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Spins on a 3D cubic lattice

  1. Oct 13, 2015 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider a system of rotors, each of them placed at one node of a 3D cubic lattice. The system is known as a Heisenberg spin model. Each rotor is represented by a vector of unit length ##\mathbf S(\mathbf r)##, where ##\mathbf r## is its position on the lattice. The Hamiltonian characterising this system is defined via $$-\beta \mathcal H = \sum_{\mathbf r} \mathbf h \cdot \mathbf S(\mathbf r) + K \sum_{\langle \mathbf r, \mathbf r' \rangle} \mathbf S(\mathbf r) \cdot \mathbf S(\mathbf r'),$$ with K>0 and ##\mathbf h## an external magnetic field, assumed to be aligned with the z axis so that ##\mathbf h = (0,0,h)## WLOG.

    Let the easier hamiltonian be ##-\beta \mathcal H_o = \sum_{\mathbf r} \mathbf H \cdot \mathbf S(\mathbf r)##, with ##\mathbf H = (0,0,H)##.

    a)Show that the associated partition function ##Z_N^{(0)}## is given by $$Z_N^{(0)} = \left(\frac{4 \pi \sinh H}{H}\right)^N$$

    b) Find the average of the magnetisation ##\langle \mathbf S \rangle = (\langle S_x \rangle, \langle S_y \rangle, \langle S_z \rangle )## with respect to this easier hamiltonian.

    2. Relevant equations

    $$Z_N = \sum_{\sigma} e^{-\beta \mathcal H_o},$$ where the sum over sigma stands for sum over all states.

    3. The attempt at a solution
    So I imagine a cubic lattice with a small vector pointing in an arbitrary direction at each vertex of the cube. ##\mathbf H \cdot \mathbf S = HS_{z,i}## is the projection of spin on node ##i## onto the z direction, where the H field is aligned. The sum over all states becomes an integral over the possible projections of each spin onto the H field. If we put the H field coinciding with the z axis in spherical polars, then $$\mathbf H \cdot \mathbf S = HS \cos \theta, \,\,\,\theta \in [0,\pi].$$ So have $$\sum_{\sigma} e^{-\beta \mathcal H_o} \rightarrow \int d^3 \mathbf r \prod_{\mathbf r} \exp(H S\cos \theta)$$ Am I thinking about this in the right way? Thanks!
     
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  3. Oct 13, 2015 #2

    TSny

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    I think you have a good start. However, I don't understand your notation in the last integral. The vector ##\mathbf r## is a position vector for a rotor in the lattice. So, it takes on ##N## discrete values. Think about how you would describe a particular state, ##\sigma##, of the system.
     
  4. Oct 13, 2015 #3

    CAF123

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    Hi TSny :),
    In a given state of the system, each rotor is pointing in some random direction relative to the H field. For each point on the lattice, can compute the corresponding projection of the spin rotor onto the H axis. So the 'sum' over all states is where we consider all possible projections of each rotor onto the H axis at each lattice point, In the easy hamiltonian, the projection of a rotor onto the H axis at any given lattice point is not dependent on that of another rotor at another lattice point. So, $$Z_N = \int d^3 \mathbf r \delta(\mathbf r = a_i \hat r_i) \prod_{i=1}^{N} \exp(H|S_i| \cos \theta),$$ where the delta function constrains the integral over all positions to lattice points (so removes the r integral), ##a_i## is the distance of the site ##i## from some origin and ##|S_i|=1## for all i. Because each projection at each lattice point is independent of the others, can write $$Z_N = \left[\int_0^{2\pi} d \phi \int_0^{\pi} \sin \theta d\theta \exp(H \cos \theta)\right]^N = \dots = \left[\frac{4 \pi \sinh H}{H}\right]^N$$

    Does it look ok?

    For the average magnetisation, is it right to say that ##S_x, S_y, S_z## are given by $$\langle S_i \rangle = \frac{1}{Z_N} \sum S_i e^{-\beta \mathcal H_o}$$ where the sum is over all possible values of ##S_i, i=1,2,3## labelling the x,y,z axes respectively? Is it right that ##S_i## is the magnetisation of the rotor on the ith axis? I would say the average magnetisation in the x and y directions are zero because there is no field in these directions so on average these components would be cancel out (there is no external field to drive the rotor to want to be on these axes so the distributions of positions about these axes is purely due to random motion). Is that right?
     
  5. Oct 13, 2015 #4

    TSny

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    I agree with your second equation for ##Z_N##. The first equation looks odd to me. How does that equation reduce to the second equation? In particular, how does the integration over ##\theta## and ##\phi## come from your first equation?

    The phase space of one rotor is the space of all orientations of the rotor; i.e., the possible values of ##\theta## and ##\phi## for the rotor. Thus, summing over states for this rotor amounts to integrating over ##\theta## and ##\phi## for the rotor. The rotors are independent, so ##Z_N = Z_1^N##, where##Z_1## is the partition function of a single rotor.

    If you wanted to express the partition function for the N-particle system in terms of a sum over states of the system, then you would consider the phase space of the system to be the the 2N-dimensional space of all ##\theta##'s and ##\phi##'s of the rotors. Summing over states would imply integrating over this 2N-dimensional space. But there would not be an integral over ##\mathbf{r}##.

    I think this is right. Show that this reduces to ##\langle S_i \rangle = \frac{1}{Z_1} \sum S_i e^{H \cos \theta}## where the sum is over all states of a single rotor.

    Yes. You can verify that ##\langle S_x \rangle## and ##\langle S_y \rangle##will be zero by calculation.
     
    Last edited: Oct 13, 2015
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