New answers tagged

1

A general Chern-Simons theory with finite group $G$ is labeled by an element $\omega$ of $H^4[G, Z]$, which for finite or compact groups is equal to $H^3[G, U(1)]$. If $\omega$ is trivial, then we just have a pure gauge theory. The Hamiltonian version of a finite group $G$ gauge theory is equivalent to Kitaev's quantum double model with group $G$, if the ...


2

No, gauge symmetry cannot be "avoided" by not specifying coordinates. Tao's analogy is supposed to say that just as we are free to choose any coordinates we want for a manifold, we are free to choose any gauge (fixing) we want for our gauge theory. The fixing of a gauge resembles a choice of origin for a coordinate system, but it is not equivalent ...


1

Gauge invariance protects the W and Z boson masses even without supersymmetry. The W and Z bosons are the massive gauge bosons from spontaneously broken electroweak symmetry. Consider the simpler case of an Abelian gauge symmetry. In this case, the symmetry acts on the gauge boson as a shift: $A_\mu(x) \to A_\mu(x) + \partial_\mu f(x)$ for some function $f(x)...


3

In chapter 74 of Srednicki's QFT book (which the previous Phys.SE post refers to) it is implicitly assumed that in Euclidean 4D $x$-space the gauge field $A_{\mu}$ vanishes sufficiently fast as $|x|\to \infty$. In particular, large gauge transformations are implicitly excluded. In this case, the only harmonic function $\Lambda$ (i.e. such that $\Box\Lambda=0$...


0

This is certainly not a good enough answer but in here https://arxiv.org/abs/1906.08616 it is shown with examples that gauge transformations that don't vanish at the spatial boundaries have a non-trivial effect on the symplectic form. Accordingly, they alter the commutation relations and have a non-trivial effect on the Hilbert space.


0

Goggle "large gauge transformation". It's not the non-compactness but the non-trivial homotopy that matters.


6

In general the Legendre transformation$^1$ from the Lagrangian to the Hamiltonian formulation may be singular, which leads to primary constraints. This is e.g. the case for gauge theories like Yang-Mills (YM) theory with or without matter, which OP mentions. However, in case of a singular Legendre transformation, by performing a so-called Dirac-Bergmann ...


1

The set of matrices is a complex manifold, i.e. locally described by $\mathbb{C}^n$. Therefore one needs to check what holomorphicity means on $\mathbb{C}^n$. Indeed, consider a function $f:\mathbb{C}^n \rightarrow \mathbb{C}^m$ and pick a point $z_0 \in \mathbb{C}^n$. Then $f$ is complex differentiable at $z_0$ if there is a a linear map $L_z : \mathbb{C}^n ...


3

To show the two formulations are equivalent, note that in Coulomb gauge, $A^0=0$ so $E_i = \partial_i A_0 - \partial_t A_i = - \dot{A}_i$. Then using $p_i = m \dot{x}_i$, and defining $H_A$ to be the Hamilonian written in terms of $A$ and $H_E$ the Hamilontian written in terms of $E$, we have \begin{eqnarray} H_A &=& -\frac{e}{m}p^i A_i \\ &=&...


Top 50 recent answers are included