How does a wave function of a $s$-band state with odd parity looks like (in real space)? To keep it simple, the restriction to a linear chain and a state at the $\vec{k}=0$-point might be useful.

My understanding is somehow like described in https://www.ensc-rennes.fr/wp-content/uploads/2015/04/JFH-Intro-Solide.pdf. When we want to construct the crystals wave function in a tight binding fashion, we start with all electrons in their atomic s-orbital. With next-neighbour interactions, we get the cases of neighbouring orbitals being in the same or oposite phase, what gives us (with two atoms in a unit cell) one bounding and one antibounding band (see p.14 in the reference).

As I have learned from topological considerations of semiconductors, each band in a crystal which has inversion symmetry (thus also our linear chain) should have a well defined parity (at least at $\vec{k}=0$). If we put the origin of this inversion into one atom, both states (bound and antibound) have positive parity. Just if we would put the origin of the inversion exactly between two atoms, the parity of the antibound state would be odd. However, as far as I know, the origin of the unit cells in crystals and thereby the origin of the symmetry operators of the point group are defined within an atom. So how do we get a s-band with an asymmetric wave function at $\vec{k}=0$?In my opinion that is just possible by a p-like atomic orbital.

  • $\begingroup$ For real tight binding you need more than one orbital, and the crystal wave functions end up being mixtures of the orbitals, varying across $k$. This is an indication that crystal wave functions are not atomic wave functions and that, while they can be found through atomic orbitals, that is only because atomic orbitals are a complete basis set. $\endgroup$
    – Jon Custer
    Feb 12, 2020 at 14:15
  • $\begingroup$ @Jon_Custer: Thanks! However, the parity should be well defined, thus in addition to s-orbitals only even orbitals like d-, g-, i-... should be mixed into the s-band?! If the wave function would have a node at the atom, then I would name the band a p-band!? $\endgroup$
    – Matthiasho
    Feb 12, 2020 at 14:30
  • $\begingroup$ Again, the Bloch functions are a mixture of atomic orbitals. Calling a crystal band an s-band or p-band does not make sense. Some good early papers include Cardona and Pollak, Phys Rev 142(2) 530 (1966) and Chaney et al., Phys Rev B3(2) 459 (1971). Various papers have indeed plotted the "s-ness" or "p-ness" of, say, the conduction band vs $k$, but no band will be uniquely s or p. $\endgroup$
    – Jon Custer
    Feb 12, 2020 at 14:39
  • $\begingroup$ @Jon_Custer: Thanks again, but as far as I understood the subject, the notation "s-band" etc is derived from group theory and does make sense: A certain band (or rather a part of the band between high symm points) can be linked to an irreducible representation. This link is exact. Since the labels of irred. reps. are badly defined, they use typical basis-states of that irred. rep. to name the bands. An s-band is therefore described (at the gamma-point) by a function transforming like a s-atomic-orbital (representation of the full rotation group) under all symmetry operators of the goup. $\endgroup$
    – Matthiasho
    Feb 12, 2020 at 14:50

1 Answer 1


While my picture of band creation and the spatial extend of the corresponding wave functions was right, the assumption that the origin/center for inversion symmetry is always on an atom CAN be wrong. While considering a linear chain, the choice of the inversion center can be on an atom or halfway between two atoms. For a diamond lattice the inversion-center must be halfway between two atoms, while all rotations etc. are defined with respect to an origin at an atom. This shift of the coordinate system regarding different symmetry-operators is named "nonsymmorphic" space group and is rather irrelevant for further group-theory ... thus this detail is rarely mentioned.


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