# Fermi velocity

1. Mar 24, 2009

### carbon9

Hi,

Is this right: Fermi velocity is the velocity of electron-wave in a conductor. Why is it called Fermi velocity, i.e. what is its relation with Fermi energy, etc..

Thanks a lot.

Cheers

2. Mar 25, 2009

### genneth

Electrons' energies should be measured relative to the Fermi surface (or the chemical potential at non-zero temperatures). Similarly, the momenta should be measured relative to the Fermi surface. To first order, the constant of proportionality is the Fermi velocity. For non-interacting Fermi gas, this can be directly related to the dispersion relation for a single particle. For an interacting Fermi liquid, it's usually best to simply measure this.

Notice that Fermi velocity is directly related to the density of states (exercise to work out how).

3. Mar 25, 2009

Thanks.

4. Apr 1, 2009

### sokrates

Electrons energies can be measured relative to whatever energy you'd like. The selection of a zero energy is arbitrary.

Fermi velocity is the velocity with which an electron would travel, if the electron has Fermi energy.

No, not all the electrons are moving with the Fermi velocity, say, in a solid. But since the current flow occurs around the electrochemical potential (interchangeable with Fermi Energy here) most of the electrons have energies at around the Fermi energy ( a few kTs above and below).

If the temperatures are small and the device is in the linear regime (small bias) and the channel is a degenerate conductor ( where Ef is at least a few kTs above the conduction band) - we can use fermi velocity as the velocity of electrons to a very good approximation.

5. Apr 1, 2009

### genneth

Indeed, if you use the renormalised/effective mass to relate momentum to velocity

6. Apr 1, 2009

### sokrates

Yes, that's right. Thanks for correcting that. "Velocity" in the semi-classical sense (the way it is used for Fermi velocity) makes sense only under the effective mass theorem.

Say your dispersion relation looks like:

$$E = \hbar^2 k^2/ 2m^* + U_0(x)$$ where U(x) = 0 arbitrarily

and since

$$\frac{d\overline{x}}{dt} = \frac{1}{\hbar}\overline{\nabla}_k E$$ (Hamiltonian Mechanics -- reminder of our semi-classical approach)

and if you take the derivative of E (and plug E=Ef and k=kf) with respect to k and divide by hbar to get the velocity you obtain:

$$v_f = \frac{\hbar k_f}{m^*}= \frac{p}{m^*}$$

where m*'s denote the "effective mass" for the bottom of the conduction band (parabolic band approximation)

as genneth properly corrected.

7. Apr 3, 2009

Thanks.