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| From: |
YONG HUANG <Y0H8797()at()ACS.TAMU.EDU> |
| Date: |
Sat, 9 Dec 1995 16:50:59 -0600 (CST) |
| Subject: |
Summary: Koopmans' Theorem and Neglect of Bond in CAChe MOPAC Input |
A week ago I asked 2 questions about Koopmans' Theorem and CAChe MOPAC.
Many thanks to Drs. J.-M.Sichel, R.Fournier, B.Duke, M.J.Ondrechen, I.Mayer,
Jack ?, E.Chamot, R.P.Mattie, J.Pollard, D.Lichtenberger, J.Schulte,
R.Winchester and Matt. The follwoing is a summary. Words in [] are mine. I
asked Drs. J.-M.Sichel and R. Fournier more questions which are attached to
the end. Lines preceeded by > are mine.
My questions are:
(1) Can anyone tell me whether ionization potential calculated based on
Koopman's Theorem is just the IP for vertical ionization? I suspect so because
I think what Koopman's Theorem misses is just the vibrational relaxation in the
ionic state after the electron is gone. Please correct me if I'm wrong.
(2) I use CAChe MOAPC to calculate the distances and bond orders between the 3
hydrogens of a NH3 right above the pai ring of ferulic acid (4-OH-3-OCH3-
cinnamic acid). After geometry optimization, NH3 shifts to the edge of the
pai-ring. I did this calculation in 3 ways: 1) no bonding is specified between
H's of NH3 and pai-ring; 2) 1 H of NH3 hydrogen-bonded to carbons #1,3,5 (if
the 6 C's are numbered 1 to 6); 3) 3 H's of NH3 hydrogen-bonded to C's #1,3,5,
respectively. My question is: Does the calculation take into account whether I
specified the bonding at all? It seems it doesn't. The results always show
small bond order between H's of NH3 and substituents of the ring AND EVEN
SMALLER bond order between those H's of NH3 and C's of the ring, regardless of
whether I draw hydrogen bond lines in the CAChe Editor (Graphic Interface for
MOPAC). Seems MOPAC Input doesn't have bond information. I guess specifying
bonds in Editor doesn't do anything at all. Can someone clarify this for me?
Thank you very much. I'll summarize.
Yong
y0h8797 %-% at %-% acs.tamu.edu
From: SMTP%"sichelj(-(at)-)Umoncton.CA" 2-DEC-1995 15:29:38.31
[to 1st question]
There are also contributions due to:
1. Relaxation of the orbitals in the ion after the electron is gone; and
2. Difference in correlation energy between molecule and ion; and
(3. Difference in relativistic correction, but this is very small for
molecules with only light atoms).
John-M. Sichel
From: SMTP%"fournier (- at -) mail.physics.unlv.edu" 2-DEC-1995 15:55:52.48
[to the 1st question]
Koopmans' theorem approximates the vertical IP because, as you
said, it does not take into account nuclei relaxation. But what's
more, it does not take into account ELECTRON relaxation either! It
assumes orbitals stay the same before and after ionization, only
orbital occupation numbers change. That's a serious approximation and
the results would be very bad if it was not partly cancelled by the
neglect of correlation which introduces an error of opposite sign
--- the stabilizing correlation energy is larger in the N-electron
system than in the (N-1) electron system. This can be schematically
represented with energy levels for the N-electron and (N-1)-electron
systems calculated in 3 ways, (the "error cancellation effect" is
exaggerated):
Koopmans' theorem with relaxation of with correlation
orbitals in the (N-1) state in both states
("exact" result)
N-1 -------
^ \
| \
| - - - - - N-1 -------
| ^ \ - - - - - - - N-1 -------
| | ^
| | |
| | |
N ------- - - - - - - N ------- |
\ |
\ |
- - - - - - - - N -------
Sincerely,
Rene Fournier.
P.S. It is Koopmans' (not Koopman's) theorem because it is due
to Koopmans (not Koopman). [This embarrassed me.]
From: SMTP%"b_duke "at@at" lacebark.ntu.edu.au" 2-DEC-1995 19:08:29.84
[to the 1st question]
Koopman's theorem states that IP = -e sub i for vertical ionisation
from the i'th MO if MO theory is correct and the same orbitals fit the
ion as fit the parent molecule. In other words it ignors orbital relaxation - the
change in the MOs on removing an electron - and it ignors electron
correlation - the error in MO theory. See our article in J Chem Ed in
the July issue, I think (I do not have the reference with me).
[to the 2nd question]
MOPAC does not use bond information. It converts the Z-matrix to
cartesians and calculates the energy and orbitals for that geometry.
Cheers, Brian. [Cheers, Brian. Thank you.]
