9

In the Lorenz gauge $\vec{\nabla}\cdot\vec{A}+\frac{1}{c^2}\frac{\partial V}{\partial t}=0$, the Maxwell equations in terms of the potentials are, for $V$: $$-\frac{1}{c^2}\frac{\partial^2V}{\partial t^2}+\nabla^2V=-\frac{\rho}{\epsilon_0},\tag{1}$$ where $\rho=\rho(\vec{x},t)$ is the charge distribution. The general solution to this is $$V(\vec{x},t)=V_0(\...


9

the time of the density corresponds to the scalar potential. This is not correct, precisely because of causality considerations. If the charge distribution is time-dependent, the potential at time $t$ will instead depend on the configuration at some retarded time, $t_r=t-\frac{|\boldsymbol{r}-\boldsymbol{r}'|}{c}$.


8

Not that much notation, but probably not less true either: $$[\partial_\mu+A_\mu,\partial_\nu+A_\nu]\psi=(\partial_\mu+A_\mu)(\partial_\nu+A_\nu)\psi-(\partial_\nu+A_\nu)(\partial_\mu+A_\mu)\psi=$$ $$=\partial_\mu\partial_\nu\psi+A_\mu\partial_\nu\psi+\partial_\mu (A_\nu\psi)+A_\mu A_\nu\psi-\partial_\nu\partial_\mu\psi-A_\nu\partial_\mu\psi-\partial_\nu (A_\...


6

You have touched on a point here which, in my experience, often gets glossed over, mentioned offhand, or mentioned indirectly (meaning it would be hard to realize it's the same point being mentioned without already knowing the answer) in first treatments of QFT. You may have heard it mentioned at some point that the modern perspective on gauge symmetry is ...


5

A gauge theory in the broadest sense is a theory with an action that has gauge symmetries parametrized by smooth functions of spacetime $\epsilon(x)$, i.e. the theory is invariant under infinitesimal changes of the coordinates/fields as $$\delta \phi = R\epsilon(x) + R^\mu \partial_\mu\epsilon(x) + R^{\mu\nu}\partial_\mu\partial_\nu \epsilon(x) + \dots$$ and ...


5

The Dirac operator $D$ is self adoint on a dense subset of the Hilbert space, it has a complete set of $V$-valued eigenfunctions $\varphi_n(x)$ with real eigenvalues $\lambda_n$. As a consequence one can define $\exp\{-tD^\dagger D\}$ on the entire Hilbert space by the integral kernel $$ K_t(x,x')=\exp\{-tD^\dagger D\}_{xx'}=\sum_n \varphi_n(x) e^{-t\...


4

As the three-form $(g^{-1} dg)^3$ is proportional to the Haar volume form on ${\rm SU}(2)$, and there is a tacit pull-back of this to $S^3$, the quantity $Q/24 \pi^2$ will be the integer winding number (Brouwer degree) of the map $g: S^3 \to {\rm SU}(2)\equiv S^3$.


4

Being a gauge theory and being an effective theory are completely orthogonal concepts. An effective quantum field theory is a theory that is only valid up to some energy scale. These theories are often non-renormalizable, but since they possess a physical meaningful energy cutoff, this is not a problem. Conversely, a (candidate for) a "fundamental" ...


4

I don't think this is true, $\pi_5(G)$ has little to do with anomalies, at least not in any direct way. The general statement is: triangle anomalies for a given symmetry $G$, in $d$ dimensions, are classified by the free part of $\Omega_{d+2}(G)$. Here $\Omega_n(G)$ denotes the cobordism group of $G$, namely the collection of $n$ dimensional manifolds $M_n$ ...


3

OK, I collect my comments since their obviousness is evidently not unassailable. You are exactly right that in my mind gauge and local are synonyms when used for symmetries, and "gauging" meant something like "localizing". ... "local symmetry", "gauge symmetry" and "local gauge symmetry" all mean the same ...


3

You cannot reuse a free index as a dummy index, so $\frac12h\eta_{\mu\nu}\equiv\frac12h^\alpha{}_\alpha\eta_{\mu\nu}\ne\frac12h_{\nu\mu}$ Verify for yourself that the residual gauge transformation acts on the polarisation tensor as $a_{\mu\nu}\to a_{\mu\nu}+i(k_\mu B_\nu+k_\nu B_\mu-k^\sigma B_\sigma\eta_{\mu\nu})$. Contracting, $$ \eta^{\mu\nu}a_{\mu\nu}\...


3

A lot of times in theoretical physics, there is a gap between the theoretical and mathematical framework, and the underlying physical assumptions. Before getting to gauge theory, let's consider the idea of curved spacetime (which is actually quite similar but involves a less abstract space so is perhaps a bit easier to think about). What physical motivation ...


3

The field strength $F$ is $\mathfrak{g}$-valued. In that case if $X^a$ is a basis of the Lie algebra we can write $$F=\dfrac{1}{2}F_{\mu\nu}^a dx^\mu\wedge dx^\nu \otimes X^a.$$ In particular, given any representation $R:G\to {\rm GL}(V)$ of $G$ on the vector space $V$, we have the derived representation $dR:\mathfrak{g}\to {\rm End}(V)$ of the Lie algebra, ...


2

The fundamental problem here is that many people, and also Pitts in his paper, are not careful about what theory they are currently talking about. "Quantization of Gauge Systems" by Henneaux and Teitelboim is actually careful about this, and their chapter 3 shows the correct resolution of this problem, even though Pitts cites it as an example for ...


2

Well you have seen the Faddeev–Popov method for gauge fields because it is designed to help with the path integral in a gauge theory; namely with summing over equivalent gauge configurations. If you don't have a gauge symmetry you don't have a problem to try to solve with Faddeev–Popov to begin with. Now, in your case you have a global $\mathrm{SO}(2)$ ...


