# Partition Function of System of Atoms in a Magnetic Field

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?

• Probably you have got a wrong result for $Z$. Also it is strange that you've got logarithm function in the result - that should disappear after differentiating: $\partial \ln Z/\partial B = \frac{1}{Z}\frac{\partial Z}{\partial B}$. Commented May 11, 2019 at 21:18

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 $$Z_N= (Z_1)^N.$$
• The thing is that the hamiltonian of each of the particles is said to depend on the $m_j$'s of the others. So, such calculation doesn't free us from the individuals $m_j$'s - $Z_1=\sum_{m_j^1} \exp{\bigg(\beta g\mu_B B\hbar \sum m_j^i \bigg)}\\ = \exp{\bigg(\beta g\mu_B B\hbar \sum_{i\neq 1} m_j^i \bigg)}\sum_{m_j^1} \exp{\bigg(\beta g\mu_B B\hbar m_j^1 \bigg)}\\ = \exp{\bigg(\beta g\mu_B B\hbar \sum_{i\neq 1} m_j^i \bigg)}\sum_{m_j^1} \exp{\bigg(\beta g\mu_B B\hbar m_j^1 \bigg)}$ And again, we face a geometric series, no? Commented May 12, 2019 at 20:17
• @Fhoenix I think youy had better write down the precise total Hamiltonian that you are usuing because I do not see any interactions between spins in your expression for H in your first equation. You do say that it is the "H" for an individual atom, but it has $N$ terms so I took it as $H_{total}$ . If it is only for atom $i$'say, then where is the degree of freedom for that particular atom? Commented May 12, 2019 at 21:01
• @mikestone Don't you have to do a sum over $j$ as well? Commented Jan 4 at 5:36
• @Dr. user44690 The original problem says that the particle have fixed spin, although the OP replaces $J\equiv j$ by $l$ for some reason. Commented Jan 4 at 14:22
• @mikestone So, when the particle doesn't have fixed spin then you have to sum over $j$ as well right? Just confirming for my clarity. Commented Jan 5 at 6:55