System of particles with spin & symmetry breaking I was solving an exercise in which I had to find how the entropy of a system of N (assuming distinguishable) particles with spin changes when the temperature $T\rightarrow0$ and $T\rightarrow \infty$.
For $T\rightarrow0$, multiplicity is $\Omega=1$, therefore from $S=kln\Omega=kln1=0$
For $T\rightarrow \infty$, multiplicity $\Omega = 2^N$ (which is the formula when we have distinguishable objects with bins of unlimited occupancy), $S=kln\Omega=kln2^N=KNln2$.
What grabbed my attention was the following commentary, which I fail to understand for multiple reasons, which I will list below. The commentary is the following:
Note: there are actually two states with perfect order, but one cannot simply go from one of these states to the other, because that requires a macroscopic change. One speaks of spontaneous symmetry breaking, an extremely important concept in condensed matter theory and quantum field theory
I have the following questions:

*

*Are these states with perfect order,microstates of the system,macrostates,eigenstates etc?

*Which are these 2 states with perfect order?

*What does it mean for a state to have a perfect order?

*What it tries to say with "but one cannot simply go from one of these states to the other, because that requires a macroscopic change". What are some of the macroscopic changes that would force this transition? Are there cases of perfect ordered states in which macroscopic change is not needed, for it to happen?

 A: If your system has $N$ particles, each of which can have two states, $+1$ and $-1$, then there are two states with perfect order: the state where every particle is in the $+1$ state, and the one where every particle is in the $-1$ state.
(Perfect order just means that they're all the same as each other.)
Now, if $+1$ and $-1$ are both of minimal energy, then at $T=0$ the system will be in one of these two states (I suppose with some assumptions about the energy of the system being lower with order). For a particle to move away from the minimum requires energy (it's at the bottom of a trough), which is in short supply at low $T$. And for the system to move into the other ordered state would require them all to go at the same time. For one particle it's unlikely, for $10^{23}$ particles it's impossible. So the system is stuck in the neighbourhood of the minimum it's in – even though, in theory, it could have ended up in the other minimum. That's the broken symmetry part. It's been forced to 'make a choice', and now it's hovering around all $+1$ or all $-1$ (rather than around $0$ as the symmetry of the problem would suggest).
At higher temperatures, the particles flip between the two states all the time, so there's nothing to stop them ending up in either state once the temperature has been lowered again.
You may wish to play around with an Ising model simulation on the Internet – e.g. one can be found here (there may be better ones, it was just the first I found). Drag the temperature down slowly (a small lattice is easiest unless you want to be waiting around for ages) and you'll see it get stuck in one ordered state (eventually).
