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1d
comment If a symmetry operator S in a QFT annihilates the vacuum, why does S preserve the space of 1-particle states?
@Void 2) Assuming Lorentz symmetry, the renormalization factors Z in ψ R =Z(ψ)ψ are Lorentz scalars, thus nothing essential changes in the analysis when the true renormalized field is used: Its transformation properties remain the same.
1d
comment If a symmetry operator S in a QFT annihilates the vacuum, why does S preserve the space of 1-particle states?
@Void 1) Please think for a moment of $Q$ as the electric charge operator, you can write it uzing the Gauss' law as a surface integral of the electric field over a very large sphere at infinity. Due to the large distance these fields do not produce singularities when multiplied by other fields, thus the only singualrities coming from the commutator are those due to the local field $\psi$, thus the commutator itself is also a local field at $x$.
May
14
revised If a symmetry operator S in a QFT annihilates the vacuum, why does S preserve the space of 1-particle states?
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May
14
answered If a symmetry operator S in a QFT annihilates the vacuum, why does S preserve the space of 1-particle states?
May
5
comment The $U(1)$ charge of a representation
@JakobH The generator of the U(1) charge $Y_{\gamma}$ needs by definition to commute with all root generators $E_{\gamma}$ of the unremoved nodes. The Cartan-Weyl generator $H_i$ corresponding to the removed node does not possess this property, but its dual (called a coweight) does. The duality transformation can be accomplished on weight space by means of the metric tensor.
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revised The Aharonov-Bohm effect is purely classical, right?
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Mar
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comment The Aharonov-Bohm effect is purely classical, right?
@levitopher you can obtain the semiclassical aproximation (including the A-B effect) from the path integral, but Tuyman's theory requires even less structure than needed for the path integral. He does not require the symplectic form to be integral, thus does not impose the Dirac's quantization condition.
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answered The Aharonov-Bohm effect is purely classical, right?
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awarded  Good Question
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comment What are orbifolds and why are they useful and interesting for physics?
@Siva This observation is based on the solution of Schrödinger equation on a two dimensional cone where the energy eigenfunctions become more concentrated around the tip as the cone's half angle becomes smaller.
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awarded  Good Answer
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comment Aharonov-Bohm Effect and Flux Quantization in superconductors
They showed that the Shrödinger wave function of a single particle acquires a phase under a translation and a boost and this is alright, but, under a sequence of transformations: translation, boost, reverse translation, reverse boost, the overall phase does not vanish even though we returned to the initial frame of reference. This is a multivalued function related to the nontrivial central extension of the Galilean group.