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Splitting Kramers degeneracy with superconducting phase difference Bernard van Heck, Shuo Mi (Leiden), Anton Akhmerov (Delft) arXiv:1408.1563 ESI, Vienna, 11 September 2014 Plan Using phase difference in a Josephson junction as a means of


  1. Splitting Kramers degeneracy with superconducting phase difference Bernard van Heck, Shuo Mi (Leiden), Anton Akhmerov (Delft) arXiv:1408.1563 ESI, Vienna, 11 September 2014

  2. Plan Using phase difference in a Josephson junction as a means of breaking time reversal symmetry. ◮ What does ‘breaking time reversal’ mean? ◮ Why it won’t work. ◮ How to make it work (and why 3 is much better than 2)?

  3. Time reversal breaking in a mesoscopic JJ Several manifestations: ◮ Splitting of Kramer’s degeneracy (Chtchelkatchev&Nazarov, B´ eri&Bardarson&Beenakker) ◮ Closing of the induced gap ◮ Protected zero energy level crossings (switches in the ground state fermion parity) P = Pf( iH ) ◮ Spectral peak in the DOS (Ivanov, Altland&Bagrets) � 1 + sin(2 π E /δ ) � ρ ( E ) = ρ 0 2 π E /δ

  4. Setup and formalism Scattering matrices of electrons and holes: S h ( − E ) = S ∗ e ( E ) Andreev reflection matrix: r A = ie i φ i Bound state condition: S e ( E ) r A S h ( E ) r ∗ A ψ = e − 2 i arccos( E / ∆) ψ

  5. Setup and formalism Scattering matrices of electrons and holes: S h ( − E ) = S ∗ e ( E ) Andreev reflection matrix: r A = ie i φ i Bound state condition: S ( E ) r A S ∗ ( − E ) r ∗ A ψ = e − 2 i arccos( E / ∆) ψ

  6. Short junction limit S ( E ) ≈ S (0) Lowest density of Andreev states, strongest effect phase difference on a single state. Due to unitarity and time reversal symmetry of S the energies are � 1 − T n sin 2 ( φ/ 2) (Beenakker) given by E n = ± ∆

  7. DOS 0 1 E/ ∆ Why it won’t work ◮ Splitting of Kramers degeneracy ◮ Closing of the gap ◮ Protected zero energy level crossings (switches in the ground state fermion parity) ◮ Spectral peak in the DOS

  8. DOS 0 1 E/ ∆ Why it won’t work ◮ Splitting of Kramers degeneracy :-( δ E ∼ E 2 / E T < ∆ 2 / E T ◮ Closing of the gap ◮ Protected zero energy level crossings (switches in the ground state fermion parity) ◮ Spectral peak in the DOS

  9. Why it won’t work ◮ Splitting of Kramers degeneracy :-( δ E ∼ E 2 / E T < ∆ 2 / E T ◮ Closing of the gap :-( | E | ≥ ∆ cos( φ/ 2) DOS 0 1 E/ ∆ ◮ Protected zero energy level crossings (switches in the ground state fermion parity) ◮ Spectral peak in the DOS

  10. Why it won’t work ◮ Splitting of Kramers degeneracy :-( δ E ∼ E 2 / E T < ∆ 2 / E T ◮ Closing of the gap :-( | E | ≥ ∆ cos( φ/ 2) DOS 0 1 E/ ∆ ◮ Protected zero energy level crossings (switches in the ground state fermion parity) :-( ◮ Spectral peak in the DOS

  11. Why it won’t work ◮ Splitting of Kramers degeneracy :-( δ E ∼ E 2 / E T < ∆ 2 / E T ◮ Closing of the gap :-( | E | ≥ ∆ cos( φ/ 2) DOS 0 1 E/ ∆ ◮ Protected zero energy level crossings (switches in the ground state fermion parity) :-( ◮ Spectral peak in the DOS :-(

  12. A big improvement All the special properties of the spectrum originate from the small number of leads!

  13. Kramers degeneracy splitting Take a Rashba quantum dot with E ∼ E SO , R � l SO , and λ � R 15 0 15 15 0 15

  14. Kramers degeneracy splitting Take a Rashba quantum dot with E ∼ E SO , R � l SO , and λ � R Calculate the splitting between the lowest two Andreev levels 15 0 15 15 0 15

  15. Kramers degeneracy splitting Take a Rashba quantum dot with E ∼ E SO , R � l SO , and λ � R Calculate the splitting between the lowest two Andreev levels 2 π 0 . 2 δE/ ∆ π φ 2 0 0 2 π π φ 1 � Kramers degeneracy is strongly broken.

