The following discussions are for isotropic quantum harmonic oscillators which have the energy eigenvalues as follows:
$$E=\left(\sum_{i}^{N}n_i+\frac{N}{2}\right)\hbar \omega$$
where $N= $ number of dimensions.
I have two interrelated doubts regarding the partition function calculation on the above system.
First
One oscillator in 3D is equivalent to three 1D independent oscillators so that if the partition function (P.F.) for a 1D harmonic oscillator is:
$$Z_{1D}=\sum_{n=0}^{\infty} e^{-\beta E_n}$$
then for the 3-D harmonic oscillator it becomes :
$$Z_{3D}=(Z_{1D})^3 \tag{1}$$
On the other hand,
$$Z_{3D}=\sum_{n=0}^{\infty}g(n) e^{-\beta E_n} \tag{2}$$
where here,
$n=n_x+n_y+n_z$
and
$g(n)=\frac{(n+1)(n+2)}{2}$
On plugging the value of degeneracy, it is not easy to show whether (1) and (2) are equal. I tried solving it but was not getting (2) same as (1).
Shouldn't (1) and (2) be the same or is there a problem with how did the above steps?
If they indeed are the same, what manipulation is needed to show that?
Second
If after calculating the partition function for $N$ independent 1D harmonic oscillators, we try to take the indistinguishability into account by dividing with a factor of $N!$ $\\ $ we get:
$$Z_{1D}^{N}=\frac{1}{N!}(Z_{1D})^N$$
If now we have to do the same for the $N$ number of 3D oscillators, we can instead consider the $3N$ number of independent 1D oscillators so that:
$$Z_{3D}^{N}=\frac{1}{(3N)!}(Z_{1D})^{3N}$$
But I have read that instead it is correctly calculated as:
$$Z_{3D}^{N}=\frac{1}{N!}((Z_{1D})^3)^N$$
which has a different denominator value.
Which one of the above is correct and how?
Best Answer
The partition function of the 1D harmonic oscillator is $$\eqalign{ {\cal Z}_{\rm 1D}&=\sum_{n=0}^{+\infty} e^{-\beta\hbar\omega\big(n+1/2\big)}\cr &=\lambda^{1/2}\sum_{n=0}^{+\infty}\lambda^n\cr &={\lambda^{1/2}\over 1-\lambda} }$$ where $\lambda=e^{-\beta\hbar\omega}$. Consider now the 3D harmonic oscillator. First, one can note that the system is equivalent to three independent 1D harmonic oscillators: $${\cal Z}_{\rm 3D}=\big({\cal Z}_{\rm 1D}\big)^3 ={\lambda^{3/2}\over (1-\lambda)^3}$$ On the other hand, using your equation (2), we get after some algebra, $$\eqalign{ {\cal Z}_{\rm 3D}&=\sum_{n=0}^{+\infty} g(n) e^{-\beta\hbar\omega\big(n+3/2\big)}\cr &=\lambda^{3/2}\sum_{n=0}^{+\infty} {(n+2)(n+1)\over 2}\lambda^n\cr &={\lambda^{3/2}\over 2}{\partial^2\over\partial\lambda^2}\Big[ \sum_{n=2}^{+\infty} \lambda^{n+2}\Big] \cr &={\lambda^{3/2}\over 2}{\partial^2\over\partial\lambda^2}\Big[ \lambda^2\sum_{n=0}^{+\infty} \lambda^n\Big] \cr &={\lambda^{3/2}\over 2}{\partial^2\over\partial\lambda^2}\Big[ {\lambda^2\over 1-\lambda}\Big] \cr &=\lambda^{3/2}\Big[{1\over 1-\lambda}+{4\lambda\over(1-\lambda)^2} +{\lambda^2\over(1-\lambda)^3}\Big]\cr &={\lambda^{3/2}\over(1-\lambda)^3}\cr }$$ i.e. the same result, as expected!
In the Einstein solid, one considers $N$ atoms oscillating around their equilibrium position. In this simple model, two atoms are not expected to exchange their position so the atoms should be considered as distinguishable. Each atom is reduced to a 3D harmonic oscillator, equivalent to three independant 1D harmonic oscillators associated to the three directions. They also should be considered as distinguishable. As a conclusion, you should not divide the partition function by $(3N)!$, nor $N!$. Note that adding or not a factor $1/N!$ to the partition function is not of great importance: in both cases, the Dulong-Petit law for the specific heat is recovered (which is the goal of this model).