Find magnetic susceptibility using partition function

[cozycoz]

Homework Statement

A certain magnetic system contains n independent molecules per unit volume, each of which has four energy levels given by 0, $Δ-gμ_B B$, $\Delta$, $\Delta +gμ_B B$. Write down the partition function, compute Helmholtz function and hence compute the magnetization $M$. Hence show that the magnetic susceptibility $χ$ is given by $$χ = \lim_{B\rightarrow 0} {\frac{μ_0 M}{B}} = \frac{2 n μ_0 g^2 {μ_B}^2}{k_B T (3+e^{Δ/k_B T})}$$

(Sorry I don't know how to type limit in mathematical form)
\lim_{n\rightarrow +\infty} {\frac{\sin(x)}{x}}

Homework Equations

$F=-k_B T \ln Z$, $m=-(\frac{∂F}{∂B})$, $m=MV$

The Attempt at a Solution

[/B]
If there are N=Vn molecules, the partition function is $$Z=(e^{0}+e^{-β(Δ-gμ_B)}+e^{-βΔ}+e^{-β(Δ+gμ_B)})^N$$.
Then Helmholtz free energy function F is
$$F=-k_B T[-βΔ+\ln (1+2\cosh {βgμ_B B})]$$

Now I have to differentiate F by B, but I've noticed βΔ part would be gone because it doesn't have B at all.
My answer is
$$χ = \frac{2 n μ_0 g^2 {μ_B}^2}{3 k_B T}$$
where βΔ part is omitted.

If you tell me what's wrong it would be lovely.

 

[Charles Link]

It is unclear how you got $F$ from $Z$. It also isn't apparent how , given the $F$ that you have presented, that a derivative w.r.t. $B$ will not be a function of the magnetic field $B$, since $cosh'(x)=sinh(x)$. Also, in writing out $Z$, you left off the $B$ dependence. You need to be more accurate in presenting it.

 

[VKint]

Your expression for the free energy is incorrect. My best guess is that you forgot about the state with 0 energy when taking the natural log of $Z$.

Let me show you how I did the problem:

Let $\zeta$ denote the one-particle partition function, i.e.,
$$\begin{eqnarray*}
\zeta &= 1 + e^{-\beta \Delta} (1 + e^{\beta g \mu_B B} + e^{-\beta g \mu_B B}) \\
&= 1 + e^{-\beta \Delta} (1 + 2 \cosh (\beta g \mu_B B))
\end{eqnarray*}$$
Then $Z = \zeta^N$, so
$$\begin{eqnarray*}
m &= -\frac{\partial}{\partial B} (F) \\
&= \frac{\partial}{\partial B} (kT \ln Z) \\
&= \frac{kTN}{\zeta} \frac{\partial \zeta}{\partial B} \\
&= \frac{(kTN) (2\beta g \mu_B) e^{-\beta \Delta} \sinh(\beta g \mu_B B) }{ 1 + e^{-\beta \Delta} (1 + 2 \cosh (\beta g \mu_B B)) }
\end{eqnarray*}$$
We then have
$$ \chi = \lim_{B \to 0} \frac{\mu_0 m}{VB} = \lim_{B \to 0} \frac{\mu_0}{B} \frac{2ng \mu_B e^{-\beta \Delta} \sinh(\beta g \mu_B B) }{ 1 + e^{-\beta \Delta} (1 + 2 \cosh(\beta g \mu_B B)) } $$
Since we're computing a limit as $B \to 0$, we need only expand the top and bottom of this fraction to order $B$. Doing so yields the correct answer.

 

[cozycoz]

Your expression for the free energy is incorrect. My best guess is that you forgot about the state with 0 energy when taking the natural log of $Z$.

Let me show you how I did the problem:

Let $\zeta$ denote the one-particle partition function, i.e.,
$$\begin{eqnarray*}
\zeta &= 1 + e^{-\beta \Delta} (1 + e^{\beta g \mu_B B} + e^{-\beta g \mu_B B}) \\
&= 1 + e^{-\beta \Delta} (1 + 2 \cosh (\beta g \mu_B B))
\end{eqnarray*}$$
Then $Z = \zeta^N$, so
$$\begin{eqnarray*}
m &= -\frac{\partial}{\partial B} (F) \\
&= \frac{\partial}{\partial B} (kT \ln Z) \\
&= \frac{kTN}{\zeta} \frac{\partial \zeta}{\partial B} \\
&= \frac{(kTN) (2\beta g \mu_B) e^{-\beta \Delta} \sinh(\beta g \mu_B B) }{ 1 + e^{-\beta \Delta} (1 + 2 \cosh (\beta g \mu_B B)) }
\end{eqnarray*}$$
We then have
$$ \chi = \lim_{B \to 0} \frac{\mu_0 m}{VB} = \lim_{B \to 0} \frac{\mu_0}{B} \frac{2ng \mu_B e^{-\beta \Delta} \sinh(\beta g \mu_B B) }{ 1 + e^{-\beta \Delta} (1 + 2 \cosh(\beta g \mu_B B)) } $$
Since we're computing a limit as $B \to 0$, we need only expand the top and bottom of this fraction to order $B$. Doing so yields the correct answer.

Yes i forgot extra 1 taking log..
Thanks a lot