Landauer Buttiker

1Landauer-Buttiker method

Current from transmissionQuantum scattering theoryLandauer-Buttiker formulasDerivation from linear response theory

D.A. Ryndyk

1.1 Current from transmission

The Landauer-Buttiker (LB) method establishes the fundamental relation be-tween the wave functions (scattering amplitudes) of a noninteracting quantumsystem (QS) and its conducting properties. The LB method can be appliedto nd the current through a noninteracting QS or through an eectivelynoninteracting QS, for example if the mean-eld description is valid and theinelastic scattering is not essential. Such type of an electron transport is calledcoherent, because there is no phase-breaking and quantum interference is pre-served during the electron motion across the System. In fact, coherence isinitially assumed in many ab initio based transport methods (DFT+NGF,and others), so that the LB method is now routinely applied to any basictransport calculation through nanostructures and single molecules. Besides,it is directly applicable in many semiconductor quantum dot systems withweak electron-electron interactions. Due to simplicity and generality of theLB method, it is now widely accepted and is in the base of our understandingof coherent transport.

In this lecture we consider only some applications of the LB method, re-lated directly to the problems of nonequilibrium transport through QS. Manyother extensions of the method are beyond our consideration. One can men-tion the Landauer formula for hybrid superconducting and magnetic systems,application to a quantum noise problem, random matrix theory, localizationtheory, the attempts to introduce the analogous scattering description for in-teracting and dissipative systems. However, in spite of such popularity of theLB method, one should remember, that in its canonical form the LB methodis applicable only for noninteracting systems.

1.1.1 Quantum junction: reservoirs, leads, scatterer

A typical considered system (quantum junction) consists of three parts: reser-voirs (contacts), quantum leads, and scattering region (Fig. 1.1). The main

16 1 Landauer-Buttiker method

property of the reservoirs is that they are equilibrium and incoherent. Themain property of the quantum leads is that they have known mode structure,incoming (from the contacts to the scatterer) +(r) and outgoing (r) modescan be dened. In the simplest case the leads are noninteracting, but can benonequilibrium. Scattering region can be as simple as one tunneling barrier,or as complex as an interacting nonequilibrium molecule.

Sometimes one of these components can be absent or they are combined.Examples: 3D tunneling or transport through molecules and quantum dotsplaced between 3D contacts, leads are combined with reservoirs; ballisticquantum wires, quantum point contact, and billiards, no scattering regionor quantum lead can be considered as scatterer.

In this lecture we consider completely coherent transport between reser-voirs, which one are, of cause, incoherent. If a particle come into reservoir, it isthermalized and any phase information is lost. Transport through a coherentregion is described by the wave function. It means that we do not considerinelastic eects inside the scattering region, and elastic scattering is describedby some potential U(r).

1.1.2 The single-channel Landauer formula

The main idea of the method was formulated by Landauer [2, 3]. He proposed,that the conductance of the elastic scatterer is determined by the quantummechanical transmission T (reection R = 1 T ) coecient. It should benoted, that Landauer considered not the resistance of a quantum system be-tween the equilibrium contacts, but the local resistance of a system itself(in fact, the zero-temperature residual resistance). As a result he got for thezero temperature one-channel (eectively one-dimensional) conductance theso-called rst Landauer formula

G =e2

h

T

1 T =e2

h

T

R, (1.1)

the result, which seems to be reasonable at least in two limiting cases. Atsmall transmission T 0, the conductance is also small. In the opposite case,

LeadsLeftreservoir(contact)

Rightreservoir(contact)

Scatteringregion

Fig. 1.1. A scattering region is connected to the reservoirs trough quantum leads.

1.1 Current from transmission 17

when T 1, there is no scattering at all, so that the conductance shouldgo to innity, in agreement with (1.1). Note, that we do not assume the spindegeneracy in these formula, to take it into account one has to multiply theconductance by 2.

