Electrical Control of the Kondo Effect at the Edge of a ...hawk.fisica.uminho.pt/ccqs/CCQS-presentations/Johannesson_Evora... · from M. König et al., Science 318, 766 (2007) Bernevig

Erik Eriksson (University of Gothenburg)

Anders Ström (University of Gothenburg)

Girish Sharma (École Polytechnique)

Henrik Johannesson (University of Gothenburg)

Electrical Control of the Kondo Effectat the Edge of a Quantum Spin Hall System

supported by the Swedish Research Council

Correlations and coherence in quantum systems Évora, Portugal, October 11 2012

Quantum spin Hall effect... some basics

At the edge: A new kind of electron liquid

Adding a magnetic impurity...

... and a Rashba spin-orbit interaction

Electrical control of the Kondo effect!

Summary and outlook

Outline

Quantum spin Hall effect... some basics

Quantum spin Hall effect

no channel for backscattering

B“skipping currents”

quantization

chiral edge states

Quantum Hall effect

ballistic transport along the edge,accounts for the quantization of the Hall conductance

M. Büttiker, PRB 38, 9375 (1988)

Integer

B. I. Halperin, PRB 25, 2185 (1982)

Can a system be stable against local perturbations without breaking time-reversal invariance?

uniformly charged cylinder with electric field

spin-orbit interaction

cf. with the IQHE in a symmetric gauge

Lorentz force

1

E = E(x, y, 0)

(E × k) · σ = Eσz(kyx− kxy)

A =B

2(y,−x, 0)

A · k ∼ eB(kyx− kxy)

G = νe2

h

Mc ≈ 100 meV

D. Grundler, Phys. Rev. Lett. 84, 6074 (2000)

W. Hausler, L. Kecke, and A. H. MacDonald, Phys.

Rev. B 65, 085104 (2002)

�α0 � 2× 10−11

eV m

Λ � 0.5 eV

vF � 1× 106

m/s

Kc + Ks � 1.8

HR

H =

�dx [Hc +Hs]

Hi =vi

2[(∂xϕi)

2+(∂xϑi)

2]−

mi

πacos(

�2πKiϕi), (1)

∆R = γ1 sin(q0a)

with vi and Ki functions of g1τ , g2τ and g4τ

Λ ∼ bandwidth (2)

ϕc ≡ (ϕ+ + ϕ−)/

√2 (3)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (4)

R†τ and L

†τ create excitations at the Fermi points of the

right- and left-moving branches with spin projection τ

(5)

L+

vF = 2a

�t2 + γ

20 and ∆R = γ1 sin(q0a)

Hτ =−ivF

�:R

†τ (x)∂xRτ (x) ::L

†τ (x)∂xLτ (x) :

�

−2∆R cos(Qx)�e−2ik0

F (x+a/2)R

†τ (x)Lτ (x)+H.c.

�,(6)

τ = ±

q0

HR =−i

�

n,µ,ν

(γ0 + γ1 cos (Qna))

�c†n,µσ

yµνcn+1,ν−H.c.

�

γj = αja−1

(j = 0, 1)

ϕc ≡ (ϕ+ + ϕ−)/

√2 (7)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (8)

Hint = g1− :R†τLτL

†−τR−τ : + g2τ :R

†+R+L

†τLτ :

+g4τ

2(:R

†+R+R

†τRτ : +R↔ L) (9)

g2τ ≡g2τ − δτ+g1τ (10)

with vi and K functions of g1τ , g2τ , g3τ (11)

1

E = E(x, y, 0)


A =B

2(y,−x, 0)


G = νe2

h

Mc ≈ 100 meV



Rev. B 65, 085104 (2002)

�α0 � 2× 10−11

eV m

Λ � 0.5 eV

vF � 1× 106

m/s

Kc + Ks � 1.8

HR

H =

�dx [Hc +Hs]

Hi =vi

2[(∂xϕi)

2+(∂xϑi)

2]−

mi

πacos(

�2πKiϕi), (1)

∆R = γ1 sin(q0a)



ϕc ≡ (ϕ+ + ϕ−)/

√2 (3)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (4)

R†τ and L



(5)

L+

vF = 2a

�t2 + γ


Hτ =−ivF

�:R

†τ (x)∂xRτ (x) ::L

†τ (x)∂xLτ (x) :

�


F (x+a/2)R


�,(6)

τ = ±

q0

HR =−i

�

n,µ,ν


�c†n,µσ

yµνcn+1,ν−H.c.

