Complicated ising model

[sanctus]

This is a quite complicated Ising model under compression with a force 2[math]\lambda[/math], 2 dimensional, with horizontal, vertical and diagonal ising interaction. After long calculations I found that this is the density of th free energy of the system.

 

[math]f(T,\lambda)\equiv\frac{F}{N}=\frac{-1}{\beta}lim_{N\to \infty}\ln Z_N = 2\lambda a_0 +\frac{1}{2\beta}\ln\left(\frac{\pi^2 m}{\varphi_2 \beta^2}\right)+2\varphi_0-\frac{\lambda^2}{\varphi_2}-\frac{j_1^2+j_3^2}{2\varphi_2}-\frac{1}{\beta}\ln \eta_1[/math]

 

Where:

[math]\eta_{1}={e^{\beta\,\left({\it \rho_3}-{\it \rho_4} \right)}}\cosh \left(2\,

\beta\,\left( {\it \rho_1}-{\it \rho_2}\right)\right)+{e^{\beta\,\left( {\it \rho_3}+{\it\rho_4} \right)}}\cosh\left(2\,\beta\,\left({\it\rho_1}+{\it\rho_2}\right)\right)+\sqrt{\gamma}[/math]

And

[math] \gamma\equiv{e^{\beta\,\left({\it2\rho_3}-{\it2\rho_4}\right)}}\left(\cosh\left(2\,\beta\,\left({\it\rho_1}-{\it \rho_2}\right)\right)\right)^{2}+4\,{e^{-\beta\,{\it2\rho_3}}}[/math]

[math]-2\,{e^{\beta\,\left({\it\rho_3}-{\it\rho_4}\right)}}{e^{\beta\,\left({\it\rho_3}+{\it\rho_4}\right)}}\cosh\left(2\,\beta\,\left({\it\rho_1}-{\it\rho_2}\right) \right)\cdot\cosh\left(2\,\beta\,\left({\it\rho_1}+{\it\rho_2}\right) \right)+{e^{\beta\,\left({\it2\rho_3}+{\it 2\rho_4}\right) }}\left(\cosh\left(2\,\beta\,\left({\it\rho_1}+{\it\rho_2}\right) \right)\right)^{2}[/math]

 

And all the [math]\rho_i[/math] and a0 are (adjustable they contain the Isning interaction) constants.

Is there a way to compute the ground state? I don't actually ask for the calculation, but just a way you might think it works (this model is analytically solvable so there has to be a way). Just straight forward calculation gets to huge even for Maple....

 

And I also need to calculate specific heat and friends...

[Qfwfq]

Well I'm no expert on the topic and I don't think I can help much, but I think there seems to be a problem or two with the LaTeX. The , seems to upset it. Perhaps Alex should check it up.

[Erasmus00]

To find the ground state energy, take the temperature to 0, and then minimize the free energy(with respect to your parameters). F = E-TS, so with T=0, minimizing F minimizes E. Hopefully the complicated free energy gets less complicated as T heads to 0.

 

As to specific heat and the like, you have the free energy, so its relatively straight forward. The derivatives will probably be awful given the complicated free energy, but I'd guess maple/mathematica could handle it.

 

[math]S=-\frac{\partial F}{\partial T} [/math]

 

And [math]C_v = T\frac{\partial S}{\partial T} [/math]

 

Essentially, you have the free energy, so you are as good as finished with the problem.

-Will

[sanctus]

Thanks for your reply Erasmus00.

Well i thought as well that it would be just putting T to zero, but then 1/beta tends to zero as well and the second term diverges to -infinity.

I agree with your last sentence, or at least I did agree until last week, it was a real struggle to get this free energy (and I know it is right, verified on the original paper where as usual there are no details) and then I thought now you just derive it for minimizing and as well to get specific heat, but...

 

Well eventuallly I managed to get the specific heat for different fixed lambdas....I guess that will already be enough to be able to hand it in and get a sufficient mark. You know I'm quite interested in finishing it quickly because so I can eventually start my diploma work (or do you say master-thesis in english?)...

[Erasmus00]

Well i thought as well that it would be just putting T to zero, but then 1/beta tends to zero as well and the second term diverges to -infinity.

 

Doesn't the second term go to 0?

 

[math] \frac{1}{2\beta}ln(\frac{1}{\beta^2}) = -\frac{1}{\beta}ln(\beta). [/math]

 

So as beta heads to infinity, the log goes to infinity slower then the over beta goes to 0.

-Will

[sanctus]

Yes you are right I saw it as well a couple of hours after I posted it, but actually for T=0 the last term tends to -infinity so there is for sure a minimum.