Magnetism - Large Spin Limit 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.
 A: 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 $2 \pi$.
