Question

# The energy E of a system of three independent harmonic oscillators is given by $E=\left(n_{x}+\frac{1}{2}\right) \hbar \omega+\left(n_{y}+\frac{1}{2}\right) \hbar \omega+\left(n_{z}+\frac{1}{2}\right) \hbar \omega$. Show that the partition function Z is given by $Z=Z_{\mathrm{SHO}}^{3}$, where ZSHO is the partition function of a simple harmonic oscillator given. Hence show that the Helmholtz function is given by $F=\frac{3}{2} \hbar \omega+3 k_{\mathrm{B}} T \ln \left(1-\mathrm{e}^{-\beta \hbar \omega}\right)$, and that the heat capacity tends to $3 k_{\mathrm{B}}$ at high temperature.

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We have a system of three independent harmonic oscillators, the energy of the system is given by,

\begin{align} E=\left(n_{x}+\frac{1}{2}\right) \hbar \omega+\left(n_{y}+\frac{1}{2}\right) \hbar \omega+\left(n_{z}+\frac{1}{2}\right) \hbar \omega\end{align}

since the three SHOs are independent of each others, then the total partition function equals the multiplication of the partition functions of each one of the oscillators, that is,

$\boxed{Z=Z_{\mathrm{SHO}}^{3}}$

where $Z_{\mathrm{SHO}}$ is the partition function of a simple harmonic oscillator, that is,

\begin{align}Z_{\mathrm{SHO}}=\frac{e^{- \beta \hbar \omega/2}}{1-e^{-\beta \hbar \omega}} \end{align}

the Helmholtz function is, therefore,

\begin{align*} F&=- k_{\mathrm{B}} T \ln\left(Z \right)\\ &=- k_{\mathrm{B}} T \ln\left(Z_{\mathrm{SHO}}^{3}\right)\\ &=- 3k_{\mathrm{B}} T \ln\left(Z_{\mathrm{SHO}}\right)\\ &=- 3k_{\mathrm{B}} T \ln\left(\frac{e^{- \beta \hbar \omega/2}}{1-e^{-\beta \hbar \omega}}\right)\\ &=- 3k_{\mathrm{B}} T \left[ \ln\left(e^{- \beta \hbar \omega/2} \right)-\ln\left(1-e^{-\beta \hbar \omega} \right) \right]\\ &=\frac{3}{2} \hbar \omega+3 k_{\mathrm{B}} T \ln \left(1-e^{-\beta \hbar \omega}\right) \end{align*}

$\boxed{F=\frac{3}{2} \hbar \omega+3 k_{\mathrm{B}} T \ln \left(1-e^{-\beta \hbar \omega}\right)}$

the mean energy is,

\begin{align*}U&=-\frac{d \ln Z}{d \beta}\\ &=-3\frac{d}{d\beta}\left[\ln \left(e^{-\beta \hbar \omega / 2}\right)-\ln \left(1-e^{-\beta \hbar \omega}\right)\right]\\ &=-3\frac{d}{d\beta}\left[-\frac{\beta \hbar \omega }{2}-\ln \left(1-e^{-\beta \hbar \omega}\right)\right]\\ &=\frac{ 3\hbar \omega }{2}+\frac{ 3\hbar \omega e^{-\beta \hbar \omega}}{1-e^{-\beta \hbar \omega}}=\frac{3 \hbar \omega }{2}+\frac{ 3\hbar \omega }{e^{\beta \hbar \omega}-1} \end{align*}

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