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.

  • 4
    $\begingroup$ What you write in point (2) is correct, since $m(N,h)$ is an analytic function of $h$ (and $\beta$) when $N$ is finite (this should be clear by looking at how $m(N,h)$ depends of $h$). This is actually the reason one has to take the thermodynamic limit first. $\endgroup$
    – Heidar
    Oct 7 '13 at 11:17
  • $\begingroup$ @ Heidar, so you mean the function $m(N,h)$ is always continuous at $h=0$ for finite $N$? But is there any physical argument or mathematical proof for this point? Thanks. $\endgroup$
    – Kai Li
    Oct 7 '13 at 11:27
  • 2
    $\begingroup$ @K-boy: You can see that at finite $N$, the partition function (or the expectation value) is just the sum of a finite number of exponentials. This is therefore analytic, and the limit $h\to 0$ is continuous. $\endgroup$
    – Adam
    Oct 7 '13 at 13:18
  • $\begingroup$ @ Adam, good point. But from the math viewpoint, even though the number of summation terms is finite, there is still possibility that some of the terms are singularity. For example, the partition function $$Z=\sum_{\alpha=1}^{d} e^{-\beta E_\alpha(h)}$$ with $d=2^N$ the dimension of the Hilbert space of $N$ spin-1/2 system and $E_\alpha(h)$ the eigenenergy of $H$, if there exist some $E_\alpha(h)$ which are not continuous functions of $h$, then the whole story may be not continuous on $h$. Is this possible? $\endgroup$
    – Kai Li
    Oct 7 '13 at 14:47
  • 2
    $\begingroup$ @K-boy: Usually, one expects that the eigenvalues are analytical functions of $h$ if the Hilbert space is finite (though I don't have a proof of that). Of course, you can always invent a model where that's not the case, but that's not typical. Where you are right is that even though the system is finite, and that the eigenvalues are all analytical, the partition function can be non-analytical in $h$ in the limit $\beta\to \infty$ (you can show that with only one quantum spin). This corresponds to a kind of quantum phase transition associated with a level-crossing. $\endgroup$
    – Adam
    Nov 24 '13 at 23:41

(1) $m(N,h)$ for finite $N$, is a continuous function of $h$. Indeed, for your Hamiltonian, $\exp(-\beta H)$ is a convergent (in any norm, since it is finite-dimensional) power series in $h$:

$$ \exp(-\beta H) = \sum_{n \geq 0} \frac{(-\beta)^n}{n!} H^n \ .$$

taking the trace of that gives

$$ Z = \sum_{n \geq 0} \frac{(-\beta)^n}{n!} \text{Tr} (H^n) \ .$$

(2) It is indeed true that

$$ \lim_{N\rightarrow \infty} \lim_{h \rightarrow 0} m(N,h) = 0 \ .$$

Note however that this does not imply that

$$ \lim_{h\rightarrow 0} \lim_{N\rightarrow \infty} m(N,h) = 0 \ .$$

We are in finite dimensions, so the trace will just be a sum of terms, with $\text{Tr} (H^n)$ being a polynomial in $h$ with at most order $n$. Hence this is an analytic function.


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