Rashba effect in the 2D system

Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton

(Date: February 26, 2019).

The Rashba effect, also called Bychkov-Rashba effect, is a momentum-dependent splitting of

spin bands in bulk crystals and low-dimensional condensed matter systems (such as

heterostructures and surface states) similar to the splitting of particles and anti-particles in the

Dirac Hamiltonian. The splitting is a combined effect of spin–orbit interaction and asymmetry of

the crystal potential, in particular in the direction perpendicular to the two-dimensional plane (as

applied to surfaces and heterostructures). This effect is named in honour of Emmanuel Rashba,

who discovered it with Valentin I. Sheka in 1959 for three-dimensional systems and afterward

with Yurii A. Bychkov in 1984 for two-dimensional systems.

https://en.wikipedia.org/wiki/Rashba_effect

The properties of a 2D electron gas in which the two-fold spin degeneracy of the spectrum has

been lifted by the Hamiltonian of the spin-orbit interaction of an electron,

1ˆ ˆ( )RH n σ k

where R is a Rashba coupling. Here we discuss the eigenvalue problem of Rashba Hamiltoniam

(and also Dresselhaus Hamiltonian) as exercises in Phys.421-521 (Quantum mechanics I).

1. Derivation of the spin-orbit interaction from Dirac theory of electron

We consider the origin of the spin-orbit interaction. In a relativistic quantum mechanics of the electron (Dirac theory), the Hamiltonian of the Dirac equation can be approximated by

4 22

0 2 2 2 2 2 2

1( )

2 8 4 8

e eH eA

m m c m c m c

pp σ E p E

ℏ ℏ

in the non-relativistic approximation (see the detail in the lecture note of Phys.524 and 525 Quantum Mechanics (graduate course), where e (<0) is the charge of electron.

02r

ZeE

r r

, (electric field)

0 0

ZeA

r (electric potential)

2

0 0( )Ze

V r eA er

(potential energy)

((Physical meaning)) Third term: relativistic correction

...8

1

2

...]8

1

21[

1

23

42

2

44

4

22

22

2

22

2222242

cmm

mccmcm

mc

mccm

mcmcccm

pp

pp

pp

The fourth term (Thomas correction)

Thomas term = )(4 22

pEσ cm

eℏ

For a central potential field,

eA0 = V(r)

00

1 dAA

r dr E r

0 01 1( )

dA dA

r dr r dr E p r p L

where L is an orbital angular momentum. Then the Thomas term is rewritten as Thomas term :

02 2 2 2

02 2

2 2

1( ) ( )

4 41

21 1

2

dAe e

m c m c r dr

dAe

m c r dr

dV

m c r dr

σ E p σ L

S L

S L

ℏ ℏ

(Spin-orbit interaction)

When

2

0

ZeV eA

r (attractive interaction)

2

2 2 2 2 3

1( )

4 2

e eZ

m c m c r σ E p S Lℏ

The spin angular momentum is defined by

σS2

ℏ

which is an automatic consequence of the Dirac theory. The last term is called the Darwin term. For a hydrogen atom,

)()3( rE .

It gives rise to an energy shift

0

2)(

22

2232)()3(

22

22

)(8

)()(8

rrrrrSchrodingerSchrodinge

cm

exd

cm

e

ℏℏ

which is non-vanishing only for the s state.

We note that the spin orbit interaction becomes stronger as the atomic number Z increases.

The fine structure factor is defined by 2 1

137.035999139

e

c ℏ

.

2. Spin-orbit interaction

2

2 2 3

1

2SO

av

ZeH

m c r S L

Using the formula

2 2

2n

nr

mZeℏ

(Bohr radius)

we can evaluate the magnitude of the spin-orbit coupling.

2

2 2 31

32 2

2 2 2

2 3 3 6

2 2 6

2 24

2

24

2

1

2

2

2

( )2

( )2

SO

ZeH

m c r

Ze mZe

m c

Ze m Z e

m c

mc eZ

c

mcZ

S L

S L

S L

S L

S L

ℏ

ℏ

ℏ ℏ

ℏ

where the fine structure factor is defined by 2 1

137.035999139

e

c ℏ

.

