Drexel Qual 2002 Modern Short

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  7. Suppose a single particle has two possible energy states: <math>-\frac{1}{2}\rho</math> and <math>\frac{1}{2}\rho</math>.
  Show that the average energy at temperature T is simply <math>\frac{1}{2}\rho\ tanh\left(\frac{\rho}{2kT}\right)</math>

We know that:

\left \langle E \right \rangle\ =\ \frac{\sum_{i} E_i\ e^{-\beta E_i}}{Q}\,

Where Q is the partition function.

Using the given values for E:

E\ =\ \pm \frac{1}{2}\rho\,

We get:

\frac{-\frac{1}{2}\rho \left(e^{\frac{1}{2}\rho \beta}-e^{-\frac{1}{2}\rho \beta} \right)}{e^{\frac{1}{2}\beta \rho}\ +\ e^{-\frac{1}{2}\beta \rho}}\,

Using trig identities:

\left \langle E \right \rangle\ =\ \frac{-\frac{1}{2}\ \rho\ sinh(\frac{1}{2}\beta \rho)}{cosh(\frac{1}{2}\beta \rho)}\ =\ - \frac{1}{2}\rho\ tanh\left(\frac{\rho}{2kT}\right)[women looking for men]
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