# 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.

Until then, enjoy your weekend!

# Non-Perturbative Effects in Quantum Field Theory

In elementary QFT we only really know how to solve free theories. These require quantizing infinitely many harmonic oscillators, which is an easy problem. But the real world is described by interacting theories with Lagrangians like

$\mathcal{L}_{\textrm{QED}} = \overline{\psi}(i\not D-m)\psi - \frac{1}{4}F^2$

in which the fields are mixed up together. The standard way of dealing with these is to write them as

$\mathcal{L} = \mathcal{L}_{\textrm{free}}+\mathcal{L}_{\textrm{int}}$

and then split the path integral as

$\exp(i\int d^4x \mathcal{L}) = \exp(i\int d^4 x \mathcal{L}_{\textrm{free}})(1+i\int d^4x\mathcal{L}_{\textrm{int}}+\dots)$

The first factor explicitly yields free theory propagators. The second factor may now be expanded as a Taylor series in some coupling constant $g$. Often this is trivial since $g$ appears in $\mathcal{L}_{\textrm{int}}$ precisely to first order. For example in QED we have

$\mathcal{L}_{\textrm{int}} = g\overline{\psi}\gamma^{\mu}\psi$

with $g$ being the electric charge. This means that the perturbation expansion is exactly the usual exponential series.

At present it looks like perturbation theory must capture all information in a QFT, provided you sum up all the terms. However, we’ve missed a crucial mathematical point. We’ll view the scattering amplitude $\Gamma$ for a given process as a complex-valued function of a real coupling $g$; that is to say $\Gamma \equiv \Gamma(e)$. We construct this function in (somewhat) trivial way from smooth functions, so it should be smooth.

However, smoothness does not guarantee analyticity! Put another way, we cannot be sure that the Taylor series for $g$ will converge for $g\neq 0$. Even if it does, it’s not guaranteed to converge to the function $\Gamma$ itself. This means that there are effects that lie outside the realm of perturbation theory, even if you could sum every term.

Let’s take an example. Suppose for instance that $\mathcal{L}_{\textrm{int}}$ has a $1/g^2$ dependence. If you’re balking at this “unphysical” choice, then rest assured that such Lagrangians do crop up. Now try to compute the Taylor expansion around $g = 0$ for $exp(-1/x^2)$. A little thought shows that this is identically zero.

Such non-perturbative effects require new approaches. One fruitful method involves solving the full theory classically then then considering quantum perturbations around such solutions. For Yang-Mills theories in particular this yields the notion of an instanton.

Finally let’s go back to QED and work out whether we should see non-perturbative effects there. Although not immediately as obvious as the simple $1/g^2$ example, one can argue that the perturbation series has $0$ radius of convergence. We follow Dyson and observe that if the radius of convergence were not zero, we’d be able to “reverse” the electric field by flipping the sign of $g$. This would render the vacuum unstable against decays into electrons and positrons, since these would repel rather than annihilate within a “Heisenberg allowed” time period. To rule out this option, we must take $g = 0$ formally.

This argument is not mathematically rigorous, and I don’t know whether one can make it so. Please comment if you know! Thankfully there’s a more modern perspective that sheds new light. We can view the sickness in QED in the context of Wilsonian renormalization. In particular, the theory has no well-defined high energy limit, as the coupling constants flow to infinite values at finite energy. This Landau pole behaviour is perhaps evidence that non-perturbative effects rule high energy QED. Again, I don’t whether this argument can be made watertight, so please take a rain-cheque on that one!