Consider an Ising model of just two elementary dipoles, whose mutual interaction energy is $\pm ϵ$. Enumerate the states of this system and write down their Boltzmann factors. Calculate the partition function. Find the probabilities of finding the dipoles parallel and antiparallel, and plot these probabilities as a function of $kT/ϵ$. Also calculate and plot the average energy of the system. At what temperatures are you more likely to find both dipoles pointing up than to find one up and one down?

The simplified model of a magnet is defined as Ising model. For an Ising model of just two elementary dipoles, the energy is $-\epsilon$when the dipoles are parallel and $+\epsilon$when the dipoles are antiparallel.

The states of the system on their Boltzmann factors are as follows:

$\uparrow \uparrow :e^{\epsilon /kT}$

$\uparrow \downarrow :e^{-\epsilon /kT}$

$\downarrow \uparrow :e^{-\epsilon /kT}$

$\downarrow \downarrow :e^{\epsilon /kT}$

Here,

$k$=Boltzmann factor

$T=$temperature.

The above four equations represent Boltzmann's factors for the given four states.

For $\epsilon$,

$Z_1=2\exp \left(\frac{\epsilon }{kT}\right)$

For $-\epsilon$,

$Z_2=2\exp \left(-\frac{\epsilon }{kT}\right)$

The partition function for the system with two elementary dipoles is as follows:

$Z=Z_1+Z_2$

Substituting the value of $2\exp \left(\frac{\epsilon }{kT}\right)\text{=}{Z}_1\text{and}2\exp \left(-\frac{\epsilon }{kT}\right)\text{=}{Z}_2\text{.}$

$Z=2\exp \left(\frac{\epsilon }{kT}\right)+2\exp \left(-\frac{\epsilon }{kT}\right)$

$=4\cos h\left(\frac{\epsilon }{kT}\right)$

The partition function for the system with two elementary dipoles =$=4\cos h\left(\frac{\epsilon }{kT}\right)$

${\mathrm{P}}_{\text{parallel}}=\frac{Z_2}{Z}$

$\text{Substitutingthevalueof}2\exp \left(\frac{\epsilon }{kT}\right)\text{=}{Z}_2\text{and}2\exp \left(\frac{\epsilon }{kT}\right)+2\exp \left(-\frac{\epsilon }{kT}\right)\text{=}Z$

${\mathrm{P}}_{\text{parallel}}=\frac{2\exp \left(\frac{\epsilon }{kT}\right)}{2\exp \left(\frac{\epsilon }{kT}\right)+2\exp \left(-\frac{\epsilon }{kT}\right)}$

$=\frac{1}{1+\exp \left(-\frac{2\epsilon }{kT}\right)}$

Thus, the probability that the dipoles are parallels is=$=\frac{1}{1+\exp \left(-\frac{2\epsilon }{kT}\right)}$

${\mathrm{P}}_{\text{antiparallel}}=\frac{Z_1}{Z}$

$\text{Substitutingthevalueof}2\exp \left(-\frac{\epsilon }{kT}\right)\text{=}{Z}_1\text{and}2\exp \left(\frac{\epsilon }{kT}\right)+2\exp \left(-\frac{\epsilon }{kT}\right)\text{=}Z\text{.}$

${\mathrm{P}}_{\text{antiparallel}}=\frac{2\exp \left(-\frac{\epsilon }{kT}\right)}{2\exp \left(\frac{\epsilon }{kT}\right)+2\exp \left(-\frac{\epsilon }{kT}\right)}$

$=\frac{1}{1+\exp \left(\frac{2\epsilon }{kT}\right)}$

Thus, the probability that the dipoles are anti-parallels =$=\frac{1}{1+\exp \left(\frac{2\epsilon }{kT}\right)}$

Average energy of the system=

$U=⟨E⟩$

$=-\frac{1}{Z}\frac{\partial Z}{\partial \beta }$

Substituting the value of $4\cos h\left(\frac{\epsilon }{kT}\right)\text{=}Z\text{.}$

$U=\frac{\epsilon \left(4\sin h\right(\epsilon /kT\left)\right)}{4\cos h(\epsilon /kT)}$

$=-\epsilon \tanh (\epsilon /kT)$

$\frac{U}{\epsilon }=-\tanh (\epsilon /kT)$