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Hot answers tagged symmetry

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The question that you have asked have some vague arguments as well as some partially true facts regarding Standard Model (SM). First, Yes SM describes physics up to some energy scale which is 14 TeV. On the other hand, if we accept Plank energy ($~10^{18}$GeV) as a fundamental energy scale, then we can possibly expect new beyond the SM energy scale. A ...

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If you define $C:q\rightarrow -q$, $P:(x,y,z)\rightarrow (-x,-y,-z)$ and $T:t\rightarrow -t$, then all the Maxwell eaquations are invariant under $C$, $P$, $T$ or any combination of them. To see this you just have to notice how the transformations act on sources and coordinates. The charge conjugation acts non trivially as \begin{align} C\rho&=-\rho,\\ ...

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This is a very broad question, so there are many ways to answer it. Here is one interpretation. A principal distinction between gauge symmetries and global symmetries is that gauge symmetries lead to long-range interactions between charged particles; the gauge symmetry demands the existence of a massless field which can propagate over arbitrarily long ...

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The first answer to such a question must always be: A gauge symmetry has no "physical" meaning, it is an artifact of our choice for the coordinates/fields with which we describe the system (cf. Gauge symmetry is not a symmetry?, What is the importance of vector potential not being unique?, "Quantization of gauge systems" by Henneaux and Teitelboim). Any ...

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You can only distinguish the sublattices in this case because you've tagged them A,B. The process of inversion only exchanges identical carbon with carbon, leaving the crystal physically unchanged. If you gave me a crystal with one orientation and I then returned it to you without telling you whether or not it's been inverted, you'd have no way of knowing. ...

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The definition of a spin liquid as a spin system "with no spontaneously broken symmetries" is out of date and no longer used, partially for the reason you describe. If you perturb as spin-liquid Hamiltonian by adding small terms that break all the symmetries, then the ground state will still be a spin liquid even though there are no longer any symmetries ...

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You are correct that Goldstone's theorem does not apply to the 1-D Heisenberg chain, or indeed to any local 1-D system, because there is no continuous symmetry breaking in 1-D for local Hamiltonians. But Goldstone modes are not the only possible kinds of gapless excitations - you don't need continuous SSB for a system to be gapless. There are plenty of ...

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For any crystal, the First Brillouin Zone is found using the Wigner-Seitz construction for the reciprocal lattice. The high-symmetry points are labeled by certain letters mainly as a convention--like you said Gamma for (0,0,0) etc. The important thing to realize as far as the group theory, is that the group of the wavevector at the Gamma point has the full ...

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