Why is the sequence of limits $\lim\limits_{V\to\infty}\lim\limits_{B\to 0}m(B,V)$ when reversed does not give the same result? For spontaneous magnetization $m$ in a sample of volume $V$, what do the limiting operations $$
\lim\limits_{V\to\infty}\lim\limits_{B\to 0}m(B,V)=0,\\
\lim\limits_{B\to 0}\lim\limits_{V\to\infty}m(B,V)\neq 0$$ physically mean and why don't they commute? $B$ denotes the applied magnetic field.
 A: The physical meaning of the two different double limits is quite obvious:
$$\lim\limits_{V\to\infty}\lim\limits_{B\to 0}m(B,V)$$ corresponds to start with a finite volume $V$, making the external magnetic field going to zero. At this point, no spontaneous magnetization is possible, due to the analytic behavior of the free energy of a finite system as a function of the thermodynamic parameter $B$, which forces $m(B,V)$ to be an odd function of $B$ ($m(-B,V)=-m(B,V)$, which implies $m(0,V)=0$).  Therefore,  magnetization density is zero for any finite $V$, and even taking the limit $V\to \infty$ will result in a zero magnetization density.
On the contrary,
$$
\lim\limits_{B\to 0}\lim\limits_{V\to\infty}m(B,V)$$ may be different from zero because, as result of the first limit, $m(B)$ for the infinite volume sample may be a non-analytic function of $B$, allowing to escape the conclusion $m=0$ at vanishing external field.
The non equivalence of the interchanged double limit is the fingerprint of a spontaneously broken symmetry: as a consequence of the thermodynamic limit, a symmetry of the hamiltonian (here the time reversal symmetry) may cease to be a symmetry of the infinite system macrostate.
