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1 vote

Why semi-simple and compact Gauge Group in YM Theory?

The compact criterion comes from the fact that the kinetic term showing up in the lagrangian is actually, strictly defined via a associative bilinear form (often the killing form). But for the theory ...
mika's user avatar
  • 56
1 vote
Accepted

Gravitational waves from metric perturbation

The reason a gauge is chosen is so that the equations of motion can be simplified and adapted to the problem at hand. Gauge transformations in this context are the same as coordinate transformations. ...
Vincent Thacker's user avatar
2 votes

The Abelian versus the non-Abelian commutator of covariant derivatives in field theory

In the case of a non-abelian gauge group, $A^\mu(x) := A^\mu_a(x) T_a$ (summation convention used!) denotes a matrix-valued field, where the matrices $T_a$ form a basis of some (faithful) ...
Hyperon's user avatar
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1 vote

Why did Peter Higgs, et al. suspect that the $W$ boson(s) had mass? All the way back in 1963, '64?

In 1961, Sheldon Lee Glashow already published his paper with the correct EW symmetry group SU(2)$\otimes$U(1) and hypothesized the W and Z bosons. The question, traced back to Julian Schwinger (...
Jon's user avatar
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0 votes

Renormalizability of Quantum Gravity

Well, Griffiths is seemingly equating gauge theory with Yang-Mills theory, but there are many other types of gauge theories, cf. e.g. this and this Phys.SE posts. Concerning non-renormalizability of ...
Qmechanic's user avatar
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1 vote

Proof of Batalin-Fradkin-Vilkovisky (BFV) theorem using BRST operator and graded Poisson bracket algebra

OP's calculation looks overall correct, except perhaps for the notion of the superdeterminant/Berezinian and supertrace. For comparison and clarity, let us provide a complete calculation, which ...
Qmechanic's user avatar
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7 votes

Why did Peter Higgs, et al. suspect that the $W$ boson(s) had mass? All the way back in 1963, '64?

Already in 1957, Julian Schwinger, Annals of Physics 2 (5) November 1957, pp 407-434, "A theory of the fundamental interactions", had inferred from the chiral structure and puny strength ...
Cosmas Zachos's user avatar
4 votes

Why the expectation value of three currents is important in the anomaly?

The expectation value of three currents is important because it's the first nontrivial expectation value in the computation of the anomaly, and it determines the whole thing. Here's an illustration ...
John Dougherty's user avatar
1 vote

How are the gauge transformations of $\epsilon(\mu)$ and $A^\mu$ related?

One issue is that you consider $A^\mu$ to be an operator, i.e. it includes creation and annihilation operators, while $\alpha (x)$ is just a function. This would imply the transformation $A^\mu \...
yaron kedem's user avatar
1 vote

What is the relation between gauge field and Levi-Civita connection?

Both are connections on a principal fibre bundle, but on different principal fibre bundles. For Yang--Mills theory with structure group $G$, there is a principal $G$-bundle $\pi:P\rightarrow M$ over ...
Bence Racskó's user avatar
1 vote
Accepted

Gauge transformation with harmonic one-form

You can add any vector field whose curl is zero (any one-form whose differential is zero). If you are working in a simply connected domain (no holes) then these vector fields can only be the gradients ...
Jahan Claes's user avatar
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1 vote

Gauge transformation with harmonic one-form

Harmonic forms are solutions of the Laplace equation $$ (\delta d +d\delta)\psi=0. $$ In infinite flat space a harmonic form would be a $C_mdx^\mu$, where the $C_\mu$ are numerically constant. With, ...
mike stone's user avatar
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1 vote

Is there an argument for using the $\theta$-vacuum for a Yang-Mills theory that works regardless of the presence of fermions?

The first argument you give, which you consider not complete as it only works in the semi-classical approach can be shown for more general cases. This is explained in Section 93 of Sredniki, as stated ...
Gabriel Ybarra Marcaida's user avatar
0 votes

Propagator in massive QED/Schwinger model

Inserting unity $\gamma_5^2=1$, we will obtain $\mathrm{det}(i\not D+m)=\mathrm{det}(\gamma_5(i\not D+m)\gamma_5) $. Therefore, the fermion determinant can be rewritten as follows $$\mathrm{det}(i\not ...
Siam's user avatar
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1 vote
Accepted

Can we prove in general that gauge fields associated with broken generators form representations of the unbroken group?

I fail to imagine the contrapositive. Assuming there are no matter fields to be integrated out in an effective theory (in a rearrangement spoiling renormalizability), in a pure gauge theory of G, all ...
Cosmas Zachos's user avatar
0 votes

What defines a large gauge transformation, really?

If you want a rather physical argument as to why you should have the extra condition $$ U(x) \rightarrow I \; \text{for} \; x \rightarrow \infty, $$ we refer to such a gauge transformation which ...
NoName's user avatar
  • 43
0 votes

Review: If true, what makes the vacuum of a local ${\rm U(1)}$ gauge theory unique?

I want to present a "counter"example to Ruben's answer --- a state that preserves local gauge symmetry but breaks the global symmetry. The issue is rooted in the thermodynamic limit. For ...
Yuan Yao's user avatar

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