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To be honest, much of this feels like very irresponsible journalism, partly on the part of the BBC and very much so on the part of Science alert.
If you're looking for an accessible resource to what the paper does, the cover piece on APS Physics and the phys.org piece are much more sedate and, I think, much more commensurate with what's actually reported ...
45
I'll give a very qualitative answer / overview.
The classification 'first-order phase transition vs. second-order phase transition' is an old one, now replaced by the classification 'first-order phase transition vs. continuous phase transition'. The difference is that the latter includes divergences in 2nd derivatives of $F$ and above - so to answer your ...
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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 ...
32
The articles are a little on the hysterical side, but I think they are just saying that violation of CP-symmetry means there must be violation of T-symmetry.
T-symmetry means that physical laws are unchanged if we reverse the direction time flows. Classical theories obey T-symmetry, and it seems intuitively obvious that quantum mechanics would as well. But ...
28
In most of the textbooks discussing this point, you should find something like : superconductors breaks the U(1)-gauge symmetry down to $\mathbb{Z}_{2}$. Fine, but what does it mean ?
To explain it, let me be a bit outside the main stream discussion. What I'll discuss below is more a personal reflexion than something clearly stated in any book.
Clearly, ...
26
It is frequently stated the Higgs mechanism involves spontaneous breaking of the gauge symmetry. This is, however, entirely wrong. In fact, gauge symmetries cannot be spontaneously broken.
A standard argument for this is that gauge symmetries are not actual symmetries, they are just a
reflection of a redundancy in our description the system; two states ...
24
The key idea is that time crystals are externally driven at a certain frequency, but they respond at a different (in fact, slower) frequency.
First of all, terminology:
what does it have in common with the usual concept of crystals, whose whole structure can be represented by spatially replicating the unit cell. Is the time crystal an addendum to the ...
24
In the case of electroweak force and electromagnetism there is an Higgs mechanism, which makes the $W^{\pm}, Z$ bosons massive, and preserves the photon $\gamma$ massless. But the symmetry relating the electric and magnetic fields is actually a Lorentz symmetry, which is global and remains unbroken. The difference between electric and magnetic fields emerges ...
answered Oct 23 '20 at 22:06
spiridon_the_sun_rotator
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23
Uniform bodies are idealizations like frictionless surfaces or no air resistance. They make the work easier. In reality there will be slight deviations in material properties (such as density, tensile strength, etc.) along various parts of the body. In your example these deviations become more and more important and extreme as the body is stretched (which ...
22
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 ...
20
In short: The spontaneous breaking of global U(1) symmetry, rather than local 'gauge symmetry', gives rise to the non-zero vacuum expectation value of Higgs field. This non-zero VEV is an essential part of the Higgs mechanism, which describes how the Higgs field gives mass to other particles, and its value is proportional to the generated mass.
To study the ...
answered Jun 22 '15 at 7:32
an offer can't refuse
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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 ...
16
In systems that break continuous translation symmetry (regular crystals,lamellar solids, smectics etc.) one also necessarily breaks rotational invariance (but note that this is not true the other way around, allowing liquid crystals to exist). As you rightly pointed out the simple goldstone mode count does not seem to work in such ordered phases.
The reason ...
16
No. Separating electric fields from magnetic fields would require the breaking of local Lorentz symmetry, not some gauge symmetry, and there is no reason why that is going to happen.
15
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 ...
15
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 $\...
15
Gapless just means that the energy of the excitation goes to zero as its momentum goes to zero. That is, with an infinitesimal (arbitrarily small) amount of energy you can always excite the system (with an excitation with very low momentum). If the excitation is gapped, you need a finite amount of energy to excite the system.
For instance, a free particle ...
15
I think you have understood it almost well.
The masses do not change, they are what they are; at least at colliders. At high energy, it is true that the impact of masses and, more generally, of any soft term, becomes negligible. The theory for $E\gg v$ becomes very well described by a theory that respects the whole symmetry group.
Notice that to do so ...
15
If I understand it correct, then in a nutshell, you are asking why is the VEV independent of spacetime. If the Higgs field had different values at different points in space i.e., if it had a spacetime variation, then the gradient term would give a positive contribution to the Hamiltonian, and hence, the total energy will not be minimized.
15
You could consider it as one more demonstration of the underlying quantum mechanical frame keeping atoms and molecules bonded together. Quantum mechanics is a probabilistic theory, and which bond will "break" depends on the square of the wavefunction describing the rod, with a probability which manifests in this one instance of breakage.
To get the ...
14
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 ...
14
An experimentalist's answer,
Our observations tell us that baryon and lepton number are conserved, within the accuracies of our experiments and observations. This means we have chosen as a standard model SU(3)xSU(2)xU(1) because in the group structure of the possible representations of all the quantum numbers assigned to the particles and resonances we ...
14
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 $U(1)...
14
Why can't fermions have a non-zero vacuum expectation value (VEV)? Lorentz invariance.
If anything other than a Lorentz scalar has a non-zero VEV, Lorentz invariance would be spontaneously broken.
For example, suppose we have a Lorentz invariant term in a Lagrangian for a vector
$$
\mathcal{L} \supset m^2 A_\mu A^\mu.
$$
Now suppose the vector obtains a ...
14
It is quite funny that people continues to cite the Mermin-Wagner theorem in a context where credit should be given to David Mermin for a paper he wrote alone, in which he derived the theorem which directly applies to the problem of crystalline order in 2D.
As an example of the ongoing citation confusion, there is a very highly cited paper by M I Katsnelson,...
14
The question has two parts: (1) What does "tree level" mean, and why are tree-level effects said to be classical in nature? (2) Is the Higgs mechanism classical in nature?
Tree level $\to$ classical
Solving linear differential equations is easy. Solving nonlinear differential equations is hard, usually too hard. If the nonlinearities are small ...
13
We have the functional of the external source $J$, which gives us v.e.v.s of field operators, by functional differentiation:
$$e^{-iE[J]} = \int {\cal{D}}\phi\, e^{iS[\phi]+iJ\phi} $$
$$\phi_{cl}=\langle\phi\rangle_J = -\frac{\delta E}{\delta J}$$
Where $\langle\phi\rangle_J$ is the v.e.v of $\phi$ in presence of external source $J$. That could be ...
13
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, E....
answered Nov 3 '13 at 9:03
Philip Gibbs - inactive
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13
The bottom line is the spontaneous symmetry breakdown from global $U(1)$ to $\mathbb{Z}_2$ and the concomitant rigidity of the omnipresent coherent phase down to which the system breaks. However, both the microscopic action and the BCS ground state (3) of a superconductor possess local $U(1)$ gauge symmetry.
By rigidity, I mean something reminiscent of a ...
13
In this answer we give a proof$^1$ of Goldstone's theorem at the physics level of rigor following Ref. 1:
We are given a spacetime-translation-covariant 4-current
$$\hat{J}^{\mu}(x)~=~e^{i(\hat{H}t-\hat{\bf P}\cdot {\bf x})}
\hat{J}^{\mu}(0)e^{i(\hat{\bf P}\cdot {\bf x}-\hat{H}t)} \tag{1}$$
that satisfies the continuity equation
$$d_{\mu}\hat{J}^{\mu}(x)~=~...
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