We note that the lowest energy for the Bohr model is

22 4 2 2 2

21 2

( )2 2 2n

Z me mc Ze mcE Z

c

ℏ ℏ

Thus we have the ratio:

2

4

22

2 221

( )2 ( )

( )2

so

n

mcZ

HZ

mcEZ

S L

S Lℏ

ℏ

So the spin orbit coupling is proportional to the factor 2( )Z . We note that the spin orbit

coupling becomes important when the atomic number Z becomes large.

Fig. The factor 2Z as a function of the atomic number Z.

3. Rashba Hamiltonian

The Rash Hamiltonian is given by

0 1ˆ ˆ ˆH H H

with

2 2

2 20

1 0ˆˆ 10 12 2

H k km m

ℏ ℏ,

and

1ˆ ˆ( ) ( )R R y x x yH k k n σ k

with

0 1ˆ ˆ[ , ] 0H H

where n is the unit vector along the z axis. The Rashba field is defined by

2

2( , ,0)

2(sin , cos ,0)

RR

B

Ry x

B

R

B

g

k kg

k

g

B k n

Fig. Rashba effect. 2 R

R

Bg

B k n . k n . n//z. The spin orbit coupling and asymmetry of

the potential in the z axis (half spaces with z>0 and z<0 are not equivalent), which is the

surface induced asymmetry.

Using this field, the Rashba Hamiltonian can be rewritten as

1ˆ ˆ( )

ˆ ( )

2ˆ( )

2

ˆ2

R

R

B R

B

BR

H

g

g

g

n k

k n

k n

B

Fig. The effective magnetic field in the Rashba Hamiltonian. 2 R

R

Bg

B k n . When k n ,

2 RR

B

kB

g

Using the Pauli matrices;

0 1

1 0x

, 0

0x

i

i

, 1 0

0 1z

1H can be expressed as

1

0 ( )ˆ

( ) 0x y

R

x y

i k ikH

i k ik

We use

( , ,0)x yk kk

With

cosx

k k , siny

k k

or

(cos sin ) i

x yk ik k i ke

(cos sin ) i

x yk ik k i ke

Using this notation, we have

1

0ˆ0

i

R i

ieH k

ie

((Eigenvalue problem)) We solve the eigenvalue problems using the Mathematica. We have

1 1 1ˆ

RH k , 1 2 2ˆ

RH k

where the eigenkets of 1H are given by

1

11

2iie

(eigenvalue: k )

2

11

2iie

(eigenvalue k )

Note that

0 1 1ˆ

RH k , 0 2 2ˆ

RH k

and

0 1ˆ ˆ[ , ] 0H H

So 1 and 2 are simultaneous eigenkets of 0H and 1H . In other words, 1 and 2 are

the eigenkets of H . Then we have

2 2

1 1 1 1ˆ ( ) ( )

2 R

kH k E k

m

ℏ,

2 2

2 2 2 2ˆ ( ) ( )

2 R

kH k E k

m

ℏ

Note that

2 2 2

21 2 2( ) ( )

2 2 2R R

R

m mkE k k k

m m

ℏ ℏ

ℏ ℏ.

2 2 22

2 2 2( ) ( )

2 2 2R R

R

m mkE k k k

m m

ℏ ℏ

ℏ ℏ

We note that

2 1( ) ( )E k E k , 1 2( ) ( )E k E k

Fig. 1( )E k for the state 1 (red) and 2( )E k for the state 2 (blue). Symmetry:

2 1( ) ( )E k E k , 1 2( ) ( )E k E k

The average values of spin operators are given by

1 1 1

1 1 1 1 1 1

ˆ2

ˆ ˆ ˆ( , , )2

(sin ,cos ,0)2

x y z

S σℏ

ℏ

ℏ

2 2 2

2 2 2 2 2 2

ˆ2

ˆ ˆ ˆ( , , )2

(sin ,cos ,0)2

x y z

S σℏ

ℏ

ℏ

We note that

2 1 S S (which is always true)

and

1

2cos(2 )R

R

B

k

g

S B

1 sin(2 )2

k S k

ℏ.

x

y

O

BR

k

S1

S2

Fig. x-y plane. k: black arrow [wave vector, (kx, ky)]. 1S : red arrow. 2S : blue arrow. RB :

green arrow.