However, the conductance through a System between equilibrium contacts,calculated by the exact linear response method [4, 5] is quite dierent

G =e2

hT. (1.2)

This conductance is nite just in the case of a perfectly transmitted junc-tion (T = 1). The contradiction between two formulas was unexpected andstimulated active discussion. Finally, it was shown that both formulas arereasonable and give the same current, but correspond to the voltages denedbetween dierent points. The key dierence between the formulas (1.1) and(1.2) is that the rst one is for the conductance inside the System (betweenpoints A and B, see Fig.1.8), while the second gives the conductane of the QSrelated to the equilibrium contacts (between points L and R in the Fig.1.8).In the section 1.3.3 we obtain both formulas and discuss the relation betweenit. The puzzle with the nite resistance at T 1 is also understood, it isclear now that for the nite number of conductance channels the conductanceis also nite, just without scattering and at zero temperature. The physicalreason is that the number of electrons going trough the System is limited,as well as the current associated with one electron state, consequently theaverage current is determined by the number of channels, their transmission,and the level population.

For the transport problems, considered in these lectures, only the secondtype of the Landauer formula is important usually. Besides, the rst typeformulas are not exact for nite-size nanostructures, because are dependenton the particular electrical potential distribution.

1.1.3 The multi-channel Landauer (Fisher-Lee, Buttiker) formulas

Now let us mention the main extensions of the formula (1.2), which will bediscussed in the lecture. Of the principal importance is the extension of thesingle-channel formula to the multi-channel case.

The zero temperature multi-channel conductance is given by the Fisher-Leeformula [5]

G =e2

hTr(tt)=

e2

h

|t |2 = e2

h

n

Tn, (1.3)

t is the transmission matrix, t is the transition amplitude between incomingand outgoing channels and , Tn are the eigenvalues of the matrix tt.

In the multi-channel case, at nite voltage and nite temperature, thefollowing general Landauer formula takes place


J(V ) =e

h

T (E, V ) [fL(E + eV ) fR(E)]dE, (1.4)

where T (E, V ) = Tr(tt)is the eective transmission function for the par-

ticles with the energy E. The most important advantage of this formula is,that the transmission function can be calculated from the quantum scatteringtheory. Thus, the kinetic problem is reduced to the pure quantum mechanicalproblem of a single particle in a static potential. The formula (1.4) is themost general two-terminal formula. All other Landauer formulas (in particu-lar mentioned above zero-temperature conductance formulas) are obtained inthe limiting cases from (1.4).

The next important contribution has been done by Buttiker, who extendedthe Landauer formula to a multi-terminal case [7, 9]. In particular, four-terminal description (Fig. 1.8) is of great importance for experiments. Thecurrent from the i-th contact to the System is

Ji =e

h

j =i

T ij(E, V ) [fi(E + eVij) fj(E)]dE, (1.5)

where Vij is the voltage between contacts i and j. We consider the multi-terminal formulas in the section 1.3.4.

1.1.4 The origin of dissipation

The important question, discussed in connection with the Landauer resistance,is the origin of dissipation in this approach. Indeed, nite dc current at nite dcvoltage means that the energy is permanently dissipated. On the other hand,we consider only elastic scattering, so that the energy can not be dissipatedin the scattering process. This problem is closely related to the phenomenaof the residual resistance at low temperature, caused by impurities. In bothcases we should introduce some thermalisation. In the case of the transportbetween the equilibrium contacts, this puzzle is resolved quite easy, the energyis dissipated in the contacts, the details of the dissipation are not relevant.More precisely, the incoming from the contacts to the System particles areequilibrium distributed, while outgoing particles propagate into the contactsand are thermalised here.

1.2 Quantum scattering theory

As we see, the main formal problem to be solved in the Landauer theory is thesingle-particle scattering problem. The conductance is determined then fromthe elements of the transmission matrix. For this reason, the LB method iscalled also a scattering method. In this section we discuss the quantum scat-tering theory, and formulate it in the form convenient for further calculations.

1.2 Quantum scattering theory 19

0 z

U(z)

U0

0 z

A+

A-

B-Transmission

Reflection

Incoming wave

Fig. 1.2. One-dimensional potential. Reection and transmission.