From: SMTP%"MARYJO { *at * } neu.edu" 3-DEC-1995 22:43:33.84
Regarding Koopmans' Theorem:
It misses two things-
1) Electronic relaxation, which is the reajustment of
the N-1 electrons upon lost of the first electron, and
2) Nuclear relaxation.
Koopmans' "Theorem" is not really a theorem, as it is
not rigorous, however it works in Hartree-Fock because
the error due to electron correlation is usually equal
and opposite of the error due to electron relaxation.
In cases where correlation is unusually small, the
relaxation correction can be significant. One is
always better off with some sort of transition
operator method.
Good luck with your project,
Mary Jo Ondrechen
From: SMTP%"ccl at.at cric.chemres.hu" 4-DEC-1995 07:09:29.59
1. Koopmans theorem is only an approximation even to vertical ionization
potential because:
a) it is pertinent to the SCF wave function, so no electron correlation
is accounted for;
b) the "relaxation" of the one-electron orbitals taking place during the
ionization is also neglected.
Effects a) and b) are often of opposite sign and partly compensate each
other.
2. A quantum chemical program like MOPAC is devoted to determine where the
bonds are, not to utilize this knowledge.
Regards,
Prof. Istvan Mayer
From: SMTP%"jas at.at medinah.atc.ucarb.com" 4-DEC-1995 07:24:14.99
[to the 1st question]
Koopmans' theorem basically says that the MO energy approximates the
vertical IP by ignoring ELECTRONIC relaxation in the ionic state. The
correction for this electronic relaxation is to perform two separate total
energy calculations for the neutral and ionics states and subtract (usually
called 'delta SCF'). This still ignores any vibrational structure due to
zero-point vibrations in both states as well the vibrational fine structure
due to Frank Condon overlaps between the vibrational amplitudes of the two
states. It also ignores any nuclear (geometric) relaxation of the ionic
state. To do it correctly, you need to fully characterize the PE's for
both the neutral and ionic states, use total energy differences, add
zero-point corrections, solve the vibrational (nuclear) problem for each
state, and compute Frank Condon overlaps.
You could also compute the dipole moment derivatives to get the relative
IR intensities (and polarizabilities for Raman intensities).
[to the 2nd question]
MOPAC is a quantum mechanical (albeit semi-empirical) method and could
care less about the bonds, atom types, atomic charges, etc, used to
construct the molecule in the Editor. Only the atomic coordinates and
total (net) charge of the molecule are used by any QM method. The bond and
atom types are used by the Editor's beautifier and MM2 calculations only.
- Jack
From: SMTP%"echamot ^at^ xnet.com" 4-DEC-1995 07:41:48.83
[to the 1st question]
Yes, you are basically correct. In your second question:
[about the 2nd question]
Your observation [MOPAC doesn't care about bonding in the input file]
is also correct. You have to remember, however, that drawing bonds (on
paper or on the computer) is really an artifact to help us chemists envision
in our own mind the more complicated (but more accurate) description of
molecules, by thinking of them simply as atoms held together by discrete
bonds: the ball and spring model. Quantum Mechanical methods (including
semi-empirical methods such as the method in Mopac) use principles from
physics to model the system more accurately as a set of positive nuclei
surrounded by a cloud of electrons, using wavefunctions to describe the
orbitals. When you model a compound semi-empirically, then, the program
optimizes the electron distribution as well as the molecular geometry to try
to find the way that the molecule really wants to exist. After the
calculations converge, Mopac calculates the bond orders you are interested
in, but this reflects the (valence) electron density between each pair of atoms.
Drawing bonds in the Editor is still useful, though. For one thing, these
bonds are used by the classical, Molecular Mechanics calculations, which do
stick with the bonding scheme that you specify. If you run a Mechanics
optimization on your system first, with each of the bonding schemes you're
interested in, you will get a good set of starting points for the various
geometries you are trying to investigate. Then use Mopac (or another
Quantum Mechanical method) to refine the structure, and see which (if any)
are the preferred geometries.
Another reason to draw the bonds, at least with CAChe Mopac, is that the
calculated bond orders (that otherwise only appear in Mopac's text output
file) will be included in the molecule file for each of the bonds shown in
the molecule. If you hadn't specified a particular bond when building the
molecule, and the atoms didn't get close enough for CAChe Mopac to suggest a
bond, then that calculated bond order for that atom pair will not show up in
the visualization.
Ernest Chamot
From: SMTP%"rpmattie "-at-" voicenet.com" 4-DEC-1995 10:04:29.71
The Koopmans Theorem for IP is just the orbital
energy of the ionized electron. Depending on how
you look at it, there are a number of types of
relaxation that are neglected.
Do not rely upon the Koopmans Theorem! It does not
get the ionization energies in the proper order. You
cannot use it to interpret ionization spectra.
