I'm new to DFT. Can you please help me to understand DOS and PDOS (projected density of states) in simple wording? I only want to know some basics of these terms and how they are linked to DFT.

  • $\begingroup$ I think the Joint density of states (JDOS) must also be included in the question to attract more detailed answers on DOS types. $\endgroup$
    – AbPhys
    Aug 16, 2022 at 14:10

2 Answers 2


From the Wiki (a good starting point):

DOS (Density of State):

In solid state physics and condensed matter physics, the density of states (DOS) of a system describes the proportion of states that are to be occupied by the system at each energy. The density of states is defined as $D ( E ) = N ( E ) / V$ , where $N ( E ) \delta E$ is the number of states in the system of volume $V$ whose energies lie in the range from $E$ to $E + \delta E$. It is mathematically represented as a distribution by a probability density function, and it is generally an average over the space and time domains of the various states occupied by the system.

PDOS (Projected Density o States)
Gives the projection of particular orbital of particular atom on the density of states. So, if you sum over all the projections, you will have the total density of state, or simple, the DOS.

As wrote at the beginning, both concepts are related with how the energy states are distributed. DFT is not directly related with those concepts. DFT if a quantum mechanic method used to determine the quantum properties of the system using a functional of the energy. You can use other method to calculate both DOS and PDOS.

As an example, the article below, is related to calculations of projected density of state with ONETEP software:

  • J. Aarons, L. G. Verga, N. D. M. Hine and C-K Skylaris. Atom-projected and angular momentum resolved density of states in the ONETEP code. Electron. Struct.1 035002 (2019). DOI: 10.1088/2516-1075/ab34f5.
  • $\begingroup$ Is this always true that "if you sum over all the projections, you will have the total density of state"? Shouldn't it be a little less (or more) due to the Wigner radius that is chosen to project them? $\endgroup$ Mar 1, 2023 at 23:08

Basically, the central job of KS-DFT is to solve self-consistently the following non-colinear KS equation:

\begin{equation} \left[ -\dfrac{1}{2}\nabla^2+v_{ks}(\vec{r}) \right]\phi_n(\vec{r})=E_n\phi_n(\vec{r}) \end{equation}

Here $v_{ks}$ represents the KS effective potential and $n=(atom, orbital, k, spin)$ the collective quantum number of the quantum state. With this minimal knowledge, we can answer your question.

  • For the density of states (DOS), we are counting the quantum number $n=(atom, orbital, k, spin)$ at the energy interval $\Delta E$, in which the information related to atom, orbital, $k$ and $spin$ are integrated.

  • For the partial density of states (PDOS), such as the atom-resolved PDOS (here I assume there are only two kinds of atoms: A and B), we are also counting the quantum number $n=(atom, orbital, k, spin)$ at the energy interval $\Delta E$. BUT at this time, we integrate the information about $orbital, k$, and $spin$ for A and B respectively. Finally, we will obtain the DOS plot for A and B. With the same logic, we can plot the orbital-resolved PDOS and spin-resolved PDOS.

In summary, the total DOS is counting the quantum state with the full consideration of the state indexes at some energy interval, and the PDOS is also counting the quantum state but with the partial consideration of the state indexes at some energy interval.

PS: When are talking about the DOS, the information related to $k$ always be integrated.

Related: Properties that can be deduced from Band structure and DOS


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