Introduction to the Scattering Matrix Formalism

Introduction to the Scattering Matrix Formalism Fabrizio Dolcini Scuola Normale Superiore di Pisa, NEST Italy Dipartimento di Fisica del Politecnico di Torino Italy Lecture Notes for XXIII Physics GradDays, Heidelberg, 5-9 October 2009

Contents Scattering Matrix Formalism 3. Introduction.................................... 3.2 Example: Ballistic Quantum Wire........................ 4.2. The case with one impurity....................... 4.3 Relation between Transfer matrix and Scattering matrix........... 7.3. General remarks............................. 7.3.2 Hybrid junction.............................. 9.4 Combining Scattering Matrices: The case with two impurities........ 2.4. Transmission Coefficient......................... 7.5 Properties of the Scattering Matrices: Unitarity and Onsager relations.... 9.6 Landauer-Büttiker formalism........................... 20.6. The single channel case.......................... 20.7 Current operator................................. 22.7. Definition................................. 22.7.2 Electrons with parabolic dispersion................... 22.7.3 Expression in the Left Lead....................... 23.7.4 Expression in the Right Lead...................... 26.8 Average Current and Non-linear Conductance................. 27.8. Zero temperature limit.......................... 28.8.2 Source of Resistance........................... 28.8.3 Example. Fabry-Pérot oscillations in carbon Nanotubes........ 30.9 Multi-channel case................................ 32.9. Example: Quantum Point Contact in a Semiconductor 2DEG.... 36 2

Chapter Scattering Matrix Formalism. Introduction Let us consider a mesoscopic sample connected to two reservoirs on the LeftL and RightR, as shown in Fig... LEFT ELECTRODE LEAD mesoscopic sample LEAD RIGHT ELECTRODE Figure.: A mesoscopic sample connected to two normal metallic electrodes. In experiments the typical size reservoirs is wide compared to the typical cross-section of the mesoscopic conductor. Consequently, the mesoscopic system represents only a small perturbation for the reservoirs, which can thus be described in terms of an equilibrium state, characterized by chemical potentials µ L and µ R and temperatures T L and T R, respectively. The distribution functions of electrons in the reservoirs are therefore Fermi distribution functions f X E = X = L/R. + e E µ X/k B T X It is worth emphasizing the differences between the reservoirs and the mesoscopic sample: While in the mesoscopic sample no inelastic processes occur, in the reservoirs inelastic processes do occur, for otherwise no equilibrium state could be established. Furthermore, electrodes are characterized by a large number of modes, which are energetically very closely spaced, whereas the mesoscopic system has a finite number of modes. 3

4 Example: Ballistic Quantum Wire The contacts between the mesoscopic conductor and the reservoirs can be schematized as ideal narrow channels widening into the large reservoirs, which are called leads, or contact regions. We suppose for simplicity that the temperatures in the two electrodes are the same T L = T R = T, whereas we apply a voltage bias V between the two electrodes V = µ L µ R /q.2 where q is the elementary electron charge q = e = e or q = e = e depending on the convention used in the literature. The voltage induces a current I through the sample and drives it into a non-equilibrium state. The purpose is to determine the transport properties average current, current fluctuations, etc. as a function of the applied bias V and of the typical parameters characterizing the mesoscopic system..2 Example: Ballistic Quantum Wire.2. The case with one impurity Let us consider free electrons with parabolic dispersion and with one δ-like impurity located at x = x 0 ; in second quantization the Hamiltonian reads + H = Ψ x 2 2 + Λδx x 2m x 2 0 Ψx dx where the electron field fulfills anti-commutation relations { Ψx, Ψ y } = δx y.3 Introducing we can also rewrite H = 2 2m + Λ = Λ 2m 2.4 Ψ 2 Ψx x x 2 The equation of motion for the field Ψ reads + Λ δx x 0 Ψx dx i Ψ = [Ψ, H].5 i.e. i Ψx, t = 2 2 2m x Ψx, t + Λδx x 2 0 Ψx, t.6

Chap.. Scattering Matrix Formalism 5 From this equation we observe that 2 xψ has a δ-like singularity, implying that x Ψ is discontinuous at x = x 0. The field Ψ itself, in contrast, is continuous. Integrating around x = x 0 we obtain that x Ψx + 0 x Ψx 0 = ΛΨx 0.7 Thus, the equation of motion.6 is equivalent to i Ψx, t = 2 Ψx, t x x 2m x 2 0 2 Ψx + 0, t = Ψx 0, t x Ψx + 0, t x Ψx 0, t = ΛΨx 0, t In order to solve these equations, we make the following Ansatz â Lk e ikx + ˆb Lk e ikx x < x 0 Ψ k x, t = e ie kt/.8 ˆbRk e ikx + â Rk e ikx x > x 0 where E k = 2 k 2 2m.9 which satisfies the first condition. The second and third conditions respectively yield â Lk e ikx 0 + ˆb Lk e ikx 0 = ˆb Rk e ikx 0 + â Rk e ikx 0 ik ˆbRk e ikx 0 â Rk e ikx 0 ik â Lk e ikx 0 ˆb Lk e ikx 0 = Λ â Lk e ikx 0 + ˆb Lk e ikx 0.0 The solution of this set of linear equations for the amplitude operators can easily be found see below and reads ˆbRk â Lk = M k. â Rk ˆbLk where the matrix M k M k = + Λ Λ ex 0 Λ e x 0 Λ.2

