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11

The simplest example in condensed matter physics that spontaneously breaks time reversal symmetry is a ferromagnet. Because spins (angular momentum) change sign under time reversal, the spontaneous magnetization in the ferromagnet breaks the symmetry. This is a macroscopic example. The chiral spin liquid (Wen-Wilczek-Zee) mentioned in the question is a ...


10

If you made the most perfect cone possible, so that its tip was a single atom, and stood it on the most perfect surface possible (a perfectly smooth, perfectly hard sheet of atoms), and completely removed all forces other than gravity, it would still topple. This is because those atoms are all jiggling around due to thermal motion. This effect fundamentally ...


7

Actually, mass and charge are only superficially similar. Yes, they both appear in inverse square force laws, namely Newton's law of gravitation and Coulomb's law of electrostatic force, but both of those are approximations. Coulomb's law ignores quantum effects, which is a very slight approximation, but Newton's law ignores all of relativity, which makes a ...


7

No, the photon is not "the non-chiral piece" of $SU(2) \times U(1)_Y$ before symmetry breaking. The photon is the $SU(2) \times U(1)_Y$ gauge boson that is invariant under $Q_{\rm elmg}$, the electric charge, which is given by $$ Q_{\rm elmg} = \frac{Y}{2} + T_3 $$ where the first term is the hypercharge, the generator of $U(1)_Y$, and the second term is the ...


7

You may notice that the equations don't pass the test of dimensional analysis. Some factors are missing. However, let me answer your question: The reason why the acceleration never exceeds $g$ is that the dome is actually finite, it is truncated at the bottom. For too high values of $r$, your initial formula for $h(r)$ will actually exceed $r$ itself, and ...


7

In non-relativistic systems both $E\sim k$ and $E\sim k^2$ are possible. Quadratic dispersion relations occur if $\langle 0|[Q_i,Q_j]|\rangle\neq 0$ for some of the generators. This occurs in a ferromagnet because rotational invariance is broken and $J_z$ has an expectation value. In terms of effective lagrangians the difference between ferromagnets and ...


7

The possibility of spontaneous Lorentz symmetry violation due to the infrared problem of the Dirac-Maxwell equation was conjectured a long time ago by Frohlich, Morchio and Strocchi, in references [1,2] mentioned in the given Balachandran and Vaidya article. In perturbative QED, we usually assume that the scattering states are free eigenstates of the number ...


6

A quick answer: "screening" currents in the superconductor are proportional to the vector potential. With an appropriate choice of gauge, the screening current appears as a mass term in the wave equation for the vector potential. From "An Informal Introduction to Gauge Field Theories": (This excerpt from Google books)


6

No, it doesn't work like that. The Higgs boson doesn't complete a set of particles that we had some theoretical reason to expect to exist. (Other particles have been predicted in roughly that way, e.g. the charm and top quarks.) So in the sense I believe you're thinking about it, physicists had no reason to predict the existence of the Higgs boson. Where it ...


5

First of all, the soft SUSY-breaking terms are just an "effective" description that replaces lots of qualitative, unknown physics by 100+ parameters for the known physics. At the end, one wants to construct a full theory. For example (an important example), if the full theory is a stabilized string theory compactification, there aren't any undetermined ...


5

I just discovered this very interesting website through Prof Wen's homepage. Thanks Prof Wen for the very interesting question. Here is my tentative "answer": The spontaneous symmetry breaking in the ground state of a quantum system can be defined as the long range entanglement between any two far-separated points in this system, in any ground state that ...


5

... I was a proponent of M-brane theory before many physicists took string theory seriously... M-theory is not the statement that strings have thickness. The brane theory is interpreting the black hole solutions of supergravity within string theory, and giving different "infinitesimally thin" descriptions of gravity which all are equivalent. In ...


5

As is easily checked, fields linear in creation and annihilation operators (and hence amenable to a particle interpretation) have zero vacuum expectation value. Thus the $\phi$ field with its nonvanishing vacuum expectation value cannot be given a particle interpretation. But the field $\psi=\phi-v$ has such an interpretation as its vacuum expectation value ...


4

This question posted by Prof. Wen is so profound that I had hasitated to response. However motivated by Jimmy's insightful answer, I eventually decided to join the discussion, and share my immature ideas. 1) Quantum SSB is a non-linear quantum dynamics beyond the description of Schordinger's equation. Regarding the transverse field Ising model mentioned in ...


4

Dear D-brane, yes, in principle, one could have theories that spontaneously break the Lorentz symmetry. Just add a vector field (or another non-scalar field) and some potential of the form $$(V_\mu V^\mu - v^2)^2 $$ which will drive $V_\mu$ towards a vector of the right length, i.e. $(v,0,0,0)$ which would pick a preferred reference frame at each point much ...


