# Why does symmetry spontaneously break? [closed]

Why does symmetry spontaneously break? For example,in the case of the mexican hat potential, as far as I can tell, a particle that is at the unstable equilibrium ought to be able to continue staying at that equilibrium. Why would the blue particle move down?

Isn't it unphsical to say that the symmetry is broken?

What are the axioms of gauge theory that allow for spontaneous symmetry breaking? • Why would the blue particle move down? Because it’s position is unstable. The smallest thermal or quantum fluctuation makes it start to move down. Can you balance a pencil on its point? Jan 23, 2020 at 22:24
• Your last question is completely unrelated to the first three. Please ask one question per post. Jan 23, 2020 at 22:26
• Whoops, sorry. I meant to ask "what are the axioms of gauge theory that allow for spontaneous symmetry breaking", I somehow filled that up with "super-symmetry". Jan 23, 2020 at 22:29
• One thing you could argue is that classically, the blue particle might potentially stay it is unstable equilibrium forever. However, if you take into account quantum fluctuations, the uncertainty principle says that $\Delta x$ is nonzero. Since the particle needs to be exactly at the equilibrium position to stay at an unstable equilibrium, even the smallest fluctuation in the position of the particle will move it away from equilibrium, at which point it must continue down the potential hill.
– d_b
Jan 23, 2020 at 22:42
• No, that is not true. The wavefunction must evolve in a way that respects the uncertainty principle regardless of whether there is measurement or not.
– d_b
Jan 23, 2020 at 22:51

There are two things happening here.

SSB refers to cases where the symmetry of a Lagrangian or Hamiltonian does not match the symmetry of its ground states.

Taking the Hamiltonian case, you imagine a unitary operator $$U$$ that expresses a symmetry of the system, such that $$U H U^\dagger = H$$ and therefore $$[U, H] = 0$$ so that $$U + U^\dagger$$ and $$i U - i U^\dagger$$ must be observables that are conserved. We say that this symmetry is spontaneously broken when there is a family of ground-states $$|g_\phi\rangle$$ such that they all have the same energy but $$U |g_\phi\rangle = |g_\xi\rangle \text{ for } \phi\ne\xi.$$

Now, such a system could be in a state $$\frac1N \sum_\phi |g_\phi\rangle$$ or $$\int \mathrm d\phi~f(\phi)~|g_\phi\rangle$$ depending on whether the index is discrete or continuous. This state is also an eigenstate of the Hamiltonian with the ground energy. In such a case the symmetry is being broken by measurement. The standard reasoning applies: there must be some other system which is weakly coupled to $$U + U^\dagger$$ or so, such that the two get correlated and some sort of information about $$\phi$$ is getting leaked to an experimental apparatus. Then the system needs to fall into one of those eigenstates in particular and you have a nonunitary collapse to one of the ground states. That's the first thing that happens.

The second thing is dissipation. Of course you might have higher-energy excited states that are not $$U$$-sensitive, $$U|e\rangle = |e\rangle.$$ This is the sort of thing you are thinking about with your particle right on the tip of the Mexican hat. As long as the system is never coupled to an external environment, this sort of thing can remain in a strange state: it might not remain precisely over the center; the wavefunction is more likely to expand and contract and then expand again and contract again, eventually reforming the singularity. But once it is being coupled to that larger environment, one starts to imagine noise creeping in and decoherence happening and the system decaying down to a lower energy state.

There is no general theorem that I know of which says that this always happens, if you are not counting, say, the second law of thermodynamics or so. You do need to have some sort of dissipation and measurement and all of that other stuff to really make sense of it, and therefore you have to decide on a form under which those terms enter your Hamiltonian, and this prevents the derivation from being truly universal.

What usually happens in these non-general derivations on the condensed matter side is something like the following:

1. You assume very weak coupling to an infinite number of fermionic or bosonic creation and annihilation operators.
2. You use the state-matrix formulation to put all of those modes into a thermal state.
3. Then you look typically for second-order effects on the overall state matrix, usually using the interaction picture.
4. These second-order effects typically take the form of some sort of integral or double-integral, so you argue that the integral can be simplified with a Born-Markov approximation.
5. If you're lucky, some terms come out at the end having that beautiful characteristic Lindblad form, $$A \rho A^\dagger - \frac12 A^\dagger A \rho - \frac12 \rho A^\dagger A,$$ often with some sort of really strange limit needed on the coupling (like you have N modes with a coupling constant that decays like $$1/\sqrt{N}$$ or something) which you just brush past and say “assume this is some constant.”
6. That Lindblad form then describes a nonunitary dynamics and you can look there for things like dissipation, and you can try to convert it to a stochastic equation and do numerical simulations. But in the spontaneous-symmetry-breaking side, that can create a flow from the center of the Mexican hat out towards the ground states.

I have also seen a bunch of other approaches to quantum mathematics but I haven't worked in the field in years so I don't remember in detail whether they apply here. So for example I can't tell you whether the folks working with random matrices and pulling out spectrums of eigenvalues have some super-easy way to do these sorts of manipulations.

• (a) In perfect isolation, if I understand correctly, the particle can remain at equilibrium, and the wave function evolves strangely (?) Can I have a worked out computation of the wavefunction evolution if possible, please? (b) How do we define a ground state? Are they those states with lowest energy, or something else? Jan 23, 2020 at 23:45