6 Example: Ballistic Quantum Wire is called the Transfer Matrix, for it describes how the amplitude operators on the left of the impurity are transfered to the ones on the right of the impurity. We notice that the Transfer Matrix exhibits the following properties of the transfer matrix: detm k = M 2 = M 2.3 the order of â, ˆb operators appearing in Eq.. is different on the left and right side. This is because typically Transfer Matrix is defined as relating corresponding velocity amplitudes. A right-moving electron is described by ˆb Rk on the right of the impurity, and by â Rk on the left of the impurity. Proof of Eq..0 Starting from Eqs..0 one can now introduce â Lk = â Lk e ikx 0 ˆb Lk = ˆb Lk e ikx 0 ˆb Rk = ˆb Rk e ikx 0 â Rk = â Rk e ikx 0.4 and reexpress the above equations in these new variables â Lk + ˆb Lk ik ˆb Rk â Rk = ˆb Rk + â Rk ik â Lk ˆb Lk = Λ â Lk + ˆb Lk i.e. â Lk + ˆb Lk â Lk + Λ ik = ˆb Rk + â Rk ˆb Lk Λ ik = ˆb Rk â Rk We have now two equations for four unknowns. In order to determine the transfer matrix we have to express the variables of the RIGHT-part as a function the the ones in the LEFT part. It is therefore straightforward to replace the above equations with their sum and difference ˆb Rk = + Λ â Lk + Λ ˆb Lk â Rk = Λ â Lk + Λ ˆb Lk

Chap.. Scattering Matrix Formalism 7 Coming now back to the original variables.4 the latter equations can be rewritten as ˆbRk = + Λ â Lk + Λ e x 0 ˆbLk â Rk = Λ ex0 â Lk + Λ ˆbLk which are simply Eqs..0 written in matrix form..3 Relation between Transfer matrix and Scattering matrix.3. General remarks The Transfer matrix expresses the coefficients of the states on the right side of scatterer in terms of the coefficients of the same states on the left side of the scatterer: ˆbR = M M 2 â L.5 â R M 2 M 22 ˆbL The relations.5 between operators are linear. It is thus easy to express the same relations also in different equivalent ways. In particular, one can expresses the outgoing state operators as a function of the incoming state operators, ˆbL ˆbR = S S 2 S 2 S 22 â L â R.6 and the matrix S is called the Scattering Matrix. It is straightforward to verify that the S-matrix can be expressed in terms of the Transfer matrix entries as follows: S = M 22 M 2 M 2.7 where we have used the property: detm =.8 Usually the Scattering matrix entries are denoted as r t S = t r.9

8 Relation between Transfer matrix and Scattering matrix where t k, t k and r k, r k are transmission and reflection amplitudes. Let us for instance assume to prepare the incoming wave packet as coming from LEFT we thus set â R = 0. Then the transmission and reflection coefficients read T = t 2 = R = r 2 = M 22 2 M 2 M 22 2 = 2 M 2 M 22.20 Exploiting the property: M 2 = M 2.2 one can observe that the same result is obtained if the incoming wave packet comes from RIGHT. Example: Scattering Matrix and Transmission Coefficient for the case with one impurity In the case described above of a quantum wire with one single impurity the Scattering matrix can be obtained using the Transfer Matrix and applying the relation.7 between Scattering and Transfer matrices S = Λ Λ ex 0 Λ Λ Λ Λ e x 0 Importantly we observe that the Scattering matrix is unitary S S = I.22 The transmission coefficient is easily computed inserting Eqs..2 into Eq..20. T = = + Λ 2 4k 2 8mE 2 8mE 2 + Λ 2 = 2 2 m E 2 2 m E + Λ2.23 We observe that the transmission coefficient does not depend on the position x 0 of the impurity.

Chap.. Scattering Matrix Formalism 9.0 0.8 λ=.0 T 0.6 λ= 0.4 0.2 λ=3.0 0.0 0 2 4 6 8 0 E Figure.2: The Transmission coefficient for a quantum wire with parabolic dispersion in the presence of one impurity..3.2 Hybrid junction Let us consider a hybrid junction between two different materials e.g. two different semiconductors, characterized by different effective masses, i.e. by different parabolic dispersion. Let us assume that the junction is ideal and that no backscattering term like the δ-like impurity in the previous example is present. In general this will not be the case, but this is unimportant for what we want to point out here. At a given value of the energy E the characteristic momenta and velocities in the two regions forming the junction are different, due to the different dispersion relations. Since the system breaks translational invariance, momentum is not conserved and is not a good quantum number, whereas energy is see Fig..3. Model: The hybrid junction can be modeled as a system with a space-dependent mass H = + Ψ x 2 x 2mx Ψx x dx.24 where the effective mass reads mx = m L θx x 0 + m R θx 0 x.25 The equation of motion for the field Ψ reads

0 Relation between Transfer matrix and Scattering matrix m L m R v L v R k L k R Figure.3: A junction between different materials. i Ψx, t = 2 x 2mx Ψx + 0, t = Ψx 0, t Ψx, t x x x 0 x Ψx + 0, t = x Ψx 0, t m R m L To solve the above relation we make the following Ansatz â LkL e iklx + ˆb LkL e ik Lx Ψx, t = e iet/ ˆbR kr e ikrx + â R kr e ik Rx x < x 0 x > x 0.26 where k L = 2mL E k R = 2mR E.27 It is straightforward to see that the solution is b LkL b RkL = S kl k R a LkL a RkL.28 with where S kl k R = v L + v R v L E = k L m L v L v R e Lx 0 2v R e ik L k R x 0 2v L e ik L k R x 0 v L v R e Rx 0.29 v R E = k R m R.30

Chap.. Scattering Matrix Formalism Here the two subscripts k L k R in S kl k R remind that, differently from the previous case of a single impurity, momentum is not conserved in this problem. Importantly, one can easily prove that this scattering matrix S kl k R is not unitary. Motivated by the remark that energy and not momentum is conserved, we try to introduce operators that are related to energy. Let us then pass from momentum label to energy label. In doing that, we also pass to the continuum We notice that with... = Ω 2π k dk... = Ω 2π de....3 ve {â Lk, â Lk } = {ˆb Lk, ˆb Lk } = δ k,k = 2π Ω δk k = 2π Ω veδe E,.32 ve = E k,.33 We can thus introduce energy amplitude operators â XE and ˆb XE, defined through the relations v â LkL = L Ω âle v ˆbLkL = L ˆb Ω LE.34 v ˆbR kr = R ˆb Ω RE ˆbR kr = v R ˆb Ω RE They obey the following anticommutation relatiosn {â LE, â LE } = {ˆb LE, ˆb LE } = δe E {â LE, ˆb LE } = {â LE, ˆb LE } = 0.35 In terms of the new operators Eqs..28 and.29 are easily rewritten as b LE = SE a LE.36 b RE a RE with a new scattering matrix v L v R e Lx 0 2 v R v L e ik L k R x 0 SE = v L + v R 2 v R v L e ik L k R x 0 v L v R e Rx 0.37 Notice that in Eq..72 the velocity is independent of the LEFT/RIGHT or IN/OUT; it only depends on the energy. This is equivalent to assuming that the band is symmetric for k k. This is also consistent with Eq..65, where only one energy appears.

