If we sum over all the PDOS (Projected/Partial Density of States), will we have the total DOS or simply the DOS? This was not the case in my DOS and PDOS results for a simple Al atom. I am wondering what could be the reason.

Is the choice of Wigner radius affecting the PDOS results? So, summing over the PDOS results might be a little less or a little more than the total DOS in my understanding because you cannot fill a square-edged box with spheres of finite radius.

Is this explanation correct? Or is there another explanation such as this is a known issue only in Quantum ESPRESSO (QE)? Or should we get the DOS exactly by summing over all the projections (which means I made some mistakes during my calculation)?

Edit: On page 24 of this tutorial by Giannozzi (author of QE), he mentioned that they are not exactly the same. He didn't explain the reason there though. So, I don't think this is a mistake. Now my question is, what could be the reason behind this discrepancy? What exactly is the role of the Wigner radius in the calculation of PDOS?

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    $\begingroup$ See the comments on this earlier answer for some explanation of why the sum of PDOS won't exactly equal the DOS: mattermodeling.stackexchange.com/a/9968/7 $\endgroup$
    – Tyberius
    Commented Mar 6, 2023 at 16:51
  • $\begingroup$ Thanks, it certainly helped. Still, it will be nice if someone wrote an answer explaining the following: "there is usually some charge which cannot be represented in the sum over atomic orbitals (called 'spilling')" $\endgroup$ Commented Mar 7, 2023 at 6:57
  • $\begingroup$ I seem to have been summoned... how different are your PDOS and DOS? What orbitals are you including in your PDOS? $\endgroup$ Commented Mar 8, 2023 at 23:51
  • $\begingroup$ @PhilHasnip, It was for Fe atom and the PP includes 4s and 3d orbital. The difference between DOS and PDOS is almost negligible in my case. I was following the official QE totorial by P. Giannozzi (hands-on pwscf). There he mentioned that DOS and the sum of the PDOS are not exactly the same. I am just curious what is the reason. So far I found two possible reasons: (1) spilling electron (2) 'due to wigner radius that is used to project the wavefunction'. It'd be helpful if I can get some related resources. I don't understand both of the reasons $\endgroup$ Commented Mar 10, 2023 at 17:48
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    $\begingroup$ OK, I'll write an answer and you can see whether it helps! $\endgroup$ Commented Mar 11, 2023 at 0:52

1 Answer 1


The fundamental problem is that the basis set used for the partial density-of-states (PDOS) does not span the plane-wave basis set space, so it is not usually possible to represent the Kohn-Sham states exactly in the atom-centred basis used for PDOS. There are two main issues:

1. Finite radial extent

The atom-centred functions usually decay quite rapidly as they move away from their centres. Exactly how rapidly they decay depends on what the functions are, and there are many different choices used (e.g. Gaussians, hydrogenic orbitals (exponential) or pseudoatomic orbitals); however, they all have a limited distance. A covalent bond could easily be long enough that neither of the bonded atoms' atom-centred states could represent it well, and the problem is even worse when electrons are delocalised (e.g. metals or aromatic rings).

2. Limited spherical harmonics

The atom-centred states are the product of a radial part and a spherical harmonic, the latter corresponding to particular orbital angular momentum quantum numbers (l, m). However, orbital angular momentum is not a good quantum number for a material or molecular system, and there is no guarantee that the actual Kohn-Sham (or whatever) states can be represented well by a sum of spherical harmonics over a small range of orbital angular momenta.


As a simple example, consider using only a hydrogenic 1s and 2s orbitals for the PDOS of bulk lithium. At first glance, you might expect this to be reasonably good, since lithium atoms only have s-states occupied in the simple hydrogenic picture, but there are some problems which arise immediately.

Firstly, the hydrogenic states ignore electron-electron repulsion, and so their radial components tend to be more compact than in real atoms; this means that the simple hydrogenic states cannot represent density a long way from the lithium atoms, because the states don't actually extend far enough from the nuclei.

Secondly, the s-states are spherically symmetric, so we cannot represent any directional dependence in the electron density near the atom at all, the only way to get any directionality at all is to mix the s-states of two or more lithium atoms.

Quantifying the problems

Since we have the original Kohn-Sham states, we can quantify the error in our PDOS projection by calculating the difference in total electron density between the total projected states and the original states. This quantity is called the "spilling", and is often expressed as a percentage of the total density.

Beware that the spilling is effectively the mean error, whereas you may care more about the worst error. For example, if the spilling is 0.5% you might think that is no problem at all, but if you have 400 electrons in your simulation, 0.5% spilling means your PDOS is missing 2 electrons' worth of density; if this is a 0.5% error in all the states it is probably fine, but what if the error is almost entirely in just one or two states? You could find that it's exactly the state you're interested in which is represented most poorly!

In principle, the spilling should reduce if you include more states in the atom-centred basis. However, in a simple Mulliken-style PDOS the assignment of charges to atom-centred orbitals can become ambiguous when the atomic states overlap, which means you have the paradoxical situation where you want the atom-centred states to cover the whole simulation volume, but not overlap; this is not possible with spherical atomic states!

Since this is only a post-processing step, it is not usually worth worrying too much about it. You will usually want your total spilling to be small (perhaps less than 1%) and I recommend always plotting the total PDOS and comparing to the DOS from the Kohn-Sham states, so you can see whether there are particular regions of energy which are poorly represented (and whether you care about those regions or not). Provided the errors are small, the results will be fine for qualitative analysis, which is the aim of this approach in the first place.

  • $\begingroup$ Thank you Dr. Hasnip for your great answer, as always. I have a couple of questions please. How/possible to calculate spilling in CASTEP? Does using a finer k-point grid for PDOS calculations reduce the spilling? After all, does this mismatching between TDOS and PDOS suggest that it is better to look at TDOS plots for quantitative considerations? $\endgroup$
    – Sha
    Commented Mar 11, 2023 at 5:35
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    $\begingroup$ @Sha CASTEP always reports the spilling in the output (.castep) file when it does the population analysis. The k-point grid doesn't affect it in any meaningful way, the issue is just that the atomic orbitals can't represent everything that the plane-waves can. The total DOS is always the most reliable, since it's the actual DOS you computed, but that doesn't mean the PDOS can't give you useful insight into why the DOS looks the way it does. $\endgroup$ Commented Mar 12, 2023 at 23:09
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    $\begingroup$ I love CASTEP! Thank you very much @Phil Hasnip. $\endgroup$
    – Sha
    Commented Mar 13, 2023 at 6:38

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