I’ve been looking at how to prove this result, so I can now sketch the main ingredients for you.

If you’re from a mathsy background you won’t be surprised to learn that the $n$-gluon Parke-Taylor identity is proved using induction. For the uninitiated there’s a simple analogy with climbing stairs. If you can get up the first one, and you can get from every one to the next one then you can get to the top!

With this in mind, our first task is to prove the simplest case, which turns out to be $n=3$. Why exactly? Well the Feynman rules for QCD have 3 and 4 point vertices at tree level, so there’s no tree level 2 point amplitudes! Turns out that the $n = 3$ case is neatly dealt with using spinor helicity formalism. Roughly speaking this takes into account the special helicity structure of the MHV amplitudes to lock in simplifications right from the start of the calculation! Add in momentum conservation and hey presto the $n=3$ Parke-Taylor identity drops right out.

So now we need to climb from one stair to the next. This is where recursion relations come in handy. Nine years ago, a group of theoreticians spotted a cunning way to break apart tree level gluon amplitudes into smaller, more manageable pieces. Mathematically they spotted that the $n$-gluon scattering amplitude factorized into the product of two distinct on-shell amplitudes, each with a complementary subset of the original external legs plus an extra leg with momentum $\hat{P}$. The only added ingredient needed was a factor of $1/P^2$ corresponding to a propagator between the two diagrams.

Woah – hold up there! What’s all this terminology all of a sudden. For the uninitiated I’m guessing that on-shell sounds a bit confusing. But it’s no cause for alarm. In general an on-shell quantity is one which obeys the equations of motion of the system involved. Here the relevant equation is the Weyl equation, which implies that $\hat{P}^2 = 0$.

Why are these recursion relations so useful? Well, they give us exactly the ingredient we needed for the induction step. And we’re done – the Parke-Taylor identity is proved, with a little bit of algebra I’ve shoved under the rug.

There’s one more point I’ve neglected to mention. How do you go about finding these mythical on-shell recursion relations? The answer comes from doing some subtle complex analysis, transforming momenta into the complex plane. It might not sound very physical to do that, but in fact the method opens up oodles of new possibilities. One reason is that complex integration is both more powerful and easier than its real counterpart, so it can be used to extract valuable identities from the world of scattering processes.

I’ll leave you with this great basic article clarifying the subtleties of the on-shell/off-shell distinction. It goes a bit deeper than that too, so is worth a read even if you’re more of an expert!

Tomorrow I hope I’ll bring you something more supersymmetrical in nature. $\mathcal{N}=4$ SUSY is a favourite playground for scattering enthusiasts because it is finite (no renormalization needed) and very simple (no free parameters). We’ll encounter its stark beauty in due course.

The Parke-Taylor Formula

Unfortunately I’m not going to have time today to give you a full post, mostly due to an abortive mission to Barking! The completion of that mission tomorrow may impact on post length again, so stay tuned for the first full PhD installment.

Nonetheless, here’s a brief tidbit from my first day. Let’s think about the theory of the strong force, which binds quarks and nuclei together. Mathematically it’s governed by quantum chromodynamics (QCD). At it’s simplest we can study QCD with no matter, so just consider the scattering interactions of the force carrying gluon particles.

It turns out that even this is pretty complicated! At tree level in Feynman diagram calculations, the simplest possible approximation, there are about 12000 terms for a four gluon scattering event. Thankfully these all cancel to give a single, closed form expression for the scattering amplitude. But why?

There’s a simpler way that makes use of some clever tricks to prove the more general Parke-Taylor formula that the maximal helicity violating $n$ gluon amplitude is simply

$\frac{\langle 12 \rangle^4}{\langle 12 \rangle \langle 23 \rangle \langle 34 \rangle \dots \langle n1 \rangle }$

What does this all mean?

Qualitatively, that there is a formalism in which these calculations come out very simply and naturally. This will be the starting point for my exploration of modern day amplitudology – a subject that ranges through twistor theory, complex analysis and high dimensional geometry!

For the real mathematics behind the formula above, I’m afraid you’ll have to wait until tomorrow or Wednesday!

New Year’s Resolution

Happy New Year everyone! I’ll be starting my PhD at Queen Mary, University of London next week and I’ve decided to pick up my blog activity. I’ll try to post a quick article every couple of days detailing something I’ve learnt, or any interesting thoughts I’ve had. Over time I might build up to longer reviews.

If there are any specific requests for theoretical physics material you’d like to see covered, either at a layman or a research level then please do comment them below!