2 Combining Scattering Matrices: The case with two impurities where k L and k R are related to energy through Eq..27. Notice that, due to the fact that velocities are different, the matrix.37 is not simply the same matrix.29 re-expressed in terms of energy E. Importantly the scattering matrix.37 for the new energy operators is unitary. One can wonder: why the previous matrix.29 was not unitary? redefinition.34 of the operators led unitarity emerge? Why a simple v dt v 2 dt Figure.4: Flux conservation. This can be understood as follows: In hybrid systems, such as heterojunctions between two different semiconductors or junctions between normal and a superconductors, density is typically not conserved across the junction. Indeed, as shown in Fig..3, at a given energy we have different k F s in the two sides of the junction, i.e. different electron densities. This is quite similar to what happens to an incompressible fluid flowing through a pipeline with inhomogeneous velocities: in order for the total number of particles to be preserved, the density is not uniform. In contrast the flux ρv is preserved: the lower the velocity the higher the density see Fig..4. In our case the flux is simply the current density, which is conserved across the junction. J = qv Ψ Ψ q v â â k âk E â = qv v E = q v h â E âe The conservation of J implies the unitarity for the scattering matrix connecting â E and ˆb E operators: the incoming flux must equal the outgoing flux. For this reason the scattering matrix is usually meant as refereed to the energy amplitude operators, and not to the â k operators..4 Combining Scattering Matrices: The case with two impurities Let us consider free electrons with parabolic dispersion, but with two δ-like impurities located at x = x and x = x 2 with x < x 2. This could be for instance a nice toy model to account for the presence of non-ideal contacts between a ballistic quantum wire and the leads notice that non-ideal means T < ; we shall see, however, that even for ideal contacts with T = there exists an intrinsic quantum

Chap.. Scattering Matrix Formalism 3 of resistance R Q. The length of the quantum wire is then The Hamiltonian then reads: H = 2 2m + L = x 2 x length of the wire.38 Ψ x 2 x 2 Ψx dx + Λ Ψ x Ψx + Ψ x 2 Ψx 2 Introducing again Eq..4, i.e. Λ = Λ 2m.39 2 and integrating the equation of motion around x = x and then around x = x 2, we obtain the following conditions i Ψx, t = 2 2 2m x Ψx, t x x, x 2 2 Ψx continuous at x = x and at x = x 2 x Ψx + x Ψx = ΛΨx x Ψx + 2 x Ψx 2 = ΛΨx 2 The following Ansatz Ψ k x, t = e ie kt/ â Lk e ikx + ˆb Lk e ikx x < x ˆγ + k eikx + ˆγ k e ikx x < x < x 2.40 ˆbRk e ikx + â Rk e ikx x > x 2 fulfills the first condition with energy: E k = 2 k 2 2m.4 The second, third and forth conditions respectively yield â Lk e ikx + ˆb Lk e ikx = ˆγ + k eikx + ˆγ k e ikx ˆγ + k eikx 2 + ˆγ k e ikx 2 = ˆb Rk e ikx 2 + â Rk e ikx 2 ik ˆγ + k eikx ˆγ k e ikx ik â Lk e ikx ˆb Lk e ikx ik ˆbRk e ikx 2 â Rk e ikx 2 = Λ â Lk e ikx + ˆb Lk e ikx ik ˆγ + k eikx 2 ˆγ k e ikx 2 = Λ ˆbRk e ikx 2 + â Rk e ikx 2.42

4 Combining Scattering Matrices: The case with two impurities This linear system of equations can be solved straightforwardly see below for explicit derivation, and one finds that ˆγ + + Λ Λ k e x â Lk = ˆγ k Λ.43 ex Λ ˆbLk }{{} M k x =Transfer Matrix at x ˆbRk â Rk = + Λ Λ e x 2 Λ ex 2 Λ }{{} ˆγ + k ˆγ k.44 M k x 2 =Transfer Matrix at x 2 where the transfer matrices M k x and M k x 2 are the single impurity transfer matrices see Eq..2 respectively at x and x 2. The whole Transfer Matrix ˆbRk = M k â Lk.45 â Rk ˆbLk is thus obtained as the product M k = M k x 2 M k x =.46 = + Λ 2 2 Λ e x 2 x Λ + Λ e x + Λ Λ e x + Λ e x 2 + Λ e x 2 Λ 2 2 Λ e x 2 x We now pass to energy amplitude operators â Lk = ˆbR k = ˆbR k = ˆbLk = v Ω âle v ˆb Ω LE v ˆb Ω RE v ˆb Ω RE.47

Chap.. Scattering Matrix Formalism 5 Since in this case the velocity v E = k E m = 2E m.48 is the same everywhere, the Trasfer Matrix for the â XE and ˆb XE operators is simply Eq..46, expressed in terms of energy E. Recalling that one rewrites Eq..46 as Λ = Λ 2m 2.49 M E = = 2 2 + Λ i v E Λ i v E e x 2 x Λ i v E Λ i v E e Ex + Λ i v E + Λ i v E e x + 2 + Λ i v E e x 2 Λ i v E Λ i v E e x 2 Λ i v E 2 e E x 2 x It is straightforward to verify that ME = M E ; moreover, since the determinant of each M E x i s equals, one also has that detm E =. Proof of Eqs..44 and.44: Consider Eq..42 and introduce â Lk = â Lk e ikx ˆb Lk = ˆb Lk e ikx ˆb Rk = ˆb Rk e ikx 2 â Rk = â Rk e ikx 2.50 and rewrite the equations as â Lk + ˆb Lk = ˆγ+ k eikx + ˆγ k e ikx ˆγ + k eikx 2 + ˆγ k e ikx 2 = ˆb Rk + â Rk ik ˆγ + k eikx ˆγ k e ikx ik â Lk ˆb Lk = Λ â Lk + ˆb Lk ik ˆb Rk â Rk ik ˆγ + k eikx 2 ˆγ k e ikx 2 = Λ ˆb Rk + â Rk