�

γj = αja−1

(j = 0, 1)

ϕc ≡ (ϕ+ + ϕ−)/

√2 (7)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (8)


†−τR−τ : + g2τ :R

†+R+L

†τLτ :

+g4τ

2(:R

†+R+R

†τRτ : +R↔ L) (9)

g2τ ≡g2τ − δτ+g1τ (10)


1

E = E(x, y, 0)


A =B

2(y,−x, 0)


G = νe2

h

Mc ≈ 100 meV



Rev. B 65, 085104 (2002)

�α0 � 2× 10−11

eV m

Λ � 0.5 eV

vF � 1× 106

m/s

Kc + Ks � 1.8

HR

H =

�dx [Hc +Hs]

Hi =vi

2[(∂xϕi)

2+(∂xϑi)

2]−

mi

πacos(

�2πKiϕi), (1)

∆R = γ1 sin(q0a)



ϕc ≡ (ϕ+ + ϕ−)/

√2 (3)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (4)

R†τ and L



(5)

L+

vF = 2a

�t2 + γ


Hτ =−ivF

�:R

†τ (x)∂xRτ (x) ::L

†τ (x)∂xLτ (x) :

�


F (x+a/2)R


�,(6)

τ = ±

q0

HR =−i

�

n,µ,ν


�c†n,µσ

yµνcn+1,ν−H.c.

�

γj = αja−1

(j = 0, 1)

ϕc ≡ (ϕ+ + ϕ−)/

√2 (7)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (8)


†−τR−τ : + g2τ :R

†+R+L

†τLτ :

+g4τ

2(:R

†+R+R

†τRτ : +R↔ L) (9)

g2τ ≡g2τ − δτ+g1τ (10)


1

E = E(x, y, 0)


A =B

2(y,−x, 0)


G = νe2

h

Mc ≈ 100 meV



Rev. B 65, 085104 (2002)

�α0 � 2× 10−11

eV m

Λ � 0.5 eV

vF � 1× 106

m/s

Kc + Ks � 1.8

HR

H =

�dx [Hc +Hs]

Hi =vi

2[(∂xϕi)

2+(∂xϑi)

2]−

mi

πacos(

�2πKiϕi), (1)

∆R = γ1 sin(q0a)



ϕc ≡ (ϕ+ + ϕ−)/

√2 (3)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (4)

R†τ and L



(5)

L+

vF = 2a

�t2 + γ


Hτ =−ivF

�:R

†τ (x)∂xRτ (x) ::L

†τ (x)∂xLτ (x) :

�


F (x+a/2)R


�,(6)

τ = ±

q0

HR =−i

�

n,µ,ν


�c†n,µσ

yµνcn+1,ν−H.c.

�

γj = αja−1

(j = 0, 1)

ϕc ≡ (ϕ+ + ϕ−)/

√2 (7)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (8)


†−τR−τ : + g2τ :R

†+R+L

†τLτ :

+g4τ

2(:R

†+R+R

†τRτ : +R↔ L) (9)

g2τ ≡g2τ − δτ+g1τ (10)


1

B = ∇×A

E = E(x, y, 0)


A =B

2(y,−x, 0)


G = νe2

h

Mc ≈ 100 meV



Rev. B 65, 085104 (2002)

�α0 � 2× 10−11

eV m

Λ � 0.5 eV

vF � 1× 106

m/s

Kc + Ks � 1.8

HR

H =

�dx [Hc +Hs]

Hi =vi

2[(∂xϕi)

2+(∂xϑi)

2]−

mi

πacos(

�2πKiϕi), (1)

∆R = γ1 sin(q0a)



ϕc ≡ (ϕ+ + ϕ−)/

√2 (3)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (4)

R†τ and L



(5)

L+

vF = 2a

�t2 + γ


Hτ =−ivF

�:R

†τ (x)∂xRτ (x) ::L

†τ (x)∂xLτ (x) :

�


F (x+a/2)R


�,(6)

τ = ±

q0

HR =−i

�

n,µ,ν


�c†n,µσ

yµνcn+1,ν−H.c.

