Partition Function and Thermodynamics of a Spin-1/2 System

Consider a system consisting of $N$ spin-$\frac{1}{2}$ particles, each of which can be in one of two quantum states, ‘up’ and ‘down’. In a magnetic field $B$, the energy of a spin in the up/down state is $\pm\mu B/2$ where $\mu$ is the magnetic moment. Show that the partition function is \[Z = 2^N \cosh^N\!\left(\frac{\beta\mu B}{2}\right).\tag{1.1}\]

Find the average energy $E$ and entropy $S$. Check that your results for both quantities make sense at $T=0$ and $T\to\infty$.

Compute the magnetisation of the system, defined by $M = N_\uparrow - N_\downarrow$, where $N_{\uparrow/\downarrow}$ are the number of up/down spins. The magnetic susceptibility is defined as $\chi \equiv \pd M/\pd B$. Derive Curie's Law, which states that at high temperatures $\chi \sim 1/T$.

First, consider the partition function of a system with a single particle, $z$: \[z = e^{-\beta\mu B/2} + e^{+\beta\mu B/2} = 2\cosh\left(\frac{\beta\mu B}{2}\right).\tag{1.2}\] Assuming the particles are distinguishable (localised on a lattice), the partition function of a $N$ particle system is, therefore, \[Z = z^N = 2^N\cosh^N\left(\frac{\beta\mu B}{2}\right).\tag{1.3}\]

The average energy is given by \[E = - \frac{\partial\ln Z}{\partial\beta} = -\frac{\mu BN}{2}\tanh\left(\frac{\beta\mu B}{2}\right),\tag{1.4}\] and the entropy is \[S = k_B(\ln Z + \beta E) = Nk_B\left[\ln 2 + \ln\cosh\!\left(\frac{\beta\mu B}{2}\right) - \frac{\beta\mu B}{2}\tanh\!\left(\frac{\beta\mu B}{2}\right)\right].\tag{1.5}\] At $T=0$, $\beta\to\infty$: $\ln\cosh(\beta\mu B/2)\approx \beta\mu B/2$ and $\tanh\to 1$, so the last two terms cancel and $S\to 0$, consistent with a unique ground state. For $T\to\infty$, $\beta\to 0^+$: $\ln\cosh\to 0$ and $\tanh\to 0$, giving \[S \to Nk_B\ln 2,\tag{1.6}\] the maximum entropy corresponding to all $2^N$ microstates being equally probable.

To compute the magnetisation, let us go back to consider the single particle system. The average magnetisation of such a system is given by \[\langle M\rangle = \frac{1}{z}(e^{-\beta\mu B/2}- e^{\beta\mu B/2}) = -\tanh\left(\frac{\beta\mu B}{2}\right).\tag{1.7}\] The magnetisation of a $N$-particle system is simply \[M = -N \tanh\left(\frac{\beta\mu B}{2}\right).\tag{1.8}\]

For large temperature, $\beta\to0^+$, which gives \[\chi \approx-\frac{N\beta\mu}{2} \sim \frac{1}{T}.\tag{1.9}\]

Partition Function and Thermodynamics of a Spin-$\frac{1}{2}$ System