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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

[itex]q_{vib} = \frac{e^{{-h \nu}/{2kT}}}{1-e^{{-h \nu}/{kT}}}[/itex]

for each frequency and then multiply them all.

Which would give

[itex]q_{vib,1} = 0,020806[/itex]

[itex]q_{vib,2} = 0,000127[/itex]

[itex]q_{vib,3} = 0,000096[/itex]

[itex]q_{vib,tot} = 2,53E-10[/itex]

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

[itex]q_{vib} = \frac{1}{1-e^{{-h \nu}/{kT}}}[/itex]

gives

[itex]q_{vib,1} = 1,000433438[/itex]

[itex]q_{vib,2} = 1,000000016[/itex]

[itex]q_{vib,3} = 1,000000009[/itex]

[itex]q_{vib,tot} = 1,000433463[/itex]

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 [itex]{h \nu}/{2kT}[/itex] be the correct zero?

Is it even correct to just plainly multiply the single vibrational partition functions?

Thanks for any help. Sorry for the long question, I'm just really confused about this.

Ciao!

Giogio

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# Vibrational partition function - Calculate from several frequencies

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