So I used quantum espresso to do bands calculation of an isolated Fe atom in a cell of 20 x 20 x 20 angstroms. I was expecting to see the 5 degenerate d orbitals but I didn't really get a proper degenerate energy level instead I got 6 values which were fairly close and then 3 values which were degenerate separately and another 1 energy level (I only calculated for nbnd=10).

Could someone explain how to look at such data and how to see the orbital degeneracy of Fe system?

[EDIT]: I wasn't able to find the original data so I had to redo the calculation, here is the input file


    nosym = .TRUE.
 Fe 55.845 Fe-pbe.UPF

 Fe 10.0 10.0 10.0

K_POINTS {automatic}
 1 1 1 0 0 0

 20.0 0.0 0.0
 0.0 20.0 0.0
 0.0 0.0 20.0

The output I got after doing the bands calculation this time was

    0.0000 -104.5764
    0.0000  -65.4071
    0.0000  -65.4069
    0.0000  -65.4068
    0.0000   -6.4181
    0.0000   -5.9636
    0.0000   -5.9635
    0.0000   -5.9634
    0.0000   -5.8398
    0.0000   -5.8397
    0.0000   -2.9550
    0.0000   -2.9550

This is the gnu file that was generated. I can see that there are 5 values similar which I should assume are the d orbitals but there is also a random hanging value of -6.4181 which I have no idea what it's for?

  • 2
    $\begingroup$ Could you please add the inputs files as a code block and the band structure that you obtained in your question? $\endgroup$ Jun 11 at 14:03
  • 2
    $\begingroup$ There is no band structure for an isolated atom, but atomic orbital levels. When you cell is so large that the orbitals become atomic-like, a plane wave code (QE) is very inefficient unless you use a very large cut-off. There are numerous atomic orbitals basis set codes for this task. This could be the reason why your data are inconsistent. $\endgroup$
    – M06-2x
    Jun 11 at 14:16
  • 2
    $\begingroup$ Band structure is only for periodic systems. One isolated atom is not a periodic system. What you get when running such calculations are the atomic orbitals. A note: you need to use a sufficient big cell in order to one cell does not interact with others image cell. If they interact, you will obatin a mix of orbitals from different cells. $\endgroup$
    – Camps
    Jun 11 at 14:18
  • $\begingroup$ Without knowing if the calculation is open or closed shell, or if you have constrained the orbital occupations in any way, it's impossible to answer this. In the above which states are occupied and which are virtual? Mapping iron's 6 3d electrons onto the 5 3d orbitals is an interesting problem. $\endgroup$
    – Ian Bush
    Jun 11 at 16:48
  • $\begingroup$ @M06-2x Could you elaborate on how to conduct calculation for getting atomic/molecular orbital energy levels then? $\endgroup$ Jun 11 at 17:43

1 Answer 1


Following the comment, a plane wave basis set is used for a slowly varying density, this means that the exponential shape of the atomic orbitals needs a lot of plane waves to match their structure. This is the main reason why the code used (QE) required a pseudopotential for atomic-like core orbitals.

The effective potential is not hydrogen-like, the repulsion $\sum_{i,j}\frac{1}{|\vec{r_i}-\vec{r_j}|}$ has also an angular character, you should not expect to have a degeneracy. However, if you are insisting using a PW code for this task, the cell needs to be very large, you also need to increase $E_{cut}$ for more precise eigenvalues.

Note that there is a self-interaction affecting the eigenvalues of a DFT calculation, they are upshifted. The Hartree-Fock eigenvalues are more meaningful, reason why a LCAO code is the best choice for this task.

  • $\begingroup$ So to account for the discrepancy in degeneracy, if I do a calculation of DOS which to get a better idea of seeing degenerate orbital would that be better? $\endgroup$ Jun 12 at 5:31
  • $\begingroup$ In the output of pw.x, you need first to count the levels below the Fermi energy, normally you should have six levels for $d$ and $s$ orbitals. As said you should not expect five degenerate $d$ levels. A discrete DOS and flat bands are not needed. the three degenerates $d$ levels (please check) are $d_{xy}$, $d_{xz}$ , $d_{yz}$ having the same symmetry. You can also try a spin polarized calculation, because of the Hund rule, there is a magnetization of the isolated atom. $\endgroup$
    – M06-2x
    Jun 12 at 11:20
  • $\begingroup$ I have edited the question to have the input and output, I didn't do spin polarized calculation yet since I didn't want any d orbital splitting (which I wasn't sure of). But I understand your point of 6 energies below the fermi energy. $\endgroup$ Jun 12 at 17:04

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