CONFORMAL TRANSFORMATIONS THAT CLICKEPISODE 8 / Where this series shakes hands with Episode 6 of the previous one

"1/137 becoming 1/128" was the sound of conformal symmetry breaking

Quantum mechanics
breaks conformal symmetry Episode 7 concluded: no mass, no problem. ── But that was the classical statement.
Quantise, and the symmetry dies even in a massless theory. And the way it dies should look familiar.

What you need: logarithms, slopes, and your memory of Episode 6 of the previous series \(T^\mu{}_\mu=\dfrac{\beta(g)}{2g}F_{\mu\nu}F^{\mu\nu}\)

For seven episodes we have treated the conformal transformation as "swapping the ruler". In Episode 7 we saw light pass straight through the swap. But — quantum mechanics will not let you swap the ruler. Merely defining a field theory smuggles in a standard of resolution. As a result a theory that was perfectly conformally invariant classically loses the symmetry the moment it is quantised. This is called the trace anomaly (or conformal anomaly). And the most important line of this episode is: that breaking is exactly the running of \(\alpha\) we watched over and over in the previous series.

01Quantum theory always imports a "resolution"

Consider a massless theory. In Episode 7's language, a theory carrying no standard of length. Yet the moment you try to compute anything in it, you run into trouble — summing all the way down to infinitely fine detail makes the answer diverge.

So in field theory you first decide "how finely will I look?". That standard is written \(\mu\) (the renormalisation scale). In the language of Episode 6 of the previous series, it is the "resolution knob".

Something fatal has just happened.

The price of quantisation

A massless theory had no standard of length.
But to define the theory, a standard \(\mu\) must be imported.
→ Conformal symmetry is lost at the moment of quantisation.

We handed a ruler to a theory that had none. In Episode 7 we wrote that light passes through a conformal transformation because it has no ruler; quantum theory forces a ruler on light as well.

02The size of the breaking is measured by the β function

Once \(\mu\) has been imported, coupling constants depend on the \(\mu\) at which they are measured. The strength of that dependence is the β function.

$$\beta(g)\equiv\frac{dg}{d\ln\mu}$$

And the breaking of conformal symmetry is expressed by exactly that \(\beta\). The quantity \(T^\mu{}_\mu\), used in Episodes 5 and 7 as the indicator of whether conformal symmetry is broken, was zero classically — but quantum mechanically:

The trace anomaly
$$T^\mu{}_\mu=\frac{\beta(g)}{2g}\,F_{\mu\nu}F^{\mu\nu}\qquad(\text{massless gauge theory})$$

If the coupling runs (\(\beta\ne0\)), conformal symmetry is broken. If it does not run (\(\beta=0\)), it is not.

This equation says that two things which look entirely unrelated are one and the same.

How a particle physicist says itHow a conformal transformation says it
The coupling runs (\(\beta\ne0\))Conformal symmetry is broken quantum mechanically
The coupling does not run (\(\beta=0\))Conformal symmetry is restored (a conformal field theory)
\(\alpha\) goes from 1/137 to 1/128The size of the breaking is being measured in the laboratory
◇ ◇ ◇

03Episode 6 of the previous series comes back here

The title of that episode's first half was "1/137 was a number that moves". Look more finely and \(\alpha\) grows, reaching about 1/128 at energies of order the Z boson mass — an established experimental fact.

In one-loop QED the running can be written like this.

Calculation — extracting the slope of the running

① One-loop running (for \(N\) species of charged particle)

$$\frac{d\alpha}{d\ln\mu}=\frac{2N}{3\pi}\alpha^2 \qquad\Longleftrightarrow\qquad \frac{d(1/\alpha)}{d\ln\mu}=-\frac{2N}{3\pi}$$

Written for the reciprocal it is simply a straight line, of slope \(-2N/3\pi=-0.212\,N\).

② From the electron mass to the Z boson mass

$$\ln\frac{M_Z}{m_e}=\ln\frac{91.19\ \mathrm{GeV}}{0.511\ \mathrm{MeV}}=12.09$$

③ With the electron alone (\(N=1\))

$$\frac{1}{\alpha(M_Z)}=137.036-0.212\times12.09=134.5$$

The measured value is 127.95, so this is not enough — the other charged particles (muon, tau, quarks…) each join the running once their own mass is passed. Effectively \(N\simeq3.5\). The slope of the running is counting what charged particles the universe contains.

