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What is the Fermi energy of (undoped) graphene?

by Izzhov
Tags: energy, fermi, graphene, undoped
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Izzhov
#1
May2-14, 06:17 PM
P: 117
All of the sources I have found for this online have been wildly unclear. Many use the phrase "Fermi energy" to refer to the "Fermi level" (which is emphatically not what I'm looking for; I want the Fermi energy as defined in this Wikipedia article: http://en.wikipedia.org/wiki/Fermi_energy ). Fermi energy is always greater than zero.

Has anyone ever measured the Fermi energy of graphene? Is there any way to calculate it from known quantities, such as the Fermi velocity, which is approx. [itex]10^6[/itex] m/s? i.e. is there a reason why the usual formula [itex]E_F = \tfrac{1}{2}m_e v_F^2[/itex] wouldn't work here? I read that the Fermi energy for undoped graphene is equal to the energy at the Dirac points, but I read elsewhere that that value is less than zero, which makes no sense because, again, Fermi energy is always greater than zero.
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UltrafastPED
#2
May3-14, 02:50 AM
Sci Advisor
Thanks
PF Gold
UltrafastPED's Avatar
P: 1,908
See http://journals.aps.org/prl/abstract...ett.108.116404
"Direct Measurement of the Fermi Energy in Graphene Using a Double-Layer Heterostructure"
Douasing
#3
May8-14, 03:55 AM
P: 33
Quote Quote by Izzhov View Post
All of the sources I have found for this online have been wildly unclear. Many use the phrase "Fermi energy" to refer to the "Fermi level" (which is emphatically not what I'm looking for; I want the Fermi energy as defined in this Wikipedia article: http://en.wikipedia.org/wiki/Fermi_energy ). Fermi energy is always greater than zero.
I think ,though many use the phrase "Fermi energy", they should know their "Fermi energy" just as the "Fermi level" in their mind. Even though wikipedia have already provided "a few key differences between the Fermi level and Fermi energy", we still need to consider them carefully for our specific system. Here,consider a simplest system (just,say,hydroden atom),we use a universal DFT tool (for example, Vasp.5.2)to examine the "Fermi level". The related energy results are as follows (which are extracted from OUTCAR):

Free energy of the ion-electron system (eV)
---------------------------------------------------
alpha Z PSCENC = 0.00238712
Ewald energy TEWEN = -2.04282400
-1/2 Hartree DENC = -5.62336435
-exchange EXHF = 0.00000000
-V(xc)+E(xc) XCENC = 1.90272023
PAW double counting = 0.87256925 -0.87765194
entropy T*S EENTRO = -0.02820948
eigenvalues EBANDS = -6.30934890
atomic energy EATOM = 12.07484263
---------------------------------------------------
free energy TOTEN = -0.02887945 eV

energy without entropy = -0.00066997 energy(sigma->0) = -0.01477471

-------------------------------------------------------------------------------------

average (electrostatic) potential at core
the test charge radii are 0.5201
(the norm of the test charge is 1.0002)
1 -41.6613


E-fermi : -6.3093 XC(G=0): -0.4452 alpha+bet : -0.0289
add alpha+bet to get absolut eigen values

k-point 1 : 0.0000 0.0000 0.0000
band No. band energies occupation
1 -6.3093 1.00000
2 -0.1793 0.00000
3 0.8620 0.00000
4 0.8620 0.00000
5 0.8620 0.00000
6 1.2780 0.00000
----------------------------------------------------------------------------------------

Where,"E-fermi:-6.3093" is just the "Fermi level".Because here is only one electron in "Hydroden atom" system, the "Fermi level" is also equal to the "eigenvalues EBANDS",but,it is not equal to the total energy as wikipedia said.
On the other hand, "Fermi energy" (not the real meaning of many people) can't be seen directly from the results above,because it is included in the total energy.
From the view of classical theory,we may understand them easier. (see that:
http://guide.ceit.metu.edu.tr/thinkquest/apndx3.htm)


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