1.2.1 Reection and transmission

To start, we simplify the problem, assuming that the motion of electrons is ef-fectively one-dimensional. For example, in planar structures the wave functionof the state with the energy E is

(r) = (x, y)(z), (1.6)

where is mode or channel index. In the 3D layered system k. Similarsituation is in the eectively 1D or 2D spatially quantized systems (electronicwaveguides), where (x, y) describes dierent transverse modes. Now weforget for a moment about the structure of transverse modes and consider only(z), the full energy is the sum of the transverse (E) and one-dimensional(Ez) energies: E = E + Ez.

This type of transport can be named mode-conserving because there is notransitions between dierent modes. More general situation with inter-modescattering will be consider in the section 1.2.5.

Now consider a one-dimensional potential (schematic in Fig. 1.2) which isconstant far from the scattering region: U(z ) = 0, U(z ) = U0.Then

(z ) =A+eikz + Aeikz, (1.7)(z +) =Beikz, (1.8)

where A+ is the incoming and A, B are the outgoing waves (Fig. 1.2). Wavevectors k and k are dened as

k =2mEzh

, (1.9)

k =

2m(Ez U0)

h. (1.10)

To determine the transmission and reection probabilities, we should useconservation of probability ux density (current)


f =ih

2m[(r)(r) (r)(r)] . (1.11)

From the Schrodinger equation it follows that

|(r, t)|2t

+ divf = 0, (1.12)

in the stationary case simplydivf = 0. (1.13)

For the plane wave Aeikz the probability ux density (current) is reducedto

f =hk

m|A|2 = v|A|2. (1.14)

We dene the transmission coecient as the ratio of the transmitted toincident probability ux

T (Ez) =ftranfinc

=v|B|2v|A+|2 =

k

k

|B|2|A+|2 , (1.15)

and the reection coecient as the ratio of the reected to incident probabilityux

R(Ez) =freffinc

=|A|2|A+|2 . (1.16)

From the probability ux conservation it follows that

T (Ez) + R(Ez) = 1. (1.17)

Note, that if we consider the incident wave from the right side of thebarrier, transmission coecient is the same at the same energy

TRL(Ez) = TLR(Ez). (1.18)

1.2.2 Single barrier

Now we consider the general scattering (or transmission) problem, assumingthat there are incoming modes from the left and from the right sides of thebarrier (Fig. 1.3).

-barrier

We start from the -potential

U(z) = (z). (1.19)

The solution is given by


(z) ={

A+eikz + Aeikz, z < 0

Beikz + B+eikz, z > 0(1.20)

where A+, B+ are incoming and A, B are outgoing waves,

k =2mEzh

.

Boundary conditions are

(0) = (0+), (1.21)(0+) (0) = 2m

h2(0). (1.22)

We can present this boundary condition using transmission matrix M(K =

h2k

m

)

(A+A

)= M

(BB+

)=(

1 + iKiK

iK 1 iK

)(BB+

), (1.23)

or, alternatively, scattering matrix S

(AB

)= S

(A+B+

)=

1iK1

iKiK1

iKiK1

1iK1

(A+

B+

). (1.24)

The scattering matrix relate the amplitudes of the outgoing waves to theamplitudes of incoming waves.

To nd transmission and reection coecients we set now B+ = 0, then

0 z

A+

A- B+

B-

Fig. 1.3. Single barrier. General scattering problem.


T (E) =[vLvR

] |B|2|A+|2 =

[vLvR

]1

|M11|2 =[vLvR

]|S21|2 = K

2

1 + K2, (1.25)

R(E) =|A|2|A+|2 =

|M21|2|M11|2 = |S11|

2 =1

1 + K2, (1.26)

T + R = 1.

The transmission matrix can be written explicitly in terms of transmissionand reection coecients

M =

1t

rt

r

t1t

, (1.27)

as well the scattering matrix in terms of the elements of M

S =

M12M11

1M11

1M11

M12M11

. (1.28)

The elements of transmission matrix are not independent, namely

M22 = M11, M21 = M12, |M11|2 |M12|2 = 1. (1.29)

Descriptions by scattering or transmission matrices are completely equiv-alent and the choice is only dependent on the convenience and the problemto be solved. Typically, in the end of a calculation the S-matrix should beobtained, because it determines the conductivity by the Landauer formula.But to calculate the scattering by the complex system, the M-matrix can beconvenient, as we shall see in the next section.

