2
$\begingroup$

Three examples in the spontaneous symmetry breaking that occurs at a phase transitions:

  • A ferromagnet below the Curie temperature chooses an axis of quantisation along which all the spins align, giving rise to a macroscopic magnetisation in that direction. Symmetry broken is $SO(3)$ (isotropy of space) to $SO(2)$ (only symmetry about magnetisation axis).

  • Bose-Einstein condensation: below $T_c$, bosons amass in the same ground state, described by the same wavefunction and with a physical (non-gaugeable) phase. Symmetry broken is $U(1)$.

  • Higgs mechanism. The complex Higgs doublet chooses a phase, a non-zero vacuum expectation value (VEV) $\propto \mu^2/\nu$. This then determines the mass of the Higgs, the $W^\pm$ and $Z^0$ bosons, and the coupling to fermions. Symmetry broken is $SU(2)_L \times U(1)_Y$ to $U(1)_{em}$


in the absence of any decoherence, coupling to environments and measurements, how is the phase chosen?
Why is not a superposition of all possible ones.

The phase controls the masses of particles, for the higgs. So it's quite important. What caused the field to choose that particular phase during the Higgs phase transition?

$\endgroup$
  • $\begingroup$ What do you mean "why is not a superposition of all possible ones"? How could the direction of the magnetization of a ferromagnet be "a superposition"? Also, an analogy: Take the mexican hat potential literally and classically and place a ball at its top. When it rolls down into the hat's brim, are you also wondering what chose the direction it rolled in? Do you consider it non-obvious that it stays at the top if you postulate unrealistically that it is placed at the exact top and that there are no influences from the environment whatsoever? $\endgroup$ – ACuriousMind Apr 13 at 11:31
  • $\begingroup$ Well it does not have to choose an axis. Why can't $M$ be n a quantum mechanical superposition of all directions? $\endgroup$ – SuperCiocia Apr 13 at 11:35
  • $\begingroup$ I understand the environment in the context of BEC and ferromagnet. But what is the environment for the Higgs transition? A field that permeates the whole of space? $\endgroup$ – SuperCiocia Apr 13 at 11:35
  • $\begingroup$ In the ferromagnet a measurement of its magnetic field is made. In the Higgs case a measurement of the mass of a particle made. I fail to see a fundamental difference here. $\endgroup$ – ACuriousMind Apr 13 at 11:38
  • $\begingroup$ Unless I make a measurement of the mass, is the system is a superposition of several mass eigenstates? $\endgroup$ – SuperCiocia Apr 13 at 12:04
0
$\begingroup$

You don't need symmetry breaking for a symmetry broken phase. What is relevant is that all ground states -- be they symmetric or symmetry broken -- have a long-range order in the relevant order parameter. This long-range order is present regardless of the presence or absence of any external fluctuations which explicitly break that symmetry.

Note that there are examples where this is arguably the case, such as superconducivity: The U(1) symmetry -- corresponding to particle number conservation -- will not be explicitly broken, since (at least at the relevant energies) electron number is in fact a conserved quantity. Rather, what is important is the long-range order in the superconducting order parameter, and the relative value of the order parameter of two coupled superconductors. However, assuming that the symmetry is broken is convenient for calculations.

A similar scenario is encountered when talking of coherent lasers with a fixed phase: What is actually fixed is the relative phase of different beams to each other, while the global phase is constantly fluctuating. However, assuming a fixed global phase simplifies calculations and has no negative consequences.

$\endgroup$
  • $\begingroup$ Remind me again why does U(1) mean particle number conservation? $\endgroup$ – SuperCiocia Apr 13 at 12:52
  • $\begingroup$ Also, I see your point of relative phases. But what about the cases where the order parameter has a direct physical meaning? Such as the magnetisation direction (ferromagnetic) or the coupling to fermions (Higgs)? $\endgroup$ – SuperCiocia Apr 13 at 12:54
  • $\begingroup$ @SuperCiocia I doubt that there is any context where there is an absolute symmetry breaking in an absolute sense. Essentially, whenever you measure you are comparing two references, or at least, you can think of it that way. Regarding U(1): The U(1) symmetry is $a\mapsto e^{i\phi}a$, i.e. a state with $N$ particles transforms as $e^{-i\phi N}$. $\endgroup$ – Norbert Schuch Apr 13 at 13:05
  • $\begingroup$ @SuperCiocia don't know too much about the Higgs mechanism, but isn't the order parameter the phase of the VEV in the mexican hat potential, while the coupling is determined by its absolute value (which is independent of the order parameter = the phase)? $\endgroup$ – Norbert Schuch Apr 13 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.