�

γj = αja−1

(j = 0, 1)

ϕc ≡ (ϕ+ + ϕ−)/

√2 (7)

ϕs ≡ (ϕ+ − ϕ−)/

√2 (8)


†−τR−τ : + g2τ :R

†+R+L

†τLτ :

+g4τ

2(:R

†+R+R

†τRτ : +R↔ L) (9)

g2τ ≡g2τ − δτ+g1τ (10)

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Consider a Gedanken experiment... B. A. Bernevig and S.-C. Zhang, PRL 96, 106802 (2006)

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First proposed by Kane and Mele for graphene (2005)

Bernevig et al. proposal for HgTe quantum wells (2006)Experimental observation by König et al. (2007)

too weak spin-orbit interaction, doesn’t quite work...

Two copies of a IQH system, bulk insulator with helical edge states

Quantum spin Hall (QSH) system single Kramers pair

from M. König et al., Science 318, 766 (2007)

Bernevig et al. proposal for HgTe quantum wells (2006)Experimental observation by König et al. (2007)

Yes! As long as the perturbationsare time-reversal invariant! Look at the band structure of an HgTe quantum well...

strong spin-orbit interactions in atomic p-orbitals create an inverted band gap (p-band on top of s-band)

Are the helical edge states stable against local perturbations?

supports a single Kramers pair of helical edge states inside the inverted gap

Kramers degeneracy at k=0 protectsthe stability of the edge states

ballistic transport

1

G =2e

2

h

L ∼ 1µm

L ∼ 20µm

g

g ∼ 1/2v

ξ < L

α(±2kF )

±2kF

z⊥ ≡ 2g

√CK

z� ≡ 4K − 2

γ = (α(2kF ) + α(−2kF ))/2

∂xθ

ψ↑ = η↑ exp�i√

π[φ + θ]�/

√2πκ

ψ↓ = η↓ exp�−i√

π[φ− θ]�/

√2πκ

H0 + Hd + Hf + HR

O(α2)

α(x) = �α(x)�+

�

n

α(kn)eikxn

Ψ↑↓(x)=ψ↑↓(x)e±ikF x

Ψ↓(x)=ψ↓(x)e−ikF x

HR = α(x)(z × k) · σ

B = ∇×A

E = E(x, y, 0)


A =B

2(y,−x, 0)


G = νe2

h

Mc ≈ 100 meV



Rev. B 65, 085104 (2002)

�α0 � 2× 10−11

eV m

Λ � 0.5 eV

vF � 1× 106

m/s

Kc + Ks � 1.8

HR

B. A. Bernevig et al., PRL 95, 066601 (2005)4-terminal measurement, equilibration in contacts


# of Kramers pair in a nontrivial (trivial) helical liquid

NK = 1 mod 2 = 0 mod 2

2D topological insulator(a.k.a. quantum spin Hall system)

is a ”Z2” topological invariant and can be calculated fromthe band structure of the bulk (”bulk-edge correspondence”)

1

ν = 0, 1

J⊥,z � ω � T

( )

K → K �

E =c

8log(

2�1�22�1 + �2

) + const.

θ

2�1 �2

γ0 ∼ (Jx + J �y)

2T 2(√K−λ/2)2−1, γ�

0 ∼ (Jx − J �y)

2T 2(√K+λ/2)2−1, γE

0 ∼ J2NCT

2K−1, γE0 ∼ J2

NCT

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

L. Fu and C. L. Kane, PRB 76, 045302 (2007)

What if time-reversal symmetry is broken...?


