Consider the Heisenberg Hamiltonian: $$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} H=-\f{J}{2} \sum_{\mean{ij}}\vec S_i \cdot \vec S_j+\mu_Bg_S \vec B \cdot \sum_i \vec S_i$$ We can write the contributions which involves a given spin as: $$ H_i=-J\vec S_i \cdot \sum_{j}\vec S_j+\mu_Bg_S \vec B \cdot \vec S_i$$ In the mean field approximation we then replace $\vec S_j$ with it's expectation value to get: $$ H_i=-J\vec S_i \cdot \sum_{j}\mean{\vec S_j}+\mu_Bg_S \vec B \cdot \vec S_i$$ Now my question is; is it enough that we have rotational symmetry about $\vec B$ to say that $\mean{\vec S_j}$ must point along $\vec B$ or do we have to make the ising approximation?

My thought is that it is the latter since $\mean{\vec S_j}$ is not necessarily zero when $\vec B$ is.


1 Answer 1


You are correct when saying that $\vec m_j = \langle\vec S_j \rangle$ is not necessarily aligned with $\vec B$.

We can write your second equation as $$ H_i = -JS_i \cdot (N\vec m + \vec B)$$ where $\vec m$ is the average of the $\vec m_j$s over $j$. It should be intuitively clear that if -- for any reason -- a majority of spins happens to be aligned along an arbitrary axis, then $|\vec m|$ is large, and the $i$-th spin described by the Hamiltonian above will align with $\vec m$, if $\vec B$ is small (even if nonzero). This means that this "arbitrary axis" solution is stable.

Your intuition did tell you that there must be something wrong, considering the system is rotationally symmetric. Indeed, in phase transitions, systems do undergo spontaneous symmetry breakings. In the case $B=0$, one could think that the total magnetisation must be zero because of symmetry, but that is not necessarily the case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.