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QuantumATK => General Questions and Answers => Topic started by: Roc on May 12, 2009, 03:16

Title: setting the quantum number to calculate the HOMO and LUMO
Post by: Roc on May 12, 2009, 03:16
Dear everyone,

I have some confusion  about setting the quantum number to calculate the HOMO and LUMO of two-probe system. for example, for the molecule line C4BPH12, there are 40 outer-shell electrons, usually based on that two electrons occupy one orbit, so the 40 outer-shell electrons will occupy 20 orbits, and the quantum number of HOMO should be 19 (the quantum number of the first one lowest energy orbit is 0 ), for LUMO is that 20. however, it seems not like that through calculating the energy spectrum, as following,

0   -4.64
1   -4.36
2   -4.14
3   -3.8
4   -3.77
5   -3.76
6   -3.65
7   -3.06
8   -2.94
9   -2.7
10   -2.45
11   -2.19
12   -2.09
13   -2.07
14   -1.86
15   -1.67
16   -1.38
17   -1.31
18   -1.24
19   -0.71
20   -0.27
21   0.77
22   1.54
23   2.04
24   2.12
25   2.26
26   2.76
27   2.84
28   3.01


is it possible that two special electrons occupy two orbits, so, the quantum number of HOMO should be setted to 20(-0.27eV) .
or, the energy -0.27 could be considered as 0, so, the quantum number of HOMO and LUMO should be setted to 19 (-0.71eV) and 21(0.77eV).
how to explain it resonablely?
Another, for the molecule owning odd outer-shell electrons, ie. Si5Li cluster. there are total 23 electron, do they should occupy 12 orbits, and which orbit is occupied with only electron, HOMO ?
Thank you!
Title: Re: setting the quantum number to calculate the HOMO and LUMO
Post by: Anders Blom on May 12, 2009, 04:01
Your way of counting assumes that the charge remain conserved on the molecule. However, in a two-probe system charge is not conserved. And, even if it usually almost conserved in the system as a whole (typical charge loss or gain in the central region is typically not more than 1e or so), the charge can easily redistribute itself onto or off the "molecule", which you have defined via the "surface_atoms" keyword.

The eigenvalues of the two-probe molecule spectrum are normalized such that zero is the (average) Fermi level. Thus states with negative energy are indeed occupied, and those with positive energy are not (well, more exactly, they are all populated according to the Fermi distribution).

Title: Re: setting the quantum number to calculate the HOMO and LUMO
Post by: Roc on May 13, 2009, 02:58
thank you Anders Blom  for your answer! ;)