# Why does Spontaneous Symmetry Breaking Depend on Energy Scale?

In the Standard Model we tend to argue that electroweak symmetry is broken below a certain (large) energy scale, yielding the $W$ and $Z$ bosons and the photon. The usual argument for spontaneously symmetry breaking relies on a Higgs potential of the form

$V(\varphi)= \varphi^4 - \mu^2\varphi^2$

and the argument follows from the degeneracy of the lowest energy state.

Remark: Strictly speaking we really ought to be talking about the effective potential, which takes into account radiative corrections. The lowest energy state would then be the vacuum expectation value of the field. I’ll treat the rigorous foundations of SSB in a future post, a topic which is often overlooked in lectures and textbooks!

Critically the standard argument makes no mention of energy scale. So why should we expect it to play a role. The answer is twofold.

Firstly we must remember that we can never observe a full theory. In fact our observations are at best approximations to our theory. This slightly backwards way of looking at things aids our understanding of spontaneous symmetry breaking. At low energies we see the “Mexican hat” clearly, and observe it’s effects in experiments. But as we go to higher energies we “zoom out” on the $V$ axis. This means that the Mexican hat effect is no longer “visible”. From an experimental perspective it’s effects are dominated by other parameters such that they are unobservable. To all intents and purposes the symmetry remains unbroken.

In some theories there’s something more fundamental which averts the SSB scenario at high energies. Recall that renormalization causes the coupling constants to run. In particular our mass pseudoparameter $\mu^2$ can actually change sign at high energies. This eliminates the vacuum degeneracy in the (effective) potential. To see an example, look at Figure 13 in Ben Allanach’s SUSY notes.

I’m grateful to Zac Kenton for a fruitful discussion over a (much-needed) coffee.

# Non-Perturbative QFT: A Loophole?

I’ve been musing on yesterday’s post, and in particular a potential loophole in my argument. Recall that the whole shebang hinges on the fact that $e^{1/g^2}$ is smooth but not analytic. We were interpreting $g$ as the coupling constant in some theory, say QED. But hang about, surely we could just do a rescaling $A_\mu \mapsto A_\mu/g^2$ to remove this behaviour? After all the theory is invariant classically under such a transformation.

But it turns out that this kind of rescaling is anomalous in (most) quantum field theories. Recall that renormalization endows quantum field theories with a $\beta$ function, which determines the evolution of coupling constants as energy changes. The rescaling will only remain a symmetry if $\beta(g)$ is globally zero. Otherwise the rescaling only superficially eliminates the non-perturbative effect – it will reappear at different energies!

This raises a natural question: can you have instantons in finite quantum field theories? By definition these have $\beta$ function zero. Naively we might expect scale invariance to kill non-perturbative physics. A popular finite theory is $\mathcal{N}=4$ SYM, which crops up in AdS/CFT. A quick google suggests that my naive thinking is wrong. There are plenty of papers on instantons in this theory!

There must be a still deeper level to non-perturbative understanding. Sadly most physics papers gloss over the details. Let’s keep half an eye out for an explanation!

# 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!

# Javascript Namespaces in Meteor

I’ve just been watching this excellent EventedMind presentation to get my head around routing for Meteor. I’ve realized that a better way to manage my code (particularly for subscriptions) is to harness the power of javascript namespaces. For example

Subscriptions = { posts: Meteor.subscribe('posts') };

Just keeps everything much cleaner, and prevents quite likely name conflicts. I think I’ll be using this from now on in the ongoing Fixer’s Dream development!