# Boltzmann Partition Function of H_2

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1. Apr 16, 2017

### transmini

1. The problem statement, all variables and given/known data
In the real world, most oscillators are not perfectly harmonic. For a quantum oscillator, this means that the spacing between energy levels is not exactly uniform. The vibration levels of an $H_2$ molecule, for example, are more accurately described by the approximate formula
$$E_n \approx \epsilon(1.03n-0.03n^2), \space\space n=0, 1, 2, ...,$$
where $\epsilon$ is the spacing between the two lowest levels. Thus, the levels get closer together with increasing energy. (This formula is reasonably accurate only up to about $n=15$; for slightly higher $n$ it would say that $E_n$ decreases with increasing $n$. In fact, the molecule dissociates and there are no more discrete levels beyond $n\approx15$.) Use a computer to calculate the partition function, average, energy, and heat capacity of a system with this set of energy levels. Include all levels through $n=15$, but check to see how the results change when you include fewer levels. Plot the heat capacity as a function of $\frac{kT}{\epsilon}$. Compare to the case of a perfectly harmonic oscillator with evenly spaced levels, and also to the vibrational portion of the graph in Figure 1.13.

2. Relevant equations
Partition Function: $Z = \sum_n \space e^\frac{-E_n}{kT}$
Average Energy: $\bar{E} = \sum_n \space E_ne^\frac{-E_n}{kT}$
Heat Capacity: $\frac{\partial\bar{E}}{\partial T}$

3. The attempt at a solution
First I tried to determine what $\epsilon$ was but just found $\epsilon=E_1-E_0=1-0=1$ but I don't feel like this could be right, because if $\epsilon$ was just $1$, why would it be there at all. But rolling with this, I wrote some code in Matlab to find $Z \approx 1$ and $\bar{E} \approx 1.61x10^{-17} eV$, but the average energy doesn't make sense either, because it's 17 orders of magnitude lower than the first nonzero energy level. Then I didn't type anything for the heat capacity yet, but tried to derive another formula:
$$C = \frac{\partial\bar{E}}{\partial T} = \frac{\partial}{\partial T}(\frac{1}{Z}\sum_n \space E_ne^{-\beta E_n})$$
$$C = \frac{-1}{Z^2}\frac{\partial Z}{\partial T} \sum_n \space E_ne^{-\beta E_n} + \frac{1}{Z}\sum_n \space (-E_n^2e^{-\beta E_n} \frac{\partial\beta}{\partial T})$$
$$\frac{\partial Z}{\partial T} = \frac{\partial}{\partial T} (\sum_n \space e^{-\beta E_n}) = \sum_n \space (-E_ne^{-\beta E_n} \frac{\partial\beta}{\partial T})$$
$$\frac{\partial\beta}{\partial T} = -kT^2$$
$$C = \frac{-1}{Z^2}kT^2(\sum_n \space E_n e^{-\beta E_n})^2 + \frac{1}{Z}kT^2 \sum_n \space E_n^2 e^{-\beta E_n}$$
I haven't tried coding this out yet, since I'm not even sure if the formula is correct.

So my questions here are:
How do I find $\epsilon$ if I do not have it correct already
If my values for $Z$ and $\bar{E}$ are incorrect, is it something with my formulas, $\epsilon$, or should I double check my code?
Is the formula I have for $C$ a correct derivation from the starting equation?

Last edited: Apr 16, 2017
2. Apr 16, 2017

### kuruman

It's probably easier to start with$$<E>=-\frac{\partial \ln(Z)}{\partial \beta}$$ then use $$C=\frac{\partial <E>}{\partial \beta}\frac{\partial \beta}{\partial T}$$
As far as ε is concerned, remember that the problem asks you to plot the specific heat as a function of kT/ε which is a dimensionless quantity. Therefore, you don't need a value for it. That should also resolve your 17 orders of magnitude issue.

Edit: Corrected erroneous expression for $<E>$.

Last edited: Apr 16, 2017
3. Apr 16, 2017

### transmini

But for the $\epsilon$, along with the plot, it also was asked to calculate the partition function. Doesn't this imply you need a specific value of $\epsilon$ and $T$? Or is it asking just for the general form of the function here, in which the values do not matter? Also, isn't $<E> = -\frac{\partial}{\partial\beta}ln(Z)$?

4. Apr 16, 2017

### kuruman

You are not given any numbers. You are expected to find expressions for Z and <E> and C in terms of parameter y = kT/ε which is dimensionless therefore works for any ε and T.
Yes, sorry, I forgot the "ln". I edited the post to correct the error. Thanks for pointing out.

5. Apr 16, 2017

### transmini

Okay, thank you. That's just poor wording in my opinion then since it says to "calculate with a computer" when there's nothing to technically calculate. I'll leave this open until next week when I get the solutions back.

6. Apr 16, 2017

### kuruman

Sure there is. Note that the partition function is dimensionless and can be cast in terms of a single parameter y = kT/ε. See if you can find an algebraic expression for the specific heat in terms of y only. You can then use a computer to tabulate C for a set of y values and plot as a function of y.