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A Why is the nearest hopping kept real in Haldane model?

  1. Oct 29, 2017 #1
    I am leaning the Haldane model :
    https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.61.2015
    Haldane imaged threading magnetic flux though a graphene sheet, and the net flux of a unit cell is zero.
    He argued that since the loop integral ##\exp [ie/\hbar \oint {A \cdot dr} ]## along a path of nearest bonds vanishes, the nearest hopping is not changed.

    However, I cannot see the connect between the vanishing loop integral and the unchanged nearest hopping, can anybody help?
     
  2. jcsd
  3. Nov 4, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Nov 5, 2017 #3
    In the Haldane model of graphene the hopping strength can be complex with a phase coming from the Aharonov-Bohm effect in the presence of a magnetic field. If one moves a particle around a closed contour, then the phase difference between the final and initial states is proportional to the magnetic flux enclosed by the contour ##\phi =\frac{e}{h} \iint \mathbf B \cdot \mathbf S = \frac{e}{h} \oint \mathbf A \cdot \mathbf {\mathcal l}##.
    Consider three sites a, b and c. The hopping strength between these three sites is ##t_{ab}##, ##t_{bc}## and ##t_{ca}##. If a particle hops from a to b and then to c, then the hopping strength around the loop is: $$t_{ab}t_{bc}t_{ca}=\left | t_{ab}t_{bc}t_{ca} \right | e^{i(\phi_{ab} + \phi_{bc} + \phi_{ca})}$$
    The phase picked up by the electron is: $$\phi_{ab} +\phi_{bc} + \phi_{ca} = \frac{e}{h} \iint \mathbf B \cdot \mathbf S$$
    If B is nonzero inside the triangle formed by these three sites, the phase for these hoppings are nonzero. On the other hand; no net field means no phase change.
     
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