# Tag Info

13

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 ...

12

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 ...

12

A Goldstone boson is a generic type of particle formed when symmetries are spontaneously broken. If you want to suggest that dark matter is a Goldstone boson then that says very little unless you suggest a specific model with a symmetry to be broken. When exact symmetries are broken you get a massless Goldstone boson (except in a few special circustances, ...

10

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 ...

10

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 ...

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 ...

10

Nice question! The short answer is that the group is not $SU(2)\times U(1)$, it is $SU(2)_L \times U(1)_{em}$. In other words the two groups act on different standard model particles differently. For example the left handed neutrino does interact weakly and so transforms under the $SU(2)_L$, but is electrically neutral so it doesn't transform under the ...

9

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 ...

8

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 ...

8

In theories with spontaneous symmetry breaking, the phase transition can usually be characterized by a local order parameter $\Delta(x)$, which is not invariant under the relevant symmetry group $G$ of the Hamiltonian. The expectation value of this field has to be zero outside the ordered phase $\langle\Delta(x)\rangle = 0$, but non-zero in the phase ...

8

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 ...

8

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 ...

8

I'll state one version of the theorem, valid for classical systems. I'll not give the most general framework, as things become messy, but this should still give you an idea of how general the result is. We need the following ingredients: Spins: to each vertex of the lattice $\mathbb{Z}^2$, we attach a spin $\phi_x$ taking values in some compact ...

8

In vacuum and with only the particles we know about the answer is no. Let's look at the symmetries we know exist in nature: $SU(3)$ colour: confined, only colourless states exist below the QCD phase transition $SU(2)\times U(1)_Y$ electroweak: Higgsed to $U(1)_{EM}$ electromagnetism $U(1)_{EM}$: Here we have opportunity. See below... $U(1)_{B-L}$: Global ...

7

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 Norton's equation of motion (5). I prefer to call the (non-negative) arc length $r$ for $s$, and the vertical height $h$ for $z$. Like Lubos Motl, I will introduce a proportionality factor $K$ for dimensional reasons, so that ...

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

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

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)

7

1） Gauge theory is a theory where we use more than one label to label the same quantum state. 2） Gauge “symmetry” is not a symmetry and can never be broken. This notion of gauge theory is quite unconventional, but true. When two different quantum states $|a\rangle$ and $|b\rangle$ (i.e. $\langle a|b\rangle=0$) have the same properties, we say that there ...

6

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 ...

6

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 ...

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 ...

6

There would probably be no life if the photon were massive because electromagnetism as the long-range force we know would be replaced by short-range forces only. They're too weak when distances are significant so in such a world, almost all interactions would only proceed by direct collisions – like in nuclear physics in the real world. One can't really ...

5

AFAIR it has two massless modes, as there are no quadratic terms around the minimum.

5

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 ...

5

No, I believe the Standard Model does not predict monopoles as a result of symmetry breaking. This is because the symmetry breaking $\mathrm{SU(2)} \times \mathrm{U(1)} \rightarrow \mathrm{U(1)}$ does not allow for topological solitons to exist. Edit: $\pi_2(\mathrm{SU(2)} \times \mathrm{U(1)}/\mathrm{U(1))}=\pi_2(S^3)=0$ Source: To be or not to be? ...

5

It's simple. The dilaton-axion (complexified) field in supergravity (and similar classical theories with a noncompact symmetry) is invariant under $SL(2,R)$ transformations $$\tau \to \frac{a\tau+b}{c\tau+d}, \quad ad-bc=1$$ However, the same transformation must also transform the charges of objects. For example, one-dimensional branes always carry the ...

5

The worldsheet Weyl anomaly in bosonic string theory is an example. More generally in any dimension you can have trace and Weyl anomalies that break scale or conformal invariance, even in systems with only bosons.

5

Actually we have the following Lie algebra isomorphism $$u(1)\oplus su(2)\cong u(2),$$ and there exists the following Lie group isomorphism $$[U(1)\times SU(2)]/\mathbb{Z}_2 ~\cong~ U(2).$$ In other words, there is a two-to-one map between $U(1)\times SU(2)$ and $U(2)$. So in that sense the Glashow-Salam-Weinberg $U(1)\times SU(2)$ model already contains ...

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