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Triangular lattice and criticality

  • Thread starter CAF123
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Homework Statement


Consider a two dimensional triangular lattice, each point of which can either contain a particle of a gas or be empty. The Hamiltonian characterising the system is defined in terms of the particle occupation numbers ##\left\{n_i \right\}_{i=1,...,N}## which can either be 0 or 1. N is the total number of points on the lattice. The Hamiltonian is $$H = \mu \sum_i n_i - J \sum_{\langle i j \rangle} n_i n_j$$ with ##\mu >0## the chemical potential and ##J \geq 0## an effective interparticle interaction.

a) In the limit J=0, find the partition function, the free energy and the average occupation number.

b) Using a mean field variational approach, the average occupation number is $$\langle n \rangle = \frac{1}{1+ \exp(3J\beta - 6 \beta J \langle n \rangle)}$$ Expand this equation around ##\langle n \rangle = 1/2## and find the critical value of J for which the mean field predicts a phase transition between a particle poor phase and a particle rich one.

Homework Equations


$$Z = \sum_{\left\{n\right\}} e^{- \beta H}$$

The Attempt at a Solution


a) In the first part I was getting ##Z = (2\cosh \beta \mu)^N##, ## F = -N/\beta \,\, \ln(2 \cosh \beta \mu)## and ##\langle n \rangle = -N \tanh \beta \mu##

b) Write the equation as $$ \langle n \rangle = (1 + \exp 3J \beta (1-2 \langle n \rangle))^{-1}$$ Write $$\exp 3J \beta (1-2 \langle n \rangle) = \exp(3J \beta) \exp(-6J \beta \langle n \rangle)$$ and expand around ##\langle n \rangle = 1/2## so set ##\langle n \rangle = 1/2 + \eta## and get $$ \exp(3J \beta) \exp(-6J \beta( 1/2 + \eta)) \approx (1-6J \beta \eta)$$ Inputting this into the equation for ##\langle n \rangle##, and replacing ##\eta = \langle n \rangle - 1/2## I get a quadratic equation for ##\langle n \rangle##? Should this happen and if so, how to interpret the two solutions? I think J is that value at which either side of it ##\langle n \rangle## goes from being less than 1/2 to greater than it. (I haven't found such an equation yet either)

Thanks!
 

Answers and Replies

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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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