Partition Function (Quantum Oscillators)

For an ensemble of quantum oscillators $\omega_i=\sqrt{k_i/m_i}$ and $E_i=\hbar\omega_i(n_i+1/2)$ evaluate the partition function and the specific heat of an oscillator. 

The partition function is

$Z=∑_n exp⁡[-(\hbar ω(n+1/2))/k T]=exp⁡(-(\hbar ω)/2kT)*∑_n exp⁡(-(n\hbar ω)/k T)$

If we define the Einstein temperature as

$k T_E=\hbar ω$ and take into account that $∑_(n=0)^∞ exp⁡(-n x)=e^x/(e^x-1)$


$Z=exp⁡(-T_E/T)*exp⁡(T_E/T)/exp⁡(T_E/T-1) =exp⁡(T_E/2T)/(exp⁡(T_E/T)-1)$

The energy is

$E=KT^2*d(ln⁡(Z) )/d T=k*exp⁡(T_E/T)/((exp⁡(T_E/T)-1))-k/2$

The specific heat is

$C v=d E/d T=k/T^2 *(exp⁡((2T_E)/T)-exp⁡(T_E/T)*(exp⁡(T_E/T)-1))/(exp⁡(T_E/T)-1)^2$

$C_v=1/T^2 *e^{1/T}/(e^{1/T}-1)^2$    taking $T_E=1$ and $k=1$

For $k=1$ and $T_E=1$  the graph of $C v$ is the following: for $T>>T_E (=1)$ one the value of $C v$ is approaching unity, which is consistent with the classical value $C v =k$   (for an average classical energy $E=k T$, the value of $C v=d E/d T =k$)

For small temperatures one can approximate

So that

$C_v=1/T^2 *e^(1/T)/(1+1/T-1)^2 =e^{1/T}$

So the function at $T<<T_E$ rises as an exponential.

By taking just the first state in the sum of the partition function one has

$Z=exp⁡(-(\hbar ω)/k T)*exp⁡(-(\hbar ω)/2kT)=exp⁡(-(3T_E)/2T)$


and $E=k T^2*d(ln⁡(Z) )/d T=(3kT_E)/2=constant$ so $C v=d E/d T=0$