Warning: jsMath requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.

SM J04 1

From QED

< PlasmaWiki(Link to this page as [[PlasmaWiki/Old Prelims/SM J04 1]])

Jump to: navigation, search

Consider N non-interacting quantized spins in a magnetic field $\vec{B}=B\hat{z}$ . The energy of the spins is $- B M z$ , where:

$M_{z}\equiv\mu\sum_{i=1}^{N}S_{z}^{(i)}$

is the total magnetization. For each spin, $S_{z}^{(i)}$ takes only 2S+1 values -S, -S+1, ..., S-1, S. Given the temperature of the system T:

a. Calculate the Gibbs partition function Z(T,B);

b. Calculate the Gibbs free energy G(T,B) and evaluate its asymptotic behavior at weak ( $\mu B\gg k_{B}T$ ) magnetic field;

c. Calculate the zero-field magnetic susceptibility

$\chi\equiv\left(\frac{\partial M_{z}}{\partial B}\right)_{B=0}$

d. Calculate the magnetic susceptibility at strong fields $\mu B\gg k_{B}T$ .

The partition function for one particle is given by:

$Z_{1}=e^{\beta E}=e^{-\beta B\mu S_{z}}$

We can get the expectation of $S z$ by:

$\langle S_{z}\rangle=\frac{\sum_{S_{z}}S_{z}e^{-\beta B\mu S_{z}}}{\sum_{S_{z}}e^{-\beta B\mu S_{z}}}$

Where the sums are over all possible values of $S z$ . Using this, we get the total partition function to be:

$Z(T,B)=\left(e^{-\beta B\mu\langle S_{z}\rangle}\right)^{N}=e^{-\beta B\mu N\langle S_{z}\rangle}$

The Gibbs free energy may be found by:

$G=-k_{B}T\ln(Z)=k_{B}T\beta B\mu N\langle S_{z}\rangle=B\mu N\langle S_{z}\rangle$

In the case that $\mu BS\ll k_{B}T$ , $e βμ B S ˜1$ , and so we find:

$\langle S_{z}\rangle\sim\frac{\sum_{S_{z}}S_{z}}{\sum_{S_{z}}1}=0$

so that the Gibbs free energy $G\rightarrow0$ .

In the other case that $e^{\beta\mu BS}\gg e^{\beta\mu B(S-1)}$ , so:

$\langle S_{z}\rangle\sim\frac{-Se^{\beta B\mu S}}{e^{\beta B\mu S}}=-S$

and the Gibbs free energy $G\rightarrow-B\mu NS$ .

The magnetic susceptibility is defined by:

$\chi\equiv\left(\frac{\partial M_{z}}{\partial B}\right)_{B=0}=\mu N\left(\frac{\partial\langle S_{z}\rangle}{\partial B}\right)_{B=0}$

Taking the derivative

$\left(\frac{\partial\langle S_{z}\rangle}{\partial B}\right)_{B=0}=\frac{\sum_{S_{z}}-\beta\mu S_{z}^{2}}{\sum_{S_{z}}1}+\frac{\sum_{S_{z}}S_{z}}{\sum_{S_{z}}-\beta\mu S_{z}}=\frac{\sum_{S_{z}}-\beta\mu S_{z}^{2}}{2S+1}-\frac{k_{B}T}{\mu}$

Plugging in:

$\chi=-\mu N\left(\beta\mu\frac{\sum_{S_{z}}S_{z}^{2}}{2S+1}+\frac{1}{\beta\mu}\right)$

1.1.4 Part d

This is similar to the last part, but we get:

$\left(\frac{\partial\langle S_{z}\rangle}{\partial B}\right)_{\mu B\gg k_{B}T}=\frac{-S^{2}\mu\beta e^{\beta B\mu S}}{e^{\beta B\mu S}}+\frac{-Se^{\beta B\mu S}}{\beta\mu Se^{\beta B\mu S}}=-S^{2}\mu\beta-\frac{1}{\beta\mu}$

So that: