0
$\begingroup$

The following pictures shows some typical examples of density of states:

enter image description here

and for superconductor:

enter image description here

Here is my interpretation: Energy level zero represents Fermi energy level. Positive energy represents energy levels for holes. Negative energy level represents energy levels for electrons. Density of states (DOS) on positive axis represents probability of electrons/holes to be found if we multiply DOS with Fermi function. For superconductor example, we have two curves separated by gap of aproximately +/- 1meV. It is a superconducting gap. Within that gap we have superconducting state.

My question is for insulator DOS example. Top unshaded peak represents electron bands and and shaded one represents hole bands. What is the physical interpretation if unshaded peak is on negative axis instead positive and shaded peak is still positive? Is there such thing as negative energy levels? (antiparticle?)

$\endgroup$
  • $\begingroup$ Just a remark on the top figure (because it is not clear what you mean by negative axis): The energy on such vertical diagrams is increasing upwards. So, in the example of an insulator the shaded DOS corresponds to lower energies than the unshaded DOS above.In general, the energy measured relative to something. $\endgroup$ – Riddler May 27 at 13:01
  • 1
    $\begingroup$ Small addition: The figure below is a cartoon of a typical result of a tunneling microscopy measurement, where the bias defines the offset relative to the Fermi level, that is why the zero of the energy axis is located exactly at the Fermi level. $\endgroup$ – Riddler May 27 at 13:08
1
$\begingroup$

The hole and electron bands in your example are, in other words, occupied and unoccupied (virtual) energy bands. If you flipped that, it means all the electrons have been exited into the conduction bands, leaving behind unoccupied states in what used to be the valence bands. Such a scenario could be realized, for instance, with powerful laser pulses, where the excitation density is high enough to momentarily excite everything from the valence bands.

The energy of some states is negative only because of the chosen reference value, which is generally the Fermi level. There is no need to invoke the antiparticle concept here.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks Raul Laasner for feedback. So for topological insulator, negative DOS peak on negative energy axis and positive DOS peak on positive energy axis implies particle and antiparticle? $\endgroup$ – Aschoolar Oct 18 '17 at 17:45
  • 1
    $\begingroup$ I haven't had any experience with topological insulators so I can't go into details. As far as I know, they don't use the particle-antiparticle concept. Normally antiparticles really only come into play when dealing with high energy physics. $\endgroup$ – Raul Laasner Oct 20 '17 at 2:01

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.