# of Kramers pair in a nontrivial (trivial) helical liquid

NK = 1 mod 2 = 0 mod 2

2D topological insulator(a.k.a. quantum spin Hall system)

is a ”Z2” topological invariant and can be calculated fromthe band structure of the bulk (”bulk-edge correspondence”)

1

ν = 0, 1

J⊥,z � ω � T

( )

K → K �

E =c

8log(

2�1�22�1 + �2

) + const.

θ

2�1 �2

γ0 ∼ (Jx + J �y)

2T 2(√K−λ/2)2−1, γ�

0 ∼ (Jx − J �y)

2T 2(√K+λ/2)2−1, γE

0 ∼ J2NCT

2K−1, γE0 ∼ J2

NCT

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

L. Fu and C. L. Kane, PRB 76, 045302 (2007)

... for example, by the presence of a magnetic impurity?

case study:

large and positive single-ion anisotropy

1

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

1

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

1

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T

anisotropic spin exchange with the edge electrons

1

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T

To model the edge electrons, we introduce the two-spinors ΨT=

�ψ↑,ψ↓

�, where ψ↑ (ψ↓) annihilates a right-moving

(left-moving) electron with spin-up (spin-down) along the growth direction of the quantum well. Neglecting e-e

interactions, the edge Hamiltonian can then be written as

HK = Ψ†(0)

�J⊥(σ

+S−+ σ−

S+) + Jzσ

zSz�Ψ(0),

1

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

1

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

R. Zitko et al., PRB 78, 224404 (2008)

from M. König et al., Science 318, 766 (2007)


The Kondo interaction is time-reversal invariant! Could it still cause a spontaneous breaking of time reversal invariance and collapse the QSH state?


Recall the Kondo effect

One-loop RG equations:

strong-coupling physics for

P. W. Anderson, J. Phys. C 3, 2436 (1970)

1

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (1)

J0 ≡ max(J⊥, Jz)D=D0

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

1

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (1)


TK = D0 exp(−const./J0)

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

1

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (1)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

1

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (1)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

formation of impurity-electron singlet (”Kondo screening”)


Pauli principle:punctured 1D lattice

T=0 insulator!


Does this really happen for the helical liquid?To find out, first add e-e interactions.... important in 1D!

1

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (1)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

1

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (1)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

1

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (1)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

bulk

local (at impurity site)

+

from Kondo

+ +

≠ 0 if U(1) spin symmetry is broken

1

( )

K → K �

E =c

8log(

2�1�22�1 + �2

) + const.

θ

2�1 �2

γ0 ∼ (Jx + J �y)

2T 2(√K−λ/2)2−1, γ�

0 ∼ (Jx − J �y)

2T 2(√K+λ/2)2−1, γE

0 ∼ J2NCT

2K−1, γE0 ∼ J2

NCT

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

1

( )

K → K �

E =c

8log(

2�1�22�1 + �2

) + const.

θ

2�1 �2

γ0 ∼ (Jx + J �y)

2T 2(√K−λ/2)2−1, γ�

0 ∼ (Jx − J �y)

2T 2(√K+λ/2)2−1, γE

0 ∼ J2NCT

2K−1, γE0 ∼ J2

NCT

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

T. L. Schmidt et al., PRL 108, 156402 (2012)


functions of Kondo couplings

”Luttinger liquid parameter”

1

K

H = (v/2)

�dx

�(∂xϕ)

2+ (∂xϑ)

2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

...perturbative RG and linear response

J. Maciejko et al., PRL 102, 256803 (2009)

Adding the kinetic energy and bosonizing...


from J. Maciejko et al., PRL 102, 256803 (2009)

”weak” e-e interaction

”strong” e-e interaction

no correction to the edge conductance at T = 0 !1

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK

Jx = Jy = Jz = 10 meV

Jx = Jy = 5 meV, Jz = 50 meV

θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Due to its topological nature, the QSH state follows the new shape of the edge.Weak coupling QSH states are robust against local breaking of time-reversal symmetry!


