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I) As a purist, I disprove of the common praxis to call the implication $$\tag{1} \{Q,H\}+\frac{\partial Q}{\partial t}~=~0 \qquad\Rightarrow\qquad \frac{dQ}{dt}~\approx~0.$$ for a 'Hamiltonian version of Noether's theorem', as explained in my Phys.SE answer here. Moreover, the implication (1) is not equivalent to the full Noether's theorem for various ...

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Whenever indices are repeated up and down and assumed to be summed over, they are always invariant under rotations because they are nothing but the scalar product between a vector and its dual (or more precisely, the action of a dual vector onto a vector). Such scalar product is invariant under orthogonal transformations, hence the eventual result will be, ...

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Actually there are analogies or generalisations of results which reduce to Noether's theorems under usual cases and which do hold for discrete (and not necesarily discretised) symmetries (including CPT-like symmetries) For example see: Anthony C L Ashton (2008) Conservation Laws and Non-Lie Symmetries for Linear PDEs, Journal of Nonlinear Mathematical ...

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I wanted to complement the answers above. For (1) $so(4) = so(3) \times so(3)$, one $so(3)$ is from the geometric 3D symmetry of the Hamiltonian, and the other $so(3)$ is from the potential term of $\frac{k}{r}$. For (2). the second $so(3)$ symmetry is a dynamic symmetry and only holds when potential term is inversely proportional to $r$. One has to do ...

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Your diagram is a representation of three dimensions on a two dimensional screen... so talking about "up" and "down" is pretty confusing. Think about it this way: if you look from the side, the coaxial cable appears to be two concentric circles right? From this perspective, the set up has ROTATIONAL symmetry(if you rotate the system around the axis, you ...

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This is just a property of Fourier transformations. If the correlation function is translational invariant then, by definition, the position space representation $D(x,y)$ transforms as $D(x+a,y+a) = D(x,y)$ for any constant $a$. Thus $D(x,y) = D(x-y,0)$ and so the correlator depends only on the difference $x-y$. For simplicity, we'll define $D(x-y) = ... 1 To start with what are identical to each other are the elementary particles of the standard model. of particle physics. When complex composites of these particles are built this complete identity starts differentiating. In interacting with each other quantum numbers enter and energy states. One proton may be indistinguishable from another proton , but a ... 1 I guess what is to be noted here is the fact that, the Correlation function (operator), commutes with the momentum operator, since $$[D,\text e^{ixp}] = 0 \implies [D,p] = 0$$ Having that to be the case, one can recollect that any operator represented in its own eigenspace is diagonal should answer your question. PS : I am not completely sure about the ... 0 To expand on the answer given by @By Symmetry, let the two point function be defined as $$f(x_1,x_2)=\langle\Omega|\, \phi(x_1)\phi(x_2)\,|\Omega\rangle.$$ In order the above to be translationally invariant one must require that $$\langle\Omega|\,\left[P, \phi(x_1)\phi(x_2)\right]\,|\Omega\rangle = 0$$ where$P$is the generator of the translations as ... 0 The fact that the system is translationally invariant implies that the translation operator commutes with the Hamiltonian. This implies that they have a basis of mutual eigenstates. Since the momentum operator generates the translations, i.e. $$T =e^{-\imath x p}$$ a state is an eigenstate of the translation operator if and only if it is an eigenstate of ... 3 First of all, my opinion is that the paper on your link is full of notational inconsistencies and therefore causes a great amount of confusion for someone who struggles to understand Noether’s theorem. So, allow me to formulate the field-theoretic version of Noether’s theorem in a more, according to my taste, charming way. Preliminaries 1: Lie Groups and ... 0 Your question is a bit fuzzy. I will do my best to answer and let you complain about what I did not understand properly. The references that I give are just papers that I know and like. There is much more to read on the topic. I will use the following sign convention for$\lambda$, $$\partial_t h + \frac{\lambda}{2}\left(\vec{\nabla}h\right)^2 = \nu ... 1 You may find the following paper useful: A Symbolic Solution of the Hubbard Model for Small Clusters, by J. Yepez. You may also want to review group theory for condensed matter physics, because your questions essentially span the basics of group and representation theory. Many texts give good overviews of the fundamentals of group theory as applied to ... 0 If I understand correctly, you're asking how to prove that symmetry of a tensor is coordinate independent, but you seem to be having trouble with the definition of a tensor. Well, you're not the first. Let me give you a definition that might help. First, suppose you have some space (it can be 3-space or spacetime or whatever) and you have a set of ... 1 To see how the components of the electromagnetic tensor transform from F to F' under a Lorentz transformation you take the tensor$$F^{\mu \nu}=\begin{pmatrix} 0 & -E_x /c & -E_y /c & - E_z /c\\ E_x /c & 0 & -B_z & B_y\\ E_y /c & B_z & 0 & -B_x\\ E_z /c & -B_y & B_x & 0 \end{pmatrix}$$and apply the ... 1 Our professor defined a rank (k,l) tensor as something that transforms like a tensor Oh dear. All too common, and all too pedagogically faulty. A tensor is nothing more or less than a linear map from (possibly multiple copies of) a vector space (and possibly copies of its dual space) into the scalar field. If I give you components T_{\mu\nu} (16 ... 3 A tensor is not particularly a concept related to relativity (see e.g. stress tensor), but is a more general concept that describes the linear relationships between objects, independent of the choice of coordinate system. This coordinate independence results in the transformation law you give where, \Lambda, is just the transformation between the ... 0 There certainly does seem to be confusion here. It is a question of symmetry, but not just the symmetry of the shape and Gaussian surface. It is electric charges that are responsible for electric fields, whether you are talking about insulators or conductors. You don't say, but I am assuming you are dealing with an insulator that has a certain amount of ... 1 There is no general algorithm for doing so, and even figuring out how many conserved quantities a system has can be difficult. A famous example is the Toda lattice, a system which was originally proposed by Toda in 1967 and was believed to be chaotic, but was in fact proven to be integrable (to have too many conserved quantities to be chaotic) in 1974 by ... 5 The submaximal dimension of the group of isometries of a Pseudo-Riemanniann manifold of dimension n with n\ge4 and n\neq5 is$$\frac{1}{2}n(n-1)+ 2 .$$However, a result proved here(Theorem 3.2) shows that a spacetimes with that amount of isometries in dimension$4\$ must be Minkoswki spacetime. Hence, the maximal number of Killing vectors you can ...

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