49

The problem In case you were not aware of this, finding a proof for confinement is one of the Millenium Problems by the Clay Mathematics Institute. You can find the (detailed) answer to your question in the official problem description by Arthur Jaffe and Edward Witten. In short: proving confinement is essentially equivalent to showing that a quantum Yang-...


27

Most of the forces induced by a point particle follows the 1/r^2 rule No, it's the forces mediated by point particles with no mass and charge that follow the the 1/r^2 rule. then why does strong force don't obey it? The inverse square law is a consequence of the particles having no mass and/or charge. Such particles have long/infinite lifetimes and can ...


21

A theory is usually denominated a 'gauge theory' if all the interactions in that theory are introduced by promoting global symmetries to gauge symmetries. Note that a gauge theory is a gauge invariant theory, but a gauge invariant theory doesn't has to be a gauge theory (for example, the Standard Model is gauge invariant, but it's not a gauge theory since ...


20

As Lubos Motl and twistor59 explain, a necessary condition for unitarity is that the Yang Mills (YM) gauge group $G$ with corresponding Lie algebra $g$ should be real and have a positive (semi)definite associative/invariant bilinear form $\kappa: g\times g \to \mathbb{R}$, cf. the kinetic part of the Yang Mills action. The bilinear form $\kappa$ is often ...


20

A theory is typically described by a Lagrangian, and varying this gives us the equations of motion of the system. The symmetries you describe are symmetries of the Lagrangian i.e. they are transformations that leave the Lagrangian unchanged. It would be nice to think that the Lagrangians that describe our leading theories of physics were derived in some ...


17

If we write the field strength in terms of "electric" and "magnetic" fields $\vec{E}$ and $\vec{B}$, the relevant expression can be written as $$\text{Tr}F_{\mu\nu}\tilde{F}^{\mu\nu}=4\,\text{Tr}\vec{E}\cdot\vec{B}.$$. Under parity transformations, $\vec{E}\rightarrow-\vec{E}$ and $\vec{B}\rightarrow\vec{B}$, while under charge conjugation, $\vec{E}\...


15

It's not a sufficient explanation. There are asymptotically free theories which are not strongly coupled in the IR. The rate at which the coupling gets strong is important. In QCD, it seems to get strong very quickly near the confinement scale, so that beyond a certain scale, you only see hadrons. It is not really understood how this works. The ...


15

Bundles and compactified spacetime A gauge theory cannot be looked at purely locally, it has inherently global features one cannot see locally. The proper mathematical formalization of a Yang-Mills gauge theory is that the gauge field $A$ is a connection on a principal bundle $P\to M$ over spacetime $M$. However, in practice, it turns out that physicists don'...


14

The possibility of spontaneous Lorentz symmetry violation due to the infrared problem of the Dirac-Maxwell equation was conjectured a long time ago by Frohlich, Morchio and Strocchi, in references [1,2] mentioned in the given Balachandran and Vaidya article. In perturbative QED, we usually assume that the scattering states are free eigenstates of the number ...


14

You've got things slightly backwards. In constructive QFT, one almost always starts in Euclidean spacetime -- where it is 'easy' to define the path integral -- and then analytically continues to get correlation functions on Minkowski spacetime. (The Osterwalder-Schrader Theorem tells you when this analytic continuation 'works', meaning when the resulting ...


13

I) Vanishing field-strength $F=0$ does not imply that the gauge potential $A$ is pure gauge. It only holds locally. There could be global obstructions. In fact, topological obstructions could happen even if the gauge group $G$ is Abelian. II) Let us sketched the proof of the local statement in a sufficiently small neighborhood $\Omega\subseteq M$ of a point ...


13

The usual action for Yang-Mills theory is, using differential forms $$S = \int \operatorname{tr} (F \wedge \star F)$$ where $\star$ is the Hodge dual. Now note that the integral of a differential form is always defined with respect to an orientation, and the Hodge dual is also defined with respect to an orientation. Parity is reversing the orientation, which ...


13

In Yang-Mills, the gauge connection plays the role of a potential and the curvature form plays the role of a "field strength". In GR, the metric tensor plays the role of a potential, and the connection plays the role of a field strength. This is why, in particular, the gravitational force is not a real force, as the connection is not gauge-covariant. Of ...


12

It seems the resolution to OP's question lies in the difference between the Levi-Civita symbol, which is not a tensor and whose values are only $0$ and $\pm 1$; and the Levi-Civita tensor, whose definition differs from the Levi-Civita symbol by a factor of $\sqrt{|\det(g_{\mu\nu})|}$.


12

The standard model Higgs particle has a weak charge, but no color charge. As a result it generates as mass gap in electroweak theory (masses for the W and Z), but not in QCD (no gluon mass). However, we know that QCD (even pure QCD, without fermions) does have a mass gap: Glueballs are heavy. The millenium prize problem asks why that is.


11

I recommend that you to read the chapter 15.2 in "The Quantum Theory of Fields" Volume 2 by Steven Weinberg, he answers precisely your question. Here a short summary In a gauge theory with algebra generators satisfying $$ [t_\alpha,t_\beta]=iC^\gamma_{\alpha\beta}t_\gamma $$ it can be checked that the field strength tensor $F^\beta_{\mu\nu}$ transforms as ...


