Does pure Yang-Mills have a scale?

Consider pure Yang-Mills (YM) in 4 dimensions. The YM mass gap problem (as described in https://www.claymath.org/wp-content/uploads/2022/06/yangmills.pdf) tells us that this is supposed to have a mass-gap $$\delta$$. The gauge coupling, $$g$$, is dimensionless. So what sets the scale of $$\delta$$? To put it another way, if, starting from the YM partition function it were possible to compute $$\delta$$, it would obviously be a function of $$g$$, so $$\delta\equiv\delta(g)$$, but from where would $$\delta$$ get its dimension of mass? Is it somehow related to the energy at which you probe the theory?

To put it even more explicitly, if somebody computes $$\delta$$ and claims it is $$1GeV$$ then we can all ask them where $$GeV$$ came from. How does this problem know anything about $$GeV$$ or $$MeV$$ or $$eV$$?

• @emirsezik Yes but where does that scale come from? Aug 6, 2023 at 9:48
• Mathematically, it can be expressed as a function of an arbitrary scale even though it is arbitrary and $\Lambda_{QCD}$ doesn't depend on this arbitrary scale. This is called dimensional transmutation. Aug 6, 2023 at 9:57
• @emirsezik So starting from pure YM why should Λ be O(MeV). There is nothing in pure YM pointing to MeV right? Why not KeV or eV? Aug 6, 2023 at 11:07
• Experimentally. We can relate one value of the coupling at one energy scale to another via the beta function. Aug 6, 2023 at 11:33
• The scale is a fundamental quantum parameter of QCD. There could be undiscovered QCD-like theories with much higher scales. You can't ask where fundamental dimensional parameters came from. They just are. Aug 6, 2023 at 13:09

If you could answer where the mass gap "really comes from", you'd have solved the millenium problem!

The formal description of the problem includes two points which are commonly conflated under "confinement" (="no color-charged free states exist"), namely quark confinement and the mass gap:

Quark confinement is the statement that you have to explain why in a theory with a quark field there are no free quark states, i.e. why quark states are confined.

The second part of confinement broadly construed is that there are no free gluon states. In the formal statement this is the "mass gap": Since gluons are massless, if free gluon states were allowed, there would be no mass gap, since there would be "soft" gluons, just like soft photons, with arbitrarily low energy.

The value of the mass gap the question calls $$\delta$$ is just the mass of the lowest-lying state of the theory that remains after the disappearance of these soft gluon states has been explained. Its precise value is not particulary interesting nor the focus of the problem - you'd expect this to be just to be the mass of the lightest color-neutral bound state that can be formed.

(here's a rephrasing of the answer) It's better to think of pure Yang-Mills theory as a zero-parameter QFT¹. In other words, the spectrum is unique, and the actual values for the energies of the states (glueballs) is completely determined by your choice of units. For example, if you look at ratios of glueball masses (say, $$\frac{m(2^{++})}{m(0^{++})} \sim 1.39$$), this is just some fixed number, there are no parameters.

How does this come about in perturbation theory? Renormalized perturbation theory works by first specifying $$g(\mu)$$, where $$\mu$$ is our renormalization scale. So, even though $$g$$ is dimensionless, to do any calculations you have to specify the value of $$g$$ at some scale, e.g. saying $$\alpha_S(m_Z) \sim 0.1176$$. By specifying this scale, you specify the units, with which you can then measure physical things.

¹ Except maybe you consider the $$N$$ in $$SU(N)$$ a free parameter!

• The renormalization scale $\mu$ has nothing to do with the existence of YM mass gap. One could have a perturbative $g(\mu)$ which does NOT blow up at any energy scale, then there is NO mass gap. Aug 7, 2023 at 18:31
• I'm not sure I agree with this answer. The running of the coupling constant with energy means there is a dimensionful parameter in the theory -- you could define this parameter as saying it's the value of energy at which the coupling takes the value 1, for example. Classically pure YM has no parameters, but quantum mechanically a scale is generated. A truly zero parameter QFT would not have a running coupling, so it would be some kind of CFT. A trivial example of a zero parameter QFT would be a single free massless scalar field. Aug 7, 2023 at 23:37
• @MadMax Yes, the scale $\mu$ has nothing to do with the existence of the mass gap. But, you agree that if (in pure yang-mills), if I set $g(\mu_0) = 0.1$, then the spectrum energies vary linearly in $\mu_0$? That's pretty much the point I was making. Nothing to do with the mass gap Aug 8, 2023 at 2:00
• @Andrew Yes I agree, when I say "zero-parameter theory" I just mean that, up to a re-definition of units, there are no parameters to pure yang-mills. For example, a super-intelligence that can solve QCD nonperturbatively would find it very natural to measure all dimensionful quantities in units of the $0^{++}$ mass. There are no parameters that change any physics (whereas for example in full QCD, there are quark masses that you can change, that actually change the physics of the theory) Aug 8, 2023 at 2:03
• @Andrew Your point about the fact that we have to pick the scale by (for example) the energy at which the coupling takes a certain value is what I was describing in my second paragraph. For example, if I say that at some scale the coupling is $0.1$, I have to specify the units with which I measure that scale (e.g. GeV, or Joules, or whatever). Specifying these units is the same data as specifying the scale! Aug 8, 2023 at 2:06

So after some help from the comments I am satisfied with the following version of events: Consider a measurement $$O$$. It can expressed as a function of the gauge coupling $$g$$ and the momentum cutoff $$\Lambda$$: $$O=f(g,\Lambda)$$ Invert this relation for $$g$$: $$g=F(O,\Lambda)$$ Since $$g$$ is dimensionless and $$\Lambda$$ is dimensionful, it must be that $$O$$ is dimensionful.