6 Combining Scattering Matrices: The case with two impurities i.e. â Lk + ˆb Lk = ˆγ+ k eikx + ˆγ k e ikx ˆb Rk + â Rk = ˆγ+ k eikx 2 + ˆγ k e ikx 2 + Λ ik Λ ik â Lk Λ ˆb ik Lk = ˆγ + k eikx ˆγ k e ikx ˆb Rk + Λ â Rk ik = ˆγ+ k eikx 2 ˆγ k e ikx 2 These are 4 equations for 6 unknowns; in order to determine the transfer matrix, we express now everything in terms of the variables in the LEFT side; thus, we proceed as follows: I e ikx I + III III e ikx I III II Λ II IV ik IV + Λ II + IV ik obtaining ˆγ + k = e ikx + Λ â Lk + ˆγ k = eikx Λ â Lk + Λ Λ ˆb Lk ˆb Lk â Rk = Λ eikx2 ˆγ + k + Λ e ikx2 ˆγ k ˆb Rk = + Λ e ikx2 ˆγ + k + Λ e ikx2 ˆγ k.5

Chap.. Scattering Matrix Formalism 7 or, in matrix form: ˆγ + k ˆγ k ˆb Rk â Rk = = e ikx e ikx e ikx 2 + Λ Λ + Λ e ikx e ikx Λ Λ e ikx 2 Λ Λ eikx 2 e ikx 2 Λ â Lk ˆb Lk By recalling the definition.50, one can see that the above equations yield.43 and.44..4. Transmission Coefficient The transmission coefficient is easily computed inserting the entry M 22 of Eq..50 into Eq..20 2 T = M 22 = 2 Λ i ve Λ 2e 2 E L i ve One has: Λ 2 2 2 Λ e EL = i ve i ve 2 2 4 [ 2 Λ Λ Λ = + + + Λ 2 e EL + + Λ 2 e EL] + = v E v E v E i v E i v E 2 4 [ 2 ] 2 Λ Λ Λ Λ = + 2 + 2 + 2 cos 2k E L + 4 Λ sin 2k E L = v E v E v E v E v E 2 4 [ 2 2 Λ Λ Λ Λ 2 = + 2 + 2 + 2 cos 2 k E L ] + 8Λ sin k E L cos k E L = v E v E v E v E v E 4 2 2 3 Λ Λ Λ Λ = + 4 + 4 cos 2 k E L + 8 sin k E L cos k E L = v E v E v E v E 4 2 3 Λ Λ Λ = + 4 sin 2 k E L + 4 cos 2 k E L + 8 sin k E L cos k E L = v E v E v E [ 2 ] 2 Λ Λ = + 4 sin 2 k E L + cos 2 k E L + 2 Λ sin k E L cos k E L = v E v E v E 2 [ Λ = + 4 cos k E L + Λ ] 2 sin k E L v E v E whence T E = 2 [ Λ + 4 v E cos k E L + ˆγ + k ˆγ k ] 2.52 Λ v E sin k E L

8 Combining Scattering Matrices: The case with two impurities We now recall that the typical energy range we are interested in temperature and applied voltage are much smaller than the equilibrium Fermi energy. We can thus write and therefore we have E = ε F + ξ ξ ε F.53 2mE 2mεF + ξ k E = = = 2m ε F + ξ = ε F = [use k F = 2m εf ] = k F + ξ = ε F k F + ξ = 2ε F and one can well approximate Eq..52 as [use v F = k F m and 2ε F = v F k F ] = k F + ξ = v F k F = k F + ξ v F.54 where we have defined and T ε F + ξ [ ] 2.55 + 4λ 2 cos[k F L + ξ E L ] + λ sin[k F L + ξ E L ] E L = ω L = v F L λ =.56 Λ v F.57 Here E L is the energy associated with the length of the wire distance between the two contact impurities, which is the natural characteristic lengthscale in this problem. We thus observe that the transmission coefficient is an oscillatory function with a typical period given by E L, as shown in Fig..5.

Chap.. Scattering Matrix Formalism 9 T.00 0.75 0.50 λ=0.2 λ=2 0.25 0.00-0 -5 0 5 0 E /E L Figure.5: The Transmission coefficient.55 for a system of electrons with parabolic dispersion in the presence of two impurities..5 Properties of the Scattering Matrices: Unitarity and Onsager relations General symmetries imply properties on the Scattering matrix: Conservation of Flux: Time Reversal Symmetry: SE = SE.58 S ij E; B = S jie; B.59

20 Landauer-Büttiker formalism.6 Landauer-Büttiker formalism.6. The single channel case We consider independent electrons in one dimension, in the presence of scatterers. For simplicity we assume that we have only one band, in which we have IN-coming and OUTgoing particles. Examples of this situation are given in sec..2 for electrons in the continuum with parabolic dispersion. We wish here to define the general framework to treat these models. Following Blanter and Büttiker, we denote: â IN operators.60 ˆb OUT Similarly we shall denote the eigenfunctions as αx IN βx OU T eigenfunctions.6 Once the location with respect to the set of scatterers is fixed i.e. whether LEFT or RIGHT, the IN and OUT states are determined: for instance, for electrons with parabolic dispersion, one has for k 0 α R kr x = β R kr x = Ω e ik Rx Ω e ik Rx α L kl x = β L kl x = Ω e ik Lx Ω e ik Lx.62 As the examples discussed in the previous sections suggest, the field operator in the LEFT and in the RIGHT lead can be respectively written as: Ψx L, t = α L kl x â L kl t + β L kl x ˆb L kl t =.63 k 0 = e ie kt/ α L kl x â L kl + β L kl x ˆb L kl =.64 k 0 = e ie kt/ e ik Lx â L kl + e ik Lx ˆbL kl.65 Ω Ψx R, t = k 0 = k 0 k 0 α Rk x â R kr t + β Rk x ˆb R kr t =.66 e ie kt/ α Rk x â R kr + β R kr x ˆb R kr.67 = e ie kt/ e ik Rx â R kr + e ik Rx ˆbR kr Ω k 0.68