  16. Protected level crossings Once again, try a random quantum dot: 1 2 π E min π φ 2 0 0 π 2 π φ 1

  17. Protected level crossings Once again, try a random quantum dot: 1 2 π E min π φ 2 0 0 π 2 π φ 1 � Level crossings are allowed.

  18. 2 π 0 . 2 P odd π φ 2 0 0 π 2 π φ 1 Protected level crossings Are level crossings allowed for any ( φ 1 , φ 2 )?

  19. Protected level crossings Are level crossings allowed for any ( φ 1 , φ 2 )? 2 π 0 . 2 P odd π φ 2 0 0 π 2 π φ 1 No: the gap may only close when all the clockwise phase differences are smaller (or larger) than π . (Note that this result holds for any junction)

  20. Proof 1. The expression for Andreev spectrum: Sr A S ∗ r ∗ A ψ = ω 2 ψ, E = ∆ Im ω

  21. Proof 1. The expression for Andreev spectrum: Sr A S ∗ r ∗ A ψ = ω 2 ψ, E = ∆ Im ω 2. The simplified expression for Andreev spectrum: ( Sr A − r A S T ) ψ = 2 e i α | E | ∆ ψ

  22. Proof 1. The expression for Andreev spectrum: Sr A S ∗ r ∗ A ψ = ω 2 ψ, E = ∆ Im ω 2. The simplified expression for Andreev spectrum: ( Sr A − r A S T ) ψ = 2 e i α | E | ∆ ψ 3. S = − S T due to time reversal.

  23. Proof 1. The expression for Andreev spectrum: Sr A S ∗ r ∗ A ψ = ω 2 ψ, E = ∆ Im ω 2. The simplified expression for Andreev spectrum: ( Sr A − r A S T ) ψ = 2 e i α | E | ∆ ψ 3. S = − S T due to time reversal. S ( r A ψ ) = 2 | E | e i α 4. This means S ψ ≡ ψ ′ , ψ − ( r A ψ ′ ) ∆

  24. Proof 1. The expression for Andreev spectrum: Sr A S ∗ r ∗ A ψ = ω 2 ψ, E = ∆ Im ω 2. The simplified expression for Andreev spectrum: ( Sr A − r A S T ) ψ = 2 e i α | E | ∆ ψ 3. S = − S T due to time reversal. S ( r A ψ ) = 2 | E | e i α 4. This means S ψ ≡ ψ ′ , ψ − ( r A ψ ′ ) ∆ 5. The necessary and sufficient condition for existence of a unitary S : ∃ ψ, ψ ′ : � ψ | r A | ψ � + � ψ ′ | r A | ψ ′ � = 2 | E | e i χ � ψ ′ | ψ � . ∆

  25. Proof II 1. We get: | E | ≥ 1 2 ∆ |� ψ | r A | ψ � + � ψ ′ | r A | ψ ′ �| .

  26. Proof II 1. We get: | E | ≥ 1 2 ∆ |� ψ | r A | ψ � + � ψ ′ | r A | ψ ′ �| . 2. Graphical solution: 3. The lower bound on the gap: i , j cos φ i − φ j E ≥ ∆ min 2

  27. Gap closing and the spectral peak Both phenomena are visible � In the ensemble (averaging over random antisymmetric S ) RMT 20 ρ ( ǫ ) × ∆ 10 0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 ǫ/ ∆

  28. Gap closing and the spectral peak Both phenomena are visible � In a single realization � In the ensemble (averaging over chemical (averaging over random potential) antisymmetric S ) Rashba dot, l so /R = 0 . 2, l/R = 0 . 4 RMT ∆ 20 ρ ( ǫ ) × ∆ ρ ( ǫ ) [a.u.] ǫ 10 0 0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 0 π 2 π ǫ/ ∆ φ 2 ≡ 2 π − φ 1

  29. Conclusions ◮ Superconducting phase difference can strongly break time reversal symmetry in a Josephson junction. ◮ This requires more than two superconducting leads. ◮ Spin degeneracy is split by a large fraction of ∆. ◮ The induced superconducting gap only closes in a a finite subregion of the phase space.

  30. Conclusions Thank you all. The end.

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