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An external magnetic field is on a 1D lattice with N sites where each site has a magnetic moment, which can rotate freely. The magnetic field at the $j^{th}$ site is, $$\mathbf{B}_j = B_0 \cos\left(\frac{2\pi j}{N}\right) \hat{z} + B_0 \sin\left(\frac{2\pi j}{N}\right) \hat{x}.$$ There is no interaction between the magnetic moments.

I need to find the partition function for this system. I have proceeded by trying to evaluate the partition function at the $j^{th} site first$,

\begin{alignat}{1} Z & = \prod_j \int d\Omega_j \, e^{-\beta H_j} \\ & = \prod_j \int_0^{2\pi} d\phi_j \int_0^\pi \sin\theta_j \ d\theta_j \ e^{-\beta H_j} \\ & = \prod_j \int_0^{2\pi} d\phi_j \int_0^\pi \sin\theta_j \ d\theta_j \ e^{\beta m B_0 \left[ \cos\left(\frac{2\pi j}{N}\right) \cos\theta_j + \sin\left(\frac{2\pi j}{N}\right) \sin\theta_j \cos\phi_j \right]} \end{alignat}

where: \begin{alignat}{1} d\Omega_j & = \sin\theta_j \ d\theta_j \ d\phi_j, \\ H_j & = -B_0 \left[ \cos\left(\frac{2\pi j}{N}\right) m_z + \sin\left(\frac{2\pi j}{N}\right) m_x \right] \end{alignat}

I am unable to integrate the partition function at the $j^{th}$ site in the above form, so one way in which I could proceed was orienting the z-axis at each site along the magnetic field at that site. This leads to the same result for the partition function if we had a uniform magnetic field like $\mathbf{B}= B_{o}\hat{z}$ . If this is the correct method, how do I interpret this result?

HypnoticZebra
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1 Answers1

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I would say the result is just a statement about the structure of the phase/configuration space, which is spherical with no preferred direction(s) in the case of a magnetic moment.

The partition function $Z$ is a function of $\beta$ and $\vec{B}$: $$Z(\beta, \vec{B}) = \int \mathrm{d}\vec{s} e^{\beta \vec{B}\cdot\vec{s}},$$ where $\vec{s}$ takes all allowed values in the configuration space. The computation indicates that $Z$ does not depend on the direction of $\vec{B}$. In other words, if one maps $\vec{B}$ to a rotated vector $\vec{B'}$, one can find a change of integration variable $\vec{s'}$ that restores the original expression for $Z$.

What does this mean? Imagine instead that the moments could only be aligned along $\pm \hat{z}$. Here, the direction of $\vec{B}$ clearly matters when finding $Z_j$ $-$ there is no reparametrization of the configuration space that maintains the expression. $$Z_j = 2\cosh\left(\beta B_0 \cos\left(\frac{2\pi j}{N}\right)\right).$$

This indicates that the structure of the configuration space is important. For a magnetic moment which is unrestricted in its orientation, the sum over allowed states (the entire spherical shell) in $Z$ ensures that $\vec{B}$'s direction doesn't matter.

It might also be helpful to look at the answers to these questions, which try to give a physical meaning to the partition function.

Crisco
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