SLIDE 1
HgTe is inconvenient because the heavy holes! Unstrained HgTe - - PowerPoint PPT Presentation
HgTe is inconvenient because the heavy holes! Unstrained HgTe - - PowerPoint PPT Presentation
HgTe is inconvenient because the heavy holes! Unstrained HgTe Strained HgTe point point L point HgTe z>0 SnTe z>0 CdTe z<0 PbTe z<0 Surface states in IV-VI heterostructures Ideal -Dirac
SLIDE 2
SLIDE 3
SnTe PbTe Ξ Ξ βππ¨π€ βπβπ€ βπ+π€ ββππ¨π€ βππ¨π€ βπβπ€ βπ+π€ ββππ¨π€ β Ξ β Ξ |πΉβ > |πΉβ > |πΌβ > |πΌβ > πΉ = Ξ2 + βππ¨π€ 2 + βππ π€ 2 Bulk solution Ξ = βπΉ1 Ξ = πΉ2
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Ξ Ξ βππ¨π€ βπβπ€ βπ+π€ ββππ¨π€ βππ¨π€ βπβπ€ βπ+π€ ββππ¨π€ β Ξ β Ξ |πΉβ > |πΉβ > |πΌβ > |πΌβ > π¨
SnTe z>0 Pb0.2Sn0.8Se PbSe z<0
Ξ = βπΉ1 π(π¦, π§, π¨) = πππππ¨π¨ππ(ππ¦π¦+ππ§π§) ππ¨ = π1 Pb0.2Sn0.8Se: πΉ = Ξ2 + βππ¨π€ 2 + βππ π€ 2 ππ
2 = ππ¦ 2 + ππ§ 2
Ξ = πΉ2 ππ¨ = π2 π1 = πΉ2 β πΉ1
2 β βππ π€ 2
βπ€ π2 = πΉ2 β πΉ2
2 β βππ π€ 2
βπ€ PbSe: πππππ(0) = πππππππ(0) α€ ππππππ πΉ2ππ¨
π¨=0
= β α€ ππππππππ πΉ1ππ¨
π¨=0
Boundary conditions:
SLIDE 5
π1
2
πΉ1
2 = π2 2
πΉ2
2
Bound Dirac state from boundary conditions
At z<0 πΉΒ± = Β± βπ€ππ π1 = πΉ2 β πΉ1
2 β βππ π€ 2
βπ€ π2 = πΉ2 β πΉ2
2 β βππ π€ 2
βπ€ π1 = βπΉ1
2
βπ€ = ππΉ1 βπ€ π2 = ππΉ2 βπ€ π(π¦, π§, π¨) = π1πβπ1π¨ππ(ππ¦π¦+ππ§π§) π2ππ2π¨ππ(ππ¦π¦+ππ§π§) π¨ π 2 PbSe
interface
PbSnSe
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Which one is topological PbSe or PbSnSe (HgTe or CdTe)
PbSe
interface
PbSnSe HgTe
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Observation of bound surface state β angle resolved photoemission
Pb0.9 Sn0.1Se Pb0.8 Sn0.2Se
Analytis et al .Phys. Rev. B 81, 205407 (2010)
- Phys. Rev. Lett. 119 106602 (2017)