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Boltzmann Partition Function of H_2

  1. Apr 16, 2017 #1
    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. jcsd
  3. Apr 16, 2017 #2

    kuruman

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    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
  4. Apr 16, 2017 #3
    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)##?
     
  5. Apr 16, 2017 #4

    kuruman

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    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.
     
  6. Apr 16, 2017 #5
    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.
     
  7. Apr 16, 2017 #6

    kuruman

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    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.
     
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