Statistical Mechanics: classical Heisenberg Model

I don't know off the top of my head. I'll have to think about it.As for the phase transition, I think you'll have to do the integral numerically and then plot M vs. T. Then you can look for where the derivative of M with respect to T changes sign to find where the phase transition occurs.
  • #1
eXorikos
284
5

Homework Statement


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].

Homework 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]

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?
 
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  • #2
Anybody?
 
  • #3
*Bump*
 
  • #4
eXorikos said:

Homework Statement


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].

Homework 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]

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?

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).
 
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  • #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.
 
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  • #6
Mute said:
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).

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.
 
  • #7
eXorikos said:
I'll write where I'm at now:
Set [tex]\vec{h}^{eff}=\frac{\beta Jq\vec{M}}{2N} + \beta \vec{h}[/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.

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).
 
  • #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!
 
  • #9
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.
 

1. What is the classical Heisenberg model in Statistical Mechanics?

The classical Heisenberg model is a mathematical model used in Statistical Mechanics to describe the behavior of magnetic materials, such as metals, at the atomic level. It takes into account the interactions between the magnetic moments of individual atoms and their neighboring atoms.

2. How does the classical Heisenberg model differ from the quantum Heisenberg model?

The classical Heisenberg model describes the behavior of magnetic materials using classical mechanics, while the quantum Heisenberg model uses quantum mechanics. In the classical model, the magnetic moments of atoms are treated as vectors, while in the quantum model, they are treated as operators.

3. What are the key assumptions of the classical Heisenberg model?

The classical Heisenberg model assumes that the magnetic moments of atoms are fixed in direction and magnitude, and that they only interact with their nearest neighbors. It also assumes that the magnetic moments are independent and do not interact with other external magnetic fields.

4. What is the significance of the classical Heisenberg model in studying magnetic materials?

The classical Heisenberg model allows scientists to predict the behavior of magnetic materials at different temperatures and magnetic fields. It also helps in understanding the phase transitions and critical phenomena of these materials.

5. How is the classical Heisenberg model used in practical applications?

The classical Heisenberg model is used in various fields, including material science, condensed matter physics, and engineering. It helps in the design and development of magnetic materials for applications such as data storage, magnetic sensors, and magnetic recording devices.

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