# Fermi energy as an energy difference?

I was traveling down the Wikipedia rabbit hole when I ran across something that caught me off guard. On the Fermi energy page, it says that the Fermi energy is:

...the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature.

Ok, so far so good. But then a bit later says that it is "an energy difference (usually corresponding to a kinetic energy)"? I got a little stuck on this, wondering how is kinetic energy related to a difference in energy? Especially when it then goes on to say "the Fermi energy is the kinetic energy of the highest occupied state."

How is the Fermi energy the difference between the highest and lowest occupied states of fermions, and how does this difference give the kinetic energy of the highest occupied state?

• A similiar question has been asked before, maybe this helps: physics.stackexchange.com/q/520184/246032
– Tera
Dec 20 '19 at 18:13
• Ok, looks like an assumption there is that the lowest state is equal to zero, therefore the difference is just equal to the max. How does this hold up if the lowest state isn't zero? Is the difference between highest and lowest state somehow going to be equal to the kinetic energy of the highest state? Dec 20 '19 at 18:23
• Can you tell me a case where it is not zero? As far as I know not.
– Tera
Dec 20 '19 at 18:24
• Ok, I think that might be where I was getting confused. I thought the zero assumption was for some sort of ideal case. Will this assumption hold for a material at absolute zero, regardless of the material? Insulator, semiconductor, metal, we can assume that all of the lowest energy states are equal to 0? Dec 20 '19 at 18:34
• @Tera, often we measure the Fermi energy relative to the vacuum energy rather than to the ground state. In this case the lowest energy bound state is at a more negative energy than the Fermi energy. Dec 20 '19 at 19:33

There are basically two questions:

1. How is the fermi energy an energy difference, and at the same time the energy of the highest occupied state (implying that the lowest state is 0)?

This comes from the confusion based on the not full or incorrect citation, "...the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature."

Now the correct statements come as follows:

• the Fermi energy is the energy difference between the highest and lowest single particle energy state at absolute zero

• in a Fermi gas, the lowest state is known to have zero kinetic energy, but in a metal, lowest occupied state is the bottom of the conduction band, thus non zero kinetic energy

So the answer to your question is that in a Fermi gas, yes, the energy difference is equal to the energy of the highest occupied state, because the lowest is zero, but in a metal, the answer is no, because the lowest energy level is non zero energy.

1. Why is the Fermi energy a kinetic energy?

• the Fermi energy is an energy difference, usually kinetic energy, whereas the Fermi level is kinetic and potential energy

As a consequence, even if we have extracted all possible energy from a Fermi gas by cooling it to near absolute zero temperature, the fermions are still moving around at a high speed. The fastest ones are moving at a velocity corresponding to a kinetic energy equal to the Fermi energy. This speed is known as the Fermi velocity.

https://en.wikipedia.org/wiki/Fermi_energy

The Fermi momentum and velocity:

$${\displaystyle p_{\mathrm {F} }={\sqrt {2m_{0}E_{\mathrm {F} }}}},$$

$${\displaystyle v_{\mathrm {F} }={\frac {p_{\mathrm {F} }}{m_{0}}}}.$$

They correspond to the Fermi energy, thus the Fermi energy is regarded as a kinetic energy (the speed of the electrons).