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Essentially you are asking why only scalars are allowed to develop a vacuum expectation value (VEV). A scalar (as the name suggests) does not point to any direction -it has spin 0- therefore it can have a VEV without breaking the Lorentz symmetry. On the other hand, a boson with higher spin, e.g. a vector (spin 1) would spontaneously break Lorentz by ...


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In order for a theory to present stable monopole solutions it has to satisfy three requirements: i) It has to have the topological conditions, generally showed as non trivial second homotopy group of the vacuum manifold. ii) It has to satisfy a quantization condition $$e^{ieQ_m}=\mathbb 1,$$ where $Q_m$ is the (non-Abelian) magnetic charge. This is a ...


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Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras. Therefore, the generators are not a subset of the original's group generators. That is not quite right. A special subalgebra is one such that their step operators do not form a subset of the algebra step operators.That is what is ...


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I think the question was already answered here, look at either my answer of the one by Adam. The punch line is that that the expectation value of a field (such as the field $\phi$ at the bottom of the mexican hat potential) is fixed by the way the source $j$ that couples to $\phi$ in the path-integral for the functional generator is sent to zero. As there ...


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Many answers discuss the transition from the unstable state to the stable one. Let me thus discuss the issue of choosing the ground state itself. I will suppose a two dimensional Mexican hat potential. As Numrock realised, it has degeneracy. There is nothing which lift this degeneracy in principle. Then you can change the ground state without energy. This ...


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In relativistic QFT, this cannot be a process in time. The unstable initial state does not exist at all: An unstable ground state is impossible in relativistic QFT at temperature T=0 (i.e., the textbook theory in which scattering calculations are done) since it would be a tachyonic state with imaginary mass, while the Kallen-Lehmann formulas require $m^2\ge ...


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When we say that a lattice has a particular symmetry we mean that the lattice is mapped onto itself by the symmetry. So if I have a (2d) material which has inversion symmetry in the bulk and which has an atom at a point $(x,y)$ then inversion symmetry tells me that there is another, identical atom at $(-x, -y)$. At the surface, however, this is no longer ...


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In fact in some sense the full state of the quantum system shouldn't break any symmetry. It's just that the overlap of the different vacuum states (in field theory) vanishes. So in other words the full quantum state is a superposition of all the different vacua, but when we make observations we "collapse the wave function" (feel free to insert your ...


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Physically, you look for the ground state of your theory in order to make a correct predictive calculus around the ground state. The "random" choice of the ground state depends on the dimension of the theory, in fact, you can obtain a "hat" shape or a simple 2-dimensional shape with just 2 possibilities for the ground state (look the scalar quantum ...


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Naively, suppose you have a field F in your theory. If you look for the mass term of the field F in the Lagrangian it appears like -L $\supset$ $m^2$ $\bar{F}F$, which is invariant under the charge operator C, i.e., -$L^c$ $\supset$ $m^2$ $F^\dagger$$C^\dagger$$\gamma^0$$C$$F$ = $m^2$ $F^\dagger$$\gamma^0$$F$ =$m^2$ $\bar{F}F$ = -L, which means that the same ...


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To a given Lie algebra $\mathfrak g$ there is a unique group $\tilde G$, called the universal covering group, with the property of being simply connected. For example, the covering group of the algebra $\mathfrak{su}(2)$ is $SU(2)$. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way ...


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The vector $(0,0,0,v)$ is left invariant by the set of matrices of the form \begin{align*} M=\begin{bmatrix} R & \vec 0 \\ \vec 0^T & 1\end{bmatrix} \end{align*} where $\det(M)=\det(R)=1$ and $M^{-1}=M^T$ implies $R^{-1}=R^T$. By definition, $SO(3)$ is the group of 3 by 3 orthogonal matrices with determinant 1. In general, you need to know the Lie ...


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One can avoid the concept of symmetry breaking in this context, to avoid "non-conservation of the particle number". People have devised way to do that, see for example http://arxiv.org/pdf/cond-mat/0105058v1.pdf. However, all these approaches gives the same results than standard Bogoliubov-like methods in the thermodynamic limit. This is not too ...



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