SLIDE 12 Why are carriers massless in graphene?
◮ In general, define ˙
◮ Since
v = ∇
pE, this yields (m∗)−1 = ∇ p∇ pE ◮ Near the Dirac point in graphene, E = vF |
k| = vF | p|
◮ If
p along x, v = ∇
pE = vF ˆ
x
◮
v follows direction of p without changing magnitude!
◮ What mass does this correspond to? ◮ Conventional explanation in literature: linear dispersion (Dirac equation)
⇒ massless particles in relativity
◮ Important: this is an analogy: vF ∼ c/400; many aspects of relativistic
particles like photons do not apply
◮ Examine mass tensor at more carefully:
(m∗)−1 = ∇
p∇ pvF |
p| = vF /p
p = pˆ x with eigenvalues 0 and vF /p, i.e. ∞ and p/vF for m∗
◮ At Dirac point, p → 0 ⇒ m∗ T → 0 while m∗ L → ∞
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