Fall 2007 18 Sm

The Hamiltonian of N noninteracting spin-1/2 particles in magnetic field H is given by

(1)
\begin{align} \mathcal{H}_0 = - H M \end{align}
(2)
\begin{align} M = \mu \sum_{i = 1}^N \sigma_i \end{align}
(3)
\begin{align} \sigma_i = \pm 1. \end{align}

(a) Calculate the average magnetization <M>, the average square of the magnetization <M2>, and the magnetic susceptibility $\chi$ = (d/dH)<M> at temperature T.

(b) Verify that your results obey the thermodynamic identity

(4)
\begin{align} \left < M^2 \right > - \left < M \right > ^ 2 = k T_{\chi} \end{align}

(c) Prove that the above identity holds even in the presence of interactions, $\mathcal{H}_0 \rightarrow \mathcal{H}_0 + \mathcal{H}_{int}$, for arbitrary $\mathcal{H}_{int}(\{\sigma_i \} )$.

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A shorter derivation can be given if we start with part (c).

(c) The partition function is

(5)
\begin{align} Q = \sum_{\{\sigma_i\}} \exp\left ( \beta H M (\{\sigma_i\} ) - \beta E_{int}(\{\sigma_i\} ) \right ) \end{align}

whence

(6)
\begin{align} \left < M \right > = \frac{1}{\beta Q}\frac{\partial Q}{\partial H} \end{align}
(7)
\begin{align} \left < M^2 \right > = \frac{1}{\beta^2 Q} \frac{\partial^2 Q}{\partial H^2} \end{align}

Now

(8)
\begin{align} \chi = \frac{\partial}{\partial H} \left < M \right > = \frac{1}{\beta Q}\frac{\partial^2 Q}{\partial H^2} - \frac{1}{\beta Q^2}\left(\frac{\partial Q}{\partial H}\right )^2 = \beta ( \left <M^2\right> - \left < M \right>^2) \end{align}

which proves the identity.

(a) We can now apply the above formulas to the problem in hand. We have

(9)
\begin{align} Q = \prod_i \sum_{\sigma_i} \exp (\beta \mu H \sigma_i) = \left [ 2 \cosh (\beta \mu H) \right]^N \end{align}

Taking the requisite derivatives, we find

(10)
\begin{align} \left < M \right > = N \mu \tanh(\beta \mu H) \end{align}
(11)
\begin{align} \left < M^2 \right> = \mu^2 [N (N - 1) \tanh^2(\beta \mu H ) + N ] \end{align}
(12)
\begin{align} \chi = \beta \mu^2 N \sech^2 (\beta \mu H) \end{align}

(b) We have

(13)
\begin{align} \mu^2 [N ( N-1 ) \tanh^2(\beta \mu H)+N]- [N \mu \tanh(\beta \mu H]^2 = \mu^2 N [1 - \tanh^2(\beta \mu H)] \end{align}
(14)
\begin{align} = \mu^2 N \sech^2(\beta\mu H) = k T_{\chi} \end{align}