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| visits | member for | 2 years, 2 months |
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| stats | profile views | 896 |
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Aug 4 |
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Discrete gauge theories In the case of continuous principal bundles, the bundle automorphism group is an extension of the base space diffeomorphism group by the group of pure gauge transformations. A semidirect product is only a special case of such an extension. Going to the discrete case, one would also expect more general extensions than the semidirect product. Since nontrivial extensions (i.e., non-semidirect product extensions) exist for finite groups, I wonder why these cases are excluded here. Do you have a reference where it is stated that the full gauge group "must" be a semidirect product. |
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Aug 3 |
answered | Can Fermionic symmetries be fully integrated into geometric deformation complexes or symplectic reduction? |
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Aug 1 |
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What is a basis for the Hilbert space of a 1-D scattering state? Overcompleteness is a well defined mathematical concept. en.wikipedia.org/wiki/Overcompleteness I don't think that I have seen use of the terminology of a spanning set in conjunction with resolution of the unity. May be you would prefer the term tight frame defined in the Wikipedia page (which is true of course for the coherent states) |
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Aug 1 |
answered | What is a basis for the Hilbert space of a 1-D scattering state? |
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Jul 28 |
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Finding the spectrum of a polynomial of the creation and annihilation operators @Piotr - sorry for the error, of course the dimension is equal to the number of ordered partitions of $n$ into at most $d$ pieces, or equivalently the dimension of the fully symmetric $n$-tensorial representation of $SU(d)$, which can be calculated for example by using the hook length formula. By the way I am trying to think occasionally on your interesting suggestion in your last comment, but I haven't reached an answer yet. |
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Jul 28 |
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Finding the spectrum of a polynomial of the creation and annihilation operators @Qmechanic - corrected thanks. |
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Jul 28 |
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Finding the spectrum of a polynomial of the creation and annihilation operators added 12 characters in body |
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Jul 26 |
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Proving that interval preserving transformations are linear The form uniqueness of the solution is dictated by the linearity of the solutions of the equation of motion of a free particle. Nevertheless, I'll try to find a more rigorous proof of the uniqueness of the separation of variables solution. |
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Jul 26 |
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Proving that interval preserving transformations are linear @Qmechanic Of course the solution is not unique because you need boundary conditions to make it unique, but the solution form (Linear in the Cartesian coordinates) is unique. The solution of the four Hamilton-Jacobi equations on the diagonal will be with different constant coefficients which have to further satisfy the of-diagonal relations to be matrix coefficients of a Lorentz transformation. I gave a reasoning that the Hamilton-Jacobi phase function is the phase of a plane wave solution of the Klein-Gordon equation. |
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Jul 25 |
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Proving that interval preserving transformations are linear I have added an update containing the required elaboration and references. I also corrected an error: Each diagonal term of the matrix equation is equivalent to a Hamilton-Jacobi equation (no need to take the trace). |
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Jul 25 |
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Proving that interval preserving transformations are linear added 2522 characters in body |
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Jul 24 |
answered | Proving that interval preserving transformations are linear |
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Jul 21 |
answered | Why muonium is unstable? |
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Jul 21 |
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Poincare group vs Galilean group The answer to your second question. You can talk about isomorphism of different group actions, even if they correspond to different representation. Group isomorphism means that the group action multiplication table is the same. In our example, the Poincare group acts on the Minkowski space irreducibly by 4 dimensional non-unitary matrices, the action on the free particle Hilbert space(of wavefunctions) is irreducible unitary and infinite dimensional. The action on the Poisson algebra of the phase space is not even irreducible, nevertheless all actions are isomorphic. |
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Jul 21 |
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Poincare group vs Galilean group The answer to your first question is no, if you carefully perform the Wigner-Inonu contraction, you will lose the right hand side momentum-boost commutator. You will retrieve this term (proportional to the mass), only upon quantization, or classical realization by means of Poisson brackets. In an answer to a related question by Arun Nanduri, I described a limiting process that the central extension term can be obtained as a nonrelativistic limit, this derivation is not correct as a part of the Wigner-Inonu contraction which when done correctly forces the commutator to be zero. |
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Jul 21 |
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Operator relation involving the logarithm of an operator? Since $[q,p] = i\hbar$ while $[p,q] = -i\hbar$, you just have to interchange $p$ and $q$ and change the sign of $\hbar$. |
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Jul 21 |
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What conservation law corresponds to Lorentz boosts? To be precise, the conserved quantity for a system of free relativistic particles is the position (not the velocity) of the center of mass, which can be obtained by the application of the Noether theorem. |
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Jul 21 |
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Operator relation involving the logarithm of an operator? I understand that you wat to express $ \ln(q) \ln(p)$, such that all dependence on q is moved to the right. I added an new update in which a close form is technically obtained. |
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Jul 21 |
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Operator relation involving the logarithm of an operator? added 1345 characters in body |
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Jul 20 |
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Poincare group vs Galilean group No, all three Poincare group actions are isomorphic. Here no extra central extension is produced because for the Poincare group $H^2(G, U(1))$ is trivial. The Heuristic way to see that is to see that the Poincare transformation of the relativistic particle Lagrnagian given for example in:en.wikipedia.org/wiki/… does not produce a total derivative. |
