Vibrational partition function - Calculate from several frequencies

1. May 19, 2013

Giogio

Hello everybody,

I registered here hoping to finally find a fundated answer about what I by myself seem not be able to figure out.

Question in short:
We have calculated a list of wavenumbers for some molecular systems. How do you get the vibrational partition function from that?

My problem in detail:
Let's take this list of wave numbers as an example:

[cm^-1]
1602,13
3710,78
3829,60
(this is for one water molecule)

Now my collegue explained to me that it's correct to calculate the vibrational partition function as
$q_{vib} = \frac{e^{{-h \nu}/{2kT}}}{1-e^{{-h \nu}/{kT}}}$
for each frequency and then multiply them all.
Which would give

$q_{vib,1} = 0,020806$
$q_{vib,2} = 0,000127$
$q_{vib,3} = 0,000096$
$q_{vib,tot} = 2,53E-10$

which seems really weird to me because the partition function becomes lower and lower with every vibrational mode. Can a partition function even be lower than 1?

One of the few things I could find about this is this one, page 5, eq. 3.21 to 3.22: http://www.chem.iitb.ac.in/~bltembe/pdfs/ch_3.pdf
But isn't the zero of energy scale kind of arbitrary?
Anyway, using
$q_{vib} = \frac{1}{1-e^{{-h \nu}/{kT}}}$
gives

$q_{vib,1} = 1,000433438$
$q_{vib,2} = 1,000000016$
$q_{vib,3} = 1,000000009$
$q_{vib,tot} = 1,000433463$

which seems way more reasonable to me. (Right? )

But it leaves open the question of the zero of energy scale for me. My collegue insists that it's arbitrary where you put it, so both solutions should be correct. But since it gives a factor into each of the single partition functions, how can it be equivalent? Also, would ${h \nu}/{2kT}$ be the correct zero?
Is it even correct to just plainly multiply the single vibrational partition functions?

Ciao!

Giogio

2. May 19, 2013

Staff: Mentor

Both the point of zero energy and the normalization of the partition function are arbitrary, they have no physical meaning.
I have never seen this way to calculate partition functions from spectroscopic lines like that... I would calculate the energy levels as intermediate result first, I think.

3. May 19, 2013

Giogio

Hey mfb,

Thank you!
How would you do that?
In my example, the energies would be

$\epsilon_1 = h\nu_1 = 3,18E-20 J$
$\epsilon_2 = h\nu_2 = 7,38E-20 J$
$\epsilon_3 = h\nu_2 = 7,61E-20 J$

and the only equation I'm finding for those would be $q_{vib} = \prod_i \frac{e^{{-\epsilon_i}/{2kT}}}{1-e^{{-\epsilon}/{kT}}}$ , which is the same as above.

What exactly do you mean?

Ciao,
Giogio

Edit:
$q_{vib} = \sum_i q_{vib,i}$ ?! Is it that simple? It... would sort of make sense, I'm only slowly beginning to understand what the partition function actually is. But then what's whith that other equation, the product one above?

Last edited: May 19, 2013
4. May 19, 2013

Staff: Mentor

Those ϵ are energy differences - but differences between what?
Based on the formulas you use, I think they correspond to the distances between adjacent states, for three different types of vibration. In that case, I would expect a product, where each factor is a sum over all excitations of that specific type of vibration (that sum leads to those e^(...)/(1-e^(...)-terms).
Therefore, I think you can use your formula with the product.

If you use a specific temperature to get numerical values, please write that down.

5. May 19, 2013

Giogio

Oh, sorry, forgot about the temperature, it was 298K.

6. May 19, 2013

Giogio

Hi,

Okay, thanks a lot for your help so far, mfb!

I've now retraced and understood that it is of no importance for the probability distribution between the modes if you use
$q_{vib} = \prod_i \frac{e^{{-\epsilon_i}/{2kT}}}{1-e^{{-\epsilon}/{kT}}}$

or

$q_{vib} = \prod_i \frac{1}{1-e^{{-\epsilon}/{kT}}}$

to calculate the partition function, it's just that in the latter case, where the lowest energy state for each mode is set to zero, partition functions close to unity indicate most particles being in the vibrational ground state.

Okay, now I just have one last problem/question:

If you calculate the total partition function:
$q_{tot} = q_{rot}q_{vib}q_{tra}q_{ele}$,

Does it then matter where you set the zero of energy for the vibration?

For instance, for the example from above:

[cm^-1]
1602,13
3710,78
3829,60

I get

$q_{vib} = \prod_i \frac{e^{{-\epsilon_i}/{2kT}}}{1-e^{{-\epsilon}/{kT}}}= 2,60E-10$

or

$q_{vib} = \prod_i \frac{1}{1-e^{{-\epsilon}/{kT}}} = 1,000433463$

whereas for a system of two water molecules with the wavenumbers

123,05
141,42
144,3
174,76
385,44
633,67
1601,73
1617,01
3559,79
3704,74
3793,53
3818,69
(more, and smaller ones too)

the results are

$q_{vib} = \prod_i \frac{e^{{-\epsilon_i}/{2kT}}}{1-e^{{-\epsilon}/{kT}}} = 4,37E-20$

or

$q_{vib} = \prod_i \frac{1}{1-e^{{-\epsilon}/{kT}}} = 19,55$

If I want to compare these two systems with respect to the partition function, I can't imagine it's arbitrary which zero of energy I use, that is, if the vibrational contribution to the total partition function becomes smaller or larger for the bigger system. Is it?

It would be great if you could give me a tip about that as well! :D

7. May 19, 2013

Staff: Mentor

I would always use 1 for the ground state, but again that is an arbitrary choice. A different choice multiplies q by a constant factor, and drops out if you calculate any physical observable (like the probability to find a molecule in a state).

8. May 20, 2013

Giogio

Ahhh, of course, when comparing the partition functions, they get exponents so that the factor for the zero of energy falls out again. Now it's clear. Thank you very much!