# Statistical Mechanics: classical Heisenberg Model

1. Jul 27, 2011

### eXorikos

1. The problem statement, all variables and given/known data
You have a latice of particles that all have spin 1, but they can change the direction of their spin so constraint $\left|S_j\right|=1$. There is only interaction with the closest neighbours so we have the following hamiltonian:

$H = -J \sum_{\left\langle ij \right\rangle} \vec{S_i} \cdot\vec{S_j} - \vec{h} \cdot \sum^{N}_{j = 1} \vec{\vec{S_j}}$

Choose a good orderparameter to treat this in the molecular field approximation. Calculate the selfconsistent equation for this order parameter and determine the spontaneous magnetisation for $T<T_c=Jq/3k_b$.

2. Relevant equations
$Z=\int_{\left|S_1\right|=1}\cdots \int_{\left|S_N\right|=1} d^3 S_1 \cdots d^3 S_N \exp{\left(-\beta H\right)}$
$M = \frac{1}{\beta} \nabla_h ln Z$

3. The attempt at a solution
As order parameter I pick $\vec{M} = \sum_j \vec{S_j}$ and than I approximate the hamiltonian with q nearest neighbors by
$H = \frac{-Jq}{2N} \left(\sum^N_{j=1} \vec{S_j}\right)^2 - \vec{h} \cdot \sum^{N}_{j = 1} \vec{\vec{S_j}}$

This gives
$Z=\int_{\left|S_1\right|=1}\cdots \int_{\left|S_N\right|=1} d^3 S_1 \cdots d^3 S_N \exp{\left(\frac{\beta Jq}{2N} \left(\sum^N_{j=1} \vec{S_j}\right)^2 + \beta \vec{h} \cdot \sum^{N}_{j = 1} \vec{\vec{S_j}} \right)}$

But I can't manage the integral. How do I calculate this integral? The rest I presume is correct?

2. Aug 2, 2011

### eXorikos

Anybody?

3. Aug 17, 2011

### eXorikos

*Bump*

4. Aug 22, 2011

### Mute

In your integral, replace one factor of $\sum_i S_i$ with your order parameter, M. Your integral is then

$$Z=\int_{\left|S_1\right|=1}\cdots \int_{\left|S_N\right|=1} d^3 S_1 \cdots d^3 S_N \exp\left[\left(\frac{\beta Jq\vec{M}}{2N} + \beta \vec{h}\right) \cdot \sum^N_{j=1} \vec{S_j}\right].$$

Treating M as independent of the S_i, the different sites are no longer coupled and you can now perform the integrals. If you then compute M by $M = \beta^{-1}\nabla_h \ln Z$, you will find the right hand side has M in it - this is your self-consistent equation for M. (Well, you have an equation for each component of M).

Analyze the self-consistent equation(s) and show that at some temperature there is a phase transition (i.e., a new solution to the self consistent equation appears).

Last edited: Aug 22, 2011
5. Sep 2, 2011

### eXorikos

Am I on the right track trying to use spherical coordinates or is there another way of calculating the integral more easy? Because then I can keep R=1 and integrate both angles respectively from 0 to pi and 0 to 2pi.

Last edited: Sep 2, 2011
6. Sep 3, 2011

### eXorikos

I'll write where I'm at now:
Set $$\vec{h}^{eff}=\frac{\beta Jq\vec{M}}{2N} + \beta \vec{h}$$
$$Z=\int_{\left|S_1\right|=1}\cdots \int_{\left|S_N\right|=1} d^3 S_1 \cdots d^3 S_N \exp\left[\vec{h}^{eff} \cdot \sum^N_{j=1} \vec{S_j}\right]$$
$$= \prod_{j=1}^{N} \int_{\left|S_j\right|=1} d^3S_jexp\left(\vec{h}^{eff} \cdot \vec{S_j}\right)$$

Now trying to calculate it seems easier to work with spherical coordinates because of the condition that $\left|S_j\right|=1$, but I'm guessing there is some kind of technique I don't know about for calculating this integral. Because doing the dot product in spherical coordinates will make life hard.