1.2.3 Transfer matrix

Consider two sequential barriers with transmission matrices M and M

(Fig. 1.4), so that(A+A

)= M

(BB+

),

(A+A

)= M

(BB+

). (1.30)

Outgoing coecients B and incoming coecients A are related by transfermatrix MW (

BB+

)= MW

(A+A

)=(

eikL 00 eikL

)(A+A

). (1.31)

To show that in the most simple way, note that Beikz and A+eikz describethe same plane wave in two dierent points z = 0 and z = L, the phasedierence is obviously kL.


Finally we can write(

A+A

)= MT

(BB+

)= MMWM

(BB+

). (1.32)

Thus, the transmission matrix for a sequence of barriers can be dened asa product of particular transmission matrices of barriers and transfer matrices

MT = M1MW1,2M2...MnMWn,n+1Mn+1...MN1M

WN1,NMN. (1.33)

1.2.4 Double barrier

Consider now the double-barrier case (Fig. 1.4). We apply the transfer matrixmethod. We need only MT11 because it determines T (see (1.25)) and R =1 T . Transfer matrix MT for a two-barrier structure is

MT = MMWM =(

M11 M12M21 M22

)(eikL 0

0 eikL

)(M 11 M

12

M 21 M22

). (1.34)

MT11 = M11M 11eikL + M12M 21e

ikL (1.35)

For transmission coecient we nd

T (E) =T 21

T 21 + 4R1 cos2(kL ), (1.36)

where is the phase of the complex M11. T1 and R1 are transmission andreection coecients of the single barrier (here we assumed that the twobarriers are the same).

0 z

A+

A- B+

B- A+

A- B+

B-

Fig. 1.4. Double barrier.


From this general expression one can see the important property of two-barrier structures: there are transmission resonances, at some specic energiesEn the transmission coecient is large (T (En) = 1 in symmetric structures),while between resonances it can be small.

When the barriers are -functions M11 = 1 +i

K, = arctan

1K

=

arctanm

h2kand the equation for resonances (T = 1) is

tan kL = h2k

m. (1.37)

Breit-Wigner formula

Close to the resonance, around one of the resonance energies En, the trans-mission coecient has a Lorentzian form

T (E) 2n

(E En)2 + 2n, (1.38)

where the width n is given for two -barriers as

n =(2h2EnT 21mL2R1

)1/2. (1.39)

1.2.5 Multi-channel scattering. S-matrix

Now we are able to consider the general multichannel case, when scatteringis possible between dierent modes. It is convenient to dene separately left(L) and right (R), incoming (+) and outgoing (-) modes (Fig. 1.5)

0 z

A1+

A1-

An+

An-

B1+

B1-

Bm+

Bm-

Fig. 1.5. Multi-channel scattering.


+Ln(r) = Ln(x, y)An+eiknz, (1.40)

Ln(r) = Ln(x, y)Aneknz, (1.41)

+Rm(r) = Rm(x, y)Bm+eikmz, (1.42)

Rm(r) = Rm(x, y)Bmeikmz. (1.43)

Scattering matrix relate all incoming with all outgoing modes, so that justin the one-channel case, considered in the rst lecture S is 22 matrix (1.24).In the general multi-channel case S-matrix is dened as

A1...

ANLB1...

BNR

= S

A1+...

ANL+B1+...

BNR+

=

S11 S12 ... S

1N

S21 S22 ... S

2N

... ... ... ...

... ... ... ...

... ... ... ...SN1 S

N2 ... S

NN

A1+...

ANL+B1+...

BNR+

, (1.44)

where NL and NR are the numbers of left and right channels.Unitarity: S is not unitary if there are dierent velocities in the leads,

because unitarity means that the norm of the vector is conserved, while inour case

n

|A2n+| =n

|An|2, (1.45)

instead one has n

vn|A2n+| =n

vn|An|2, (1.46)

where vn = hkn/m are velocities.Note that these velocities are dierent, because the energy E = En + Ez

is conserved, scattering from the mode n into mode m means that transverseenergy En is changed and dierent modes coupled by (1.44) have dierent Ezand consequently dierent wave vectors kn and velocities vn.

