# Tag Info

## Hot answers tagged yang-mills

50

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-...

28

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 ...

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 ...

18

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, $\vec{E}\... 18 Rant on notation I hate the physics convention of defining transformations by writing something like "$S \to S$,$\phi \to 2 \phi$,$x \to x + 1$, apples$\to$oranges,$1 \to -1$,$\pi \to e$, true$\to$false". The problem is that while this might be a succinct description of what to mechanically do to carry out a transformation, it's ambiguous ... 18 Color charge is a general term that describes how a particle transforms under$SU(3)$transformations, i.e. what is its$SU(3)$representation. The terms red, green and blue refer to the fundamental or defining representation of$SU(3)$which is 3 dimensional. Red, green and blue refer to the three basis vectors in this representation, denoted by$| r \...

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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 ...

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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'...

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 ...

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TL;DR: 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. More details: The starting point is a connected (but not necessarily simply connected) gauge Lie group $G$ and a ...

15

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 ...

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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 ...

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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.

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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 ...

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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 $j,k=1,... 11 Certainly I admit that power counting law is violated, but why does violation of power-counting have relation with renormalizablity? As per the Dyson-Weinberg power-counting theorem (see Ref.1, chapter 12 and Ref.2, chapter 8-1), a diagram is convergent if and only if each of its subdiagrams has a negative superficial degree of divergence$\omega$. The ... 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 ... 10 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$is the degeneracy of the level), the ... 10 If you don't impose power-counting renormalizability, there are a host of other possibilities, since higher order derivatives or higher order interactions can be introduced. For example, terms$(Tr(F^2)^m)^n$and are gauge invariant but for$m>1$or$n>1$not renormalizable. If you impose power-counting renormalizability, uniqueness is fairly ... 10 The difference between the two cases is the nature of the vacua. In the case of spontaneous symmetry breaking, you find that the tunneling amplitude between them is proportional to the volume, so that in the infinite volume limit of QFT there is no tunneling at all between the different sectors of these vacua - they are effectively superselection sectors. ... 9 OK, blathering it is. The key thing to understand in Yang-Mills theory (whether or not you're trying to be mathematically rigorous) is that gauge symmetry is not a physical symmetry. (Its job is to keep track of the overcounting in a redundant description.) Classically, this means that the observables -- the numerical quantities which we can measure -- ... 9 Consider the finite dimensional unitary representations$\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$.... 9 OP is considering Yang-Mills theory over a curved base space$(M,g)$. If the base space connection is the Levi-Civita connection$\nabla^{LC}=\partial+\Gamma$, then it doesn't matter whether one uses the gauge-covariant derivative$D=\partial+A$or the full covariant derivative$\nabla=D+\Gamma$since the Christoffel symbols$\Gammadrops out of the Yang-... 9 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 likeA_\mu\partial_\nu$. 9 The structure of standard model$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 ... 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 ... 9 Background. Indeed, the quark bilinears have an extra exact symmetry$U(1)_B$(baryon number) and the lepton ones an extra symmetry$U(1)_L$(lepton number). These two commute with each other and the 12 gauged ones of the SM. You are presumably asking why these are not gauged. In the SM, nonperturbative ("topological") EW anomalies break the$U(1)_{B+L}$... 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 $$[J^a_n,J^b_m]=i{f^{ab}}_cJ^c_{n+m}+mkd^{ab}\... 8 This is nicely answered in Jaffe-Witten's "problem description" of the problem of quantization of Yang-Mills theory: By the 1950s, when Yang–Mills theory was discovered, it was already known that the quantum version of Maxwell theory – known as Quantum Electrodynamics or QED – gives an extremely accurate account of electromagnetic fields and forces. ... 8 It's not particularly strange or unusual for a term to seemingly not appear in the Hamiltonian, but still have a physical effect. For example, the Hamiltonian for a free particle is$$H = \frac{p^2}{2m} = \frac12 m v^2, \quad p = mv.$$On the other hand, the Hamiltonian for a particle in a magnetic field is$$H = \frac{(p-eA)^2}{2m} = \frac12 m v^2, \quad p ...

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