# Tag Info

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You are right that the argumentation for the existence of ortho and para hydrogen is identical to that for the existence of ortho and para helium, that is, applying the permutation operator on a wave function for two identical particles should result in an eigenvalue of $+1$ or $-1$ if one considers bosons or fermions, respectively. Whereas in helium you ...

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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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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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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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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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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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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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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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In the absence of Yukawa couplings (only kinetic terms), the SM has the global flavor symmetry: $$G_{y=0} = U(N_f)^5=U(3)^5$$ Because there are 5 distinct representations in the SM (3 for quarks: $u_R$, $d_R$, $Q_L$; and 2 for leptons: $e_R$, $L_L$). However, $U(N) \sim SU(N)\times U(1)$, so the group can also be written as: $$G_{y=0} = SU(3)^5 \times ... 0 I often see \mathrm{SU}{(3)}_\text{flavor}. However, I have seen$$\mathrm{U}(3)_\mathrm L \times \mathrm{U}(3)_\mathrm R = \mathrm{SU}(3)_\mathrm L \times \mathrm{SU}(3)_\mathrm R \times \mathrm{U}(1)_\text{vector} \times \mathrm{U}(1)_\text{axial} where the last one is broken by the quantum anomaly. See slide 14 in this lecture summary of Theoretical ...

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A very late answer, but for symmetry-protected topological (SPT) phases, I believe it is true (certainly, no counterexamples are known) that the boundary is "non-trivial" if and only if the bulk is a non-trivial SPT phase. Here "non-trivial" boundary has a very specific meaning. A boundary is "non-trivial" if there is NO symmetry-respecting terms that we can ...

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It helps to remember that invariant quantities are seen as scalars to the transformation (they have no indices in the target space). In the other hand, covariant quantities are objects that transform in a certain way. Example: Vectors in $R^{2}$, under rotation $R_{ij}$, transform covariantly since $v'_{i}=R_{ij}v_{j}$, but it's length is invariant since ...

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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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