# 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

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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.
$$χ = \frac{2 n μ_0 g^2 {μ_B}^2}{3 k_B T}$$
where βΔ part is omitted.

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

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Homework Helper
Gold Member
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.

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