Pumping 1

Example taken from P.W.Brouwer Phys. Rev.B 1998 Two parameter pumping in 1d wire back to phase 1 length along wire Uniform conductor: no bias, no current some charge shifted to left and right

Berry phase associated to two-parameter pumping

There is a clear connection with the Berry phase (see e.g. Di Xiao,Ming-Che-Chang, Qian Niu cond-mat 12 Jul 2009). Bouwer formulation for Two parameter pumping assuming linear response to parameters X 1, X 2 But this is not the only kind of pumping discovered so far. A circuit is not enough: one needs singularities inside. The magnetic charge that produces the Berry magnetic field is made of quantized Dirac monopoles arising from degeneracy. The pumping is quantized (charge per cycle= integer). Circuit in parameter space

Mono-parametric quantum charge pumping ( Luis E.F. Foa Torres PRB 2005) quantum charge pumping in an open ring with a dot embedded in one of its arms. The cyclic driving of the dot levels by a single parameter leads to a pumped current when a static magnetic flux is simultaneously applied to the ring. The direction of the pumped current can be reversed by changing the applied magnetic field (imagine going to the other side of blackboard). The response to the time-periodic gate voltage is nonlinear. time-periodic gate voltage The pumping is not adiabatic.No pumping at zero frequency. The pumping is not quantized.

6 6 See also: Cini-Perfetto-Stefanucci,PHYSICAL REVIEW B 81, (2010)

7 7 Another view of same quantum effect described above Bias U  current in wires  vortex  magnetic moment of ring  Interaction with magnetic field proportional to U^3 This is Magnetic pumping It must be possible to make all in reverse! Interaction with B  current vortex->magnetic moment of ring  current in wires  Bias

8 Model: laterally connected ring, same phase drop on all red bonds Different distribustions of the phase drop among the bonds are equivalent in the static case, but not here. This choice is simplest.

9 Half flux in and then out. Charging of ring with no net pumping We may avoid leaving the ring excited by letting it swallow integer fluxons this is emf first clockwise then counterclockwise the ring remains excitedthe ring remains charged charge is sent to left wire similar charge is sent to right wire

10 Pumping by an hexagonal ring – insertion of 6 fluxons (B  chirality) emf always same way ring returns to ground state pumping is achieved

11 If the switching time grows the charge decreases. It is not adiabatic and not quantized! Rebound due to finite leads Pumping by an hexagonal ring – insertion of 6 fluxons (B  chirality) effect of 6 fluxons in 100 time units effect of 6 fluxons in 200 time units effect of 6 fluxons in 300 time units

12 What happened? We got 1-parameter pumping (only flux varies) Charge not quantized- no adiabatic result Linearity assumption fails and one may have nonadiabatic 1 parameter pumping We got a strikingly simple and general case where linearity assumption that holds in the classical case fails due to quantum effects. In the present time-dependent problem the roles of cause and effect are interchanged.

Memory storage Insertion of 3.5 flux quanta into a ring with 17 sides connected to a junction (left wire atoms have energy level 2 in units of the hopping integral t h, right wire atoms have energy level 0). The figure shows the phase pulse and the geometry. Time is in units of the inverse of the hopping integral. Right: expectation value of the ring Hamiltonian. The ring remains excited long after the pulse. It remembers.

Charge on the ring. The ring remains charged after the pulse. It remembers. Fine! But memory devices must be erasable. How can we erase the memory?

Same calculation as before performed in the 17-sided ring, but now with the A – B bond cut between times t = 30and t = 70. The ring energy and occupation tend to return to the values they had at the beginning, and the memory of the flux is thereby erased.

16 Graphene Unit cell a a=1.42 Angstrom Lower resistivity than silver-Ideal for spintronics (no nuclear moment, little spin-orbit) and breaking strength = 200 times greater than steel.

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18 Corriere della sera 15 febbraio 2012

19 = basis the lattice is bipartite b a

20 Primitive vectors

21 Reciprocal lattice vectors To obtain the BZ draw the smallest G vectors and the straight lines through the centres of all the G vectors: the interior of the hexagon is the BZ.

K’  K K K BZ and important points. M

23 M K K’  M K K BZ and important points.

24 K K’  M K K BZ and important points. M

25 Tight-binding model for the  bands: denoting by a and b the two kinds of sites the main hoppings are : ba Jean Baptiste Joseph Fourier

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27 Why 2 component? It is the amplitude of being in sublattice a or b.

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29 (upper band)

30 Band Structure of graphene

31 Note the cones at K and K’ points

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33 no gap

34 Expansion of band structure around K and K’ points

35 Expansion of band structure around K and K’ points But the 2 components are for the 2 sublattices

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