10

It's because you want the kinetic part of the Yang Mills action $$ \int Tr({\bf{F^2}}) dV$$ to be positive definite. To guarantee this, the Lie algebra inner product you're using (Killing form) needs to be positive definite. This is guaranteed if the gauge group is compact and semi-simple. (I'm not sure if it's only if G is compact and semi simple though. ...


10

I) Bosonic part: When we Wick-rotate, it is more natural to use sign convention $$\tag{1} \eta_{\mu\nu}~=~{\rm diag}(-1,+1,+1,+1)$$ for the Minkowski (M) metric, and $$\tag{2} \delta_{\mu\nu}~=~{\rm diag}(+1,+1,+1,+1)$$ for the Euclidean (E) metric. Here we will use Greek indices $\mu,\nu=0,1,2,3$, to denote spacetime indices, and Roman indices $j,k=1,...


10

Yes, one traditional alternative to the path integral formalism is the operator formalism. For QED with abelian gauge group, the old quantization formulation is the Gupta-Bleuler formulation. For QCD/Yang-Mills theory with non-abelian gauge group, the Gupta-Bleuler formulation is replaced by the BRST formulation. The BRST formulation exists in at least 3 ...


9

By definition, Lagrange multipliers are only coefficients that enter the extremized quantity (action) etc. linearly – and that multiply constraints. In some exceptional cases, an auxiliary field could enter in this way. However, they typically appear in a more complicated way and bilinear terms in the auxiliary fields are a rule rather than an exception. So ...


9

First impressions based on a quick read of the preprint: I'm out of my depth on this! I couldn't tell you if their derivation is correct, but assuming that it is: They don't treat real QCD. They study SU(2) YM without quarks. The authors claim they can do real QCD and get the same result, but this is not demonstrated in the paper (they defer this to a later ...


9

Gravity isn't Yang-Mills theory in the narrow sense – well, except for equivalences such as AdS/CFT or Matrix theory that imply that a quantum gravitational theory is fully equivalent to a gauge theory living in a different space (e.g. in AdS/CFT, on the boundary of the AdS space). However, gravity is a gauge theory in the broader sense because it's ...


9

OK, blathering it is. The key thing to understand in Yang-Mills theory (whether or not you're trying to be mathematically rigorous) is that gauge symmetry is not a physical symmetry. (Its job is to keep track of the overcounting in a redundant description.) Classically, this means that the observables -- the numerical quantities which we can measure -- ...


9

Consider the finite dimensional unitary representations $\alpha,\beta,\gamma$ of the given compact group $G$ on corresponding vector spaces $V_1,V_2,V_3$. Let $|i\rangle_j,i=1,\dots,n_j$ be an orthnormal basis of $V_j$ where $dim V_j=n_j$. Then $\{|i\rangle_1\otimes|j\rangle_2\otimes|k\rangle_3\}$ forms an orthonormal basis of $V=V_1\otimes V_2 \otimes V_3$....


9

The short answer is that only if the Berry curvature is defined by: (in matrix notation): $$F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} + [A_{\mu}, A_{\nu}]$$ it becomes gauge covariant, i.e., for a gauge transformation: $$A_{\mu} \rightarrow g^{-1}A_{\mu} g+g^{-1}\partial_{\mu}g$$ $g \in U(N)$ ($N$ is the degeneracy of the level), the ...


9

If you don't impose power-counting renormalizability, there are a host of other possibilities, since higher order derivatives or higher order interactions can be introduced. For example, terms $(Tr(F^2)^m)^n$ and are gauge invariant but for $m>1$ or $n>1$ not renormalizable. If you impose power-counting renormalizability, uniqueness is fairly ...


9

It may be worth looking at this formula as the infinitesimal limit of an exponentiated relation. We have $$ \exp(- a^\mu D_\mu) \exp(- b^\nu D_\nu) \exp( a^\mu D_\mu) \exp( b^\nu D_\nu) \approx 1 + a^\mu b^\nu [D_\mu, D_\nu] + \dots $$ Just like $\exp( a^\mu \partial_\mu)$ is an operator of translation in a flat space, $\exp( a^\mu D_\mu)$ is an operator ...


8

In this context, a "current" is an object obeying an affine Lie algebra, also called current algebra and a special case of a Kac-Moody algebra. It is an algebra formed by unit weight operators: take for example a current $J^a(z)$, where $a$ is a label and $z$ is a complex coordinate. The algebra is given by $$[J^a_n,J^b_m]=i{f^{ab}}_cJ^c_{n+m}+mkd^{ab}\...


8

Note that, for example, \begin{align} [A_\mu,\partial_\nu]f&=A_\mu\partial_\nu f-\partial_\nu(A_\mu f)\\ &=A_\mu\partial_\nu f-\partial_\nu(A_\mu)f-A_\mu\partial_\nu f\\ &=-f\partial_\nu A_\mu\,. \end{align} So you don't get terms like $A_\mu\partial_\nu$.


8

The structure of standard model $SU(3)\times SU(2)\times U(1)$ is chiral which basically tells you the necessity of chiral fermions. If left-handed fermions transform under a representation $R$ of the symmetry group then due to charge-conjugation relating left-handed and right-handed fermions as $$\psi_{Right}=C(\bar{\psi^C})^T_{Left}$$ and so, right handed ...


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