Chap.. Scattering Matrix Formalism 2 where the operators â Xk s and ˆb Xk s X = L/R fulfill {â XkX, â Y k } Y = δ XY δ k,k.69 {ˆb XkX, ˆb Y k } = δ X,Y δ k,k.70 {â XkX, ˆb Xk X } = {â XkX, ˆb Xk = 0 X}.7 Since we have explicitly separated IN and OU T states, the sum over the momentum only runs on positive k s. Explicit realizations of Eqs..63-.68 have been provided in the previous sections; in particular i in.8 and.40 for electrons with parabolic dispersion in the presence of one and two impurities. We have also seen that it is customary to introduce energy amplitude operators â XE = 2π Ω v XE âxk X ˆbXE = ˆbXkX 2π Ω v XE X = L/R.72 which obey anticommutations relations {â X kx, â Xk } = {ˆb X X kx, ˆb Xk = δ X} kx,k X = 2π Ω δk X k X = 2π Ω veδe E,.73 Similarly we also relabel α L kl x = α L EL x β L kl x = β L EL x.74 where k E 0 denotes the positive momentum corresponding to the energy E mostra in figura. We thus obtain that Eqs..65 and.68 can be rewritten as Ψx L, t = ET E B Ψx R, t = ET E B de vl E e iet/ e ik LEx â LE + e ik LEx ˆbLE de vr E e iet/ e ik REx â RE + e ik REx ˆbRE.75.76 where E B and E T are respectively the bottom and top values of the energy band.

22 Current operator.7 Current operator.7. Definition The current operator is defined as Îx, t = q N i= 2 δx x iˆv i + ˆv i δx x i.77 where ˆv i is the velocity operator of the i-th particle, and the symmetrized form is required, because in general ˆv i and δx x i do not commute. It is important to realize that the form of the velocity operator is not unique, it depends on the context. As a rule, in k-space the velocity is given by: v k = E k k.78.7.2 Electrons with parabolic dispersion In the case of fermions with parabolic dispersion one has E k = 2 k 2 /2m one gets v k = k/m, which yields, in x-space ˆv = ˆp m.79 so, in second quantized formalism, one has: Îx, t = i q 2m Ψ x, t Ψ x Ψ x, t x, tψx, t x.80 The current is typically measured in the leads. Replacing Eq..75 and Eq..76 into the general expression.80 we have that the current in the LEFT and RIGHT leads respectively.

Chap.. Scattering Matrix Formalism 23.7.3 Expression in the Left Lead Replacing Eq..75 into Eq..80 we obtain the current in the left lead. Îx L, t = i q de de t/ 2m veve eie E. = q Recalling now that one obtains Îx L, t Linearization. = q { ik E + k E e ik E k E x â LEâLE ik E + k E e ik E k E x ˆb LEˆb LE + ik E k E e ik E+k E â LEˆb LE + ik E k E e ik E+k E x ˆb LEâLE } = de de veve eie E t/ { ke + k E [ e ik E k E x â 2m LEâLE + k E k E 2m ve = k E m ] eik E k E x ˆb LEˆb LE + [ e ik E+k E x â LEˆb LE + e ik E+k E x ˆb LEâLE ] }.8 de de e ie E t/.82 { ve + ve [ ] 2 e ik E k E x â veve LEâLE eik E k E x ˆb LEˆb LE + + ve ve [ ] } 2 e ik E+k E x â LEˆb veve LE + e ik E+k E x ˆb LEâLE We now observe that typically the energies which are relevant to most of the physical quantities are those in a narrow range with respect to the Fermi energy. We can thus linearize around the Fermi energy and approximate and so that Eq..82 simplifies to. Îx L, t = q de de e ie E t/ v L E v LF.83 k LE k LE E E v LF.84 { e ie E x/ vf â LEâLE } x/ v F eie E ˆb LEˆb.85 LE Notice that the linearized form.85 looks much simpler than the general expression.82, although linearization is not a necessary condition to apply Landauer-Büttiker formalism.

24 Current operator The linearization procedure is equivalent to treating Dirac electrons, which are characterized by a linear dispersion relation E k = v F k, for which the velocity is operator v k = v F.86 i.e. a constant. The situation is in fact slightly more complicated, because one has two branches, i.e. right and left movers Ψ and Ψ, with velocities v k = ±v F respectively. Thus, the form of the current operator in the second quantized form for Dirac fermions reads Îx, t = qv F Ψ x, tψ x, t Ψ x, tψ x, t.87 The definition.87 of the current, together with the expression for the density: ρx, t = q Ψ x, tψ x, t + Ψ x, tψ x, t ensures the continuity equation t ρ = x J for the solutions of the Dirac equation. Use of the Scattering Matrix Suppose that we now want to compute the average current in the say left lead. We would have. I L = Îx L, t = = q de de e ie E t/ Since energy is conserved we can write { } e ie E x/ v F â LEâLE x/ v F eie E ˆb LEˆb LE â LEâLE = δe E â LEâLE.88 ˆb LEˆb LE = δe E ˆb LEˆb LE.89 and we can conclude that I L = q de { } â LEâLE ˆb LEˆb LE.90 In evaluating these average values we are facing a problem. We know the the statistical properties of the incoming electrons, because we can experimentally control them. However, the distribution of the outgoing electrons is quite complicated. Consider for instance a right-moving electron traveling in the left lead towards the mesoscopic system. This electron can only originate from the left electrode and therefore are in thermal

