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Homework Help: Statistical Mechanics: classical Heisenberg Model

  1. Jul 27, 2011 #1
    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 [itex]\left|S_j\right|=1[/itex]. There is only interaction with the closest neighbours so we have the following hamiltonian:

    [itex]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}}[/itex]

    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 [itex]T<T_c=Jq/3k_b[/itex].

    2. Relevant equations
    [itex]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)}[/itex]
    [itex]M = \frac{1}{\beta} \nabla_h ln Z[/itex]

    3. The attempt at a solution
    As order parameter I pick [itex]\vec{M} = \sum_j \vec{S_j}[/itex] and than I approximate the hamiltonian with q nearest neighbors by
    [itex]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}}[/itex]

    This gives
    [itex]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)}[/itex]

    But I can't manage the integral. How do I calculate this integral? The rest I presume is correct?
  2. jcsd
  3. Aug 2, 2011 #2
  4. Aug 17, 2011 #3
  5. Aug 22, 2011 #4


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    In your integral, replace one factor of [itex]\sum_i S_i[/itex] with your order parameter, M. Your integral is then

    [tex]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].[/tex]

    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 [itex]M = \beta^{-1}\nabla_h \ln Z[/itex], 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
  6. Sep 2, 2011 #5
    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
  7. Sep 3, 2011 #6
    I'll write where I'm at now:
    Set [tex]\vec{h}^{eff}=\frac{\beta Jq\vec{M}}{2N} + \beta \vec{h}[/tex]
    [tex]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][/tex]
    [tex]= \prod_{j=1}^{N} \int_{\left|S_j\right|=1} d^3S_jexp\left(\vec{h}^{eff} \cdot \vec{S_j}\right)[/tex]

    Now trying to calculate it seems easier to work with spherical coordinates because of the condition that [itex]\left|S_j\right|=1[/itex], 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.
  8. Sep 4, 2011 #7


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

    [tex]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][/tex]

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

    I think what may be easier to due is use [itex]h^{eff} \cdot S = |h^{eff}||S|\cos\gamma = |h^{eff}|\cos\gamma[/itex], where [itex]\gamma[/itex] is the angle between the vector [itex]h^{eff}[/itex] and [itex]S[/itex]. [itex]\gamma[/itex] runs from 0 to [itex]\pi[/itex]. There's still another angle [itex]\phi[/itex] 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.

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

    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).
  9. Sep 4, 2011 #8
    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!
  10. Sep 6, 2011 #9


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    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 [itex]\mathbf{h} = h_z \hat{z}[/itex]. 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

    [tex]Z_j= 2\pi \int_0^{\pi} d\theta~\sin^2\theta \exp\left[h^{eff}_z\cos\theta\right][/tex]
    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

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

    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

    [tex]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[/tex]
    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 [itex]M = (1/N)\sum_j S_j[/itex]. That might get rid of N in your equations.
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