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$F=\frac{a}{2}m^2+\frac{u}{4}m^4+\frac{v}{6}m^6-hm$, where F is the free energy, m is the order parameter, h is the external field, $a=a_0(T-T_c)$, and $a_0>0,u>0$ and $v>0$.We know this free energy expansion describes a second order phase transition. How to write down a free energy such that the transition is a third-order phase transition?

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Why can the higher order terms be neglected in the vicinity of the critical point? –  Jeremy Mar 30 '13 at 19:01
    
Because in the vicinity of the critical point, the order parameter is weak $m\to 0$, so higher order terms can be neglected. –  Everett You Mar 30 '13 at 20:49
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Although third-order transitions are rare to see, but it is not hard to write down Landau theory for third-order transitions. Let $m$ be the order parameter, the free energy simply reads $$F=a m^4+ b m^6+\cdots$$ with $b>0$, and $a$ is the driving parameter, which triggers a third-order transition at $a_c=0$. To verify, just calculate the saddle point from $\partial_mF=0$ so that $$m=\left\{\begin{array}{ll}\sqrt{-2a/3b} & a<0,\\ 0 & a>0.\end{array}\right.,$$ and hence $$F=\left\{\begin{array}{ll} 4a^3/9b^2 & a<0,\\ 0 & a>0.\end{array}\right.,$$ which obviously becomes singular in the third-order derivative $\partial_a^3 F$ at $a=0$. Following this line of thought, one can play with Landau theory, and write down $$F=a m^{2(n-1)}+b m^{2n}+\cdots$$ for $n$th-order phase transitions. But I personally do not consider this meaningful, because the vanishing lower order terms require fine-tuning of the model, and can hardly be realized physically.

However beyond Landau paradigm, third-order transitions can happen in topological quantum phase transitions. One known example is transition between 2D Chern-insulators, described by the following Hamiltonian $$ H =\sum_{k} c_k^\dagger (k_x\sigma_x+k_y\sigma_y+m\sigma_z)c_k,$$ where $c_k$ is the electron operator of the momentum $k$. $m$ is the driving parameter, and $m=0$ is critical. It is easy to show that the free energy (at zero temperature) of the fermion system follows $F\simeq\int\mathrm{d}^2k\sqrt{k^2+m^2}\sim -|m|^3$, so $\partial_m^3 F$ is singular at $m=0$, and hence a third-order transition. During this transition, the gap of the Dirac cone closes and reopens, leading to $\pm1$ change in the Chern number of the occupied band, which reflects the underlying change of the band topology. However there is neither symmetry broken nor local order parameters involved in this transition, so it is an example of third-order transition that can not be described by the Landau theory.

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Thanks for the answer. The Chern insulator cannot be described by Landau theory why do you still use the free energy and order parameter? –  Jeremy Mar 31 '13 at 4:45
    
@Jeremy Note that the Dirac mass $m$ is NOT an order parameter, but the driving parameter, just like $a$ in Landau theory. All phase transitions, no matter topological or not, can be described by free energy and driving parameter, but only for those symmetry breaking phase transitions, the free energy can be associated to order parameters. –  Everett You Mar 31 '13 at 4:51
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