The dispersion relation of electrons in, for example, graphene exhibits Dirac cones. The dispersion relation of a Dirac point at $\mathbf{k}=\mathbf{K}$ is linear in the momentum magnitude: $$E=\pm v_F |{\bf k-K}|$$ where $v_F$ the Fermi velocity.

The Fermi velocity can be measured directly by ARPES, or by any momentum-resolved spectroscopy. Are there measurable properties of the system which directly depends on the Fermi velocity in graphene, or in other systems where Dirac cones are present?

Addendum: To exemplify my question imagine that you could change in a gedanken experiment the velocity $v_F$ in the equation above, without changing the Fermi level or any other property of the system. Which physical quantities would be affected?

  • $\begingroup$ The most characteristic effect which depends on the Dirac cones/"relativistic" electrons, to my knowledge, is the $\sqrt{B}$ dependence of the Landau levels induced by a magnetic field. For classical particles such levels are evenly spaced. $\endgroup$
    – CR Drost
    Aug 16, 2015 at 19:32
  • $\begingroup$ @ChrisDrost Landau levels have energies proportional to $v_F$ in graphene. However, this is true only in the limit of small magnetic fields (in strong magnetic fields the situation is more complicated, i.e., Hofstadter butterfly). Are there others quantities which depend on $v_F$? $\endgroup$
    – sintetico
    Aug 16, 2015 at 19:48

2 Answers 2


To change $v_F$ is to change the dispersion and density of states, which you generally expect to have an affect on every electronic and optoelectronic property.

A particularly straightforward example: When you "gate" graphene (as you would a transistor), the fermi level moves, which you can measure most clearly by a cutting-off of IR absorption above a certain wavelength. The relation between gate voltage and IR absorption edge wavelength is given by a straightforward formula that involves $v_F$.

Gating also affects conductivity in a way related to $v_F$ for the same reason. (Actually conductivity and the IR absorption edge are related to each other by a "sum rule", ref.)

Apart from gate-based measurements: A high $v_F$ means that "the electrons move faster" which tends to increase electron mobility, other things equal. ...But mobility is also affected by defect scattering etc. So that's not an especially straightforward relationship.

If you could make a ballistic graphene electronic device (i.e., a device small enough and high-quality enough that there's no scattering), I bet $v_F$ would show up more directly in those measurements.

  • $\begingroup$ If you gate graphene, the Fermi energy moves away from the Dirac point. Without gating, the Fermi surface is a collection of points (Dirac points, 0 dimensional) but if you gate, the Fermi surface is 1 dimensional. In this case it is obvious that everything changes. However, I am not interested in this case. I am interested to the situation where the system still has Dirac points $\endgroup$
    – sintetico
    Aug 19, 2015 at 16:55
  • $\begingroup$ (1) Away from absolute zero the states above and below the fermi surface are all relevant; (2) Even with no (intentional) gating, the fermi level will not be EXACTLY at the dirac point in practice; (3) Even if it is exactly at the dirac point somewhere, it won't be exactly at the dirac point elsewhere, because of inhomogeneities like random dangling bonds in the substrate; (4) If you apply a voltage than the fermi level is tilted and cannot be at the dirac point everywhere. So again, every electronic and optoelectronic property is affected. $\endgroup$ Aug 19, 2015 at 17:30
  • $\begingroup$ You consider how a change in the Fermi level affects the bulk properties of graphene, but not, as I asked, how these properties are affected by the Fermi velocity of the Dirac point. $\endgroup$
    – sintetico
    Aug 19, 2015 at 17:37
  • $\begingroup$ As soon as the Fermi level is away from the dirac point, vF has an obvious importance--density of states in particular. I guess I'm saying, it's easy to say in theory "I assume Fermi level is exactly at the dirac point", but it's not only impossible in practice, but stops being true as soon as you do something as simple as apply a voltage, even in theory. $\endgroup$ Aug 19, 2015 at 17:53
  • $\begingroup$ I agree with your arguments. And your answer is correct, but it does not address my question. You describe a situation where the Fermi level changes, but not the Fermi velocity. For small perturbations the Fermi velocity is in fact constant (the dispersion is linear at the Dirac point). $\endgroup$
    – sintetico
    Aug 19, 2015 at 18:14

In a 2deg this is straightforward to measure from the temperature dependence of the shubnikov de haas oscillations, which depends directly on vf.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.