To make the Smatrix unitary, we introduce new normalization of incomingand outgoing modes

+Ln(r) =1kn

Ln(x, y)An+eiknz, (1.47)

Ln(r) =1kn

Ln(x, y)Aneknz, (1.48)

+Rm(r) =1km

Rm(x, y)Bm+eikmz, (1.49)

Rm(r) =1km

Rm(x, y)Bmeikmz. (1.50)

and a new matrix S with elements

Snm =

knkm

Snm. (1.51)


It can be written in the form

S =(r t

t r

)(1.52)

where matrices r (NL NL) and t (NL NR) describe transmission andreection of states incoming from the left, matrices r (NR NR) and t(NR NL) describe transmission and reection of states incoming from theright, NL (NR) is the number of left (right) channels. S is a square matrixN N , N = NL + NR.

T = |t|2 are probabilities of transmission from the left mode intothe right mode , R = |r|2 are probabilities of reection from the leftmode into the left mode , etc.

1.3 Landauer-Buttiker formulas

In this section we obtain all results in a physically transparent way. Theexample of a rigorous calculation of the conductance using the linear responsetheory, is presented in the section 1.4.

1.3.1 Mode-conserving current

Now we want to show, how the transmission coecient can be used to calculatethe current through a quantum system. We start from the mode-conservingscattering. The main assumption is that all the particles coming from theleft to the barrier and having nonzero transmission coecient are transmit-ted through the barrier with probability T (Ez) and after that their excessenergy, phase coherence, and the memory of their previous state are lost inthe right contact. The same takes place for all particles coming from the rightand transmitted to the left. Transport through the barrier is coherent in thismodel, energy and transverse quantum number (e.g. momentum k) areconserved. Irreversibility is introduced by the after-transmission relaxation.

The current from the left to the right is determined by the distributionfunction only of the left contact

JLR = e

0

TLR(kz)vLz(kz)fL(kz, )dkz2

, (1.53)

where vLz(kz) is the velocity of the particle with momentum kz, fL(kz, ) isthe distribution function, the form of this function is considered below. Notethat it is not necessary to multiply this expression additionally by the factorlike (1 fR((kz, )) as in the tunneling or golden rule theory, because thisfactor describes the number of empty states in the right contact and shouldbe included when left and right states are considered as independent. Instead,

1.3 Landauer-Buttiker formulas 27

in our approach we consider scattering states, and transmission from theleft to the right is simply probability to nd particle in the right or left thepart of this state.

Taking into account that

vz(kz) =Ez(kz)hkz

,

we obtainJLR =

e

h

UL

TLR(Ez)fL(Ez, )dEz, (1.54)

and similar expression for the current from the right to the left

JRL =e

h

UR

TRL(Ez)fR(Ez, )dEz, (1.55)

note that integration in this expressions is done from the bottom of conduc-tion band UL(R). If the zero energy level is taken the same for left and rightcontacts, the energy Ez is the same one-dimensional energy in the both ex-pressions, and taking into account the symmetry of transmission coecients(1.18) we get nal expression for the current

J =e

h

T (Ez) [fL(Ez, ) fR(Ez, )] dEz. (1.56)

Integration over Ez should be actually performed from the maximum of twoconduction band bottoms, at lower energies T (Ez) = 0.

Now in this expression distribution functions should be discussed. Thereare two dierent cases. When there is a dierence in chemical potentials (Fig. 1.6, left), distributions functions are

fL(Ez, ) =1

exp(

Ez+ELT

)+ 1

, fR(Ez, ) =1

exp(

Ez+ERT

)+ 1

.

(1.57)This case, however, is quite dicult to realize in nanostructures, because

any change of particle density causes the change in the electric eld, so that infact a dierence in electro-chemical potentials = + e with approximatelythe same chemical potentials (L = R = ), e.g. voltage dierence (LR =eV ) takes place (Fig. 1.6, right).

