Almost in every textbook of condensed matter physics, the standard description of SSB could be formulated as follows:

Consider the lattice Heisenberg model in an external magnetic field $H=\sum_{ij}J_{ij}\mathbf{S}_i\cdot\mathbf{S}_j+hS_z$, where $h$ is the magnitude of magnetic field and $S_z=\sum_iS_i^z$. Now the average magnetization per site is a function of both magnetic field $h$ and number of lattice sites $N$, say $m\equiv \sum_i\left \langle S_i^z \right \rangle/N=m(N,h)$, where $\left \langle S_i^z \right \rangle\equiv tr(\hat{\rho }S_i^z)$ with $\hat{\rho }=e^{-\beta H}/tr(e^{-\beta H})$ the density operator. Then if $$\lim_{h\rightarrow 0}\lim_{N\rightarrow \infty }m(N,h)\neq 0$$, we say the system has SSB at temperature $T$. Now I get some questions:

(1)We know at finite $N$ and zero $h$, $m(N,h=0)=0$ due to spin-rotation symmetry. But **there is no reason** for that $$\lim_{h\rightarrow 0}m(N,h)=m(N,h=0)—[1]$$, right? Since the function $m(N,h)$ may *not be continuous* at $h=0$, from the math viewpoint.

(2)If Eq.[1] is correct, and hence $\lim_{h\rightarrow 0}m(N,h)=0$, then $\lim_{N\rightarrow \infty }\lim_{h\rightarrow 0}m(N,h)=0$, right?

(3)If Eq.[1] is wrong, say $\lim_{h\rightarrow 0}m(N,h)\neq m(N,h=0)$ and hence $\lim_{h\rightarrow 0}m(N,h)\neq0$, then what about $$\lim_{N\rightarrow \infty }\lim_{h\rightarrow 0}m(N,h)?$$ And why don't we use this identity to define SSB?

Thank you very much.

This post imported from StackExchange Physics at 2014-03-09 08:42 (UCT), posted by SE-user K-boy