Chap.. Scattering Matrix Formalism 25 equilibrium with it, even when the voltage bias is applied. The statistical properties of this electron are thus characterized by a chemical potential µ L and a temperature T L : a LE a LE = f L E The same thing can be said for left moving electrons traveling on the right lead towards the mesoscopic sample: they can only originate from the right electrode and are thus characterized by a chemical potential µ R and a temperature T R : a RE a RE = f R E These remarks cannot hold for ougoing electrons. Indeed left moving electrons traveling in the left contact may originate either from the right electrode or from the left electrode, possibly after scattering processes through the mesoscopic sample. As a consequence, their distribution is strongly non-equilibrium distribution, which is not known a priori and is in general not experimentally controllable : b LE b LE =? b RE b RE =? The underlying idea of the Landauer-Büttiker formalism is to exploit the Scattering Matrix to express the outgoing operators in terms of the incoming operators only: ˆbLE ˆbRE re t = E t E r E }{{} SE âle â RE.9 The S-matrix can easily be obtained by the transfer matrix M, as described in sec..3. examples of transfer matrices have been given in Eqs..2 and.50 for fermions with parabolic dispersion. In particular, for the current in the left lead we can use the first line of Eq..9 ˆbLE = r E â LE + t EâRE ˆb LE = r Eâ LE + t Eâ RE.92 and insert them into Eq..93 to obtain. Îx L, t = q de de e ie E t/.93 { e ie E x/ vf â LEâLE + } e +ie E x/ v F rer E â LEâLE + r Et E â LEâRE + t Er E â REâLE + t Et E â REâRE

26 Current operator We can now collect together all the terms that involve the same operator product Îx L, t = q de de e ie E t/ [ e ie E x/ v F e +ie E x/ v F rer E â LEâLE e +ie E x/ v F t Et E â REâRE + e +ie E x/ v F ret E â LEâRE + ] e +ie E x/ v F t Er E â REâLE Introducing now the following dimensionless quantities: A LL L E, E ; x = e ie E x/ v F e +ie E x/ v F rer E.94 we can rewrite Îx L, t = A RR L E, E ; x = e +ie E x/ v F t Et E A LR L E, E ; x = e +ie E x/ v F r Et E A RL L E, E ; x = e +ie E x/ v F t Er E q X,Y =L/R.7.4 Expression in the Right Lead.95 dede e ie E t/ A XY L E, E ; x â XEâY E.96 Similarly we can proceed to evaluate the current in the right lead. Replacing Eq..76 into Eq..80, exploiting the second line of Eq..9 ˆbRE = t E â LE + r EâRE ˆb RE = t Eâ LE + r and introducing the following dimensionless quantities: we obtain Îx R, t = A LL R E, E ; x = e ie E x/ v F t Et E Eâ RE A RR R E, E ; x = e +ie E x/ v F + e ie E x/ v F r Er E A LR R E, E ; x = e ie E x/ v F t Er E A RL R E, E ; x = e ie E x/ v F r Et E q X,Y =L/R.97.98 dede e ie E t/ A XY R E, E ; x â XEâY E.99

Chap.. Scattering Matrix Formalism 27.8 Average Current and Non-linear Conductance We now compute the average value of the expressions.27 and.99 for the current operator in the left and right lead, respectively. We exploit the fact that incoming electrons are statistically independent, and that in each lead electrons obey Fermi statistics â XEâY E = δ XY δe E f X E X = L/R.00 where f X is the Fermi distribution in the X-th electrode. Left Lead Using Eq..27 one obtains I L = Îx L, t = q ET E B de A LL L E, E; x f L E + A RR L E, E; x f R E.0 Recalling the expression for the coefficients.95, one has that the diagonal entries E = E read A LL L E, E; x = r E 2 = t E 2 = T E and therefore I L = Îx L, t = q A RR L E, E; x = t E 2 = T E.02 ET de T E f L E f R E.03 E B Right Lead Similarly, using the expression.99 for the current operator in the right lead one obtains I R = Îx R, t = q ET de A LL R E, E; x f L E + A RR R E, E; x f R E.04 E B and, using the expression for the coefficients.98 one has A LL R E, E ; x = t E 2 = T E A RR R E, E ; x = + r E 2 = t E 2 = T E.05 and therefore I R = Îx R, t = q ET de T E f L E f R E.06 E B

28 Average Current and Non-linear Conductance Comparing Eqs..03 and.06 we observe that the average current is actually independent of time and of position x continuity of current, and equals I = I L = I R = q We define the electrochemical potentials of the leads as µ L = ε F + qv where ε F is the common equilibrium Fermi level, whereas de T E f L E f R E.07 µ R = ε F.08 V = µ L µ R /q.09 is the applied bias voltage. From the expression.0 we observe that, with the above definition of V, the current has the same sign as V, independently of the sign of the charge q..8. Zero temperature limit Let us consider the case of zero temperature T = 0. Then the Fermi functions appearing in Eq..0 become Heaviside functions and one obtains I = q εf +qv T E de.0 ε F Let us now consider the Non-linear Conductance: differentiating Eq..0, we obtain that GV = di dv = e2 h T ε F + ev. where we have used the fact that q = ±e, depending on the convention adopted. In particular, if the transmission coefficient can be considered as energy-independent close to the Fermi energy one obtains I = e2 h T V.2 which is a linear dependence, just like Ohm s law, where the resistance is given by R = h/t e 2..8.2 Source of Resistance One can see that even when the mesoscopic sample has a perfectly ideal transmission T =, the conductance is not infinite. There exist therefore an upper bound to the conductance or

Chap.. Scattering Matrix Formalism 29 equivalently a lower bound to the resistance, which is known as the elementary conductance quantum G 0 = e2 h R Q = h 26kΩ resistance quantum per spin channel.3 e2 We have neglected spin so far. If we include spin degeneracy the quantum of conductance is twice as much G 0 = 2e 2 /h. The natural question arises Where does this resistance originate from? Furthermore, resistance is usually associated with dissipative effects, i.e. irreversible processes of energy loss. Here we have used Hamiltonian formalism which ensures reversibility to compute the Scattering Matrix, and we have emphasized that in a mesoscopic system energy is conserved. Thus another question arises: 2 Where are the irreversible processes? To understand this point, one should remember there are two types of chemical potential. On the one side we have the electrochemical potentials µ L and µ R of the electrodes, which are large reservoirs where inelastic scattering processes ensure thermalization. Thus µ L and µ R are the ones corresponding to equilibrium Fermi distribution. On the other side, we have the chemical potentials at the leads here modeled as ideal D channels. As observed when introducing the Scattering Matrix formalism, in the say left lead one has both electrons thermally equilibrated with the left electrode and electrons equilibrated with the right electrode. At the left of the sample there are no equilibration processes, and the electron distribution is not a simple Fermi function. Thus, strictly speaking, it is not even obvious that such non-equilibrium chemical potential can be defined. However, in the linear response regime, one can formally define a chemical potential µ A associated to the average electron density on the left of the sample. Similarly one can introduce a chemical potential µ B on the right of the mesoscopic sample. These chemical potentials are in general different from µ L and µ R see as depicted in Fig..6. The difference V AB = µ A µ B /q between these chemical potentials is called the voltage drop across the mesoscopic system, and is different from the voltage bias µ L µ R. Typically one has µ A µ B = T µ L µ R.4 As one can see the voltage drop is indeed vanishing if the sample has ideal transmission T =. Expressing the current in terms of the voltage drop, one would obtain I = e2 h T T V AB