In this case expression (1.56) should be used with care! First of all, thepotential U(z) is now a function of applied voltage, and consequently thetransmission coecient is a function of voltage too. And the energy shift in adistribution function should be taken into account. One obtains

J(V ) =e

h

T (Ez, V ) [fL(Ez, , V ) fR(Ez, )] dEz, (1.58)


0 z

L

R

U(z)

0 z

L+eV

R

U(z)

Fig. 1.6. Energy diagrams for chemical potential dierence (left) and voltage dif-ference (right).

with

fL(Ez, , V ) =1

exp(

Ez+EeVT

)+ 1

, fR(Ez, ) =1

exp(

Ez+ET

)+ 1

.

(1.59)

1.3.2 Conductance of a perfect wire

First of all, let us consider the conductance of a perfect wire placed between tworeservoirs. Perfect wire means that there are several reectionless channels(with transmission coecient T (E) = 1). Besides, it is assumed that there isno reection for electrons transmitting from the wire into the contact, thusright going electrons are populated only by the left reservoir and left going


electrons are populated only by the right reservoir (Fig. 1.7). We can say thatright moving electrons have the (pseudo-) electro-chemical potential L, whileleft moving electrons R. Of course, the state of electrons inside the wireis not equilibrium, and these left and right chemical potentials give thenumber and energy of corresponding particles in the channel, but they arenot usual thermodynamic potentials.

Now we simply use the expression for the current. Distribution functionsin the contacts at zero-temperature are step-functions

fL(Ez, , V ) = ( + eV Ez E), (1.60)fR(Ez, ) = ( Ez E), (1.61)

the current is

J(V ) =e

h

T(Ez, V ) [( + eV Ez E) ( Ez E)] dEz =

=e

h

E+eVE

T(Ez, V )dEz =e2

hNV,

(1.62)

where we used T (Ez, V ) = 1, and N is the number of open channels betweenL = + eV and R = . For the conductance one has

G =e2

hN. (1.63)

Where does the resistance of a perfect wire come from? The origin ofthis resistance is in the mismatch between the large number of modes in thecontacts and a few channels in the wire. So this is NOT the resistance of aperfect wire, but rather the resistance of the interface between contacts andwire.

1.3.3 Back to the single-channel Landauer formula

Consider now the single-channel case with imperfect transmission T (Ez, V ) =T = 1, repeating the same calculation as in (1.62) we obtain

J =e

hT (L R) = e

2

hTV, (1.64)

G =e2

hT. (1.65)

This is conductance between reservoirs, e.g. between some two points Land R inside the contacts (see Fig. 1.8). Now consider two other pointsA and B inside quantum wire. Distribution functions and corresponding


0 z

L+eV

R

Fig. 1.7. Left and right particles in a perfect wire (energy diagram).

LeadsLeftreservoir(contact)

Rightreservoir(contact)

Scatteringregion

A B

L R

Fig. 1.8. Four-point schema.

electro-chemical potentials (these potentials are not true potentials, butgive only the correct number of electrons, as we discussed before) are dierentalso for left and right moving electrons. Now, however, these potentials aredierent at dierent sides of the barrier (Fig. 1.9). Potential L of the rightmoving electrons is equal to L only in the left part of the wire, as well asR = R in the right part. In the other parts electro-chemical potentials aremodied by the reection from the barrier. For example, only the part ofright moving electrons is transmitted through the barrier and correspondingpotential should be T L, but additionally (1 T )R are reected and moveback. Finally, we obtain


L = L, R = T L + (1 T )R, (1.66)

R = R, L = T R + (1 T )L. (1.67)

The dierence of both left moving and right moving chemical poten-tials across the barrier is the same

L R = L R = (1 T )(L R). (1.68)

We can consider eVAB = (1 T )(L R) as the actual potential dropacross the barrier, because if one makes additional contacts to points A andB, the external voltage VAB should be applied to prevent current through thisnew circuit (4-point measurement). Thus we can dene the conductance (withthe current (1.64))

G =J

VAB=

e2

h

T

1 T =e2

h

T

R. (1.69)

which is the self-conductance of a wire. Not surprising that for perfect wirewith T = 1 and R = 0 this conductance is innite.