When you are looking at experimental IP energies, make
sure the authors have specified whether these are
adiabatic or vertical ionizations.
Renee Peloquin Mattie
From: SMTP%"jpollard -x- at -x- U.Arizona.EDU" 4-DEC-1995 10:11:40.83
You are correct about your thoughts on Koopman's theorem. This is why
computational chemists had such a hard time assigning the photoelectron
spectrum of dinitrogen and ferrocene (relaxation energy) JRP, UofA
From: SMTP%"DLICHTEN at.at XRAY0.CHEM.ARIZONA.EDU" 4-DEC-1995 10:53:25.98
In regard to your recent question, Koopmans' (note where apostrophe is)
Theorem refers to the vertical ionization because no nuclear relaxation is
included. This is the correct assumption for vertical ips measured by
electron spectroscopy. Koopmans' Theorem is still an approximation to ips
because, most importantly, it does not take into account electron relaxation
energies (a theoretical consequence of the frozen orbital assumption implicit
in the theory) and electron correlation (a consequence of the Hartree-Fock
approximation). You have an expert on this at TAMU, Mike Hall in chemistry.
He will be able to explain this very clearly to you. [Thanks Dr. Lichtenberger]
Dennis Lichtenberger
From: SMTP%"schulte -x- at -x- ws09.pc.chemie.th-darmstadt.de" 4-DEC-1995
12:32:53.06
Dear Dr. Huang [should be "Dear Pre-dr. Huang"]
According to Koopman's theorem the measured IP for a
certain state corresponds to the orbital
energy of the ionized electron.
But for quite a lot of molecules this relation is only a rather crude
approximation, because several electronic relaxation and correlation
processes remain unconsidered in Koopmans theorem.
1. SCF-relaxation energy
After the electron has been removed, the remaining electrons of the cation
reorganize. This relaxation energy is a pure electronic effect. Theoretically
you can describe this effect by a Delta SCF calculation: one calculation for
the neutral state and one calculation for the cationic state. Then you obtain
the difference in the total energies of both systems, which corresponds to
the IP corrected by those relaxation effects which are described in the SCF
picture. Very often this is sufficient, if Koopmans theorem itself fails.
2. correlation effects
These effects can be considered in a perturbational Greens function approach,
which can be extended to different orders.
a. If you go from the N-electron- to the N-1-electron system you also have a
loss in ground-state correlation energy. In second order it is called ''pair
removal energy''
b. Due to the reorganization of the electrons in the cationic state also
the correlation energy changes. This energy is called ''pair relaxation
energy''
All the calculations listed in 1 and 2 are done within the Born-Oppenheimer
approximation, which states that the nuclear coordinates do not change
during electronic processes. Vibrational effects therefore still are
unconsidered in this picture.
I think MOPAC 7 contains a GF package written by D. Danovich. Since I
use another GF package I cannot tell you much about its functionality.
Here in Darmstadt we have an INDO program written by M.C. Boehm, which
is capable of the above listed calculations.
Sincerely Yours,
Joachim Schulte
From: SMTP%"randy_winchester -8 at 8- msmtp.iddw.saci.org" 4-DEC-1995
08:31:23.37
I'm not too sure about the answer to your question about Koopman's theorem,
but I think you are right.
Your conclusions about MOPAC are also correct. There is (as far as I know)
no transfer of information from the editor to the program MOPAC about bond
order or even prescence of bonds. This is usually a plus, since the
calculation is not biased by your expectations.
I hope this is helpful.
Randy Winchester
From: SMTP%"/G=Matthew/S=Harbowy/OU=LIPTONUS-EC02/O=TMUS.TJL/ (+ at +)
LANGATE.gb.sprint.com" 5-DEC-1995 06:30:58.83
[to the 2nd question]
One of the big flaws of a graphical interface is the thinking that
the graphics themselves have meaning. This is especially true for
semiempirical and ab-initio schemes.
Unless you do something to lock a particular conformation, risking
locking it into a non-minimum conformation, all the atoms are
'free' to do whatever they want in a minimization. There's no such
thing as a bond, especially a hydrogen bond, outside of the
'bonding state' generated by a particular agreeable overlap of
orbitals. Try drawing a molecule with every atom bonded to every
other atom, and you'll get the same result as bonding no atoms to
any other atom in the graphical interface for MOPAC.
Molecular mechanics, on the other hand, is a whole different story.
matt
[I asked Dr. Sichel more questions who kindly answered in the following]
From: SMTP%"sichelj -AatT- Umoncton.CA" 6-DEC-1995 09:30:50.50
On Wed, 6 Dec 1995, YONG HUANG wrote:
> 1. How big is the orbital relaxation in the ion compared to atomic vibrational
> relaxation? I assume your "orbital relaxation" means electron relaxation. Its'
> time scale should femtosec while atomix relaxation is picosec., right?
