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There is no better definition than what Wikipedia offers - in general, a topological excitation is a (field) state, i.e. a localized quantity since fields depend on spacetime, whose integral is a topological invariant. One prime example are Yang-Mills theories in 4D, where the integral $\int \mathrm{Tr}(F\wedge F)$, as essentially the second Chern class of ...

6

The Lagrangian provided is Maxwell's Lagrangian, supplemented by a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The equations of motion are, $$\partial_\mu F^{\mu\nu} + \partial^\nu (\partial_\mu A^\mu) = \partial_\mu \partial^\mu A^\nu = 0$$ Instead of making a gauge fixing procedure a ...

4

The extra term, in general $$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{2 \xi}(\partial_{\rho}A^{\rho})^2$$ is called gauge fixing term. This term is needed in order to be able to quantize the field $A_\mu$. Without this extra term the photon propagator is ill defined $$D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu ... 2 Conservation of helicity requires that$$ \sum_{i=1}^n h_i = 0 $$in any amplitude. If the above equation is not satisfied, then conservation of helicity requires that the corresponding amplitude vanish. Maximally violating helicity conservation implies that the above sum take its maximum possible value with its corresponding amplitude being non-zero. Now, ... 6 In normal usage, a gauge is a particular choice, or specification, of vector and scalar potentials \mathbf A and \phi which will generate a given set of physical force fields \mathbf E and \mathbf B. More specifically, a physical situation is specified by the electric and magnetic fields, \mathbf E and \mathbf B. A set of potentials \mathbf A ... 2 Continuous symmetries of the action of a system which are global, that is, do not depend on where they act, give rise through Noether's theorem to conserved quantities. For example, a translation in time t \to t+\epsilon for \epsilon \in \mathbb{R} is a global transformation, and leads to energy conservation. On the other hand, if an action is invariant ... 1 Good to see you solved the paradox yourself! Another way to find the same result is to compute the number of orbits which is possible to stack in a surface equal to the area of the system. Let's first remind basic things on Landau levels. The hamiltonian of a particle moving in 2D \{x,y\}\equiv\{r,\theta\} plan through a static magnetic field reads :$$ ...

1

The problem with my thought is that I mix the "Canonical angular momentum" with the physical one. Since $\hat p_x,\hat p_y$ is canonical momentum, they are explicit gauge dependent. See What is canonical momentum? for a little reference. The $\hat l_z$ in my question admit no physical meaning, so the bounded condition for it is just invalid. The true ...

2

The trace is just the inner product for the Lie algebra. The field strengths are Lie algebra valued, i.e., $\mathbf{F}_{\mu\nu}$ is an element of the Lie algebra, and can be written as a linear combination of generators: $\mathbf{F}_{\mu\nu} = \sum_a F^a_{\mu\nu} t^a$. One usually normalizes the generators such that $\left \langle t^a, t^b\right\rangle = ... 1 Actually, they are one and the same thing. Before I delve into the question you asked, let me quickly describe a closely related analogy - the covariant derivative in GR. This is a quantity$\nabla_\mu$that acts differently on different objects. In particular$$\nabla_\mu \phi = \partial_\mu \phi,~~~ (\nabla_\mu V)_\nu = \partial_\mu V_\nu - ... 1 (I am dropping the bothersome factors of$\mathrm{i}$in this answer, they contribute nothing to understanding what is going on) The gauge covariant derivative exists for all forms on the spacetime manifold$\mathcal{M}$taking value in a representation of the gauge group. (Formally, these are sections of associated vector bundles to the gauge principal ... 3 There is no contradiction since you should not be doing complex linear combination of generators that are already hermitian (indeed, you want the group transformation to be unitary). Hence, your linear combination with a complex coefficient (that brings you away from the$SU(2)$group) doesn't imply a massless excitation. 0 After reading the section put forth by @gcsantucci I think I kind of understand what was happening, but I'm eager to hear feedback on this. If you break$ SU(2) $using a doublet you indeed break all the generators and get massive gauge bosons. What was confusing is that a linear combination of generators leaves the vacuum invariant (namely,$ \sigma _1 ...

0

I guess instead of me trying to explain, it is way better if you read section "Non Abelian Examples" of Peskin and Schroeder, in chapter 20. It explains exactly what you are asking for. It's actually the predecessor of the Standard model. Georgi and Glashow proposed this model before the SM, because they didn't know about the Z boson. So it is exactly what ...

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