But... one important thing is missing from the analysis!

z

d

zd

Rashba spin-orbit interaction!

2DEG

semiconductor heterostructure

Spatial asymmetry of band edges mimics an E-field in the z-direction

1

HR = α(kxσy − kyσ

x)

HSO = λcrystal(∇V × k) · σ

λcrystal ≈ �2/4m

∗Eg ≈ 106

λvac

λvac = �2/4m

2

0c2 ≈ 3.7× 10−6

A2

HSO = λvac(∇V × k) · σ

Kc ≈ 2.2TK

1

2

K(R)

R

K(R) ∝ (J2/D) cos(kF R)

TK ∝ D exp(−1/ρF J)

K(R) > TK

V

U

Vg

J ∝ V2/U

�d

�F

�d+U

Cimp ∝ (T/TK) ln(TK/T ), Simp = k ln(√

2), ...

Simp = k ln(√

2)

T

K(R)critical

δ = π/2

δ = 0

−ψ(r)

ψ(r)

K(R)→∞

K(R)→ −∞

Hint = J1S1 · σ1 + J2S2 · σ2 + K(R)S1 · S2

J1 �= J2

Yu. A. Bychkov and E. I. Rashba, J. Phys. C 17, 6039 (1984)

z

d

zd

Rashba spin-orbit interaction!

semiconductor heterostructure

1

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2+ (∂xϑ)

2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

1

x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2+ (∂xϑ)

2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

x

doesn’t conserve spin

Adding the Rashba interaction...... breaks the locking of spin to momentum. However, there is still a single Kramers pair on the QSH edge, and this is all that matters!

1

H = vF

�dx Ψ†

(x) [−iσz∂x]Ψ(x) + α

�dx Ψ†

(x) [−iσy∂x]Ψ(x)

x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2+ (∂xϑ)

2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

Electrical control of the Kondo effect in a helical edge liquid

Erik Eriksson,1Anders Strom,

1Girish Sharma,

2and Henrik Johannesson

1

1Department of Physics, University of Gothenburg, SE 412 96 Gothenburg, Sweden2Centre de Physique Theorique, Ecole Polytechnique, 91128 Palaiseau Cedex, France

Magnetic impurities affect the transport properties of the helical edge states of quantum spin

Hall insulators by causing single-electron backscattering. We study such a system in the presence

of a Rashba spin-orbit interaction induced by an external electric field, showing that this can be

used to control the Kondo temperature, as well as the correction to the conductance due to the

impurity. Surprisingly, for a strongly anisotropic electron-impurity spin exchange, Kondo screening

may get obstructed by the presence of a non-collinear spin interaction mediated by the Rashba

coupling. This challenges the expectation that the Kondo effect is stable against time-reversal

invariant perturbations.

PACS numbers: 71.10.Pm, 72.10.Fk, 85.75.-d

Introduction. The discovery that HgTe quantum wells

support a quantum spin Hall (QSH) state?

has set offan avalanche of studies addressing the properties of this

novel phase of matter?. A key issue has been to deter-

mine the conditions for stability of the current-carrying

states at the edge of the sample as this is the feature

that most directly impacts prospects for future applica-

tions in electronics/spintronics. In the simplest picture

of a QSH system the edge states are helical, with counter-

propagating electrons carrying opposite spins. By time-

reversal invariance electron transport then becomes bal-

listic, provided that the electron-electron (e-e) interaction

is sufficiently well-screened so that higher-order scatter-

ing processes do not come into play? ?

.

The picture gets an added twist when including effectsfrom magnetic impurities, contributed by dopant ions or

electrons trapped by potential inhomogeneities. Since an

edge electron can backscatter from an impurity via spin

exchange, time-reversal invariance no longer protects the

helical states from mixing. In addition, correlated two-

electron?

and inelastic single-electron processes? ?

must

now also be accounted for. As a result, at high temper-

atures T electron scattering off the impurity leads to a

ln(T ) correction?

of the conductance at low frequencies

ω, which, however, vanishes? in the dc limit ω → 0. At

low T , for weak e-e interactions, the quantized edge con-

ductance G0 = e2/h is restored as T → 0 with power

laws distinctive of a helical edge liquid. For strong inter-

actions the edge liquid freezes into an insulator at T = 0,

with thermally induced transport via tunneling of frac-

tionalized charge excitations through the impurity?.

