Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Zero point energy and IR spectroscopy

  1. Jul 12, 2012 #1
    I made some frequency calculations in Gaussian (using various method/basis set combos) and I have to find which method/basis set combo is the most accurate by comparing the results to literature values. I got the calculated wavenumbers along with their intensities but my professor tells me I need to be corrected for zero point energy. I struggled in his spectroscopy class last year so I don't know what zero point energy is, let alone how to correct these values for it. If I just google zero point energy, I'm a visual thinker so quantum chemistry isn't my strongest area. Can anyone explain how this zero point energy relates to IR spectra? I'm googling zero-point energy so I'll learn as much about it as I can but I suspect it will be a long time before I find an explanation as to how it relates to IR absorbances.
  2. jcsd
  3. Jul 12, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Which is to say, you want to find out which model gives results closest to observation?

    Sounds like you are expected to do a literature search...?
    We probably should not help you there since diving in to the lit is how you learn how to deal with it. If we just throw up some papers that would defeat the purpose.

    Uh oh - you will need to find a way to cope with abstract math I'm afraid, or go into a different field. Lets see what I can do:

    In IR spectroscopy, though, it is common to model the systems being observed as harmonic oscillators. In a quantum harmonic oscillator, the minimum energy is not zero: this is the zero-point energy for the system.

    Here's your visual:
    (scroll down to "zero point energy".)

    Notice there is a non-zero minimum energy at [itex]E=h\nu_0/2[/itex]?
  4. Jul 12, 2012 #3
    The energy levels of a harmonic oscillator are determined by:
    where E_n is the energy of the level, n is the excitation number of the level (n=0,1,2,3,...), h is Plancks constant, and f is the fundamental frequency of that energy level.
    The above formula is used a lot in IR spectroscopy, both absorption spectroscopy and Raman spectroscopy.
    The term 1/2 is the contribution of zero point energy.
  5. Jul 15, 2012 #4
    Yeah. Cr(CO)6 is the compound so its a relatively simple compound to work with since all the bond angles are the same. I found that the most effective method/basis set combo for optimising the structure is B3LYP/6-311G. Now I need to find out what the best method/basis set combo is for frequency and TD calculations.

    Yeah, I've found literature values to compare the predicted results to. For IR spectrum of gaseous Cr(CO)6, the CO peak lands right on 2000 cm-1.

    Visual thinking gives me an edge in many areas in chemistry but other areas, such as this one, I have to really work like mad to gain a good understanding. I make abstract visual representations of the concepts but it takes a serious amount of time and effort sometimes. Thanks for the help, I'll read all this now and see if I can get a grasp of it all.

    Thanks for the diagram but I'm not entirely sure what it represents. I'm guessing v=0 is the ground state vibrational frequency. So lets use Cr(CO)6 as an example. The C-O bond produces an intense peak at around 2000 cm-1. If I'm not mistaken, 2000 cm-1 corresponds to the energy differece between v=0 and v=1 for the C-O bond.

    I'm still trying to get my head around what exactly zero point energy is. In the article you linked, it says zero-point is the energy of a system when T=0. So lets say my system is a can of butane and the temperature is absolute zero. Zero-point energy is the energy thats left over? As for harmonic oscillators, I'll use a spring with a metal ball at the end of it as an example of a macroscopic harmonic (lets pretend its harmonic i.e. forget about gravity and air resistance etc.) oscillator. I can see that at the very moment the metal ball changes direction, its kinetic is zero but its potential energy is at its highest. How does this concept apply to a quantum harmonic oscillator?
    Last edited: Jul 15, 2012
  6. Jul 15, 2012 #5
    BTW, heres the output from one of my frequency calculations:

    74.4629, 98.1250, 116.5055, 328.6017, 337.3187, 413.5514, 527.1941, 558.8680, 666.7709, 2157.3618, 2160.9906,
    0.0000, 0.0000, 5.6900, 0.0000, 0.0000, 83.7658, 0.0000, 0.0000, 304.6480, 0.0000, 2480.6823

