# Tag Info

## New answers tagged gauge-theory

0

There is no sacred difference between the way a gauge boson and a Goldstone boson couple to a matter field. Goldstone bosons are just ordinary bosons that are massless due to the breaking of a global symmetry. To avoid Lorentz violating these have to be scalar fields so that does turn out to restrict the possible coupling that they could have, but the fact ...

2

Let us start with a simple analogy that helps grasping the concept. Say, we need to calculate a electric field of a charged cylinder. What is the first thing you do? You choose your $z$-axis along the axis of the cylinder. Why won't you choose your axes differently, like pointing whatever they want? Well, you could, but that will make the problem ...

6

The advantage of unitary gauge is that it completely removes unphysical fields, while adding additional degrees of freedom to the gauge bosons, which consequently become massive. This gauge works well for tree-level calculations, but complications arise when considering loops: The propagators of gauge fields and ghosts (which are needed to impose the ...

1

A small comment first: usually people call this theory as a Chern-Simons theory in (2+1)d, while the BF theory usually refers to a similar theory in (3+1)d. But anyways, this naming is not important. The U(1) Chern-Simons theory in (2+1)d is always formulated in the following general form $$S=\int\frac{\mathrm{i}}{4\pi}K^{IJ} a^I\wedge \mathrm{d}a^J.$$ The ...

0

The Lagrangian must be a gauge invariant and Lorentz invariant object that can be integrated over the entire spacetime $\Sigma$. So, we must first obtain an $n$-form (for $n$ the dimension of spacetime), and all that we have for that is the gauge field $A$, which itself transforms in an ugly way under gauge transformation. The only object we can build out ...

1

$R$ Parity is a discrete $Z_2$ symmetry while an $R$ symmetry is a global continuous symmetry. If you use a $Z_2$ symmetry to build your model then each field can just be either odd and even, that's it. If you impose a continuous symmetry then there are an infinite number of possible choices of $R$ charges. From a model building perspective, a continuous ...

2

No, the gauge current need not be gauge invariant, since it carries a group index in non-abelian theories. You should recall that both sides of the Yang-Mills equation (and therefore the current itself) are Lie-algebra valued and therefore transform in the adjoint representation. Not even the field strength $F^a_{\mu\nu}$ is gauge invariant, but transforms ...

2

Analyzing the spectrum of the strings, one finds that it contains $N^2$ massless vector states, which is precisely the number of gauge fields corresponding to a $U(N)$ group. Note that this is only true for massless oriented open strings; the unoriented case yields $SO(N)$ or $Sp(N)$. As is described in the same chapter of the book, open string states can ...

1

It seems OP's main question concerns the systematics of gauge-fixing. We interpret/reformulate OP's questions as essentially the following. The original gauge-invariant action $S_0$ is unsuitable for quantization, so we add a non-gauge invariant gauge-fixing term to the action. Obviously we cannot add any non-gauge invariant term to the action. ...

4

It's Stokes's theorem. Consider a field $F = dA + A \wedge A$ such that $A$ is pure gauge at infinity, that is, $\lim_{x\to\infty} A(x) = \omega\, d \omega^{-1}$ for some $\omega : S^3 \to SU(2) \sim S^3$ where $\omega$ is a function on the 3-sphere because the limit can depend on the direction out to infinity. In differential forms the first expression is ...

2

In general, the covariant derivative acting upon a field in a representation $\rho : G \to V_\rho$, is given by $$D = \mathrm{d} - \mathrm{i} g \rho(A) \wedge$$ or, in coordinates/generators, indeed $$D_\mu = \partial_\mu - \mathrm{i} g A_\mu^a \rho(T^a)$$ where $\rho(T^a)$ are the generators of $G$ in the chosen representation. Now, for the triplet ...

3

Comments to the question (v3): In contrast to QED with fermionic matter, in QED with bosonic matter, the full Noether current ${\cal J}^{\mu}$ (for global gauge transformations) tends to depend explicitly on the gauge potential $A^{\mu}$, see e.g. Refs. 1-2 and this Phys.SE post. The reason for this difference is because the QED Lagrangian for fermionic ...

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