4

Once again, I am way out of my league in answering this. I may be wrong about many things here, comments appreciated That was just a definition of mass. The Higgs explains where rest mass (but not gravity) comes from in a mathematically rigorous manner. One of the attempts to explain how our universe works in a mathematically rigorous manner is the ...


4

In addition to Lubos Motl's correct answer, I would like to make two comments related to Norton's dome: First a brief derivation of the equation of motion. I prefer to call the (non-negative) arc length $r$ for $s$, and the height $h$ for $z$. Like Lubos Motl, I will introduce a proportionality factor $K$ for dimensional reasons, so that the equation for ...


4

If you look at the diagrams of what they actually created, essentially it consists of several ions maintaining consistent spacing moving in a superconducting ring. As long as the temperature of the superconductor is maintained, the persistent rotational motion will be maintained. Since no work is being extracted this can be understood as a constant entropy ...


4

By the "noncompact $U(1)$ group", we mean a group that is isomorphic to $({\mathbb R},+)$. In other words, the elements of $U(1)$ are formally $\exp(i\phi)$ but the identification $\phi\sim \phi+2\pi k$ isn't imposed. When it's not imposed, it also means that the dual variable ("momentum") to $\phi$, the charge, isn't quantized. One may allow fields with ...


4

In simple language we do not have a Theory Of Everything (TOE) therefore any answer about the ultimate existence of specific laws is a tentative one. What we do have is a set of nested mathematical theories that fit observations mainly in the study of particle physics. These theories extrapolated to the extremely high energies at the beginning of the Big ...


3

Not quite. The Higgs mechanism actually applies at low energies. Don't think of it as an event that happens once and bestows mass upon all particles for the rest of time; instead, the Higgs mechanism is a continuous effect that explains how particles are able to have mass at low energies. For a full(er) explanation, I'll point you to another answer of mine, ...


3

The Higgs gives the W's, Z's, quarks, leptons, and (probably) the neutrinos a mass, but it doesn't have hardly anything to do with the proton mass, it only contributes the tiny amount roughly equal to the proton neutron mass difference. Its own mass is fundamental parameter in the standard model, and the natural value is the Planck mass. Physicists call the ...


3

The mass of the Higgs boson is a free parameter of the standard model and not (only) due to the interactions with a non-zero Higgs field. If the Higgs field were zero, the standard model predicts four massive Higgs bosons, which are the only massive particles. In case of a non-zero Higgs field, only one of them gains some extra mass via Higgs field ...


3

You have to read a bit about the Higgs mechanism. In particle physics, the Higgs mechanism (also called the Brout–Englert–Higgs mechanism, Englert–Brout–Higgs–Guralnik–Hagen–Kibble mechanism,1 and Anderson–Higgs mechanism) is the process that gives mass to elementary particles. The particles gain mass by interacting with the Higgs field that permeates ...


3

This is a good example of when the theoretical rubber meets the proverbial experimental road. The two issues you bring up are actually completely independent. The chemical potential problem is purely kinematic, and may be solved by simply introducing a harmonic trapping potential or any other way to modify the density of states. Mermin-Wagner is more ...


3

1) The axial vector current $j^{\mu 5}$ is a pseudovector $$j^{\mu 5}~:=~\overline{\psi}\gamma^{\mu}\gamma^5\psi~=~j^{\mu}_R-j^{\mu}_L,\qquad j^{\mu}_{R,L}~:=~ \overline{\psi}_{R,L}\gamma^{\mu}\psi_{R,L}, $$ $$\psi_{R,L}~:=~P_{R,L}\psi,\qquad P_{R,L} ~:=~\frac{1\pm\gamma^5}{2} . $$ The $4$-divergence $d_{\mu}j^{\mu 5}$ is a pseudoscalar. That the axial ...


3

The simplest $SU(5)$ GUT Higgs transforms as ${\bf 10}$ under the gauge group, an antisymmetric tensor $5\times 4/2\times 1$ with two indices of the same kind (without complex conjugation). The 2-dimensional representation of $SU(2)$ has an antisymmetric invariant $\epsilon_{ab}$ and if you extend this antisymmetric tensor to 5-valued indices of $SU(5)$ and ...


3

A key difference between spontaneously broken symmetries and "emergent symmetries" is that emergent symmetries are never exact while spontaneously broken symmetries are backed by exact maths although the ground state isn't invariant. In most cases, the "emergent symmetries" only emerge if some parameters are fine-tuned, and even if it is so, they are only ...


3

My understanding is that as you state, these states do not belong to the same hilbert space but are formally connected by the symmetry transformation. The essence is that the degeneracy of the vaccum is expressed in the notion that there are multiple equally well suited states (from separate hilbert spaces) from which one can build up excited states. Nature ...


3

Quantum field theories usually have a unique ground state – by the ground state, I mean the Hamiltonian eigenstate corresponding to the lowest energy eigenvalue. This may be demonstrated in various ways, depending on the assumptions we're allowed to make. For example, if a quantum field theory is a free field theory, the ground state may be constructed ...



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