7. Sep 4, 2011

### Mute

The dot product isn't hard - it's just $h_x^{eff}S_x + h_y^{eff}S_y + h_z^{eff}S_z$. You then switch to spherical coordinates by the relations $S_x = \cos\phi\sin\theta$, $S_y = \sin\phi\sin\theta$ and $S_z = \cos\theta$. So, your integral is then

$$Z_j= \int_0^\pi d\theta \int_0^{2\pi} d\phi~\sin^2\theta \exp\left[ h_x^{eff}\cos\phi\sin\theta + h_y^{eff}\sin\phi\sin\theta + h_z^{eff}\cos\theta\right]$$

This integral, however, is still rather hard to do.

I think what may be easier to due is use $h^{eff} \cdot S = |h^{eff}||S|\cos\gamma = |h^{eff}|\cos\gamma$, where $\gamma$ is the angle between the vector $h^{eff}$ and $S$. $\gamma$ runs from 0 to $\pi$. There's still another angle $\phi$ which runs over 0 to 2*Pi, but the integrand no longer depends on it so you just get a factor of 2*Pi. Now, I'm afraid I don't remember the Jacobian off the top of my head, so you'll have to work that out.

$$Z_j= 2\pi \int_0^{\pi} d\gamma~\mathcal J(\gamma) \exp\left[|h^{eff}|\cos\gamma\right]$$

I don't know if the resulting integral will be doable. It may be expressible in terms of some sort of Bessel function (or a modified Bessel function).

8. Sep 4, 2011

### eXorikos

That's all way to complicated to solve in an exam of 4 ours together with two other questions. This was a question of a previous exam so it shouldn't be so convoluted. Thanks for your help!

9. Sep 6, 2011

### Mute

If I were in an exam situation and I just wanted to find the critical temperature, I would probably try assuming that there is a disordered phase in which the total magnetization was zero, and then choose $\mathbf{h} = h_z \hat{z}$. Near the transition the z component of the magnetization is probably the one that would acquire a non-zero value, so the integral you would have to do is

$$Z_j= 2\pi \int_0^{\pi} d\theta~\sin^2\theta \exp\left[h^{eff}_z\cos\theta\right]$$
This apparently can be evaluated in terms of a modified Bessel function, but unless you've memorized that it probably won't be too handy on an exam. So, what I would do now is take a derivative with respect to h_z. This brings down a factor of cos(theta) and a beta, so you get

$$M_z = 2\pi \int_0^{\pi} d\theta~\sin^2\theta \cos\theta\exp\left[h^{eff}_z\cos\theta\right],$$

which still isn't doable. However, again, you just want the transition, so we can do two things: 1) set h_z = 0 now. From the Ising model, you should have some intuition that there is no transition for finite h. The transition you're looking for is a zero-field transition. 2) Assume M_z is small and expand the exponential.

You'll find you have to expand the exponential to 3rd order in M_z, as the zeroth and second order terms vanish, and you can't have just the first order term or you'd have M = stuff*M.

Doing the expansion (quickly - you'll have to check my work) yields something that looks like

$$M_z = \frac{2\pi\beta Jq}{2N}I_1 M_z + 2\pi\left(\frac{\beta Jq}{2N}\right)^3I_2 M_z^3$$
where the I's are integrals that you could do because they're just trigonometric. Anywho, we don't even care about I2. Cancelling a factor of M_z you get something like

stuff*M_z^2 = 1 - other stuff.

Now, the 1- other stuff obviously has to be positive, otherwise you have no solution other than M_z = 0 (which we divided out). So, solve (1 - other stuff) = 0 for T to find the transition temperature. I'm don't think I got the exact result you quoted in your first post, but it'll be close, and for a time-constrained test it would hopefully get you most of the points for the problem.

By the way, I might redefine your order parameter as $M = (1/N)\sum_j S_j$. That might get rid of N in your equations.