# Tag Info

1

Can we simply comment that the $\text{Tr}[T^a_rT^b_r]\equiv C(r) \delta^{ab}$ depends on the representation. For the case of SU(2) and SO(3), we can relate this to the spin-S representation. By the manner that for SU(2) group is in a spin 1/2 representation and SO(3) group is in a spin 1 representation. One can write down the relation of spin operators as: ...

2

This vector potential can be written in every point on the plane except the origin as: $$A = d\phi$$ where $\phi$ is the polar angle ($\phi = \mathrm{tan}^{-1}\frac{y}{x}$). This does not mean that $A$ is exact, because $\phi$ is singular at the origin. But this means that the magnetic field is zero at every point except the origin. At the origin ...

2

It's just not true that local gauge symmetry implies causality. That's a false statement. And, from what I can tell, the textbook you reference does not make this assertion. It only says that local gauge symmetry preserves causality, meaning that the two concepts are compatible.

1

A rough sketch: Suppose you start with $A_1$ and $A_2$ being in some configuration. Then by means of a gauge transformation, you can gauge away one component, $A_1 \to A_1' = A_1 + h\, \partial_1 h^{-1}$ where $h^{-1} = \mathcal{P} e^{\int_{-\infty}^{x_1} dx_1 A_1}$ is the holonomy of $A_1$ along a line of constant $x_2$ The point is you can choose $h$ so ...

1

A quick answer, if I may. You need $\theta$ to be smooth since you want to derive it. So mathematics imposes you to choose $\theta$ smooth. Now the trick: choosing $\theta$ to be smooth means you can always impose $\mathbf{A}$ to be smooth, and use several patches related to each other by a gauge transform. Then you should always discuss smooth vector ...

2

I'm going to wave my hands wildly here, but I think the point is that you are treating the particle semi-classically, and the amplitude for going from point a to point b is equal to the 1-particle Feynman path integral, which is dominated by the classical action. The kinetic part contributes a simple phase, while the minimal coupling with the field $A$ ...

2

Again, thanks to the $SU(2)$ PSG proposed by prof.Wen, I can answer my question now, $THT^{-1}$ is in fact $SU(2)$ gauge equivalent to $H$, and the statement "$H$ is also not SU(2) gauge equivalent to the time-reversal transformed Hamiltonian $THT^{-1}$" in my question is wrong. Let's rewrite the Hamiltonian as ...

1

To test whether $|0\rangle$ is physical, you would apply $\partial^\mu A_\mu^+$ to it. So $|0\rangle \in V$. Note that 7.48 does not mean that $\partial_\mu A^\mu = 0$ as an operator identity in this space. It means that all it's matrix elements in this space are zero. You have noticed that the state you created, $\partial_\mu A^\mu |0\rangle$, lives not ...

2

I don't know the article you refer to, but I believe the Hamiltonian you discuss should get a $\pi$-phase shift after one turn around a (2D) lattice cell. So I guess it should read $H=F^{\dagger}\cdot H_{\pi}\cdot F$ with H_{\pi}=t\left(\begin{array}{cccc} 0 & e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4}\\ e^{-\mathbf{i}\pi/4} & 0 & ...

8

Comments to the question: First it should be stressed, as OP does, that the Euler-Lagrange equations (= classical equations of motion = Maxwell's equations) are unaffected by scaling the action $S[A]$ with an overall (non-zero) constant. So classically, one may choose any overall normalization that one would like. As Frederic Brünner mentions a ...

11

The factor is there so that once you add a source term, i.e. $J^\mu A_\mu,$ you get the correct equations of motion, namely Maxwell's equations: $\partial_\nu F^{\mu\nu}=J^\mu.$

Top 50 recent answers are included