A more complete description of edge transport in a

QSH system must include also the presence of a Rashba

spin-orbit interaction. This interaction, which can be

tuned by an external gate voltage, is a built-in feature

of a quantum well?. In fact, HgTe quantum wells ex-

hibit some of the largest known Rashba couplings of

any semiconductor heterostructures?. As a consequence,

spin is no longer conserved, contrary to what is assumed

in the minimal model of a QSH system?. However,

since the Rashba interaction preserves time-reversal in-

variance, Kramers’ theorem guarantees that the edge

states are still connected via a time-reversal transforma-

tion (”Kramers pair”)?. Provided that the Rashba inter-

action is spatially uniform and the e-e interaction is not

too strong, this ensures the robustness of the helical edge

liquid?.

What is the physics with both Kondo and Rashba in-

teractions present? In this Letter we address this ques-

tion with a renormalization group (RG) analysis as well

as a linear-response and rate-equation approach. Specif-

ically, we predict that the Kondo temperature TK −which sets the scale below which the electrons screen the

impurity − can be controlled by varying the strength

of the Rashba interaction. Surprisingly, for a strongly

anisotropic Kondo exchange, a non-collinear spin inter-

action mediated by the Rashba coupling becomes rel-

evant (in the sense of RG) and competes with the

Kondo screening. This challenges the expectation that

the Kondo effect is stable against time-reversal invari-

ant perturbations?. Moreover, we show that the im-

purity contribution to the dc conductance at tempera-

tures T > TK can be switched on and off by adjusting

the Rashba coupling. With the Rashba coupling being

tunable by a gate voltage, this suggests a new inroad to

control charge transport at the edge of a QSH device.

Model. To model the edge electrons, we introduce

the two-spinors ΨT=

�ψ↑,ψ↓

�, where ψ↑ (ψ↓) annihi-

lates a right-moving (left-moving) electron with spin-up

(spin-down) along the growth direction of the quantum

well. Neglecting e-e interactions, the edge Hamiltonian

can then be written as

H = vF

�dx Ψ†

(x) [−iσz∂x]Ψ(x) +

+α

�dx Ψ†

(x) [−iσy∂x]Ψ(x) + (1)

+Ψ†(0) [Jxσ

xSx+ Jyσ

ySy+ Jzσ

zSz]Ψ(0),

with vF the Fermi velocity parameterizing the linear ki-

netic energy. The second term encodes the Rashba inter-

action of strength α, with the third term being an anti-

ferromagnetic Kondo interaction between electrons (with

Pauli matrices σi, i = x, y, z) and a spin-1/2 magnetic im-

1

Ψ� = e−iσxθ/2Ψ

H = vF

�dx Ψ†(x) [−iσz∂x]Ψ(x) + α

�dx Ψ†(x) [−iσy∂x]Ψ(x)

x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



T << TK

Mn2+

kinetic term Rashba

1

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



T << TK

1

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



1

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)

1

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)

zz’

The ”Rashba-rotated” spinorstill defines a Kramers pair

1

θ

γ0 ∼ (Jx + J �y)

2T 2(√K−λ/2)2−1, γ�

0 ∼ (Jx − J �y)

2T 2(√K+λ/2)2−1, γE

0 ∼ J2NCT

2K−1, γE0 ∼ J2

NCT

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK

C. L. Kane and E. J. Mele, PRL 95, 226801 (2005)

Adding the Rashba interaction...

e-e interaction is invariant under

1

Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

Kondo interaction

2

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



T << TK

Mn2+

(Sz)2

S = 5/2 −→ Seff = 1/2

low T


�ψ↑,ψ↓




HK = Ψ†(0)

�J⊥(σ

+S−eff + σ−

S+eff) + Jzσ

zSzeff

�Ψ(0),

1

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)

S� = e−iSxθ/2SeiSxθ/2

Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

1


H = vF



x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

k kF − kF

∂J⊥∂D

= −νJ⊥Jz + ...

