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I have a question about this paper http://arxiv.org/abs/1003.0010 In their model, when consider holographic paramagnetic-ferromagnetic phase transition, they need Yang-Mills field itself to condensate. In bulk the Yang-Mills field which is dual to spin wave has the following form $$ A^3_t=\mu \alpha(r),~~\alpha(r\rightarrow\infty)=1 $$ where the $\mu$ is dual to boundary magnetic field.

When consider holographic paramagnetic-antiferromagnetic phase transition, they focus on the adjoint representation scalar field $\Phi$ which is dual to order parameter of field theory. Near the boundary, the scalar has the following form $$ \Phi=A r^{\Delta-3}+B r^{-\Delta},~~ \Delta=\frac{3}{2}+\sqrt{m^2R^2+\frac{9}{4}} $$ When considering holographic paramagnetic-antiferromagnetic phase transition, the authors choose the standard quantization condition where $A=0$ and $B\neq 0$.

My questions are: 1) If $A\neq0$, is this condition dual to paramagnetic-antiferromagnetic phase transition with external field? why do people general not care such case? 2) Also in holographic superconductor models, why do people always require standard/alternatve quantization? why not consider cases with classic current, that is both components are not zero?

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Depending on the mass region of $\Phi$, either A or B can be taken as source and the corresponding response (vev). If $B\neq 0$ when $A=0$, it means that the system can spontaneously have a nontrivial vev even without any source. That indicates a phase transition. In the case both $A\neq 0$ and $B\neq 0$, it doesn't mean any phase transition. If we treat $\Phi$ as a fluctuation, $G_R= B/A$ means some kinds of susceptibility.

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how about the system has a vev with some source, that is $A \neq 0 $ and $B \neq 0$ ? how to conclude that "it doesn't mean any phase transition."? Naively from the duality, $B$ corresponds to the order parameter in standard quantization. That $B$ varies from zero to nonzero means a phase transition in field theory. –  Craig Thone Jun 12 '12 at 3:24
    
From linear response theory, the vev of an operator is always nontrivial if we turned on some source, no matter which phase are the system in. Just like in CMT, for a metal in the normal phase, if we turned on the source for the paring order parameter, the vev is not zero. However, the system is in the normal phase. Of course we can put the system with some external source turning on and study the phase transition, e.g. 1111.2606. Right now I am not sure which order parameter should it be in this system. –  user9510 Jun 12 '12 at 18:31
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