30 Average Current and Non-linear Conductance If the voltage drop V AB were controllable one would conclude that a finite voltage drop across a sample with perfect transmission T = would yield an infinite current. To some extent, this would seem to solve the problem for a system with no resistance is expected to have infinite current for a finite bias. However, it would be physically incorrect. T µ L µ A µ B µ R Figure.6: Different chemical potentials The crucial point is that the conductance G computed above is the one measured between the two external reservoirs V = µ L µ R /q, and not the one measured with respect to the voltage drop µ A µ B across the sample. Indeed the leads are narrow channels, widening into large reservoirs, where eventually inelastic processes lead to thermalization. The reservoirs act as sources of carriers determined by the Fermi distribution, but also act as perfect sinks of carriers irrespective of the energy of the carrier that is leaving the mesoscopic system. When reaching the reservoirs, electrons are immediately thermalized. This is the reason for the difference between µ A and µ L or between µ R and µ B. These sinks thus represent a source of an irreversible process, which dissipates energy giving rise to a contact resistance. The energy dissipation takes place over a lengthscale given by the energy relaxation length of the leads. In fact, the whole sample can be schematized as a series of 3 resistances in series, the one of the sample itself and the two contact resistances. Thus, the whole system exhibits voltage drops at the contacts and across the mesoscopic sample..8.3 Example. Fabry-Pérot oscillations in carbon Nanotubes Carbon Nanotubes are shown in Fig..7. Their energy band is linear up to ev. Let us adopt a simplified model, where we neglect the valley quantum number, and deal with just one single channel. We then model the non-ideal contacts with two impurities at the ends of the nanotube. We can thus use the model discussed above for a quantum wire with two impurities. Furthermore, due to the fact that the spectrum of nanotube

Chap.. Scattering Matrix Formalism 3 Figure.7: Structure of Carbon Nanotubes. is linear up to ev, the approximation of linearized band is quite appropriate, and the transmission coefficients were found to be [see Eq.8.55] T ε F +ξ 2 [ ] 2 ξ = E ε F Λ + 4 v F cos[k F L + ξ E L ] + Λ v E sin[k F L + ξ E L ].5 Replacing Eq..5 into the general expression Eq..6, and setting V = 0, one can then find the linear conductance GV = 0 GV = 0 = di dv = e2 h = e2 V =0 h T ε F = + 4λ 2 [cos[k F L] + λ sin[k F L]] 2.6 Experiments by Liang. The position of the Fermi level and therefore k F can be tuned by a gate voltage V g. Indeed by varying the gate voltage one fills or depletes charge into the nanotube, changing the position of the Fermi level and, as a consequence, of the Fermi momentum. Oscillations were found in G, as shown in Fig..8 These oscillations demonstrate the coherence of transport in carbon nanotubes. Indeed they originate from the interference between backscattering at the two contacts, which is possible only if coherence is preserved along the nanotube. The nanotube represents the electronic version of a waveguide with two mirrors the barriers. For this reason, these oscillations are called Fabry-Pérot oscillations. In an actual nanotube the situation is more complicated because of -the presence of the valley channels -the impurity strengths may not be identical for the two contacts

32 Multi-channel case Figure.8: Fabry-Pérot conductance oscillations as a function of gate bias in a carbon nanotube, after Ref.[8]. -there may be other impurities inside the wire -there may be charging effects Nevertheless, the main physics of Liang s experiments is already captured by our simple model.9 Multi-channel case So far we have considered a purely D problem, i.e. one single spinless channel. We now want to generalize the previous results taking into account the fact that electrons have a spin, and that a mesoscopic system is in general 3D structure. Since we have some degree of freedom in describing the leads, we can assume without losing in generality that the electron motion in the leads is separable in longitudinal x and transverse y and z directions In the longitudinal direction the system is open, and is characterized by the continuous wave vector k l. Transverse motion is quantized and described by the discrete index n corresponding to transverse energies E n which can in principle be different for the Left and Right leads, and shall therefore be denoted as E Xn. These states are referred to as transverse quantum channels. We write thus E = E n + E l.7 As an example consider a 2D case, where a wire has a transverse width W. If we model such width with hard-wall potential we obtain Then we have E l = 2 k 2 l 2m E n = 2 π 2 n 2 2mW 2 n =, 2,....8 The energies E Xn can be viewed as the band bottom of the longitudinal band for the n-th quantum channel in lead X = L/R, as shown in FIg..9 Since E l 0, for a given total energy E only a finite number of channels exists. The number of incoming channels is denoted N L E in the Left and N R E in the Right lead, respectively.