Figure: the running of \(1/\alpha\). At \(N=0\) (\(\beta=0\)) the line is horizontal — that is conformal invariance. Raise \(N\) and it tilts, and that tilt is the trace anomaly
N = 1.0 β ≠ 0 → conformal symmetry is broken 1/α(M_Z) = 134.47 (measured 127.95)
Running 1/α (one loop) The case β = 0 (conformally invariant) Measured 1/α(M_Z) = 127.95

Set the knob to 0 and the line goes perfectly flat. That is what a conformally invariant world looks like — \(\alpha\) is the same at every resolution, so the marks on the ruler have no influence on physics. Our universe is not like that. The line is tilted. That tilt is the size of the breaking of conformal symmetry.

Collecting bonus episode 6 of the previous series There was an episode asking which kind of motion "1/137 becoming 1/128" is. The answer was "the energy axis, not the time axis". Restated in the language of conformal transformations it is sharper still — motion in time is a choice of gauge (Episode 4); running in energy is an anomaly (this episode). The former is bookkeeping, the latter is physics. The two axes the previous series separated as "with freedom / without freedom" were exactly these two.

04The price of the breaking is 99% of your body weight

"A symmetry was broken" sounds like a loss, but this breaking does an extraordinary amount of work.

A massless theory has no standard of length. But once a coupling runs, a standard is born in the form of "the resolution at which the coupling takes some particular value". A scale wells up out of nothing — this is called dimensional transmutation. For the strong interaction that standard is an energy of a few hundred MeV, written \(\Lambda_{\rm QCD}\).

Now the point. Let us estimate the proton's mass from the masses of the quarks inside it.

Check — where does the proton's mass come from?

A proton is three quarks: u, u, d

$$2m_u+m_d=2(2.16)+4.67=8.99\ \mathrm{MeV}$$

Compared with the proton mass

$$\frac{8.99\ \mathrm{MeV}}{938.27\ \mathrm{MeV}}=0.96\%$$

The quark masses (which come from the Higgs) are only 1% of the proton's weight. The remaining 99% comes from the scale \(\Lambda_{\rm QCD}\) — a scale created by the quantum breaking of conformal symmetry.

What this episode most wants to say

About 99% of your body weight comes not from the Higgs
but from conformal symmetry breaking quantum mechanically.

A massless theory manufactured mass. This is not a by-product of symmetry breaking — it is the very reason there is such a thing as size in our world. Episode 7 said that mass imports a scale; the thing that first made that scale was broken conformal symmetry.

05Is there anywhere the breaking stops?

If \(\beta(g)=0\), conformal invariance is restored at the quantum level. Such a point is called a fixed point, and the theory realised there is a conformal field theory (CFT).

The trivial fixed point \(g=0\)With zero coupling there is nothing to run. But it is a dull theory with no interactions.
Non-trivial fixed pointsSpecial values at which \(\beta\) vanishes even though interactions are present. Strongly interacting yet conformally invariant — the physics of critical points at phase transitions, and \(\mathcal{N}=4\) supersymmetric Yang–Mills, belong here.
The specialness of two dimensionsAs we saw in an exercise in Episode 4, the conformal coupling \(\xi=(D-2)/4(D-1)=0\) in two dimensions. On top of that, conformal symmetry becomes infinite dimensional and the theory becomes exactly solvable. That string theory and critical phenomena in statistical physics are settled by two-dimensional CFT is a consequence of this.

Our world (the four-dimensional Standard Model) sits, unfortunately, at no fixed point. So \(\alpha\) goes on running.

06Even without interactions, curved spacetime breaks it

There is worse news. Switch off every interaction, keep only free massless fields, and conformal symmetry still breaks if spacetime is curved.

$$T^\mu{}_\mu=\frac{1}{16\pi^2}\Big(c\,C_{\mu\nu\alpha\beta}C^{\mu\nu\alpha\beta}-a\,E_4\Big)$$

The right-hand side is built from curvature squared. \(a\) and \(c\) are pure numbers fixed by the kinds of field present (scalar, fermion, vector). The curvature of spacetime counts how many fields there are. This matters in practice — in calculations of Hawking radiation and of fluctuations during inflation.

So the breaking of conformal symmetry cannot be avoided. Cut the interactions and it still leaks, as long as curvature remains.

07So Episodes 4 and 5 were about classical physics

We must now annotate the earlier episodes.

In Episode 4 we counted "one field added, one symmetry added, net zero". In Episode 5 we wrote "this is a rewriting, not physics". Both were classical statements.

Quantum mechanically, a conformal transformation is no longer a mere rewriting. Swapping frames generates anomaly terms and the calculations in the two frames stop agreeing exactly. The question "which frame is physical?", meaningless classically, can acquire meaning quantum mechanically — and that dispute is still unsettled, as we shall see in the finale (Episode 10).

Being straight with you

The straight line in the figure is a one-loop estimate that ignores thresholds. The actual value \(1/\alpha(M_Z)=127.952\) is obtained by summing the effect of each particle joining the running once its own mass is passed (thresholds), together with the hadronic contribution from the strong interaction, and cannot be represented by a single \(N\). The \(N\) in the figure is an effective value — "the slope that would pass through the measured point" — not the actual number of particles.

