  Relativistic density-functional-theory calculations of Bi(111) thin films are
performed to revisit their band structure and that of macroscopic samples. The
band structure of a our 39-bilayer film ($\sim$~15~nm) shows that (1)
$\sim$9-nm films are enough to describe that of Bi(111), (2) The two split
surface-state metallic branches along the $\overline{\Gamma M}$ direction do
not overlap with the bulk band at the zone boundary but lie within the
A7-distortion-induced conduction-valence band gap, and (3) Neither the
existence of the metallic surface states nor their observed splitting is
related to inversion \emph{asymmetry}. Thus, the spin texture observed in such
states is not caused by the lifting of the Kramers degeneracy and their
splitting is not of the Rashba-type. We instead propose that (1) the large
splitting of the metallic branches is a $m_j=\pm1/2$-$m_j=\pm3/2$ splitting and
(2) the spin texture observed for the metallic branches may only occur because
the almost unaltered strong covalent bonds retained by Bi(111) surface atoms
cannot afford magnetic polarization. We emphasize that degeneracy at the
$M$-point of the SBZ of Bi(111) -- implied by the translational symmetry of the
surface -- is satisfied irrespectively of the presence of inversion symmetry
centers. We show that the magnetic-moment discontinuity at $M$ does not exist,
which also explains why the measured spin-polarization of the metallic branches
vanishes near $M$. We induce the Rashba effect on the band structure of Bi(111)
via different structural/electronic perturbations to reveal the actual lifting
of the Kramers degeneracy and find that the magnitude of the perturbation
imposed on a film correlates with the magnitude of the splitting and the
localization of the Rashba-split states.