    The first number is the wavenumber, the number below it is the intensity. As you can see, the most intense peak is the one at 2160.9906 cm-1, which has an intensity of 2480.6823. Thats clearly the C-O peak. I have about 30 output files which all contain a calculation for this carbonyl peak. I need to determine which calculation is the most accurate. Those calculations above were done by HCTH/6-311G. My professor tells me I need to correct these calculations for zero point energy before I can compare them to literature values so thats what I'm trying to figure out now.
    Last edited: Jul 15, 2012
  7. Jul 15, 2012 #6
    Why not just save yourself some effort, and ask your professor how to do the ZPE correction?
  8. Jul 15, 2012 #7
    I emailed him earlier and he said:
    so by the sounds of it, hes suggesting I just find a correlation factor to apply to my calculations. I emailed him again, asking where to find this correlation factor but I didn't really think it through. I can probably use a correlation online that was derived for a different class of molecule and use that. In other words, if an accurate correlation factor was determined for tetrahedral carbon compounds, it probably applies to octahedral chromium complexes. Have I got the right idea or am I mistaken?
  9. Jul 15, 2012 #8
    Right - so I know what he means.

    It turns out for various reasons, ab-initio calculations for vibrational frequencies (like you do in Gaussian) turn out to be slightly incorrect when compared with experiment. It's standard practice to simply multiply the calculated values by some empirical factor to correct the problem. I've been told that this factor (which depends on the particular method you use for your calculation) tends to be fairly universal across a large set of molecules.

    The factors can be looked up in the literature (experimental chemists use them all the time to analyze their IR spectra).

    Do you really understand what your project is?
  10. Jul 15, 2012 #9
    Ah right. The gaussian site:
    http://www.gaussian.com/g_tech/g_ur/k_freq.htm [Broken]
    gives the following correction factor:
    Zero-point correction= .023261 (Hartree/Particle)
    so I suppose thats all I need to use.

    You're last question: No, I don't really know what I'm supposed to be doing for this research project. I was new to Gaussian when I started so the terminology he was using had me baffled. I have a vague idea though. My professor is researching substituted octahedral chromium carbonyl complexes and the project he assigned me is to determine if any of the newer density functional methods are more effective at making calculations on octahedral carbonyl complexes than Hartree-Fock methods. I happen to be a programmer so I've been doing all kinds of side projects for him by programming scripts to extract data from the output files etc. but I think I'm pretty much finished this research project.
    Last edited by a moderator: May 6, 2017
  11. Jul 15, 2012 #10
    While quantum chemistry isn't my strongest field, I'm glad I got this particular project because I finally got to implement my programming skills in chemistry. You'd be amazed at the power programming gives you when it comes to dealing with data outputted by Gaussian.
  12. Jul 15, 2012 #11
    I could be wrong, but I think the correction factor above is to the total energy.
    I don't think the Gaussian output file has the correction factor you want in it.
    I take it you'll either need to look it up in the literature, or find out what the real IR spectra look like for the compounds in question to compare with your results.
  13. Jul 16, 2012 #12
    I have a literature value:
    the carbonyl peaks wavelength lands right on 5 microns which corresponds to 2000 cm-1. Any idea where I'd find a zero-point correction factor to apply to my calculations? Heres my lecturers original reply:
    I have no idea how to get this correction factor. I asked him about it but he didn't answer the question.
    Last edited: Jul 16, 2012
  14. Jul 16, 2012 #13

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The correction factor is empirical - you have to use experimental data to work it out.
    Think of it as finding a systematic error in a measurement - only here it is in the calculation. The same correction factor should apply to every calculation - which tells you if the theory is good.
  15. Jul 17, 2012 #14
    Ah, I think I get it now. The experimental CO absorption for Cr(CO)6 is at 1988 cm-1 in alkane solvent. If I calculate its zero-point energy, then I can subtract it from 1988 cm-1 and get a Born-Oppenheimer approximation value. From those two values I can derive the correction factor. Have I got the right idea?
  16. Jul 17, 2012 #15
    So lets say the experimental wavenumber is 1988 cm-1, which corresponds to a wavelength of 5.03 microns and thus a frequency of 59601 GHz and an energy of 0.247 eV. I know that this is the energy difference between n=0 and n=1. I see from the equation [itex]E_n=hf(n + \frac{1}{2})[/itex] that the energy difference between n=0 and n=1 is 3/2 - 1/2 hf, in other words its just hf. So that means the energy of n=0 must be 0.247 eV / 2 = 0.1245 eV. That can't be right. That would shift the wavenumber to around 1000 cm-1.
    Last edited: Jul 17, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Zero point energy and IR spectroscopy
  1. Point Defect Energies (Replies: 3)