∂Jz∂D

= −νJ2⊥ + ... (2)



T << TK

Mn2+

1

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

HR = α kxσy

K

H = (v/2)

�dx

�(∂xϕ)

2 + (∂xϑ)2�

+A

κcos(

√4πKϕ)+

B

κsin(

√4πKϕ)+

C√K

∂xϑ

+gU

2(πκ)2cos(

√16πKϕ)

+gie

2π2√K

: (∂2xϑ) cos(

√4πKϕ) : (1)

XYZ Kondo Non-Collinear term

depend on the Rashba coupling controllable by a gate voltage

1

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

HR = α kxσy

...bosonization and perturbative RG

E. Eriksson et al., PRB 86, 161103(R) (2012)

Electrical control of the Kondo temperaturevia the ”Rashba angle” ~ gate voltage

easy-axis Kondo

1

θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


H = vF



x

1



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


Region where dominates the RG flow obstruction of Kondo screening!

1



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


challenges the Meir-Wingreen

conjecture that the Kondo effect is

blind to time-reversal invariant

perturbations

Y. Meir and N.S. Wingreen, PRB 50, 4947 (1994)

easy-plane Kondo

2

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

Low-temperature transport, (away from the ”dome”)

1

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)


1

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)

1

G =2e

2

h

L ∼ 1µm

L ∼ 20µm

g

g ∼ 1/2v

ξ < L

α(±2kF )

±2kF

z⊥ ≡ 2g

√CK

z� ≡ 4K − 2

γ = (α(2kF ) + α(−2kF ))/2

∂xθ

ψ↑ = η↑ exp�i√

π[φ + θ]�/

√2πκ

ψ↓ = η↓ exp�−i√

π[φ− θ]�/

√2πκ

H0 + Hd + Hf + HR

O(α2)

α(x) = �α(x)�+

�

n

α(kn)eikxn

Ψ↑↓(x)=ψ↑↓(x)e±ikF x

Ψ↓(x)=ψ↓(x)e−ikF x

HR = α(x)(z × k) · σ

B = ∇×A

E = E(x, y, 0)


A =B

2(y,−x, 0)


G = νe2

h

Mc ≈ 100 meV



Rev. B 65, 085104 (2002)

�α0 � 2× 10−11

eV m

Λ � 0.5 eV

vF � 1× 106

m/s

Kc + Ks � 1.8

HR

1

T = 0

K>1/4

T � TK

Jx= Jy

= Jz= 10 meV

Jx= Jy

= 5 meV,Jz

= 50 meV

θ

J�y, J

�z, JNC

displaym

ath

α

H�K= Ψ

�† (0)[Jxσ

x Sx + J

�yσy�Sy� + J

�zσz�Sz� + JNC(σ

y�Sz� + σ

z�Sy�)]Ψ

� (0)

S� = e

−iSx θ/2Se

iSx θ/2

Ψ → Ψ�

vα=

�v2F+ α2

cos θ= vF

/vα

sin θ= α/vα

Ψ�T =

�ψ↑�,

ψ↓��

1

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα


1

1/4 < K < 2/3 K > 2/3

(T/TK)8K−2 (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

1

1/4 < K < 2/3 K > 2/3

(T/TK)8K−2 (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

1

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

1

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�


1

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)

signature of

Rashba!

