# Spontaneous symmetry breaking in phase transitions

I don't really understand the concept of spontaneous symmetry breaking in phase transitions. From my understanding of how spontaneous symmetry breaking works I need to find the ground states of a system and see if these states have the same symmetry of the hamiltonian. If not we can speak of spontaneous symmetry breaking.

Let me make a concrete example.

Consider a $$O(2)$$ model. There is a spin in every lattice point that can rotate from $$0$$ to $$2\pi$$. The hamiltonian has a continuous rotational symmetry (i.e. if I rotate every spin by the same angle the energy is the same).

Let's consider first $$T = 0$$. The ground states are all those states with all the spins pointing in the same direction. Let's say that my system decides to have all the spins aligned to the x-axis. If now I rotate the spins of the same angle I can recognize that the state has changed (i.e. there is no longer the symmetry and we have SSB).

Now suppose we are at high $$T$$. The spins now can have random directions in the ground state and they are not correlated among each other. But anyway I would say that even now if I do a rotation I could in principle still see that the state has changed.

I get that intuitively in the case $$T=0$$ if I do a rotation it's easier to see the different states while in the $$T>0$$ case it would look like nothing changed (because the magnetization would still be $$0$$ and I can't actually look inside the magnet), but in principle they are different states!

So why in the case $$T>0$$ we speak of the system being symmetric while in the case $$T=0$$ we speak of SSB? Is my bold definition of spontaneous symmetry breaking correct?

• A small comment: The model you refer to is not the Ising model but the $O(2)$ model. The ising model only has a discrete $Z_2$ symmetry and a big part of the discussion about symmetry breaking does not apply to discrete symmetries at all.
– mika
Commented Apr 22 at 14:09

Now suppose we are at high T. The spins now can have random directions in the ground state and they are not correlated among each other. But anyway I would say that even now if I do a rotation I could in principle still see that the state has changed.

The above statement is where you got it wrong.

The physical system in thermal equilibrium is a weighted summation of all possible states. Therefore, at high T, even if you see a certain fixed random direction in one snapshot, the physical system is still symmetrical after you sum up all the possible states/snapshots.

You may wonder: does the above reasoning apply to the $$T = 0$$ symmetry-breaking phase as well? Is the $$T = 0$$ case also a summation of all the lowest energy (ground) states? Since the space of possible lowest energy (ground) states is symmetrical with equal summation weights, does that mean there is no SSB even for $$T = 0$$ case? Actually this the real hard question that is at the core of the SSB problem.

The answer to the hard question hinges on the thermodynamic limit ($$N\rightarrow \infty$$, $$V\rightarrow \infty$$, and $${N}/{V}\rightarrow$$ constant). The detailed answer would be appropriate for a different post. Suffice it to say that if you get this point, you would understand why there is no SSB in the usual quantum mechanics with finite of number of particles. SSB can only take place in quantum field theory where there are infinite number of particles.

• I would actually be very interested in the explanation of the "hard question". Do you have any books where I can read this? Or I can open another post.
– Alex
Commented Apr 23 at 8:21
• You may refer to this post: physics.stackexchange.com/questions/69289/… Commented Apr 23 at 13:29