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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.

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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.

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    $\begingroup$ 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. $\endgroup$
    – Andras Deak -- Слава Україні
    Commented May 18, 2022 at 9:47
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    $\begingroup$ @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. $\endgroup$
    – Gregor Michalicek
    Commented May 18, 2022 at 13:23
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    $\begingroup$ 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. $\endgroup$
    – Andras Deak -- Слава Україні
    Commented May 18, 2022 at 19:57
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    $\begingroup$ 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. $\endgroup$
    – Gregor Michalicek
    Commented May 19, 2022 at 8:25

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