Subject: Ionization Potential: Summary of Answers

From: "Fernando De La Vega" <[email-masked]>
Date: Mon, 08 Apr 2002 22:50:06 -0700

Hello,
  I would like to thank Li Zhenhua, Brian Salter-Duke,  Alan Shusterman, 
Jens Spanget-Larsen,  N.O.J. Malcolm, Marcin Makowski,  Ulrike Salzner, 
Pascal Boulet, Teodorico Ramalho, E. Lewars, David Shobe, Pierre Vitorge 
and, Robert Ponec for their response to my inquiry. I reproduce below my 
original questions and their answers. Thanks again.

My question:
I want to calculate the Ionization Potential (IP) for a particular system. I 
know I can get an approximate value through Koopman's theorem but I would 
like to be more exact. I understand that I can calculate the IP from the 
energy difference between the molecule and its +1 cation. However, what I am 
not sure about is whether I should calculate the energy for the +1 cation 
using the geometry of the of the neutral molecule or whether I should 
optimize the geometry of the cation and then calculate the difference. Your 
comments on the subject are really appreciated. I will summarize if there 
are any responses. Thanks.

Answers:
----------------------------------------------
Li Zhenhua:
The usual procedure is to optimize the geometry of the cation and calculate 
the energy difference.

--------------------------------------
Brian Salter-Duke:
It depends whether you want the Vertical Ionization Energy as measured say 
by Photoelectron spectroscopy, in which case you want the cation to be at 
the same geometry as the neutral molecule, or whether you want the adiapatic 
ionization energy where both the neutral and the cation should be at a 
geometric minimum energy.
--------------------------------------
Alan Shusterman:
The correct choice depends on what use you intend to make of your result. If 
you leave the geometry the same, you can estimate the vertical IP. If you 
optimize the cation's geometry, you can estimate the adiabatic IP.

---------------------------------------
JENS SPANGET-LARSEN:
if you use the neutral groundstate molecular geometry in the calculation for 
the cation, you end up with so-called vertical ionization energies 
(according to the Franck-Condon principle). These ionization energies 
generally correspond to the maximum, or the center of gravity, of the 
photoelectron (PE) band. If you optimize the geometries of the cationic 
states, you obtain the so-called adiabatic ionization energies that would 
correspond to the onset op the PE bands.

---------------------------------
Dr. N.O.J. Malcolm:
It really depends what you want to know, if you use teh neutral system 
geometry for teh cation then you will have calculated a vertical IP (VIP) 
otherwise using the optimised cation geometry you will have teh adiabatic IP 
(AIP). Both the VIP and AIP can be determined via experiments so check what 
you want/have. By the way you may want to have a look at Greens function 
methods for calculating IPs as these are generally a bit more accurate than 
energy difference approximations.

---------------------------------
Marcin Makowski:
Computing energy of the cation at the same geometry will give you vertical 
IP, while relaxing geometry will result in adiabatic IP. I dont know which 
of these you are interested in, because both of them might be measured in 
experiment (but different type).

---------------------------------
Ulrike Salzner:
if you use the geometry of the neutral, you get the verticel ionization
energy, if you re-optimize, you get the adiabatic ionization energy. Now you 
have to decide what you need. In experiment, the vertical IP
isusually determined, because the nuclear motion is slow compared to the 
movement of the electrons (Franck-Condon principle).

--------------------------------------
Pascal Boulet:
There is an alternative way to calculate vertical ionization potential : use 
the geometry of the neutral molecule, put half an electron on the HOMO and 
perform a single point calculation. The energy of the HOMO should give you a 
good idea of the IP. It's working well within the DFT formalism. I've not 
tried for ab initio methods.

------------------------------------------------------
Teodorico Ramalho:
Computing energy of the cation at the same geometry will give you vertical 
IP, Experimentaaly, the IP vertical is most difultof be obtained. Thi is, 
generally, obtained by spectroscopic studies. There is one option in 
Gaussian98 to calculation vertical IP, the keyword is OVGF. If you relax 
geometry will result in adiabatic IP. This is IP, generaly, obtained to 
electrochemistry studies.
------------------------------------------------
E. Lewars:

Hello, Fernando De La Vega asked about ionization energies. Whether you use 
for your calculation (1) the cation at the geometry of the neutral or (2) 
the optimized geometry of the cation, depends on which ionization energy 
(ionization potential) you want:

(1) gives the _vertical_ ionization energy,
(2) gives the _adiabatic_ IE

The adiabatic values seem to be of more interest to chemists, since they 
represent the energy difference between two "stable" species (potential 
energy surface relative minima), but compilations of IE do not always state 
clearly whether their listed IEs are vertical or adiabatic; an exception is 
R. D. levin and S. G. Lias, "Ionization Potential and and Appearance 
Potential Measurements, 1971-1981", National Bureau of Standards, Washington 
DC, 1982.

Vertical IEs seem to be slightly (up to 0.2 eV) higher than adiabatic (e.g. 
experimental CH3OH, 11.0 and 10.9 eV). Koopman' IEs are about 1.0 eV too 
high (CH3OH 12.06 by HF/6-31G*, 12.12 by MP2(fc)/6-31G*, HOMO energy). The 
difference of the calculated cation and neutral energies can give accurate 
IEs if post-HF calculations are used; for CH3OH (eV):

vertical HF/6-31G* 9.66, MP2(fc)/6-31G* 10.82 adiabatic
HF/6-31G* 9.38, MP2(fc)/6-31G* 10.6

Note that experimental and calculated vertical IEs are for electronically 
but not geometrically relaxed cations; adiabatic IEs are for completely 
relaxed cations; and Koopman' IEs refer to cations that are neither 
electronically nor geometrically relaxed.
-------------------------------------------------
David Shobe:

There's a reason for IE-vert being > IE-adia. Actually it's implicit in the 
rest of E. Lewars' message, but it's worth stating explicitly:

1. The IE-vert is the energy required to go from neutral at its geometric 
energy minimum to cation at the *neutral's* geometric energy minimum;
2. whereas the IE-adia is the energy required to go from neutral at its 
geometric energy minimum to the *cation's* geometric energy minimum.
Note that the starting point is in both cases the same: the neutral at the 
neutral's geometric minimum.

3. And by definition, the cation's energy will be lower at *its* geometric 
minimum than at any other geometry, including the neutral's geometric energy 
minimum. QED.

-----------------------------------------------------------------------
Pierre Vitorge:

optimize the geometry of the cation and then calculate the difference

------------------------------------------------------------------------
Robert Ponec:
you can calculate both, the IP calculated for the geometry of +1 ion with 
the fixed geometry of the neutral molecule is the so-called "vertical IP", 
the other possibility is to relax the geometry of +1 ion to its optimum, in 
this case the IP is called "adiabatic", both values can be in principle 
determined fron experiment.

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