# Tag Info

9

The original papers by Gerard 't Hooft himself are quite readable. On the Phase Transition Towards Permanent Quark Confinement A Property of Electric and Magnetic Flux in Nonabelian Gauge Theories Topology of the Gauge Condition and New Confinement Phases in Nonabelian Gauge Theories Whenever I open these papers, I'm always awestruck.

7

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 $V=V_1\otimes V_2 \otimes V_3$....

4

It is simple to describe mathematically. First I will recall what the equation for the Aharonov-Bohm phase means, and then I will describe (without proof) its relationship to parallel transport for $G$-bundles, which I define. The gauge potential $A$ is a connection on some principal $G$-bundle, where $G$ is the gauge group. Principal $G$-bundles over a ...

4

For SU(N), the adjoint representation can be obtained from a tensor product of the fundamental representation with its dual and projecting out the scalar. Thus we can replace the adjoint representation indices $a, ...$ by double indices $i \bar{j}$ and write this relation as: $$W^{i \bar{j}}_{k \bar{l}} = U^i_k U^{\dagger j}_l - \frac{1}{N} \delta^i_k \... 3 There exists an extensive literature for discretization of the abelian and the non-abelian gauge theories, known as lattice QED and lattice QCD, respectively. Here we will only sketch the main idea. Let us for simplicity use Euclidean signature (+,+,+,+). A small Wilson-loop$$\tag{1} W~=~{\rm Tr}{\cal P}e^{ig\int_{\gamma}A}$$lies approximately in a ... 3 If you put a non-Abelian anyon and its anti particle on a sphere, then moving the non-Abelian anyon around its anti particle only induces an Abelian phase. Also, twisting a non-Abelian anyon by 360^\circ only induces an Abelian phase as well, which define the (fractional) spin of the non-Abelian anyon. 3 Non trivial holonomies have been proposed for quantum computation, see this article http://arxiv.org/pdf/quant-ph/0007110v2.pdf The basic idea is this: Suppose you have a sistem prepared in the ground state of an Hamiltonian H(\lambda), where \lambda is a set of parameters. If you slowly change this parameters the state evolves remaining in the ground ... 2 Possible answer to your first bullet: Stokes theorem. If G=U(1), then F=dA (not dA+[A,A]) and so \int_{\partial D}A=\int_{D}dA=\int_D F. Further, if memory serves, U(1) is the only Abelian Lie algebra (thinking of the Cartan classification) so G is Abelian means G=U(1) anyway. EDIT: Coordinate expression for the Holonomy. Hol(A,C) is defined ... 2 1) Let us write the Wilson-line of a simple open curve \gamma: [s_i,s_f]\to \mathbb{R}^4 as$$\tag{1} U(s_f,s_i) ~=~ \mathcal{P}\exp \left[ i\int_{\gamma} A_{\mu}~ dx^{\mu} \right]. $$2) The path-ordering \mathcal{P} becomes important if the gauge potential$$\tag{2}A_{\mu}~=~A^a_{\mu} T_a$$is non-abelian. Here T_a are the generators of the ... 2 The Aharonov-Bohm effect for nonAbelian gauge fields is subtle, even the definition of flux and charge is more complicated than Abelian cases. Both charge and flux can be nonAbelian. A flux is defined as a conjugation class of the gauge group G, and a charge is a (irreducible) representation of (subgroup of) the gauge group. However, (in 2D) a general ... 2 I) We start with a non-Abelian Wilson line ^1$$\tag{7.1} \Psi(C)~:=~ P e^{\int_{C} \!A} $$over a parametrized (possibly open) curve C. Here P denotes path-ordering. II) We now make an infinitesimal variation of the curve C to a new curve C^{\prime}. We may define an infinitesimally thin 2-surface \Sigma with oriented boundary ^2$$\...

2

I'm going to wave my hands wildly here, but I think the point is that you are treating the particle semi-classically, and the amplitude for going from point a to point b is equal to the 1-particle Feynman path integral, which is dominated by the classical action. The kinetic part contributes a simple phase, while the minimal coupling with the field $A$ ...

2

Perhaps the resources you have seen are confusing "force" with "energy", which is a common misunderstanding that frequently leads to mistakes in terminology. Energy must be applied to overcome the color force and increase the distance between quarks. The formation of a new quark/anti-quark pair when the gluon field becomes too stretched may provide an ...

2

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 ...

1

This just classical mechanics. We have a Lagrangian ${\cal L}(U,\partial_x U)$. Since the Lagrangian does not depend on time (here: $x$), the Hamiltonian $${\cal H} = \frac{\partial {\cal L}}{\partial (\partial_x U)} (\partial _x U) -{\cal L}$$ is conserved. Presto.

1

On the same page of the paper, $\mathcal{N}$ is defined as a connection on a bundle $\sigma^* E$. The claim that its BRST variation is $$\delta_{\text{BRST}} \,\mathcal{N} = \mathrm{d} \alpha + [\mathcal{N},\alpha],~~~ (\alpha = \omega \eta + T),$$ means that $\delta_\text{BRST} \mathcal{N}$ is just a gauge transformation of $\mathcal{N}$, $\delta_\alpha \... 1 If the comparator was not unitary, you could not expand it in terms of Hermitean generators of$SU(2)$, which is required in order to construct the non-Abelian covariant derivative, as it is done in Peskin and Schroeder. 1 Edit: Sorry, my first answer was not the one expected. I let it below the correct answer I detail now: It's always tricky to discuss exponentials of operators. The good thing is that it's always the same trick: you use$e^{A}\approx1+A$in both directions, valid for small$A$. To get exact results you also use that$e^{At}=\prod_{i}\left(1+A\Delta t_{i}\...

1

I just found that the solution of $SU(2)$ matrices is really simple. When there is no hopping term, the projected spin state of the above $d+id$ mean-field Hamiltonian indeed has TR symmetry. Because there exist global $SU(2)$ matrices $G_i$ which implement the TR transformation, say $G_i=i\tau ^x$.

1

Consider the non-abelian phase factor around a closed path $C$, $$\psi(C) = \mathrm{P} e^{\oint A_\mu dx^\mu} = \mathrm{P} e^{\int_0^{2\pi} dt \, A_\nu(x(t)) \dot{x}^\nu(t) }$$ Let us take the functional derivative with respect to $x^\mu(s)$ \begin{align} \frac{\delta}{\delta x^\mu(s)} \psi(C) %& = \int_0^{2\pi} dt \, \...

Only top voted, non community-wiki answers of a minimum length are eligible