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The image below shows the shape of a Fermi surface of Cr metal measured by Angle resolved photoemission spectroscopy. (not important) The question is why does the Fermi surface shrink for the electrons and expand for the holes? This leads me to think that the holes have negative energy and electrons have positive energy... so the smaller the energy, the smaller the Fermi surface for electron, and the larger the energy for holes (being less negative?)

I am really not sure what happens here!
enter image description here

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2 Answers 2

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This question may have more to do with definitions (although I am not an expert, only a graduate student, so I may be wrong). Let me try to explain what I am thinking:

The Fermi Surface is a construction that signifies an equipotential surface whose energy is the Fermi Energy. Beneath the Fermi energy the states are defined to be occupied whereas above the fermi energy the states are defined to be unoccupied.

A hole is a quasiparticle symbolizing the "absence" of an electron. It is not a real particle, but rather a convenient alternative method of describing the motion of electrons. The filled and unfilled states of an energy diagram where the charge carriers are designated to be electrons would therefore correspond to the unfilled and filled states of an energy diagram where the charge carriers are designated to be holes.

The difference in size of the fermi surface corresponding to electrons vs. holes might therefore have to do with "reversing" the definition of what states are occupied vs. unoccupied.

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If I correctly understand the caption of the picture the surfaces in (b) are not proper Fermi surfaces, because they are taken

0.3 eV below the Fermi energy

We can imagine the Fermi surface as the surface of the 3D shape that we would build in the $k$-space (or reciprocal space) if we filled with conduction electrons the available states, by proceeding in order of increasing energy and continuing until we finish all the electrons. The electrons on the Fermi surface would all be at the Fermi energy $E_F$, that is the energy of the highest occupied state (at $0K$ to be precise).

In the case of figure (b) though we did not "place" all the available electrons but we left out the ones with energy higher than $E_F-0.3eV$: the shape we built is then smaller.

As for the holes instead we must consider that lowering the energy for the electrons correspond to increasing the potential (given that $\Delta E=-e\Delta V$ for the electrons) which means higher energies for the holes. So if we build in the same way as above a surface for the holes it would be bigger, as we would have more holes to place.

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