Department of Electronics Nanoelectronics 07 Atsufumi Hirohata Department of Electronics 13:00 Monday, 4/February/2019 (P/T 006)

Quick Review over the Last Lecture Schrödinger equation : ( de Broglie wave ) ( observed results ) ( operator ) For example, ( Eigen value ) ( Eigen function )  H : ( Hermite operator ) Ground state still holds a minimum energy :  ( Zero-point motion )

Contents of Nanoelectonics I. Introduction to Nanoelectronics (01) 01 Micro- or nano-electronics ? II. Electromagnetism (02 & 03) 02 Maxwell equations 03 Scholar and vector potentials III. Basics of quantum mechanics (04 ~ 06) 04 History of quantum mechanics 1 05 History of quantum mechanics 2 06 Schrödinger equation IV. Applications of quantum mechanics (07, 10, 11, 13 & 14) 07 Quantum well V. Nanodevices (08, 09, 12, 15 ~ 18)

05 Quantum Well 1D quantum well Quantum tunnelling

Classical Dynamics / Quantum Mechanics Major parameters : Quantum mechanics Classical dynamics Schrödinger equation Equation of motion  : wave function A : amplitude ||2 : probability A2 : energy

1D Quantum Well Potential A de Broglie wave (particle with mass m0) confined in a square well : x a V0 m0 -a E C D General answers for the corresponding regions are Since the particle is confined in the well, For E < V0,

1D Quantum Well Potential (Cont'd) Boundary conditions : At x = -a, to satisfy 1 = 2, 1’ = 2’, At x = a, to satisfy 2 = 3, 2’ = 3’, For A  0, D - C  0 : For B  0, D + C  0 : For both A  0 and B  0 :   : imaginary number Therefore, either A  0 or B  0.

1D Quantum Well Potential (Cont'd) (i) For A = 0 and B  0, C = D and hence, (1) (ii) For A  0 and B = 0, C = - D and hence, (2) /2  3/2 2 5/2   Here, (3) Therefore, the answers for  and  are crossings of the Eqs. (1) / (2) and (3). Energy eigenvalues are also obtained as  Discrete states * C. Kittel, Introduction to Solid State Physics (John Wiley & Sons, New York, 1986).

Quantum Tunnelling In classical theory, E Particle with smaller energy than the potential barrier cannot pass through the barrier. In quantum mechanics, such a particle have probability to tunnel. For a particle with energy E (< V0) and mass m0, Schrödinger equations are x a V0 E m0 C1 A1 A2 Substituting general answers

Quantum Tunnelling (Cont'd) Now, boundary conditions are Now, transmittance T and reflectance R are  T  0 (tunneling occurs) !  T + R = 1 !

Quantum Tunnelling (Cont'd) For  T exponentially decrease with increasing a and (V0 - E) x a V0 E m0 For V0 < E, as k2 becomes an imaginary number, k2 should be substituted with  R  0 !

Quantum Tunnelling - Animation Animation of quantum tunnelling through a potential barrier jt ji jr x a * http://www.wikipedia.org/

Absorption Coefficient Absorption fraction A is defined as jt ji jr Here, jr = Rji, and therefore (1 - R) ji is injected. Assuming j at x becomes j - dj at x + dx, ( : absorption coefficient) x a With the boundary condition : at x = 0, j = (1 - R) ji, With the boundary condition : x = a, j = (1 - R) jie -a, part of which is reflected ; R (1 - R) ji e -a and the rest is transmitted ; jt = [1 - R - R (1 - R)] ji e -a