3

Spontaneous symmetry breaking refers to the solution of a system loses some symmetry in its Lagrangian. Consider a Simple Harmonic Oscillator, its lagrangian is time translationally invariant but its solution is periodic in time, thus not time-translational invariant. Is this Spontaneous symmetry breaking?

Shadumu
  • 1,211

1 Answers1

2

I definitely wouldn't say this. Spontaneous symmetry breaking is when the Hamiltonian is symmetric under some transformations, but the ground state of the system is not. The ground state of the classical SHO is that the particle sits at the minimum of the potential, which is a time-independent state. In the quantum case the quantum ground state has the overall time dependence $e^{-i\omega t}$, but that drops out from all observables: all expectation values are time independent.

G_Almasi
  • 43
  • 4
  • So you say, SSB is only applied to ground state? – Shadumu Nov 25 '14 at 19:45
  • Yes. You can read more about SSB on Wikipedia as well: http://en.wikipedia.org/wiki/Spontaneous_symmetry_breaking A system having a non-symmetric state is nothing special. The special property is when the ground state does not share a certain symmetry with the dynamics, that we call SSB. – G_Almasi Nov 25 '14 at 20:04
  • but in some examples like the ferromagnet, SSB occurs when the temperature drops below the Critical Curie Temperature, but not necessarily the ground state. Although wiki does mention often the ground state SSB is considered, but I do see generally why it is restrained to ground state and the general definition of SSB does not have this restriction as well. – Shadumu Nov 26 '14 at 01:14
  • The ferromagnet breaks rotational symmetry at finite temperature as well, that is true. But this follows from the non-uniqueness of the ground state. At $T=0$ the system chooses one ground state, and when you increase the temperature, fluctuations around that ground state get important too. The probability of a transition to the neighbourhood of an other ground state has still small probability (which needs flipping of MANY spins), so you can still have symmetry breaking magnetization. – G_Almasi Nov 26 '14 at 20:48
  • Imagine a 1D double well potential (w shaped) and a ball moving in it. The ground state is the ball sitting in one of the minima, which breaks mirror symmetry. For small temperatures your ball starts slowly moving around your chosen minimum. For higher temperatures however the ball will have enough energy to get to the other part of the potential as well, so the symmetry gets restored. – G_Almasi Nov 26 '14 at 20:53
  • Then back to Simple Harmonic Oscillator, any excited state solution certainly breaks time transnational symmetry. It is certainly not explicitly symmetry breaking. If it is not spontaneous, what symmetry breaking could it be? Why we must have the ground state to be asymmetric in order to have SSB, why not any excited state solution? – Shadumu Nov 29 '14 at 17:16
  • I don't really understand your question. Let's take the classical SHO. The motion is determined by the dynamics and the initial conditions. Let's take the spatial inversion symmetry. If your initial condition is a stationary particle at the minimum of the potential, the solution will be symmetryc under spatial inversion: the particle will stay at the minimum. However if your initial condition breaks the inversion symmetry (you start the particle from somewhere else), you will get oscillation, so at any latter time the position of the particle need not be symmetric to inversion. – G_Almasi Nov 30 '14 at 17:49
  • I mean time translational symmetry breaking in any non-ground state solution in SHO, i know it is trivial but it does break it right?Why SSB has to be associated with ground state anyway? – Shadumu Dec 02 '14 at 22:19
  • I would say yes, any time dependent solution is not invariant under time translation. What breaks this (in my opinion) are the initial conditions. Take any time dependent solution. If you apply time translation, you get an other equivalently good, but different solution. What decides which solution will actually happen are the initial conditions. They break time translation explicitely, you specify at a given, chosen time the velocity/location. – G_Almasi Dec 08 '14 at 11:07