4 Effect of external magnetic field: Zeeman effect

In the presence of an external magnetic field

1ˆ ˆ ( )

2

ˆ ˆ( )2

BR

BR

gH

g

σ B B

n σ k σ B

or

1ˆ ˆ( )

2B

R y x x y

gH k k

σ B

Here we define a new parameter x as

2 2B

RR R

B

g BB Bx

kB k

g

where

2 RR

B

kB

g

.

We consider the two cases for the direction of B.

(a) / /zB

We solve this eigenvalue problem using the Mathematica.

1ˆ ( )

2

1 00

0 120

BR y x x y z

i

BR i

gH k k B

ie gk B

ie

Note that

R

Bx

B

The eigenvalue: 21Rk x

2

2 2

1

2 2

1

2 1 1

2 1 1

i

x x

x x x

ie

x x x

The eigenvalue: 21Rk x

2

2 2

2

2 2

1

2 1 1

2 1 1

i

x x

x x x

ie

x x x

The average values of spin operators are given by

1 1 1

1 1 1 1 1 1

2 2 2

ˆ2

ˆ ˆ ˆ( , , )2

sin cos( , , )

2 1 1 1

x y z

x

x x x

S σℏ

ℏ

ℏ

and

2 2 2

2 2 2 2 2 2

2 2 2

ˆ2

ˆ ˆ ˆ( , , )2

sin cos( , , )

2 1 1 1

x y z

x

x x x

S σℏ

ℏ

ℏ

Thus we have

1 2 S S

Thus we have

1 1 1H E , 2 2 2H E

with

2 2

21 1

2 R

kE k x

m

ℏ,

2 22

2 12 R

kE k x

m

ℏ

Fig. 0B . B//z. 3D plane. k: black arrow [wave vector, (kx, ky)]. 1S : red arrow. 2S : blue

arrow. RB : green arrow.

Fig. 0B for comparison. 3D plane. k: black arrow [wave vector, (kx, ky)]. 1S : red arrow.

2S : blue arrow. RB : green arrow.

Since

2B

R

g Bx

k

we have

2 22 2 2 2

21 1

2 2 2 2B B

R R

R R

g B g Bk kE k k

m k m

ℏ ℏ,

2 22 2 2 2

22 1

2 2 2 2B B

R R

R R

g B g Bk kE k k

m k m

ℏ ℏ

We assume that these energies can be expressed by

2 2 2

1y x a x b , 2 2 22y x a x b

Fig. Energy dispersion E vs k for B//z, where B is perpendicular to BR.

(b) / / RB B

1ˆ ˆ [ ( )

2

ˆ( )( )2

ˆ( )( )2

BR

BR

BR

g BH

k

g B

k

g Bk

k

σ B k n

k n σ

kn σ

where

ˆ ˆ( ) ( )R R n σ k k n σ

The energy eigenvalues are obtained as

2 2

1

2 2

( ) ( )2 2

2 2

BR

R

BR

g BkE k k

m

g Bkk

m

ℏ

ℏ

2 2

2

2 2

( ) ( )2 2

2 2

BR

R

BR

g BkE k k

m

g Bkk

m

ℏ

ℏ

Fig. Energy dispersion E vs k for B//BR.

k

E k

O

B 0

B 0

B 0

B 0

3. Dresselhaus Hamiltonian

We consider the eigenvalue problem for the Dresselhaus Hamiltonian

ˆ ˆ ˆ( )

ˆ ˆ(sin cos )

0

0

DS DS y y x x

DS y x

i

DS i

H k k

k

ek

e

where DS is a Dresselhaus coupling. Eigenvalue problem

1

11

2i

e

, (eigenvalue: k )

2

11

2i

e

(eigenvalue: k )

1 1 1 1 1 1 1ˆ ˆ ˆ{ , , }2

( cos ,sin ,0}2

x y z

Sℏ

ℏ

2 2 2 2 2 2 2ˆ ˆ ˆ{ , , }2

(cos , sin ,0)2

x y z

Sℏ

ℏ

2 1 S S

1 cos(2 )2

S kℏ

, 2 cos(2 )2

S kℏ

Fig. Dresselhaus interaction. x-y plane. k: black arrow [wave vector, (kx, ky)]. 1S : red arrow.

