I am new to solid state calculation. I have a question about the pseudopotential in terms of the relativistic effect. In solid-state calculations, ultrasoft pseudopotential using PBE functionals are widely used. I opened my ultrasoft pseudopotential file, I saw it says

'The Pseudo was generated with a Scalar-Relativistic Calculation'.

My question is, what is the definition of Scalar-Relativistic Calculation? I have a hard time finding the exact definition from a textbook. Are there any resources I can read about the scalar-relativistic effect?

Also, in order to generate the pseudopotential, the all-electron wavefunction needs to be calculated first, in ultrasoft pseudopotential, what kind of method do people use to calculate the all-electron wavefunction?

I really appreciate any comments on this. Thank you.


1 Answer 1


There is no scalar-relativistic effect. "Scalar-relativistic" is an approximation to the fully relativistic treatment. The main simplification in this technique is the neglection of spin-orbit coupling. As a consequence, instead of a 4-component wavefunction, in this approach one deals with two-component wavefunctions for each spin. In a nonrelativistic treatment you only have a single component.

Here are two papers that introduce and discuss this approximation:

  1. Koelling, D D; Harmon, B N (28 August 1977). "A technique for relativistic spin-polarised calculations". Journal of Physics C: Solid State Physics. 10 (16): 3107–3114.
  2. Takeda, T. (March 1978). "The scalar relativistic approximation". Zeitschrift für Physik B. 32 (1): 43–48.

I hope you have access to these journals. I'm sure there are also nice chapters on this in several text books but for that someone else has to give suggestions.

  • 1
    $\begingroup$ Good answer. Since most readers are unlikely to look up the references, I'd extend with one remark (since people also often confuse "spin-orbit coupling" with "relativistic effects"): in the 1/c^2 expansion of the Dirac equation we get three terms: the mass velocity term, the Darwin shift and the spin-orbit coupling. Scalar-relativistic calculations (as opposed to non-relativistic ones) include the former two terms which are often very important. Spin-orbit coupling on the other hand can often be neglected. And fully-relavistic calculations use the Dirac equation, so they go beyond spin-orbit. $\endgroup$ May 18 at 9:47
  • 1
    $\begingroup$ @AndrasDeak--СлаваУкраїні: Thank you for the additional clarification. I have to admit that I am not so sure whether one can think the scalar-relativistic approximation in terms of such a 1/c^2 expansion. Up to the order where mass-velocity-, Darwin-, and spin-orbit-coupling terms appear it is correct to think in this way. But to my understanding the scalar-relativistic approximation also covers many higher-order terms. But I have to admit that I also never studied the details. $\endgroup$ May 18 at 13:23
  • 1
    $\begingroup$ Fair enough, you indeed don't need an actual 1/c^2 expansion to isolate the spin-orbit terms based on the angular momentum dependence. But hopefully if you do the 1/c^2 expansion on what you get this way would recover the aforementioned "named" terms plus some small (as in 1/c^2 small) corrections to those. So yeah, while I like to think of the remaining effects in terms of the 1/c^2 expansion, I don't know for a fact that these are the main contributions being picked up. Thanks. $\endgroup$ May 18 at 19:57
  • 1
    $\begingroup$ I think it is pretty clear that the named terms are the main contributions. But it is also clear that the papers on the scalar-relativistic approximation (SRA) don't take the path over such a 1/c^2 expansion. One may wonder how an approximation on this basis would actually differ from the SRA. As a sidenote, there are several relativistic approximations similar to the SRA. One example is the zero-th order regular approximation (ZORA). I also don't know how these alternatives behave in comparison to the SRA. $\endgroup$ May 19 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.