New answers tagged gauge-theory
0
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Field strength tensor written as commutator of covariant derivatives in QED
Your mistake is after the third line of your equations.
Bear in mind that the operators in $(...)$ act on the $\psi$ on the right,
and then apply the product rule carefully.
$$\begin{align}
&... \\...
- 40.1k
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Field strength tensor written as commutator of covariant derivatives in QED
Just use the product rule of differentiation to evaluate commutators:
$$
[\partial_\mu,f(x)]=\partial_\mu f(x)-f(x)\partial_\mu=(\partial_\mu f(x))+f(x)\partial_\mu-f(x)\partial_\mu=(\partial_\mu f(x))...
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Field strength tensor written as commutator of covariant derivatives in QED
The derivative operators in the second commutator in the second line of Eq. 1 1/2 also act on the wave function via the chain rule. This cancels the third commutator.
- 25.3k
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Independence of $S$-matrix of $\xi$-gauge in QED
As far as I know, the rigorous proof of the LSZ formula fails in QED because of the photon cloud. Nevertheless, it still seems to work! We just have to physically assume that at $t\to\pm\infty$ the ...
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How to expand $(D_\mu\Phi)^\dagger(D^\mu\Phi)$ in $SU(2)$?
You appear to not appreciate your expression as a row vector dotted on a column vector (possibly sandwiching operators). I corrected your expression to
$$\partial_\mu\Phi^\dagger \partial^\mu \Phi - \...
- 64.2k
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Gauge theory of Electomagnetic Potentials - 2nd order derivatives
The Gauge fixing condition must be (1) accessible (This is addressed in Gabriel's answer), and (2) it must uniquely fix the gauge. In E&M one way to see that you get the gauge degree of freedom is ...
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Gauge theory of Electomagnetic Potentials - 2nd order derivatives
This is only a partial answer to your question.
What you say about X and Y being linear is not necessarily true, as I have seen other guage choices like the Weul gauge, $A_t=0$, used in QFT, or the ...
- 1,611
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Gauge transformation rule for $dA$, where $A$ is the gauge field
Given
$$
A \to g A g^{-1} - i dg g^{-1}
$$
where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$ and $d$ is exterior derivative, thus
$$
dA\\
\to d(g A g^{-1} - i dg g^{-1}) \\...
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What kind of object is a function in the context of gauge theory?
The terminology used in physics texts is often a bit imprecise, and thus the word "scalar" is somewhat ambiguous. Often if a function $\phi(x)$ of the coordinates is a "scalar", it ...
- 11.3k
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vote
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Writing gauge transformation of the gauge fields explicitly in terms of coordinates
Assume a field $\psi(x)$ transforming with respect to some arbitrary representation $$U(x)=e^{-i \alpha_a(x) T_a}, \qquad [T_a,T_b]= i f_{abc}T_c$$ of your gauge group. Defining the matrix-valued ...
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Writing gauge transformation of the gauge fields explicitly in terms of coordinates
Your derivative term should be
$$
(\partial_\mu g) g^{-1}.
$$
This combination is an element of the Lie algebra,and so expressible as a sum of the $t^\alpha$, unlike the group element $g$ itself.
It'...
- 54.5k
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Geometrical interpretation of gauge fields of spin other than 2
``Geometry'' is a very vague term. The usual gauge theories (Yang-Mills theory) is about connections on vector bundles, see https://en.wikipedia.org/wiki/Gauge_theory_(mathematics). If you include ...
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2+1-dimensional $SU(N)$ Yang-Mills Theory
From the point of view of perturbation theory, QCD in lower dimensions is essentially identical to QCD in $d=3+1$. Namely, the beta function is negative, which means that the theory is UV complete, ...
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Causality for gauge dependent operators in quantum field theories
Correct. As an example of how causality cannot be imposed on the electromagnetic potentials, consider the Coulomb gauge. In this gauge the propagation is instantaneous.
- 25.3k
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Why are there no Goldstone modes in superconductor?
A neutral fermionic superluid has gapless soundwaves that can be regarded as a Goldstone mode due to the spontaneous breakdown of translational symmetry, but the fluctuating quantity in the sound ...
- 54.5k
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Why are there no Goldstone modes in superconductor?
The Goldstone modes of a superconductor (considered as an electron gas with attractive potential) are physically the oscillations of electron density. As electrons are intrinsically charged, this will ...
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2+1-dimensional $SU(N)$ Yang-Mills Theory
Two-dimensional Yang-Mills is solvable non-perturbatively. Atiyah and Bott used equivariant cohomology to study the theory. The theory is topological and in the small-coupling regime it is classically ...
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2+1-dimensional $SU(N)$ Yang-Mills Theory
I don't know much about the 3-dimensional case however, I can maybe provide some insight into the 1+1 dimensional version. In 1+1 dimensions the number of gauge constraints is high enough to go far ...
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How to find a covariant gauge derivative from a field transformation
I'll give the example for SU(N):
Consider a field $\Phi$ in the fundamental representation of SU(N) which means that $\Phi \to U \Phi$, with $U \in SU(N)$. Now consider how its ordinary derivative ...
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