A Half-Semiconductor is a material with a narrow band-gap for one spin channel and a wide band-gap for the other channel. Knowing that for this material :

$E_{gap}(up)$ = 0.9609 eV

$E_{gap}(down)$ = 2.4974 eV

Can I consider the spin-up gap as narrow, and the spin-down as wide and call it a Half-Semiconductor material ?

  • 1
    $\begingroup$ I would say to find an actual source for this cutoff, but Wikipedia suggests 1.11 eV or lower as a narrow bandgap. $\endgroup$
    – Tyberius
    Aug 29, 2021 at 18:31
  • $\begingroup$ Thanks, that's a good starting source but as you have said I must find a real scientific source. $\endgroup$
    – Chi Kou
    Aug 30, 2021 at 3:16

2 Answers 2

  • For the up-spin channel, you can absolutely consider it as a semiconductor; for the down-spin channel, you can consider it as an insulator (may someone also consider it as a semiconductor, just different conventions). Thus, you can say your material is a half-semiconductor (HSC). (HSC is a semiconductor in one spin channel but an insulator in the other spin channel.)

  • One more thing you should check is the band alignment (don't mix the HSG and bipolar magnetic semiconductors (BMS)), seeing (d) and (f) below:

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  • Ref: https://academic.oup.com/nsr/article/3/3/365/2236578?login=true

  • PS: As @Tristan Maxson said, it's hard to identify insulators (with a large bandgap) and semiconductors (with a moderate or small bandgap) with a definitive gap value.

  • Finally how you obtained these gaps listed above? DFT, HSE06, or GW? I suggest that you calculate them with HSE06 or GW. DFT usually underestimates the bandgap of the semiconductors. If the GW gap of the up-spin channel is below 1.5eV and the GW gap of the down-spin channel is above 3eV, then it must be classified into HSCs.

  • $\begingroup$ These results were obtained using GGA+U. $\endgroup$
    – Chi Kou
    Sep 1, 2021 at 4:52

I think you will be hard pressed to find a definitive answer, however I would say you are within reason to call it a half semiconductor. Just be aware that this doesn't seem to be a well defined term and you may need to give the audience / reader a prompting on what you mean by half semiconductor.

  • $\begingroup$ Yes, you are totally right. $\endgroup$
    – Chi Kou
    Aug 30, 2021 at 3:15

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