In problem 6.20 you computed the partition function for a quantum harmonic oscillator :$Z_{\mathrm{h}.\mathrm{o}.}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left(1-e^{-\beta \epsilon }\right)$}\right.$, where $\epsilon =hf$is the spacing between energy levels.

(a) Find an expression for the Helmholtz free energy of a system of $N$harmonic oscillators.

(b) Find an expression for the entropy of this system as a function of temperature. (Don't worry, the result is fairly complicated.)

The partition function of a single harmonic oscillator is given by

The formula to calculate the Helmholtz free energy for th single harmonic oscillator is given by

$F=-kT\ln Z.....................\left(2\right)$

Substitute the value of $Z$from equation (1) into equation (2) to calculate the free energy for single harmonic oscillator.

$\begin{array}{rcl}F& =& -kT\ln \left[\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left(1-e^{-\beta \epsilon }\right)$}\right.\right]\\ & =& kT\ln \left(1-e^{-\beta \epsilon }\right).......................\left(3\right)\end{array}$

Since, Helmholtz free energy is an extensive property, multiply both sides of equation (3) by $N$to obtain the required free energy for $N$harmonic oscillators.

role="math" localid="1647054790522" $\begin{array}{rcl}{F}_N& =& NF\\ & =& NkT\ln \left(1-e^{-\beta \epsilon }\right)\end{array}$

Here,$F_N$is the Helmholtz free energy for$N$harmonic oscillator.

The formula to calculate the entropy$\left(S\right)$of the system is given by

$S=-{\left(\frac{\partial F}{\partial T}\right)}_N..........................\left(4\right)$

Substitute the formula for Helmholtz free energy from equation (3) into equation (4) and simplify to obtain the required entropy of the system.

role="math" localid="1647054991723" $\begin{array}{rcl}S& =& -\frac{\partial }{\partial T}\left[NkT\ln \left(1-e^{-\beta \epsilon }\right)\right]\\ & =& -Nk\ln \left(1-e^{-\beta \epsilon }\right)-NkT{\left(1-e^{-\beta \epsilon }\right)}^{-1}\epsilon e^{-\beta \epsilon }\left(\frac{\partial \beta }{\partial \epsilon }\right)\\ & =& -Nk\ln \left(1-e^{-\beta \epsilon }\right)+Nk\frac{{\displaystyle \raisebox{1ex}{$\epsilon $}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}{e^{\beta \epsilon }-1}\end{array}$