2S .

____________________________________________________________________________

APPENDIX

Mathematica

x

y

O

kS1

S2

Dresselhaus Hamiltonian

Clear "Global` " ;

x PauliMatrix 1 ; y PauliMatrix 2 ;

z PauliMatrix 3 ;

exp : exp . Complex re , im Complex re, im ;

A0

0;

Eigensystem A Simplify , 0 &

1, 1 , , 1 , , 1

11

2

1, ; 21

2

1, ;

1 . x. 1, 1 . y. 1, 1 . z. 1 FullSimplify

Sin , Cos , 0

2 . x. 2, 2 . y. 2, 2 . z. 2 FullSimplify

Sin , Cos , 0

P1 8;

R1 : Cos , Sin ;

BR1 : Sin , Cos ;

S1 : 0.2 Sin , Cos ;

S2 : 0.2 Sin , Cos ;

s1 Graphics Red, Thick, Arrowheads 0.03 ,

Table Arrow R1 , R1 S1 ,

, 0, 2 , P1 , PointSize 0.03 ,

Table Point R1 , , 0, 2 , P1 ;

s2 Graphics Blue, Thick, Arrowheads 0.03 ,

Table Arrow R1 , R1 S2 ,

, 0, 2 , P1 , PointSize 0.015 ,

Table Point R1 , , 0, 2 , P1 ;

s3 Graphics Black, Thick, Arrowheads 0.03 ,

Table Arrow R1 , 1.2 R1 , , 0, 2 , P1 ,

Green, Table Arrow R1 , R1 0.2 BR1 ,

, 0, 2 , P1 ;

s4 Graphics Black, Thin, Line 1, 0 , 1, 0 ,

Line 0, 1 , 0, 1 ,

Text Style "x", Black, 15, Italic , 1.3, 0 ,

Text Style "y", Black, 15, Italic , 0, 1.3 ,

Text Style "O", Black, 15 , 0, 0.05 ,

Text Style "BR", Black, Bold, 12, Italic ,

0.29, 1.03 ,

Text Style "k", Black, Bold, 12, Italic ,

0.46, 1.16 ,

Text Style " S1 ", Black, Bold, 12, Italic ,

0.14, 0.8 ,

Text Style " S2 ", Black, Bold, 12, Italic ,

0.65, 1.03 ;

s5 Graphics Black, Thin, Circle 0, 0 , 1 ;

Show s1, s2, s3, s4, s5

REFERENCES

Yu.A. Bychkov and E.I. Rashba, JETP 39, 66-69 (1984).

G. Dresselhaus, Phys. Rev. 100, 580 (1955).

R. Winkler, Spin-Orbit Coupling Effects in Two-Dimensional Electron and Hole Systems

(Springer 2003).

_____________________________________________________________________________

APPENDIX

Spin-orbit interaction

A.1. Biot-Savart law

x

y

O

BR

k

S1

S2

The electron has an orbital motion around the nucleus. This also implies that the nucleus has an orbital motion around the electron. The motion of nucleus produces an orbital current. From the Biot-Savart’s law, it generates a magnetic field on the electron.

The current I due to the movement of nucleus (charge Ze, e>0) is given by

NZeId vl ,

where Nv�

is the velocity of the nucleus and Ndt

dv

l . Note that

NZedt

dqd

t

qId v

lll

.

The effective magnetic field at the electron at the origin is

3

1

1

r

rlB

d

c

Ieff , ev vN

where v is the velocity of the electron. Then we have

3

1

13

1

1

r

re

r

rvB

v

c

Ze

c

Ze Neff .

