MIT OpenCourseWare http://ocw.mit.edu 3.23 Electrical, Optical, and Magnetic Properties of MaterialsFall 2007For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Homework # 1 September 12, 2007 Homework is due on Wednesday September 19th, 5pm 1 Linear operators, hermitian operators and more 1. What does it mean for an operator Oˆto be linear and hermitian? 2. Prove which of the following operators is linear, and which is hermitian: d 1 d 2 , , xdx i dφ 3. Let’s consider a particle of mass m trapped in a one dimensional well of length a. The wavefunction of the particle is given by ψ(x) = Ax(a − x). The wavefunction is zero outside the well. a) Find a value for A such that the wavefunction is normalized over the interval [0, a]. b) Calculate the expectation value of the position x of the particle in the well. c) What is the expectation value of the kinetic energy of the particle in the well? 2 The 2D electron gas 2D electron gases are observed at the interface between a semiconductor and a metal. One of the easiest model used to describe the main properties of the 2D electron gas is to consider non-interacting free electrons. In this model, we consider that the Coulomb repulsion between electrons is zero and that they do not feel any forces inside the box. This problem is concerned with the proper derivation of the eigenfunctions and the energy eigenvalues of a free electron in a 2D geometry. 2.1 General solution for the 3D case Let’s consider a box of dimensions Lx and Ly in respectively the x and y direc-tions, and w in the z direction. An electron is enclosed in this box and cannot escape from it, but the electron is free to move inside it. This means that the potential energy of the electron is zero inside the box and infinite outside it. 1 Nicolas Poilvert & Nicola Marzari� � � � � Since the electron cannot escape, then the electron’s wavefunction has to vanish on the sides of the box. So we will write that: ψ(0, y, z) = ψ(Lx, y, z) = 0 for the x direction ψ(x, 0, z) = ψ(x, Ly, z) = 0 for the y direction ψ(x, y, 0) = ψ(x, y, w) = 0 for the z direction Those constitute the boundary conditions for our problem. 1) Write down the Schrodinger equation for an electron moving inside the box. Now we will make the ansatz that you have seen in class and use the method of separation of variables to solve the Schrodinger equation. The wavefunction can be written as: ψ(x, y, z) = X(x)Y (y)Z(z). 2) Translate the boundary conditions for ψ in the x, y and z directions in boundary conditions for respectively the X, Y and Z fonctions. 3) By using the ansatz for the wavefunction ψ, write the left hand side of the Schrodinger equation as a sum of 3 terms that are respectively fonctions of x, y and z only. The right hand side is reduced to a c onstant E. 4) The only mathematical solution for the Schrodinger equation expressed in question 3) is that each term of the equation is equal to a constant and that those constants add up to E. Let’s call Ex, Ey and Ez those constants. Write down the 3 separate equations for the X, Y and Z functions. The general solution for those three equations should be: 2mEx • X(x) = Ax sin(kxx) + Bx cos(kxx) with kx = h¯2 2mEy • Y (y) = Ay sin(ky y) + By cos(kyy) with ky = h¯2 Z(z) = Az sin(kz z) + Bz cos(kz z) with kz = 2mEz • h¯2 5) Use the boundary conditions found in question 2) to find the allowed values for Ex, Ey and Ez in terms of integers nx, ny and nz. Find which constants in the set (Ax, Ay, Az , Bx, By, Bz) are zero and simplify the general solutions for X, Y and Z. Finally write down the allowed total energies (the eigenen-ergies) E(nx, ny, nz ) and the associated allowed wavefunctions (the eigenfunc-tions) ψnx,ny ,nz (x, y, z). At this stage we see that we still have some unknowns in the problem, namely the normalization constants in front of the eigenfunctions. Let’s denote by Anx,ny ,nz the constants in front of ψnx,ny ,nz (x, y, z). We know from physical intuition that the probability of finding the electron inside the box is 1. So for any eigenfunction, we must have: � Lx � Ly � w ψ∗ (x, y, z)ψnx,ny ,nz (x, y, z)dxdydz = 1 0 0 0 nx,ny ,nz This is the normalization condition for the eigenfunctions. 6) Using the analytical expressions for the eigenfunctions obtained in ques-tion 5), prove that: Anx,ny ,nz = LxL8 y w = V 8 , where V is the volume of the box 22.2 When can we talk about a 2D gas? Now that we have solved our problem mathematically, let’s have a closer look at the energy eigenvalues: E(nx, ny, nz ) = h¯2π2 ( n x 2 + n y 2 + n z 22 ) = h2 ( n x 2 + n y 2 + n z 22 )2m L2 L2 w 8m L2 L2 wx y x y 1) What is the energy difference ΔEx between state E(nx +1, 0, 0) and state E(nx, 0, 0)? In the same way, calculate ΔEy = E(0, ny + 1, 0) − E(0, ny, 0) and ΔEz = E(0, 0, nz + 1) − E(0, 0, nz ). 2) Calculate the ratios ΔEz and ΔEz .ΔEx ΔEy 3) Now imagine that the box is very elongated in the x and y directions, i.e Lx and Ly are macroscopic, and very thin in the z direction. To make things clearer, let’s take w ∼ 10nm and Lx = Ly ∼ 1µm, which are typical dimensions 2nz +1 2nz +1in semiconductor/metal junctions. In real devices the ratios 2nx+1 and 2ny +1 are of order unity. Knowing this, give a typical value for the ratios calculated in question 2). 4) From the numerical result of question 3), what can you deduce in terms of level spacings in x and y directions compared to the level spacing in the z direction? In which direction(s) of space can you consider that the allowed energy states form a quasi-continuum and in which direction(s) the allowed energy levels are quite well separated? When talking about real devices (that are always 3-dimensional), we can speak of a 2D gas of electrons when the motion of each electron is ”frozen” in one direction (let’s call it the z direction) and not in the two others (x and y). By ”frozen”, we mean that the energy difference between the energy states caracterized by nz = 1 and nz = 2 is bigger than any typical thermal energy kB T . 5) Calculate the energy difference ΔEz = E(0, 0, 2) − E(0, 0, 1) in a GaAs device for which the width w is 10nm and the mass of the electron is m∗ = 0.067m, …