Also, the value of \(\Lambda_{\rm QCD}\) depends on scheme and order and is not a unique number beyond "a few hundred MeV". The conclusion that 99% of the proton's mass comes from something other than quark masses is established by lattice QCD, but saying it "came out of a single number \(\Lambda_{\rm QCD}\)" is a simplification.

Exercises (solvable with this episode's equations alone)
  1. What is \(T^\mu{}_\mu\) in a theory with \(\beta(g)=0\)? What does that mean?
    Show answer
    \(T^\mu{}_\mu=(\beta/2g)F^2=0\). Conformal symmetry holds even at the quantum level — a conformal field theory. The coupling does not run, so it is "the same theory at every resolution".
  2. What is \(1/\alpha(M_Z)\) for \(N=2\)? Use \(\ln(M_Z/m_e)=12.09\) and slope \(-0.212N\).
    Show answer
    \(137.036-0.212\times2\times12.09=137.036-5.13=131.9\). Still short of the measured 127.95; agreement needs the equivalent of \(N\simeq3.5\).
  3. Of the proton's 938 MeV, what percentage comes from the Higgs (the quark masses)?
    Show answer
    \((2\times2.16+4.67)/938.27=8.99/938.27\approx0.96\%\) — about 1%. The remaining 99% comes from the scale created by the quantum breaking of conformal symmetry.
  4. (Harder) Episode 5 said "only radiation leaves conformal symmetry alone". How does this episode amend that?
    Show answer
    Episode 5's statement was classical. Quantum mechanically, even massless radiation (a system of light and charged particles) has \(T^\mu{}_\mu\ne0\) whenever \(\beta\ne0\). Furthermore, if spacetime is curved, a term proportional to curvature squared survives even without interactions. Only a theory sitting at a fixed point is strictly conformally invariant, and our universe is not there.

SUMMARYRunning and breaking were the same thing

Quantising requires importing a standard of resolution \(\mu\), and conformal symmetry dies at that instant. The size of the breaking is measured exactly by the β function: \(T^\mu{}_\mu=(\beta/2g)F_{\mu\nu}F^{\mu\nu}\). So "the coupling runs" and "conformal symmetry is broken" are two ways of saying one fact. The \(1/137\to1/128\) of Episode 6 of the previous series was, all along, a measurement of how badly it is broken.

And the breaking is no loss. The running creates a scale out of nothing (dimensional transmutation), and that \(\Lambda_{\rm QCD}\) supplies 99% of the proton's mass — most of your body weight exists thanks to conformal symmetry being broken. The breaking stops only at a fixed point \(\beta=0\) (a conformal field theory), and the Standard Model is not at one. On top of that, cut the interactions and it still leaks in the form of curvature squared whenever spacetime is curved. So Episodes 4 and 5's "a rewriting, hence safe" was, all along, a statement about classical physics.

This document is Episode 8 of the series "Conformal Transformations That Click", written for physics-minded high-school and university students. That the trace (conformal, Weyl) anomaly is proportional to the β function, \(T^\mu{}_\mu=(\beta(g)/2g)F_{\mu\nu}F^{\mu\nu}\) for a massless gauge theory, and that conformal invariance is restored at a fixed point \(\beta(g^*)=0\), are standard results. The one-loop QED running \(d\alpha/d\ln\mu=(2N/3\pi)\alpha^2\), \(\alpha^{-1}(m_e)=137.035999\) and \(\alpha^{-1}(M_Z)=127.952\) (\(\overline{\rm MS}\)) are likewise standard. The straight line in the text and figure is a one-loop estimate that ignores thresholds, and \(N\) is an effective value chosen to pass through the measured point, not a particle count. In the curved-space conformal anomaly \(T^\mu{}_\mu=(16\pi^2)^{-1}(cC^2-aE_4)\), \(a\) and \(c\) are constants fixed by the field content. The fraction of the proton mass carried by current quark masses (about 1%) is estimated with the PDG values \(m_u=2.16\) MeV and \(m_d=4.67\) MeV (\(\overline{\rm MS}\), 2 GeV); that the remainder originates in QCD dynamics is established by lattice QCD, though the numerical value of \(\Lambda_{\rm QCD}\) is scheme and order dependent. The quantum non-equivalence (frame dependence) of conformal transformations is treated in Episode 10. — To print, use your browser's Print → Save as PDF (in the print version the slider is frozen and answers are hidden).

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen the slider changes the slope of the running and shows the line going flat — conformally invariant — at β = 0. "Show answer" opens the solutions.