I know that, in the band structure of a solid, the Fermi level is the energy level at which there is a 50% probability of finding an electron (at thermodynamic equilibrium), and that it doesn't have to correspond to an actual energy level (ie it's in the band gap in an insulator). But Atkins and Shriver's Inorganic Chemistry doesn't mention this and instead defines it as the highest occupied energy level at 0 K. (page 103, 5th edition) I was wondering if these definitions are actually equivalent. If so, can you explain how? Given the fact that in an insulator the Fermi level is in the band gap, it seems like at best the definition in Atkins is oversimplified.

  • $\begingroup$ It is not a straightforward concept at all, so I am not surprised that an inorganic textbook simplifies it a bit. You might look at aapt.scitation.org/doi/10.1119/1.1629090 for the case of an intrinsic semiconductor. $\endgroup$
    – Jon Custer
    Jan 6 '20 at 13:56

Be aware of Fermi level versus Fermi energy, as it seems to me the book speaks about the latter.

The quotes below are from both Wikipedia links.

In band structure theory, used in solid state physics to analyze the energy levels in a solid, the Fermi level can be considered to be a hypothetical energy level of an electron, such that at thermodynamic equilibrium this energy level would have a 50% probability of being occupied at any given time. 

The Fermi energy is a concept in quantum mechanics usually referring to the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature. In a Fermi gas, the lowest occupied state is taken to have zero kinetic energy, whereas in a metal, the lowest occupied state is typically taken to mean the bottom of the conduction band.

  • The Fermi energy is only defined at absolute zero, while the Fermi level is defined for any temperature.

  • The Fermi energy is an energy difference (usually corresponding to a kinetic energy), whereas the Fermi level is a total energy level including kinetic energy and potential energy.

  • The Fermi energy can only be defined for non-interacting fermions (where the potential energy or band edge is a static, well defined quantity), whereas the Fermi level remains well defined even in complex interacting systems, at thermodynamic equilibrium.

Since the Fermi level in a metal at absolute zero is the energy of the highest occupied single particle state, then the Fermi energy in a metal is the energy difference between the Fermi level and lowest unoccupied single-particle state, at zero-temperature.

A collection of non interacting fermions, often called the Fermi gas is a simplified model useful in the solid state physics. There is supposed the considered fermions do not interact, so the energies of these fermion quantum states do not depend on the distribution of these fermion quantum states.

A typical case would be the classical hydrogen atom model, applied on multielectron atoms. As electrons in the same atom interact quite a lot, it is not a good model.

For solid states, electrons and electron vacancies are often taken as quasi fermions following Fermi-Dirac statistics for the Fermi gas, as they are separated enough to be considered as non interacting.

  • $\begingroup$ Thank you, that does help. The book calls it the Fermi level but from your answer it seem like they are in some sense mixing the two concepts. When do systems with non-interacting fermions actually arise? One would suppose that electrons in a solid are inevitably going to interact with each other. Just to be fully clear, is there no particular link between Fermi energy and Fermi level (beyond the fact that presumably Fermi worked on both!)? $\endgroup$
    – atbm
    Jan 7 '20 at 15:57
  • $\begingroup$ Just added the paragraph about the relation of the quantities in a metal context. $\endgroup$
    – Poutnik
    Jan 7 '20 at 16:37
  • $\begingroup$ Added Fermi gas and non interacting fermions note. $\endgroup$
    – Poutnik
    Jan 8 '20 at 6:20
  • $\begingroup$ Thank you - that's so helpful! $\endgroup$
    – atbm
    Jan 8 '20 at 10:01

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.