Problem:
Consider a system of N atoms in a magnetic field $B$ pointing along the z-axis. Each atom has angular momentum J and the Hamitonian of each atom is
$$H=-MB=-g\mu_B B\sum_i^N J_z^i$$
where $J_z^i$ is the angular momentum in the $z$ direction of the $i$-th particle, $g$ is the Lande factor, and $\mu_B$ is the Bohr magneton.
a) Find the partition function as a function of $\beta$ and $B$.
b) Find the total Magnetization.
Problems I encountered:
So when trying to solve this problem I'm stuck in trying to find the partition function. So, I believe that the Canonical Ensemble is the best choice since $N$ is fixed and since I don't see a good way to count the states. Therefore, I'd have that
$$Z=\sum_{states}\exp{\bigg(-\beta H\bigg)}$$
I know that the $J_z^i=m_j^i\hbar$ and that these $m_j^i$ will range from $-(l+1)$ to $(l+1)$ in integer steps, so I create a $m_{j_{T}}$ that will range from $-N(l+1)$ to $N(l+1)$ so that
$$Z=\sum_{m_{j_{T}}}\exp{\bigg(\beta g\mu_B B m_{j_{T}}\hbar \bigg)}$$
I don't believe this is correct, because of what we get for the magnetization as I will show. Assuming it's, I believe we would then calculate the magnetization to be:
$$\langle M \rangle = \frac{1}{\beta}\frac{\partial}{\partial B}\big( \ln{Z}\big)$$
where I just used that $\langle M \rangle = \frac{1}{Z}\sum_{m_{j_{T}}} g\mu_B m_{j_{T}}\hbar \exp{\bigg(\beta g\mu_B B m_{j_{T}}\hbar \bigg)}$.
All good until we insert what $Z$ is and calculate it via regular geometric series calculations, use $N>>1$ and taylor expansion of the $\exp$ function:
$$\langle M \rangle\approx \bigg(-N(l+1)\beta g\mu_B B\hbar\bigg) + \ln{\Bigg(\frac{\exp{\bigg(\beta g\mu_B B\hbar\bigg)}^{2N(l+1)}}{-\beta g\mu_B B\hbar}\Bigg)}$$
We would be taking the log of a negative argument, which doesn't make sense.
What am I missing?
Best Answer
Why the geometric progression calculation? I don't see where that comes from.
If you find $Z_1$ for one atom, then the partition function for $N$ of them is just $Z_1^N$. For example for spin 1/2 we have $$ Z_1= e^{\beta \mu B/2}+ e^{-\beta \mu B/2}= 2 \cosh(\beta \mu B/2) $$ and $$ Z_N = (2 \cosh(\beta \mu B/2)^N $$ giving $$ M= \frac 12 N \mu \tanh (\beta\mu B/2). $$ For spin $j$, you do need a bit of a GP for each atom:
$$ Z_1= e^{\beta \mu B j}+ e^{\beta \mu B (j-1)}+\ldots + e^{-\beta \mu B (j-1)}+e^{-\beta \mu j}= \frac{\sinh( \beta (2j+1)\mu B/2)}{\sinh (\beta j \mu B/2)} $$ but again $$ Z_N= (Z_1)^N. $$