Note thatG1 G1 = h

e2(1.70)

is the quantum resistance of the perfect wire (connected to reservoirs). G1

can be considered as the full resistance of two sequential resistances of thebarrier (G1) and the perfect wire (h/2e2).

1.3.4 Multi-channel Landauer formula

0 z

L+eV

R

L

L

R

R

Fig. 1.9. Left and right particles in an imperfect wire (energy diagram).


At zero temperature

Calculation of the current is now straightforward, following (1.62) we obtainfor the current from the left mode into the right mode

J =2eh

T (L R) , (1.71)

summing contributions from all incoming left modes we obtain current inoutgoing mode

J =e

h

NL=1

T (L R) , (1.72)

and, nally summing all outgoing modes, the full current

J =e

h

NL=1

NR=1

T (L R) . (1.73)

After the obvious mathematical transformation

NL=1

NR=1

T =NL=1

NR=1

|t|2 =NL=1

NR=1

tt =

NR=1

(tt)

= Tr(tt)

(1.74)we obtain Landauer multichannel current and conductance

J =e

hTr(tt)(L R) = e

2

hTr(tt)V, (1.75)

G =e2

hTr(tt). (1.76)

General expression

If we repeat this summation procedure at nite temperature and voltage, weobtain generalized form of equation (1.56)

J(V ) =2eh

T(E, V ) [fL(E,, V ) fR(E, )]dE, (1.77)

here we use integration over energy E = En +Ez instead of Ez because Ez isnot conserves in multimode transmission.

Equilibrium distribution functions of the contacts are

fL(E,, V ) =1

exp(

EeVT

)+ 1

, fR(E, ) =1

exp(

ET

)+ 1

. (1.78)

1.4 Derivation from linear response theory 33

These functions are the same for all modes and we obtain the following generaltwo-terminal Landauer formula

J(V ) =2eh

T (E, V ) [fL(E + eV ) fR(E)]dE, (1.79)

withT (E, V ) =

T(E, V ) = Tr(tt)(E, V ). (1.80)

1.3.5 Multi-terminal formula of Buttiker

In the 80-th the rapidly developing mesoscopic experiments, and also thetheoretical discussion about two-probe and four-probe measurements, calledfor a new formalism describing the QS with several electrodes, so that voltagesand currents can be applied and measured independently for any contact. Thisnew formulation was given by Buttiker [7, 9]. The only dierence from theconsidered before two-terminal system is that now there are several contactswith independent electrochemical potentials.

All developed scattering theory can be applied in this case in exactly thesame way, as in the multi-channel case. But the transmission functions aredetermined for any pair of contacts.

For example, the multi-channel formulas (1.75) and (1.76) can be rewrittenas

Ji =e

h

j =i

T ij (i j) = e2

h

j =i

T ijVij , (1.81)

Gij =e2

hT ij =

e2

hTr(tij t

ij

), (1.82)

where Ji is the current from the i-th contact to the System.From the time-reversal invariance in the presence of a magnetic eld H

j =i

T ij(H) =j =i

T ji(H). (1.83)

The current from the i-th contact to the System is

Ji =e

h

j =i

T ij(E, V ) [fi(E + eVij) fj(E)]dE, (1.84)

where Vij is the voltage between contacts i and j.

1.4 Derivation from linear response theory

The linear response theory (Kubo formula) can be applied to calculate con-ductance (Bruus and Flensberg, section 6.3)


G = lim0

e2

Re 0

dtei(+i)t[

J(t), J(0)]

eq

, (1.85)

where the matrix elements of the current operator I in the Hilbert space ofsingle particle eigenfunctions n(r) = n(r)n(z) are

Jnm(z) = i2m

dr

(n

mz

m n

z

), (1.86)

which should be independent of z in the stationary case because of currentconservation, and can be evaluated at any cross-section.

