In class we were computing the partition function given the following Hamiltonian:

$$H = - g\mu_B \sum_{i} \vec{h}.\vec{S_i}$$

where $\vec{h}$ is the external magnetic field, and $\vec{S_i}$ is the direction of the spin of $i-th$ particle. I understand how to exactly solve this partition function.

However, my teacher also said the following statement regarding the classical limit. In this limit the spin can be replaced by a classical vector of length $S$. I don't understand this statement properly. Also a consequence of this is that now the partition function can be written in the classical limit to be:

$$Z=\Big[2\pi\int_{0}^{\pi} d\theta sin\theta e^{-\beta g \mu_B hScos\theta}\Big]^N$$

Can somebody shed some light? Thanks.


1 Answer 1


In the case of quantum mechanics, the spin values $S_i$ can take two values: up (value is +1/2) and down (value is -1/2). Therefore you have a discrete configuration space (discreteness also for particle in the box or the harmonic oscillator in quantum statistics).

In classical mechanics however, you have a continuous phase space and you assume spins as simple rotating rods with equal length. The configuration space here for the i-th spin is

$(\phi_i,\theta_i), \phi_i \in [0, 2\pi], \theta_i \in [0, \pi]$. (spherical coordinates)

Now you can calculate the dot product of $\vec{h}$ (constant direction; it is frequently assumes that it points in the cartesian z-direction) and $\vec{S_i}$; this is, by elementary vector algebra:

$\vec{h} \vec{S_i} = hS_i cos \theta$. ($h$ and $S_i$ are all constant; $S_i = S$, i.e. every "spinning rod" has equal length)

Now, the partition function must be a multi-dimensional integral over $d \phi_i d \theta_i sin \theta_i$ (the surface element of the sphere) for all spins i. There occur equal factors for all $N$ spins and therefore you will have the partition function for one single spin but to the power $N$. Because integrand does not depend on $\phi$, integration of it factors out to the factor $2 \pi$.

  • $\begingroup$ Ohh, so the idea here is that in the large spin limit, the allowed quantized spin values tend to become continuous, so we get the classical limit? $\endgroup$ Apr 28, 2018 at 15:19
  • $\begingroup$ So here in the expression for the partition function, if the max value of $S_z$ is given by $j\hbar$, then $S$ is $\hbar^2 j(j+1)$, right? $\endgroup$ Apr 28, 2018 at 15:29
  • $\begingroup$ yes, this is right. $\endgroup$
    – kryomaxim
    Apr 28, 2018 at 19:30

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