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It is very important to distinguish whether the symmetry is broken explicitly or spontaneously. I think that the sentence "Now when I break this symmetry spontaneously (or explicitly)" indicates that its author isn't quite distinguishing these things. An explicit symmetry breaking generally lifts the degeneracy because the different parts of the multiplets ...

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It is easiest to directly derive the form of the vector fields from the boundary conditions for asymptotically flat spacetimes. See for example this paper http://arxiv.org/abs/1001.1541 or this one http://arxiv.org/abs/1106.0213 Infinitesimally diffeomorphisms act on the metric via lie derivative $\mathcal{L}_{\xi}g_{\mu \nu}$. In the case of BMS, you ...

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Other answers being good, i'll try to give a different perspective. What is a vector? As Feynman used to say ("Feynman lectures on physics") not every bunch of numbers (i.e $\left( a_1, a_2,..,a_n\right)$) makes a vector simply because it has $n$ components. Why? Because vectors have a specific relation (or more correctly transformational relation) with the ...

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All terms in a sum or both sides of an equality should be of the same kind, either vector or pseudo vector. Otherwise the expression will break reflection symmetry. This is a useful check of formulas and possible physical explains of different phenomena.

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First of all, I would suggest you to read the comments I have made in the Danu's answer to check whether I have understood your question or not. See, $\oint B\cdot d\ell~=~ \mu_0I$ has been derived only on the basis of $\vec{\nabla}\times \vec{B}=\mu_0 \vec{J}$. But actually the Maxwell equation is $\vec{\nabla}\times \vec{B}=\mu_0 (\vec{J}+\epsilon_0 ... 5 We use the idealized case of an infinitely long current to be able to justify (by symmetry) that the strength of the field will only depend on the radial coordinate$r$, so that it can be taken out of the integral, since we are only integrating over the angle which parametrizes a circle around the wire: $$\oint B\cdot d\ell = B \int_0^{2\pi}r\ d\theta = ... 2 Interesting difference for theoretical physics is that n-dimensional generalizations of quantities which are vectors have n components, while n-dimensional generalizations of quantities which are pseudovectors, such as angular momentum, have \frac{1}{2}n(n-1) components. This coincides for 3 dimensions, which is why the same vector notation is ... 2 (1) Since u(\textbf{r}) = u(\textbf{r}+\textbf{R}), we can expand this part in terms of reciprocal lattice vectors, u_k(\textbf{r}) = \sum_\textbf{G}{e^{i\textbf{G}\cdot \textbf{r}}u_\textbf{k-G}}. We can therefore write: \psi_{\textbf k+\textbf K} = e^{i(\textbf k + \textbf K)\cdot \textbf r}\sum_\textbf{G'}{e^{i\textbf{G'}\cdot ... 1 Tong is alluding to the standard trick in the derivation of Noether's theorem by promoting the (infinitesimal) x-independent parameter \epsilon to become x-dependent, see e.g. this Phys.SE post. 8 Not only can you do physics "in a mirror", but I've been part of an experiment involving exactly that. The weak interaction is, well, weak. And that makes it very hard to get access to in any physical process which can also proceed through other interactions. So, you can see the weak interaction at work in beta decay, but to leading order you can't see it ... 34 [Disclaimer: I'm not providing an argument where the distinction would be useful. I am providing an argument that pseudovectors and vectors describe intrinsically different geometrical concepts, and should, for clarify of argument, never be conflated just because they look so similar] The point is that pseudovectors, by their very nature, are not the same ... 0 Possibly (for this purpose) the simplest expression of the Wigner-Eckhart theorem in plain language is "what else could it be?" The angular motion of the nucleus is described by the spin. The spin operator is a vector. We need a second rank tensor for the quadrapole interaction. From the spin operator, you can only make one second rank traceless ... 0 Will attempt an answer. The theorem as you may know is based on representation theory. Representation theory for Lie groups plays an important part because it states that observables can be constructed from an algebra of generators of the group. The angular momentum operators are the generators of the spherical group (if i may say) So each angular ... 0 For each continuous symmetry, infinitesimal transformations may be expressed, by a bracket involving the conserved charge operator associated to the symmetry :$$\delta_\epsilon \phi(x) = i\epsilon [Q, \phi(x)] \tag{1}$$In our case, we must have :$$\epsilon \theta = i\epsilon [Q_\theta, \phi(x)] \tag{2}$$A solution is then :$$Q_\theta = \theta ... 0$T, W, P\toT, O, IC_n \to C_nD'_n \to D_n$1 The Lie algebras and$\mathfrak{so}(3)\mathrm{su}(2)$are isomorphic, but the Lie groups$\mathrm{SO}(3)$and$\mathrm{SU}(2)$are not. In fact$\mathrm{SU}(2)$is the double cover of$\mathrm{SO}(3)$; there is a 2-1 homomorphism from the former to the latter. How is this possible when every so(3) irrep can get raised to one of SO(3) as described ... 5 You already received several answers. However the fundamental physical reason is elementary: In classical, quantum and relativistic physics the physical laws describing an isolated physical system in an inertial reference frame are the same (are invariant) if you rotate (with an element of$SO(3)$) the system (there are many other symmetries depending on the ... 2 One can make sense of the introduction of$\mathrm{SO}(3)$into quantum mechanics as follows: Consider a physical system in three spatial dimensions which we'll think of as residing in a box sitting on a table, like a table-top experiment. Suppose that we prepare the system in a particular way, so that the system is measured to be in a (pure) state ... 0 The very important group in physics is the Lorentz group$SO(3,1)$. It is built from the$SO(3)$group and Lorentz boosts. The algebra of the Lorentz group generators (boosts$L_{i}$and 3-rotations$R_{i}$) doesn't have the separation on$SO(3)$and boosts' parts. But by introducing the "new" generators$J^{i}_{\pm} = \frac{1}{2}(R_{i} \pm iL_{i})$we may ... 1 Spin in (non-relativistic) QM is fairly ad hoc, it has no deep reason. The underlying reason is that, in a relativistic setting for QM/QFT our states/fields must transform in some representation of the Lorentz group$\mathrm{SO}(1,3)\$ if we want the amplitudes/Lagrangians to be Lorentz invariant (since something cannot be invariant if there's no rule how its ...

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