At zero temperature this formula can be written as [4]

G = lim0

e2h

4m2

nm

|Wnm|2( Em + En), (1.87)

where Wnm is

Wnm = nmz

m n

z, (1.88)

the sum is taken with the condition En < EF < Em.The simplest way to proceed is to use the so-called scattering states. Using

incoming and outgoing modes in the right and left leads (1.47)-(1.50), we candene the scattering states as

Ln(E, x, y, z) =

Ln(x, y)eikn(E)z +m

knkm

rmnLm(x, y)eikn(E)z, z L,m

knkm

tmnRn(x, y)eikm(E)z, z R.(1.89)

The physical sense of this state is quite transparent. It describes a particlemoving from the left and splitting into reected and transmitted parts. Weestablished that it is important for Landauer transport, that only these statesare populated from the left reservoir, so that one can accept the distributionof left scattering states to be equilibrium with the left electro-chemical po-tential. The other right states, populated by the right reservoir, are denedas

Rn(E, x, y, z) =

Rn(x, y)eikn(E)z +m

knkm

rmnRm(x, y)eikn(E)z, z R,m

knkm

tmnLn(x, y)eikm(E)z, z L.(1.90)

Now, using the scattering states (1.89) and (1.90), it is straightforward toobtain the Landauer and Fisher-Lee formulars. We leave that as an exercise!

1.5 Problems 35

1.5 Problems

1.5.1 Transmission though a symmetric rectangular barrier

Consider the rectangular barrier

U(z) =

0, z < a,U0, a < z < a,0, z > a.

(1.91)

Wave function is given by

(z) =

A+eikz + Aeikz, z < a

C1ez + C2ez, a < z < a

Beikz + B+eikz, z > a(1.92)

where

k =2mEzh

, (1.93)

=

2m(U0 Ez)

h. (1.94)

After straightforward calculation we nd

T (Ez < U0) =4k22

(k2 + 2)2 sinh2(a) + 4k22. (1.95)

1.5.2 4-point system. Derivation of the rst Landauer formulafrom the Buttiker formula

1.5.3 Linear response derivation of the Fisher-Lee formula

Additional reading

A. D. Stone and A. Szafer, What is measured when you measure a resis-tance? - The Landauer formula revisited, IBM J. Res. Develop. 32, 384(1988).

H. Bruus and K. Flensberg, Many-body quantum theory in condensedmatter physics, chapters 6,7.

D.K. Ferry and S.M. Goodnick, Transport in nanostructures, chapter 3. S. Datta, Quantum transport: atom to transistor, chapter 9. S. Datta, Electronic transport in mesoscopic systems, chapters 2,3. Y. Imry, Introduction to mesoscopic physics, chapter 5.


Bibliography

1. R. Landauer, Spatial variation of currents and elds due to lodcalizedscatterers in metallic conduction, IBM J. Res. Develop. 1, 223 (1957).

2. R. Landauer, Electrical resistance of disordered one-dimentional lat-tices, Phil. Mag. 21, 863 (1970).

3. E. N. Economou and C. M. Soukoulis, Static Conductance and Scal-ing Theory of Localization in One Dimension, Phys. Rev. Lett. 46, 618(1981).

4. D. S. Fisher and P. A. Lee, Relation between conductivity and transmis-sion matrix, Phys. Rev. B 23, 6851 (1981).

5. M. Buttiker, Y. Imry, R. Landauer, and S. Pinhas, Generalized many-channel conductance formula with application to small rings, Phys. Rev.B 31, 6207 (1985).

6. M. Buttiker, Four-Terminal Phase-Coherent Conductance, Phys. Rev.Lett. 57, 1761 (1986).

7. R. Landauer, Spatial variation of currents and elds due to lodcalizedscatterers in metallic conduction, IBM J. Res. Develop. 32, 306 (1988).

8. M. Buttiker, Symmetry of electrical condution, IBM J. Res. Develop.32, 317 (1988).

9. A. D. Stone and A. Szafer, What is measured when you measure a resis-tance? - The Landauer formula revisited, IBM J. Res. Develop. 32, 384(1988).

10. H. U. Baranger and A. D. Stone, Electrical linear-response theory in anarbitrary magnetic eld: A new Fermi-surface formation, Phys. Rev. B40, 8169 (1989).

Landauer Buttiker

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