1

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα


1

1/4 < K < 2/3 K > 2/3

(T/TK)8K−2 (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

1

1/4 < K < 2/3 K > 2/3

(T/TK)8K−2 (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2


1

G = 0

1/4 < K < 2/3 K > 2/3

(T/TK)8K−2 (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

1

T = 0

K>1/4

T � TK

Jx= Jy

= Jz= 10 meV

Jx= Jy

= 5 meV,Jz

= 50 meV

θ

J�y, J

�z, JNC

displaym

ath

α

H�K= Ψ

�† (0)[Jxσ

x Sx + J

�yσy�Sy� + J

�zσz�Sz� + JNC(σ

y�Sz� + σ

z�Sy�)]Ψ

� (0)

S� = e

−iSx θ/2Se

iSx θ/2

Ψ → Ψ�

vα=

�v2F+ α2

cos θ= vF

/vα

sin θ= α/vα

Ψ�T =

�ψ↑�,

ψ↓��

1

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

1

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

1

K < 1/4

G ∼ (T/Tbs)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J �y, J

�z, JNC

displaymath


1

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)

signature of

Rashba!

1

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα


1

1/4 < K < 2/3 K > 2/3

(T/TK)8K−2 (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

1

1/4 < K < 2/3 K > 2/3

(T/TK)8K−2 (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2


1

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

1

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

1

K < 1/4

G ∼ (T/Tbs)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

J �y, J

�z, JNC

displaymath

from instanton processes


1

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ


1

T � TK



θ

J�y, J

�z, JNC

displaymath

α

H�K = Ψ�†(0)[Jxσ

xSx + J

�yσ

y�Sy�+ J

�zσ

z�Sz�+ JNC(σ

y�Sz�+ σz�

Sy�)]Ψ�(0)


Ψ → Ψ�

vα =�

v2F + α2

cos θ = vF /vα

sin θ = α/vα

Ψ�T =�ψ↑�,ψ↓�

�

H� = vα

�dx Ψ�†(x)

�−iσz�

∂x�Ψ�(x)

signature of

Rashba!

blind to

Rashba

1

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



1

T � TK

K < 1/4

G ∼ (T/Tbs)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

”High-temperature” transport, 1

J⊥,z � ω � T

( )

K → K �

E =c

8log(

2�1�22�1 + �2

) + const.

θ

2�1 �2

γ0 ∼ (Jx + J �y)

2T 2(√K−λ/2)2−1, γ�

0 ∼ (Jx − J �y)

2T 2(√K+λ/2)2−1, γE

0 ∼ J2NCT

2K−1, γE0 ∼ J2

NCT

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

1

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



conductance correction in dc limit:1

γ0 ∼ (Jx + J �y)

2T 2(√K−λ/2)2−1, γ�

0 ∼ (Jx − J �y)

2T 2(√K+λ/2)2−1, γE

0 ∼ J2NCT

2K−1, γE0 ∼ J2

NCT

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK


1

T � TK

K < 1/4

G ∼ (T/Tbs)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

∼ (T/TK)8K−2 ∼ (T/TK)2K+2

G =2e2

h− δG

T = 0

K > 1/4

T � TK



θ

”High-temperature” transport, 1

J⊥,z � ω � T

( )

K → K �

E =c

8log(

2�1�22�1 + �2

) + const.

θ

2�1 �2

γ0 ∼ (Jx + J �y)

2T 2(√K−λ/2)2−1, γ�

0 ∼ (Jx − J �y)

2T 2(√K+λ/2)2−1, γE

0 ∼ J2NCT

2K−1, γE0 ∼ J2

NCT

I = G0V − δI

T � TK

K < 1/4

G ∼ (T/TK)2(1/4K−1)

G = 2e2/h

G = 0

1/4 < K < 2/3 K > 2/3

Summary and outlook

Magnetic impurity in a helical edge liquid:Rashba coupling allows electrical control of Kondo temperature and IV-characteristics

blocking of Kondo screening!?

Rashba-induced impurity correction accessible in experiment?

Interesting open problems:Two-loop RG, thermal transport, effects from Dresselhaus spin-orbit interactions, Kondo lattice in a helical liquid, higher-spin impurities... and more!

E. Eriksson et al., PRB 86, 161103(R) (2012)

Documents

Electrical Control of the Kondo Effect at the Edge of a ...hawk.fisica.uminho.pt/ccqs/CCQS-presentations/Johannesson_Evora... · from M. König et al., Science 318, 766 (2007) Bernevig