Deriving the partition function for a harmonic oscillator

Question

This question is in reference to an answer I posted here yesterday.

In it I derived the partition function for a harmonic oscillator as follows

$$q = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$

For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$

Note that $\epsilon_0 \neq 0$ there exists a zero point energy.

Let's write out a few terms $$q = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +..... $$

factoring out $e^{-\frac{\hbar \omega/2}{kT}}$

$$q = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right) $$

The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)} $$ herein, $ x \equiv e^{-\frac{\hbar\omega}{kT}} $

Putting all of this together

$$q = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$

I happened to check Atkins Physical chemistry (10th edition) for the same derivation earlier today.

On page 620, the vibrational partition function using the harmonic oscillator approximation is given as $$q = \frac{1}{1-e^{-\beta h c \nu'}}$$, $\beta$ is $\frac{1}{kT}$ and $\nu'$ is wave number

This result was derived in brief illustration 15B.1 on page 613 using a uniform ladder. However, in that illustration the uniform ladder starts at 0, but the harmonic oscillator has a zero point energy (which I accounted for in my derivation).

I discussed this with my instructor and he pointed out that as $T \rightarrow 0$ $q \rightarrow 1 $ (since only one state is thermally accessible.

The result derived in Atkins' does that indeed, but mine goes $q \rightarrow 0$.

Now, in the formalism developed in Atkins the do set the ground state energies to zero (on page 605), and basically add non-zero ground state energies to a calculated $\langle \epsilon \rangle$, which is fine.

Anyway, I would love it if someone could weigh in on this and help me resolve this conceptual mess in my head.

Answer 1

I'm confused why you're interpreting the partition function as a count of states. It can't be a count; it's continuous.

The zero point energy doesn't actually matter because you can just shift the energy scale so that it starts at zero. The main point of zero point energy is that the ground state of the harmonic oscillator is such that there is energy, and the system is not stationary. I'm going to use it below anyway because you are.

You can get the answer you want, but you'll want to look at the probability $P_i$ of being in state $i$, where the partition function is used to normalize:

$$P_{i} = \frac{\epsilon_{i}}{q} = \frac{\mathrm{e}^{-\beta\hbar\omega (i + 1)/2}}{\sum_{i}\mathrm{e}^{-\beta\hbar\omega (i+1)/2}}$$

Substitution with your convergent sum:

$$P_{i} = \mathrm{e}^{-\beta\hbar\omega (i+1)/2} \frac{1 - \mathrm{e}^{-\beta\hbar\omega}}{\mathrm{e}^{-\beta\hbar\omega/2}} = \mathrm{e}^{-\beta\hbar\omega i}(1- e^{-\beta\hbar\omega})$$

For $T\rightarrow 0$, $P_{i} = \delta_{i0}$, which is exactly what you are looking for.