2

There is no relation at all between the particle fields $\phi$ that are sections of associated bundles and the gauge field $A$. $A$ is the field of the gauge bosons, $\phi$ is a field of a particle charged under the interaction with that boson. The Standard Model does not contain particles charged in the adjoint of some gauge group, so there is no analogue ...


2

If you compacify to a torus then, in Cartesian coordinates, the spin connection vanishes and so is irrelevent. If you compactify to a sphere, as Fujikawa suggests, it is far less obvious that $\hat A(TM)$ is not needed. That it plays no role is because $\hat A(TM)$ is a genus and so cobordism invariant. This means that the curvature contribution to the ...


2

If you insist on not breaking one type of terms in the Hamiltonian - say, the plaquette constraints - then the remaining theory is a $Z_2$ gauge theory with the plaquette constraints as the gauge constraints. Thus, within this subspace where the plaquette constraints are not violated, the theory behaves just as a gauge theory. But this is no longer true when ...


2

The reason for this is that Gauß' law is not an equation of motion, but a constraint. The Lagrangian $$ L = -\frac{1}{4}\int F^{\mu\nu}F_{\mu\nu}\mathrm{d}^4 x $$ gives us the canonical momenta $\pi^\mu = F^{\mu0}$ and hence a primary constraint $\pi^0\approx 0$ ($\approx$ means an equality that hold on the constraint surface/upon use of the equations of ...


2

Sure, the Ward-Takashi identities hold for global symmetries too, although the trick of "promoting" a global symmetry to a local one (as is done for using Noether's theorem in a classical field theory) and dropping spacetime dependence right at the end is a quick and easy way to derive the identity for a given theory. $$ \delta\phi_i=\varepsilon_i \...


2

Classically speaking, gauge fields are connections on principle $G$ bundles. Matter fields, on the other hand, are sections of associated bundles. To construct an associated bundle, you must choose a representation of $G$. For instance, if $G = U(1)$, then this amounts to choosing an integer. When you construct the associated bundle, this integer is the ...


1

For adjoint representation $Z_N$ invariance, you demonstrated that for adjoint generator action $[Y,\phi]=\lambda \phi$, $$\phi= e^{2\pi iY}\phi e^{-2\pi i Y}=\phi+ (2\pi i)[Y,\phi]+\frac{1}{2!} (2\pi i)^2[Y,[Y,\phi]]+ ... =e^{2\pi i\lambda/N}\phi ,$$ so $\lambda$ must be an integer. So for all tensor products of this adjoint $\phi$, the over-all composite ...


1

I will merely sketch the derivation for the standard proof that the quotient of the weight lattice by the root lattice produces the centre of an (arbitrary semisimple Lie) group: prove that the centre of the group is the intersection of its maximal tori $\{T_i\}$ use the restriction of the exponential map $\exp: \mathfrak t\to T$ to define another abelian ...


1

A gauge transformation is any transformation on $A = (\phi, \vec{A})$ that does not change any physical observable, c.q. $F$. So any transformation $A \to A + d \lambda$ with $\lambda$ any scalar field. A gauge fix is, within this gauge freedom, a particular choice for $A$. This choice can be either complete ($A$ completely fixed, no further gauge ...


1

The first notation hides the Lie algebra generators by writing $A_\mu = A_m^a {\boldsymbol \lambda_a}$ where the implied generators obey $$ [{\boldsymbol \lambda}_a, {\boldsymbol \lambda}_b]= i {f_{ab}}^c {\boldsymbol \lambda}_c. $$ The second notation makes the generators explicit and writes $$ {\bf A}_\mu\cdot {\boldsymbol \lambda}. $$ For SU(2), ...


1

While writing the question, I found the answer. The part of $\delta S_{\mathrm{CS}}$ that should be made to vanish is: $$\delta S_{\mathrm{CS}} = 3\theta^M{f_{LM}}^IC_{IJK}\int A^L \wedge F^J \wedge F^K = 0$$ The wedge product on the right is symmetric in $J, K, L$ (since $F$ is a $4$ form), so the previous equation constrains only the part of ${f_{LM}}^IC_{...


1

You are probably confused because of the muddled notation such as $|\psi(x)\rangle$. Indeed I remember being confused when I first read Fujikawa's account. Let's do it with proper Dirac notation: Firstly the LHS $$ \sum_n \langle \psi_n(x)|O(\psi_n(x)\rangle $$ is to be interpreteted as $$ \sum_n\langle \psi_n|x\rangle \langle x|\hat O|\psi_n\rangle $$ ...


1

Let $P\to M$ be a principal $G$-bundle with a representation $\rho\colon G\to\mathrm{GL}(V)$. The "extra term" is the endomorphism $O\colon C^{\infty}(M,\Delta\otimes V)\to C^{\infty}(M,\Delta\otimes V)$ defined by \begin{equation} O(\psi)=[\gamma^k,\gamma^l]\cdot\{\nabla_k,\nabla_l\}(\psi)=\sum_{k,l}[\gamma^k,\gamma^l]\cdot\{\nabla_k,\nabla_l\}(\...


1

It seems you are essentially asking two questions which aren't really related. One of the two can be phrased as: given a Lie algebra-valued 1-form $A$ (whether it's a connection or not is irrelevant to this question), does there exist a spacetime vector field $v$ such that on some patch $U$ of spacetime we may obtain $A(v)=\phi$ for an arbitrary Lie algebra-...


1

The trace in those formulae is simply the trace over the group indices of the field strength tensor $F$, you are free to compute it in whichever basis you want, since the trace is an invariant, however the extra structure of an inner-product is not needed only matricial algebra. So once you have a representation $\rho$ and a basis, $\{T_a\}$ for the vector ...


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