Substitute $$g=F(O,\Lambda)$$ into the theory. Now the theory is expressed entirely in terms of $$O$$ and $$\Lambda$$. Now take $$\Lambda\to\infty$$. The theory is now expressed only in terms of $$O$$. But $$O$$ is dimensionful so we now have a scale in the theory. In particular, $$\delta$$ will be a function of $$O$$, ie $$\delta\equiv \delta(O)$$, and roughly speaking, $$O$$ sets the scale of $$\delta$$.

• Yes, 0 GeVs equal 0 eVs, and likewise for infinite ones…. Aug 7, 2023 at 10:06
• By linking to the Clay problem, you electrified the readers to go there, instead of the arbitrariness of the dimensional transmutation scale, determined by an experimental boundary condition. Perhaps you might emphasize that in your title and question... Aug 8, 2023 at 14:05

The detail mechanism of how a Yang-Mills (YM) theory produces a mass gap is obvious not known yet. However, there is a mechanism for a theory without an intrinsic mass/energy scale to produce such a mass scale. It is one of the most amazing phenomena predicted by quantum field theory.

Note that all YM theories contain at least one free parameter: the gauge coupling. However, it is dimensionless. Yet, due to radiative corrections, it runs. That is to say, the strength of the gauge coupling varies as a function of the energy scale. In non-abelian YM theories, the gauge coupling becomes stronger toward lower energy scales. These theories are said to be asymptotic free.

Now we borrow a phenomenon from solid state physics: a phase transition. The gauge coupling acts as the control parameter, which will cause some order parameter (typically some vacuum expectation value) to undergo a non-analytic change at some value of the gauge coupling, which represents the phase transition.

In the context for YM theories, the phase transition (which is associated with confinement and the formation of the mass gap) would appear at a specific energy scale where the running coupling becomes strong enough. As a result, an energy scale is produced even though the fundamental theory does not contain any fundamental mass parameters.

Although the analytical proof for the existence of such a phase transition in YM theories does not exist yet, it is believed that this mechanism is responsible for the appearance of the mass scale.

To theoretically prove (thus winning the Clay award) the existence of none-zero YM mass gap $$\delta \neq 0$$ is a wild goose chase. Theoretically, YM allows either zero or non-zero $$\delta$$.

OP's own answer is a very roundabout way to say that $$\delta$$ is a free parameter. Any quantity which can not be calculated theoretically and can only be determined by measurement is just a free parameter. Therefore, $$\delta$$ is merely a free parameter, one should not over-interpret it.

And for that mater, nothing prevent $$\delta$$ to be zero $$\delta=0$$, which means there is NO YM mass gap. That is to say, none-zero $$\delta \neq 0$$ is just an experimentally verified fact of QCD, which can NOT be derived theoretically. Hence to theoretically prove the existence of none-zero YM mass gap $$\delta \neq 0$$ is a misguided enterprise.

The situation is analogues to that of the none-zero Higgs VEV (and its specific scale), which is also an accidental fact of our universe. Theoretically, a zero Higgs VEV works just fine.

• I'm not 100% sure $\delta=0$ is really consistent with what we know about YM, since asymptotic freedom implies the coupling should blow up at some finite energy scale as you run from high energies to low energies, but let's assume that it is. Even so that would not make the Clay problem to prove the existence of a non-zero YM mass gap trivial, since the problem statement is to show that there exists a YM theory (for any appropriate gauge group) with a mass gap. That's a statement that's apparently true based on experiments but no one can rigorously prove. Aug 7, 2023 at 23:46
• @Andrew, asymptotic freedom only implies that the coupling should approach zero at the high energy end. However, at the low energy end we only know that the coupling should blow up, but we DON'T KNOW at which energy scale $\delta$ the coupling should blow up. Nothing prevents $\delta$ to be zero, that is $\delta=0$. Non-zero $\delta$ (and thus finite energy gap) is ONLY an experimentally verified fact of QCD, it's NOT a corollary of asymptotic freedom. Aug 30, 2023 at 16:39
• While it is possible to have theories that are asymptotically free and non-confining, there's not really any evidence that QCD is one of them. Lattice computations (with no need for physical experiment) strongly suggest that it's confining. Aug 30, 2023 at 20:10
• @user1504, the QCD Lagrangian has no specific scale. Any claim to theoretically derive a confinement scale from a Lagrangian with no specific scale seems fishy to me. At some point of your purported "theoretical derivation" (lattice computation or not), you must put in by hand some scale and that scale is then confirmed by experimentally measurable quantities. Do you see the circular logic here? The "no need for physical experiment" argument is lost in the bloody detail. Aug 31, 2023 at 15:23
• Right. But QCD isn't defined by the Lagrangian. The classical theory is, but the quantum theory is defined by the path integral, which includes regularization and limits. That's where the scales come from. Aug 31, 2023 at 18:22