Finding the Partition Function in Paramagnetism

[Ang Han Wei]

Homework Statement

I was given a Hamiltonian H = -$\mu$B$\sum$cos$\alpha_{i}$
where the sum is over i from i = 1 to i = N

I need the partition function given this Hamiltonian.

Homework Equations

The Attempt at a Solution

I tried using the classical approach where $Z_{N}$ = $\frac{1}{h^{3}}$$\int$$d^{3}$p$d^{3}$q exp[$\beta$$\mu$B$\sum$cos$\alpha_{i}$], but I am stuck at what to do with the cos summation.

Can anyone help?

Thanks!

 

[mark726]

I would like to provide some guidance and suggest a possible solution to this problem. First, let's break down the Hamiltonian into its individual components:

H = -\muB\sumcos\alpha_{i}

= -\muB(cos\alpha_{1} + cos\alpha_{2} + ... + cos\alpha_{N})

= -\muB\sum_{i=1}^{N} cos\alpha_{i}

Now, let's rewrite the partition function in terms of this new form of the Hamiltonian:

Z_{N} = \frac{1}{h^{3N}}\int e^{-\beta H} d^{3N}p d^{3N}q

= \frac{1}{h^{3N}}\int e^{\beta\muB\sum_{i=1}^{N} cos\alpha_{i}} d^{3N}p d^{3N}q

= \frac{1}{h^{3N}}\int e^{\beta\muB(cos\alpha_{1} + cos\alpha_{2} + ... + cos\alpha_{N})} d^{3N}p d^{3N}q

= \frac{1}{h^{3N}}\int e^{\beta\muB cos\alpha_{1}} e^{\beta\muB cos\alpha_{2}} ... e^{\beta\muB cos\alpha_{N}} d^{3N}p d^{3N}q

= \frac{1}{h^{3N}}\int e^{\beta\muB cos\alpha_{1}} d^{3N}p d^{3N}q \times e^{\beta\muB cos\alpha_{2}} d^{3N}p d^{3N}q \times ... \times e^{\beta\muB cos\alpha_{N}} d^{3N}p d^{3N}q

= \left(\frac{1}{h^{3}}\int e^{\beta\muB cos\alpha_{i}} d^{3}p d^{3}q \right)^{N}

= Z_{1}^{N}

where Z_{1} is the partition function for a single particle.

Therefore, the partition function for N particles can be written as the N-th power