Fall 2007 18 Sm

Fall 2007

Fall 2007 1 Mech
Fall 2007 2 Mech
Fall 2007 3 EM
Fall 2007 4 EM
Fall 2007 5 QM
Fall 2007 6 QM
Fall 2007 7 SM
Fall 2007 8 SM
Fall 2007 9 Misc
Fall 2007 10 Misc
Fall 2007 11 Mech
Fall 2007 12 Mech
Fall 2007 13 EM
Fall 2007 14 EM
Fall 2007 15 QM
Fall 2007 16 QM
Fall 2007 17 SM
Fall 2007 18 SM
Fall 2007 19 Misc
Fall 2007 20 Misc

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 $\mathcal{H}_{int}(\{\sigma_i \} )$.

***

Answer

***

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}

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