Qestion on MPSH of two probe system

[zhangguangping]

Dear all, I now use MolecularEnergySpectrum class to do MPSH anaylsis in two probe system. The system is a benchmark model that contains a C6H4S2 molecule. And the projection of MPSH analysis is to the bare molecule, that is, C6H4S2 (strictly speaking, it is not a molecule but a benzenethiolate). According to the manual http://www.quantumwise.com/documents/manuals/latest/ReferenceManual/index.html/chap.atomicdata.html#sect1.atomicdata.periodictable, there are 40 electrons (4*6+1*4+6*2=40) for this projection part. So, I would expect the HOMO orbital would has an index of 19 since ATK count from 0, and LUMO 20. But the MolecularEnergySpectrum reports that there are 42 electrons in the projection region. Please refer to the attached input file for more details if possible. The output for molecular eigenstates are as follows.

Code

+----------------------------------------------------------+
+------------------------------------------------------------------------------+
| Molecular Energy Spectrum Report                                             |
| ---------------------------------------------------------------------------- |
| Fermi level = -2.163589e+00                                                  |
| Number of electrons = 42.000000                                              |
| Unit = eV                                                                    |
| Eigenenergies given relative to the Fermi Level                              |
+------------------------------------------------------------------------------+

    0  -1.799324e+01   2.000000e+00
    1  -1.650674e+01   2.000000e+00
    2  -1.487298e+01   2.000000e+00
    3  -1.480179e+01   2.000000e+00
    4  -1.295187e+01   2.000000e+00
    5  -1.146399e+01   2.000000e+00
    6  -1.065004e+01   2.000000e+00
    7  -9.021545e+00   2.000000e+00
    8  -7.843826e+00   2.000000e+00
    9  -7.693816e+00   2.000000e+00
   10  -6.878082e+00   2.000000e+00
   11  -6.599574e+00   2.000000e+00
   12  -5.959106e+00   2.000000e+00
   13  -5.247004e+00   2.000000e+00
   14  -4.881207e+00   2.000000e+00
   15  -4.745161e+00   2.000000e+00
   16  -3.415383e+00   2.000000e+00
   17  -3.410642e+00   2.000000e+00
   18  -3.266553e+00   2.000000e+00
   19  -2.651504e+00   2.000000e+00
   20  -1.869709e+00   2.000000e+00
   21   2.357217e+00   5.029450e-40
   22   2.567025e+00   1.502792e-43
   23   2.927772e+00   7.440152e-44
 ....
 ....

So, what is wrong? With my best regards, Guangping

[Anders Blom]

The two extra electrons are basically those that complete the bonds to the gold, so they are borrowed from the gold to saturate the dangling thiolate bond (the missing H atoms).

[zhangguangping]

Dear Anders, As I understand, to do MPSH, one just use the Hamilitonian subspace subject to the selection atoms. And directly diagionalized the subspace Hamilitonian obtaining the MPSH eigenvalues. If as you have mentioned, two electrons have to be borrowed from the electrodes, then what the subspace Hamilitonian looks like? Or maybe the same subspace Hamilitonian is used as the above procedure. Just fill two extra electrons to the obtained MPSH eigenlevels. So, does this two electrons shift the Fermi level of the projecting atoms? Ah, I find the

Code

Fermi level = -2.163589e+00 

which is lower than the hightes occupied energy level

Code

20  -1.869709e+00   2.000000e+00

Therefore, the LUMO and HOMO energy levels for C6H4S2 benzenethiolate in the two probe system is energy level  of index 20 (although its occupation is 2.0) and 19. That is to say, to dertermine which energy level in MPSH anaylsis is HOMO and LUMO, do not use the occupation number but the Fermi level. Am I right? Guangping

[Anders Blom]

No, the numbers are fooling you. Actually the Fermi level reported in the output cannot really be used for anything, since you don't know what it's given relative to.

Note that the eigenenergies are given relative the Fermi level, so in the eigenspectrum, the Fermi level is zero. Thus it's fully consistent that the HOMO is the smallest negative energy, and LUMO the smallest positive.

The reason you have 42 electrons is that you are not projecting on the isolated molecule, but the molecule coupled to the Au surfaces.

[zhangguangping]

Dear Anders,

Thanks very much for your reply.

So, for my example, it is really hard to image that the C6H4S2 has 42 electrons because there is no such large electron numbers transfered from Au electrodes to the molecule. I have to think, there are 21 eigenenergy levels under Fermi level due to the shift of the eigenenergy levels arising from the interaction of molecule and Au electrode.

With best regards,

Guangping

[Anders Blom]

Isn't that the same thing?