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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? – Timothy 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
up vote 9 down vote accepted

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 ($n\geq2$) phase transitions. But I don't think such theoretical construction really 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 $$ H =\sum_{k} c_k^\dagger (\sin k_x\sigma_x+\sin k_y\sigma_y+(\cos k_x+\cos k_y-2+m)\sigma_z)c_k,$$ where $c_k$ is the electron operator of the momentum $k$. The topological mass $m$ is the driving parameter: $m>0$ ($m<0$) corresponds to the topological (trivial) phase, while $m=0$ is critical. It is easy to show that the free energy (at zero temperature) of the fermion system reads $F\simeq\int\mathrm{d}^2k\sqrt{k^2+m^2}\sim -|m|^3$, 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 (at $k=0$) 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? – Timothy 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
@qfzklm Yes, the mass is a physical controlled parameter. Its physical meaning depends on the system. For example, for the quantum spin Hall insulator, $m$ controls the band inversion, and is tuned by the thickness of the quantum well. – Everett You Jan 18 '15 at 21:10
@Hamurabi Free boson BEC transition is not described by a Landau theory (because there is no order parameter to start with), so the 3rd order transition in that case is not related to the missing $m^2$ term. The superfluid transition of interacting bosons is described by a Landau theory, and does have a $m^2$ term. – Everett You Oct 29 '15 at 2:37
@Hamurabi The local order parameter exists only if the phase transition is a spontaneous symmetry breaking (SSB) transition. Landau theory only applies to SSB transitions. The free boson BEC is not a SSB transition, because it breaks no symmetry. Without the notion of symmetry, local order parameters can not be defined. $N_0/N$ (occupation number of lowest level) is an indicator of the BEC transition, but it is not an order parameter, because order parameter must transform under symmetry non-trivially, but there is no symmetry to transform $N_0/N$. – Everett You Oct 29 '15 at 16:54

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