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?

  • $\begingroup$ 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: $\partial \ln Z/\partial B = \frac{1}{Z}\frac{\partial Z}{\partial B}$. $\endgroup$ Commented May 11, 2019 at 21:18

1 Answer 1


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 $$ Z_N= (Z_1)^N. $$

  • $\begingroup$ 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? $\endgroup$
    – Fhoenix
    Commented May 12, 2019 at 20:17
  • $\begingroup$ @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? $\endgroup$
    – mike stone
    Commented May 12, 2019 at 21:01
  • $\begingroup$ @mikestone Don't you have to do a sum over $j$ as well? $\endgroup$ Commented Jan 4 at 5:36
  • $\begingroup$ @Dr. user44690 The original problem says that the particle have fixed spin, although the OP replaces $J\equiv j$ by $l$ for some reason. $\endgroup$
    – mike stone
    Commented Jan 4 at 14:22
  • $\begingroup$ @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. $\endgroup$ Commented Jan 5 at 6:55

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