# Tag Info

7

You are right, it is wrong to think that in gauge theory "gauge transformations are just a redundancy". This becomes true only if one abandons locality, ignores all boundary effects, all instanton effects, hence most of what is interesting about gauge theory. Of course forming gauge equivalence classes (say of observables) is something one wants to do every ...

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$SU(N)$ has $N^2 -1$ gluons, one corresponding to each generator (this is called the adjoint representation). That "quantum number" (which could be called an index for the adjoint rep) could be tensored together with the Lorentz representation for a massless vector particle, which should have 2 polarizations (just like QED). So the theory will have $2(N^2 ... 4 To be honest, I think that the route you describe (and which is also used in many textbooks) is not physically well motivated at all. You have begun with a theory of a fermion with a global symmetry which maps physical states to different physical states. This theory has the property that specifying initial conditions on a spacelike surface completely ... 4 A good analogy for the difference between the two can be given in terms of two other examples of anomalies, that are possibly more familiar. Consider a field theory with a global symmetry, take$U(1)$for simplicity. At the classical level, the equations of motion lead to the existence of a conserved current (Noether's theorem). At the quantum level, the ... 4 The reason that the gauge particle must be a spin 1 gauge boson is because there aren't any renormalizable alternatives. To see this consider the Dirac Lagrangian: $$\bar{\psi} i \gamma ^\mu \partial _\mu \psi$$ This term is not gauge invariant under the transformation,$ \psi \rightarrow e ^{ i T ^a \theta ^a (x) } \psi $, ... 3 Yes, there is a physical significance. The longitudinal mode$A^0$is pure gauge, it does not propagate (in other words, the equation of motion for$A^0$is a constraint [Gauss Law], not an equation of motion and it's canonical momenta is identically 0 , meaning we cannot impose canonical commutation relations on it). Some of the spatial modes do propagate, ... 3 Again assuming it only has a zero$x^i=x_0^i$what you have is $$\delta(f(x^i)) = \frac{\delta(x^1-x_0^1)}{\left|\frac{\partial f}{\partial x^1}\right|_{x^i=x_0^i}} \frac{\delta(x^2-x_0^2)}{\left|\frac{\partial f}{\partial x^2}\right|_{x^i=x_0^i}}\cdots \frac{\delta(x^n-x_0^n)}{\left|\frac{\partial f}{\partial x^n}\right|_{x^i=x_0^i}} = \prod_{j=1}^n ... 3 Since I am new here and I cannot comment above I just write my comment as a small answer. I think in your question there is a small mistake. You ask how many degrees of freedom are there in a Yang-Mills theory although you want to ask how many polarizations does the gluon have. First of all, the number of generators of the gauge group will be the number of ... 3 (1) Classifying "Phase Structure of (Quantum) Gauge Theory" (with a gap) is about the same as classifying phase structure of topologically ordered states. Some topologically ordered states are described by a group and can be related to a gauge theory. Some other topologically ordered states are not related to gauge theory. (2) One way to classify "Phase ... 2 You should also specify the Representation. The Representation requires SU(N) Lie group with N×N matrix is called Fundamental Representation. Which is used in Standard model U(1) x SU(2) x SU(3). You can surely have SU(N) Lie groups with other Representation, such as Adjoint Representation, then in this case SU(N) are represented by a matrix with a rank of ... 2 I) The un-gauge-fixed QED Lagrangian density reads$$\tag{1} {\cal L}_0~:=~-\frac{1}{4}F_{\mu\nu}^2 + \bar{\psi}(iD\!\!\!\!/ \ \ -m)\psi.$$The gauge-fixed QED Lagrangian density in the R_{\xi}-gauge reads$$\tag{2} {\cal L}~=~ {\cal L}_0 +{\cal L}_{FP}-\frac{1}{2\xi}\chi^2 , $$where the Faddeev-Popov term is$$\tag{3} {\cal L}_{FP}~=~ ... 2 I have to admit that I have no idea about the model you are working on, but the standard way to determine whether a gauge theory is confining or not is to calculate the vacuum expectation value expectation value of Wilson loops. The latter are gauge invariant operators that describe parallel transport around a closed loop in spacetime. If the vacuum ... 2 Under a Wick rotation, which is what you do in order to go from Minkowski to Euclidean space, both the partial derivative$\partial_0$with respect to time and the zero-component of the gauge field transform as $$\partial_0\rightarrow i\partial_\tau$$ $$A_0\rightarrow iA_0.$$ This defines the covariant derivative for statistical field theory. 2 Leibniz rule holds for covariant derivatives, both in gauge theories and gravity. Mathematically, a derivation is one for which the Leibniz rule holds. How does it work for non-abelian covariant derivatives. I will give you an example. Let$\Phi^\dagger \Phi\$ be invariant under local non-abelian gauge transformations. Then \partial_\mu (\Phi^\dagger ... 2 You are always allowed to introduce a new integration variable as long as its not its already being summed over. This might be more clear in discrete form: \begin{align} \int d x \, f (x) & \rightarrow \Delta x\sum _i \,f ( x _i ) \\ & = \big( N \Delta y\sum _j g ( y _j ) \big) \Delta x \sum _i f ( x _i ) \\ \end{align} where  N \Delta ... 2 The lagrangian of the gauge field is independent of that of the scalar field. You have to "guess" it. The reason we pick this one is twofold: 1-it is the one which gives the Maxwell equations, so when you try to describe E&M, that looks like a good guess; 2- if you think of all the terms that are both Lorentz invariant, parity invariant and gauge ... 2 The notation $$\frac{ \partial f_i}{ \partial x ^i }$$ means the diagonal elements of the matrix: $$J _{ ij} = \frac{ \partial f _i }{ \partial x ^j }$$ where f_i is the component of the vector \vec{f} (x). I found this very confusing a few weeks ago so. Here is the proof I wrote up for the ... 1 This is how I understand this issue. First, I believe you may agree that imposing gauge invariance is a sensible thing to do. If we want our fields to be invariant under some kind of transformation it better be local, since two separate space-time points shouldn't be related in any unnecessary way, otherwise we may violate causality. A different issue is ... 1 What confused me was the explanation from the tangentbundle homepage (second yellow box in OP). The generalization is straightforward, for simple zeros we have:\bigg\vert\frac{\mathrm{d}f(x)}{\mathrm{d}x}\bigg\vert_{x_0}\delta[f(x)] = \delta(x-x_0)$$integrate$$\int \mathrm{d}x\,\bigg\vert\frac{\mathrm{d}f(x)}{\mathrm{d}x}\bigg\vert_{x_0}\delta[f(x)] ...

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The statement you cited does not imply that a complex representation of a gauge group implies a chiral gauge theory in general. This only holds true if the gauge group corresponds to a chiral symmetry in the first place. A chirally symmetric theory contains massless fermions. Regarding your counterexample: it is true that QCD contains fermions in the ...

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