Electronic (or orbital) relaxation is often 1-2 eV and occurs on the
timescale of photoelectron spectroscopy, so it shifts the peaks to lower
ionization energies. Vibrational relaxation would indeed be slower, but
vibrational transitions may accompany photoionization and lead to
splitting of an orbital peak into a vibrational multiplet. This is important
for bonding and antibonding electrons, but much less so for nonbonding
electrons whose removal does not affect bond length.
> 2. I'm not very clear about your difference in correlation energy between a
> molecule and an ion. To be honest, I'm not a calculation guy.
Correlation energy is the correction to the Hartree-Fock energy due to the
tendency of electrons to avoid each other by correlating their motions. It
is negative (stabilizing) and the most important contribution is due to
correlation of pairs of electrons in the same MO. For a closed-shell
molecule, ionization removes (one electron of) one pair so the ion will
have less correlation energy.
This effect increases the ionization energy and may roughly cancel the
orbital relaxation effect, but of course the cancellation is not
guaranteed.
An early article which explains all these points was by W Graham Richards
in the late 60's or early 70's, possibly in Journal of Mass Spec + Ion
Physics.
John-M. Sichel
[Also to Dr. Fournier who kindly answered as follows]
From: SMTP%"fournier #*at*# mail.physics.unlv.edu" 6-DEC-1995 14:39:12.33
>What's the energy of electron relaxation approximately compared to atomic
>vibrational relaxation energy?
I don't know exactly, but for ionization of a valence orbital I would
guess that the electron relaxation energy is on the order of 1 eV, but it
could be 2 or 3 eV, maybe more, in some cases. The difference in correlation
energy between the N-electron and (N-1)-electron systems is also on the order
of 1 eV, but of opposite sign: for atoms Li to K, it varies between 0.1 eV (Li)
and 1.7 eV (Ne) ["Density functional theory of atoms and molecules", R. G. Parr
and W. Yang, Oxford University Press, 1989; table 8.1 on page 178].
Generally speaking, the vertical transition corresponds to the peak
maximum of an electronic transition (excitation or ionization) and I
suppose that what you mean by "vibrational relaxation" is the difference
between peak maximum and peak origin (0-->0 transition). This difference
(the vibrational relaxation energy) depends very much on the type of
system and transition, but just to give an example it is roughly 0.1 eV,
0.5 eV and 0.5 eV for the 3 lowest ionization potentials of formaldehyde.
Roughly speaking, it is a small number (say, between 0 and 5) times the
frequency of the vibrational mode most directly affected by the
ionization. I guess it should be on the order of 0.2 eV or less for
non-bonding or delocalized orbitals because there is little change in bond
orders and bond lengths, the most intense transition would be the 0-->0,
and the only change are small distortions in all bond lengths and angles.
In an extreme case, H_2 --> H_2+ + electron, the vibrational relaxation
energy is about 1.0 eV and I guess this is one of the largest value.
In general the effect of electron relaxation and electron correlation
on calculated IPs are larger (sometimes much larger) than the vibrational
relaxation energy.
>Also, I don't quite understand the correlation (means relaxation?
I just meant the usual definition of correlation energy. The
difference between the exact energy of a system "E(exact)" and the limit
Hartree-Fock calculation, or in other words, any error remaining in a
Hartree-Fock calculation with a complete (infinite) basis set:
E(corr.) = E(exact) - E(HF) < 0
see for instance Parr and Yang (reference above) page 13-14.
Sincerely,
Rene Fournier.
From: SMTP%"sichelj -AatT- Umoncton.CA" 6-DEC-1995 15:23:40.86
> Is there any way to calculate vertical ionization potential which takes into
> account all electron effects (orbital relaxation and electron correlation) but
> not atomic vibrational relaxation? I talked to my boss about a new idea which
Many methods include correlation, e.g. CI etc. etc. To exclude nuclear
(vibrational) relaxation, just calculate the ion energy at the geometry
found for the neutral molecule, without re-optimizing geometry.
> modifies the traditional energy diagram (E vs. Rxn Coord). I want to add one
> more dimension to it---electron coord. In this new 3-D diagram, "vertical"
This would be invalid. It means calculating the energy for fixed values of
an electron co-ordinate. But an electron cannot be fixed according to the
quantum-mechanical uncertainty principle; you must consider it as a wave
with a probability distribution.
Note that the Born-Oppenheimer approximation does allow one to fix nuclei
because they are heavy. But this does not work for electrons, which is
why their co-ordinates are never included in a potential surface.
John-M. Sichel
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