Does pure Yang-Mills have a scale?

Question

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 $eV$?

Answer 1

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.

Answer 2

(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.

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

Answer 3

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 $\delta$.

Answer 4

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.

Answer 5

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.