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8 votes

Is non-temperature related Symmetry Breaking possible?

Temperature is not needed to define symmetry breaking. Symmetry breaking simply refers to the loss of some symmetry during the evolution of a system. Given that a system is in a state $A$ that is ...
Javi's user avatar
  • 1,091
4 votes

Why are there no Goldstone modes in superconductor?

A neutral fermionic superluid has gapless soundwaves that can be regarded as a Goldstone mode due to the spontaneous breakdown of translational symmetry, but the fluctuating quantity in the sound ...
mike stone's user avatar
  • 54.5k
4 votes

Why are there no Goldstone modes in superconductor?

The Goldstone modes of a superconductor (considered as an electron gas with attractive potential) are physically the oscillations of electron density. As electrons are intrinsically charged, this will ...
E. Anikin's user avatar
  • 1,011
3 votes

Is non-temperature related Symmetry Breaking possible?

In physics. Symmetry Breaking is related to temperature [...] Symmetry breaking in classical physics More correct way of saying this is that in (classical) physics symmetry breaking is related to a ...
Roger V.'s user avatar
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2 votes

Are there universes where "symmetry breaking" went differently?

In classical physics there is only one history and symmetry breaking is understood in terms of saying that any particular system is very unlikely to perfectly respect the symmetry even if the ...
alanf's user avatar
  • 8,512
1 vote

Phase Coherence in the BCS wavefunction and the Cooper Pair Wavefunction

First answer. This is a quantum theory: every amplitude is a priori a complex number. The most general expression would have all $u_{k}$ and $v_{k}$ complex. What you do to get that expression here ...
T.P. Ho's user avatar
  • 91
1 vote

Does all symmetry breaking have corresponding unitary group?

In general you can have broken symmetry groups different from $U(n)$. For example the quantum Ising model has a discrete symmetry (enacted by $P=\prod_j \sigma^z_j$) that breaks spontaneously in the ...
lcv's user avatar
  • 2,474
1 vote

Does all symmetry breaking have corresponding unitary group?

As in this SE post, you can have the Lagrangian with $SO(N)$ symmetry $${\mathcal{L}} = \frac{1}{2}(\partial_\mu \Phi)^T (\partial^\mu \Phi) - \left(\frac{1}{2}\mu^2 \Phi^T \Phi + \frac{1}{4}\lambda (...
Gabriel Ybarra Marcaida's user avatar
1 vote

Are there universes where "symmetry breaking" went differently?

I think, since we can only study the physics of this universe, this question counts as meta-physics. Our current theories allow us to describe the reality that we see, and that's all that we need. As ...
Chaddyfynn's user avatar
1 vote
Accepted

Are there universes where "symmetry breaking" went differently?

I'm not an expert in quantum field theory or string theory, but as I understand it, these "other universes" have different laws of physics (in the low-energy limit). At some very high energy ...
Allure's user avatar
  • 21.4k
1 vote
Accepted

Alternative potentials in the context of spontaneous symmetry breaking

Let $V(\phi)$ be the potential for a scalar field $\phi(x)$. The scalar can be real or complex, it can have an arbitrary number of components, the important thing is just that it is a scalar under the ...
11zaq's user avatar
  • 1,014
1 vote

Spontaneous symmetry breaking in the quantum 1D XX model?

It is not. Because of (generalisations) of the Mermin-Wagner theorem, you can only break discrete symmetries in 1D. In both Ising and XXZ the symmetry that is broken is a discrete Z2 symmetry (...
sav's user avatar
  • 11

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