Chap.. Scattering Matrix Formalism 33 4 E = µ 3 2 n = Figure.9: The subbands of a mesoscopic system. We now introduce creation and annihilation operators â XEn and â XEn, which create and annihilate electrons with total energy E in the transverse channel n in the X-th lead X = L/R, which are incident upon the sample. In the same way, the creation and annihilation ˆb LEn and ˆb LEn operators describe elec- trons in the outgoing states. They obey anticommutation relations { } â XEn, â Y E m = δ XY δ nm δe E.9 The outgoing and incoming operators are connected through the Scattering Matrix ˆbLE ˆbLE ˆbLENL ˆbRE ˆbRE ˆbRENR = re t E t E r E }{{} =SE â LE â LE â LENL â RE â RE â RENR The S-matrix has dimension N L + N R N L + N R, with.20 r E block has dimension N L N L r E N R N R.2 t E N R N L t E N L N R The electron field operator can be expressed, quite similarly to Eqs..75 and

34 Multi-channel case.76, as where Ψr L, t = Ψr R, t = N L E de e iet/ n= N R E de e iet/ n= χ Ln r e ik Enx â LEn + e ik Enx ˆbLEn vn E.22 χ Rn r e ik Enx â REn + e ik Enx ˆbREn vn E.23 k LEn = 2mE E Xn / χ Xn r s are the transversal wave-function in lead X = L/R, and r = y, z is the transversal coordinate. The electron current operator along the longitudinal direction is obtained by integrating over the transversal coordinates In the Left lead: Îx, t = i q 2m dr Ψ r, t Ψ Ψ r, t r, tψr, t x x.24 Replacing Eq..22 into Eq..24, and exploiting the normalization of transversal wavefunctions one straightforwardly generalizes Eq..93 to Îx L, t = i q dr Ψ r, t Ψ Ψ r, t r, tψr, t = 2m x x = q N L E de de e ie E t/.25 n= { vn E + v n E [ ] 2 e ik En k E nx â v LEn v LEnâLE n e ik LEn k LE n x ˆb LEnˆb LE n+ LnE + v ne v n E [ ] } 2 e ik LEn+k E x â LEnˆb v n Ev n E LE n + e ik En+k E n x ˆb LEnâLE n Adopting the linearization approximation one obtains Îx L, t = q de de N L E e ie E t/ n=.26 {[ ]} e ik En k E n x â LEnâLE n e ik LEn k LE n x ˆb LEnˆb LE n+

Chap.. Scattering Matrix Formalism 35 Using the Scattering matrix one expresses the outgoing operators as a function of the incoming ones, obtaining Îx L, t = q X,Y =L/R where we have introduced dede e ie E t/ n,m A Xn,Y m L E, E ; x â XnEâY me.27 A =???????.28 The average current can be expressed as I L = Îx L, t = q de Tr [ t EtE ] f L E f R E.29 where te is the off-diagonal block of the Scattering Matrix. Notice that the trace implies that the expression does not depend on the basis adopted for the scattering states. For every energy E the matrix t EtE can be diagonalized, and its eigenvalues will be denoted by T n E Tr [ t EtE ] = n T n E.30 so that the current reads I L = q de n T n E f L E f R E.3

36 Multi-channel case In the Right Lead We proceed similarly and obtain the same expression I R = q de T n E f L E f R E.32 n In conclusion, the Landauer-Büttiker expression for the average current reads I = I L = I R = q de T n E f L E f R E.33 n The linear conductance At zero temperature, the linear conductance is easily obtained from Eq..33 as GV = 0 = e2 h T n ε F.34 n.9. Example: Quantum Point Contact in a Semiconductor 2DEG A Quantum Point Contact is a narrow constriction in a 2DEG 2-Dimensional Electron Gas. The constriction is created by applying a potential to some top gates, as shown in Figs..0 and. split gate In electrode y split gate x In electrode Figure.0: Top view of a Quantum Point Contact. A nice model for a Quantum Point Contact has been proposed by Büttiker [7], who described the potential inducved by the split gates as V x, y = V 0 2 ω2 xx 2 + 2 ω2 yy 2.35 where V 0 is the electrostatic potential at the saddle, and ω x,y the strength of the curvatures. E Ln = E Rn = E n = V 0 + ω y n n =, 2,....36 2

Chap.. Scattering Matrix Formalism 37 gate gate Ga As cap Al0.3 Ga0.7 As undoped Al0.3 Ga0.7 As n-doped Al0.3 Ga0.7 As undoped Ga As carrier supplier barrier 2DEG Figure.: Side view of the sample see Ref.[4] p.3-34. and it can be shown that the transmission coefficients read T n E V 0 = + e be En b = 2π/ ω x.37 which exhibit a Fermi-like shape, where the inverse temperature b is given by the energy spacing ω x related to the longitudinal confinement strength. The behavior of T n E is shown in Fig..3: each coefficient changes from 0 to within a smoothing energy of ω x, as a function of V 0. Replacing Eq..37 into the expression.38 for the conductance, we obtain Gε F V 0 = e2 h n= + e bε F V 0 ω yn 2.38 As one can see G exhibits a staircase dependence on V 0, where the steps occurs whenever V 0 matches the values V 0 = ε F ω y n /2. The height of the steps plateauto-plateau is given by the quantum of conductance. This quantization of conductance has been experimentally observed in a series of famous experiments by Van Wees et al. [6] see Fig..4. These experiments represent a striking proof of the existence of contact resistance. Furthermore it pointed out the role of transverse modes in transport properties of narrow conductor. The current is carried by a discrete number of transverse modes. This discreteness is not evident if the conductor is extremely wide W λ F, for a small change in W changes the number of transverse channels by a large amount

38 Multi-channel case V x, y x y Figure.2: Potential profile.35 for a Quantum Point Contact. Figure.3: Transmission coefficient.37 for a Quantum Point Contact.

Chap.. Scattering Matrix Formalism 39 Figure.4: Experimental results for conductance quantization in a Quantum Point Contact, after Ref.[6].

Bibliography [] S. Datta, Electronic Transport in Mesoscopic Systems, Cambridge University Press, Cambridge 995. [2] S. Datta, Quantum Transport: Atom to Transistor, Cambridge University Press, Cambridge 2005. [3] Y. Imry, Introduction to Mesoscopic Physics, Oxford University Press, New York 997. [4] D. K. Ferry, and S. M. Goodnick, Transport in Nanostructures, Cambridge University Press, Cambridge 2009. [5] Y. Blanter and M. Büttiker, Shot Noise in Mesoscopic Conductors, [ArXiv version cond-mat/99058] Articles [6] B.J. van Wees et al., Phys. Rev. Lett. 60, 848 988; B.J. van Wees et al., Phys. Rev. B 43, 243 99. [7] M. Büttiker, Phys. Rev. B 4, 7906 990. [8] W. Liang et al., Nature 4, 665 200. 40