Since rr 1 , effB can be rewritten as

z

e

ezeff

cm

vZem

c

Zev

c

Zevv

c

Zee

rr

e

r

er

r

reB 2233

�

where me is the mass of electron. The Coulomb potential energy is given by

Vc(r) Ze2

r,

dVc(r)

dr

Ze2

r2 .

Thus we have

zzc

e

ze

e

zeff Ldr

rdV

cermrvm

r

Ze

cermcer

rvZeeeeB

)(112

2

3

2

,

or

zzc

e

eff Ldr

rdV

rcemeB

)(11 ,

where Lz is the z-component of the orbital angular momentum, vrmL ez .

A.2. Derivation of the expression for the spin-orbit interaction

Fig. Electron in the proton frame, and proton in the electron frame. The direction of magnetic field B produced by the proton is the same as that of the orbital angular momentum L of the electron (the z axis in this figure).

We consider the circular motion of the nucleus (e) around an electron at the center. The nucleus rotates around the electron at the uniform velocity v. Note that the velocity of the proton in the electron frame is the same as the electron in the proton frame. The magnetic field (in cgs) at the center of the circle along with the current I flow.

cr

IB

2 .

according to the Biot-Savart law. The current I is given by

r

ev

v

r

e

T

eI

22 ,

where T is the period. Then the magnetic field B at the site of the electron (the proton frame) can be expressed as

33 crm

e

crm

rvem

e

z

e

Ne LeB ,

where L is the orbital angular momentum of the electron (the electron frame). Note that the direction of B is the same as that of L. The spin magnetic moment is given by

Sμℏ

Bs

2 ,

where S is the spin angular momentum and B is the Bohr magneton,

cm

e

e

B 2

ℏ .

Then the Zeeman energy of the electron is given by

SLL

SBμ

322

2

3 2)

2(

2

1

2

1

rcm

e

crm

eH

ee

Bs

ℏ

,

where the factor (1/2) is the Thomas correction. In quantum mechanics we use the notation

SL ˆˆ1

2ˆ

322

2

ave

sorcm

eH .

A.3. Thomas correction

The spin magnetic moment is defined by

Sμℏ

Bs

2,

where S is the spin angular momentum. Then the Zeeman energy is given by

)(

)(1

2

1

)(112

2

1

2

1

22

LS

LS

LSBμ

dr

rdV

rcm

dr

rdV

rcemH

c

e

c

e

Beffsso

ℏ

where the factor 1/2 is the Thomas correction and the Bohr magneton B is given by

cm

e

e

B2

ℏ .

Thomas factor 1/2, which represents an additional relativistic effect due to the acceleration of

the electron. The electron spin, magnetic moment, and spin-orbit interaction can be derived directly from the Dirac relativistic electron theory. The Thomas factor is built in the expression.

LS soH ,

with

ave

c

ercm

e

dr

rdV

rcm3

2

22

1

2

1)(1

2

1

.

When we use the formula

)1)(2/1(

133

3

lllanr

B

,

the spin-orbit interaction constant is described by

)1)(2/1(2)1)(2/1(2 632

8

3322

2

lllnc

em

lllancm

e e

Beℏ

where

2

2

ema

e

B

ℏ = 0.53 Å (Bohr radius).

The energy level (negative) is given by

20

2

2

2 nan

eE

B

n

,

where

B

e

a

eem

22

2

2

4

0 ℏ

.

The ratio nE/2ℏ is

)1)(2/1()1)(2/1( 2

2

222

42

lllnlllnc

e

En

ℏ

ℏ

with

037.137

12

c

e

ℏ

((Note)) For l = 0 the spin-orbit interaction vanishes and therefore = 0 in this case. ((Summary))

The spin-orbit interaction serves to remove the l degeneracy of the eigenenergies of hydrogen atom. If the spin-orbit interaction is neglected, energies are dependent only on n (principal quantum number). In the presence of spin-orbit interaction (n, l, s = 1/2; j, m) are good quantum numbers. Energies are dependent only on (n, l, j).