# [Pw_forum] fermi level (XQ Wang)

Stefano Baroni baroni at sissa.it
Fri Aug 15 15:30:59 CEST 2008

```OK. Let me try to throw my own 5 cents into the discussion. Let me
comment later on some of the statements made by others. The concept of
Fermi energy is a very simple one and is best explained thinking of a
system of non-interacting electrons (of which Fermi liquids are
generalizations I will not comment on).

1) The eigenstates of the Hamiltonian of a system of non-interacting
electrons can ALWAYS be chosen as antisymmetrized products ("Slater
deteminants") of one-particle wavefunctions ("molecular orbitals" in
quantum chemistry, "Bloch states" in solid-state physics) which are
eigenfunctions of a one-electron Hamiltonian. Pay attention to the
conceptually simple, trivial, but often overlooked, difference between
the many-body Hamiltonian (with its-own eigenfunctions and
eigenstates) and the one-particle Hamiltonian. The many-body energy is
simply the sum of the one-electron energies of all the molecular
orbitals whose product is the many-body eigenstate.

2) The antisymmetric nature of the many-body wavefunctions is such
that, if you construct  a product of a set of functions where two of
them are equal, the result will vanish. This is the PAULI PRINCIPLE.
No two electrons can occupy the same one-electron state.

3) Because of (2), the lowest possible many-body energy (ground state)
is the sum of the lowest N one-particle energy eigenvalues (N being
the number of electrons). The highest occupied one-electron energy
level is the difference between the ground state energy of the system
with N electrons and that with N-1 electrons (ionization potential).
The lowest unoccupied energy is the difference between the ground
states with N+1 and N electrons (electron affinity).

4) For any finite system (as well as for insulating infinite ones) the
electron affinity is different from the ionization potential. For
(infinite) metals, they coincide and the define the Fermi energy: by
definition, the energy necessary to add or to remove an electron from
the system (in classical thermodynamics this same quantity is called
the chemical potential). For insulators, it is not that the Fermi
energy "does not exist". Only, it is ill-defined it the zero-
temperature limit (it can be assumed to take any value between the
electron affinity and the ionization potential). At any finite
temperature, thermodynamic considerations remove this indeterminacy.

On Aug 15, 2008, at 11:47 AM, wangxinquan wrote:

> Dear Eyvaz,
>
>> 1. Scf calculations do not give the Fermi level, because you use
>> fixed occupations used by default for semiconductors (insulators).
>> I.e. if nothing specified as occupations, >your system is
>> considered as a semiconductor
>> (insulator). In this case the number of bands treated is exactly
>> half of the number of valence electrons, i.e. 4 in your case.
> Do you mean that the occupations of electrons were constrained
> during iterations? To my limited experiences,
> the occupations should be changed by setting the
> "starting_ns_eigenvalue" parameters (lda+u calculation) and
> the occupation matrix will be tuned through the iterations process.

the electron occupations are usually fixed following the "filling
principle": states with lower energies are filled first. in an
insulator, this is easily done because the number of filled states is
independent of the wavevector in the Brilloion zone. in metals, this
is more tricky because the number of occupied states will depend on
the key point, and in order to sort out this dependence one has to
know in advance the electronic structure (i.e. the sequence of energy
levels) at ALL the k points in the Brillouin zone

> I'm sorry for that I confused "electron density" with "occupation".
> In my opinion, both
> concepts describe the probabilty of electrons locating at lattic site.

I am sorry but your opinion is wrong. The "occupation" does not refer
to the probability of finding an electron at any particular lattice
site, but rather in any of the molecular orbitals (which, being
generally delocalized, cannot be attached at any specific lattice site).

> The "electron density"
> which is derived from wavefunctions belongs to quantum mechanics
> field, while "occupation"
> belongs to condensed matter physics field with respect to band
> structure.

I do not quite see the difference between "belonging to quantum
bechanics" as opposed to "belonging to condensed matter", but for the
fact that you may have learnt one concept in one textbook and another
in a different book. As a rule of thumb, what you learn tends to be
rather useless up the time when you can forget where/when you learnt
it and you can pretend that the concepts you are using are YOURS ...

> The electron density
> should be changed but the occupation of bands should be fixed in scf
> calculation. right?

if so you like to think, this is not wrong ... (which indicates that
you could have seen by yourself why your previous statement is wrong)

> The electron states(eigenvalue of the density matrix)

what a mess, here! electron states, if ever, may be eigenSTATES of a
quantum operator, not eigenVALUES. it is true that for independent
electrons the "electron states" (i.e. the eigensSTATES of the one-
particle Hamiltonian) are also eigenstates of the one-particle density
matrix, but the viceversa is not true. Being an eigenstate of the
density matrix is not a sufficient condition for being a legitamate
"electron state" (in the sense of being an eigenstate of the
Hamiltonian). this is so because the density matrix is a projector,
whose eigenvalues (0 and 1) are highly degenerate ...

> which construct the band structures will not
> be occupated entirely. For insulator, the valence band is full while
> the conduct band is empty.
> In the output file of scf I have no idea what the "fermi energy"
> results are. Does it not mean the
> "real" fermi energy? E.J.Yoffa and D.Adler (Phy. Rev. B, 12, 2260)
> have talked about the calculation
> of fermi energy for Mott insulator. I'm afraid the definition of
> fermi energy were misunderstood.

if you mean misunderstood by you, this is quite probable.

> My brain doesn't work.

take it easy. you are probably one of the many victims of the modern
having properly understood the fundamentals. as trivial as these
fundamentals may be, it takes time to master them. I am sure it is not

> Any help to remove my understanding to fermi energy will be deeply
> appreciated.

hope I did help, a little bit - stay tuned

Stefano B

---
Stefano Baroni - SISSA  &  DEMOCRITOS National Simulation Center -
Trieste
http://www.sissa.it/~baroni / [+39] 040 3787 406 (tel) -528 (fax) /
stefanobaroni (skype)

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de la pensée - Jean Piaget

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