I ntroduction to Nanoelectronics Nanoelectronics I ntroduction to - PowerPoint PPT Presentation

I ntroduction to Nanoelectronics Nanoelectronics I ntroduction to Prof. Supriyo Datta ECE 453 Purdue University 09.15.2004 Lecture 9: Schrödinger Equation Ref. Chapter 2.1 Net work f or Comput at ional Nanot echnology

00: 03 1D Schrödinger Equation, Dispersion Relation ∂ ∂ � 2 2 ? = − + • Substitution of this solution into the � i ? U ( x ) ? ∂ ∂ 2 equation leads to an E-k relationship. t 2 m x • (2) in (1) => • We’ve been talking about solutions to 1D − � 2 Schrödinger equation. A simple example is iE ( ) ⋅ Ψ = − Ψ + Ψ � 2 i ik U the case of constant potential. For harder � 0 2 m cases, we we’ll learn a numerical method � 2 2 k = + that would help us to solve for the energy n E U Dispersion Relation n 0 levels of a material with arbitrary potential. 2 m First we’ll consider the 1D case and then we’ll get into 3D. E • The simplest case is: ∂ ∂ � 2 2 ? = − + � i ? U ? (1) ∂ ∂ 0 2 t 2 m x • The solutions to a differential equation U with constant coefficients (like above) can 0 k be solved by plane waves: ( ) − Ψ = / � ikx e iEt x , t Ae (2)

06: 19 Vibrating String • Schrödinger equation is a wave equation. • Again by substituting the solution in the A common example of a wave equation is wave equation we can get the dispersion the acoustic waves on a string. There the relation. (1) in (2) => ( ) ( ) quantity that is used to describe the wave − ω = 2 2 2 i u v ik is the displacement of each point on a ω = 2 2 2 string from an equilibrium point at a v k Dispersion Relation particular time. ω Harmonically oscillating string U ( x , t ) x = 0 x = L k • The equation describing such waves is: • You can see the analogy between the two ∂ ∂ 2 2 u u wave equations. = 2 v (1) ∂ ∂ • One point is in order and that is: there are 2 2 t u times where people start from a dispersion • Solutions can be written as: relation and deduce form it a differential − ω ∝ ikx e i t U ( x , t ) e (2) equation; reveres of what we’ve done here.

12: 25 Waves in a Box - Allowed Values of k • This way we can have solutions like: • The next thing to understand is solving for the energy levels having more 8 8 Infinite complicated potential functions. The first Square example that was presented was that of a Well particle in a box. 8 E 3 8 U(x) E 2 x = 0 x = L E 1 U = 0 • This allows only certain values of k which x = 0 x = L leads to discreteness in the energy levels shown in the left figure. The reason is that • Assuming that the potential is very high the wavelength must be such that the wave at the two ends, one can show that the could fit inside the box. (k is related to the wave function has to go to 0 at the two = π λ k 2 Ψ = = Ψ = = ends: wavelength by ) ( x 0 ) 0 & ( x L ) 0

15: 58 Discrete Energy Levels • The corresponding energy also • What we saw last day was that for the wave becomes quantized: that goes to 0 at the two ends we cannot use: ( ) π � 2 2 Ψ = − � ikx e iEt / x , t Ae = + 2 E n U n 0 2 2 m L because this is never 0. Notice that there are two • This shows that whenever an values of k for a given energy. We can use a electron is confined in a certain superposition of solutions with k and –k to write region, the allowed energy values the proper solution. become quantized. E k + − = • We have: ikx ikx e e 2 cos kx − − = ikx ikx e e 2 i sin kx • Since in this problem the wave has to go to 0 at x=0, we choose the “sinkx”. Further, the = π boundary condition at x=L requires: k n L n

18: 50 Particle in a Finite Box • Suppose we have potential box like: • In region b we can write the solution as: − Ψ = + ik x ik x Ae Be 1 1 = = 1 U ( x ) U U ( x ) U • This is because we get two possible 2 2 values of k for a given U from equation (1) (a) (c) (b) so we can use a linear combination as a general solution. = U ( x ) U • In regions b & c we can write the solution 1 − Ψ = + ik x ik x as: Ae Be x = 0 2 2 x = L 2 = − � k 2 m ( E U ) •If E<U2, then 2 2 • We are trying to find a solution with a Gives an imaginary value for k2 and we get given E. A solution like a decaying exponential in region a & c: ( ) Ψ = − � ikx e iEt / x , t Ae De γ + Ce γ − Ψ = Ψ = x x a c will not satisfy the equation because we • For region (c), exp( ? x) is not allowed have two different values for the potential; because the wavefunction cannot go to hence two different values for k. To see this infinity for large x. consider: • For region (a), exp(- ? x) is not allowed � 2 2 k because the wavefunction cannot go to = + ⇒ = − � E U k 2 m ( E U ) (1) infinity for large negative x. 2 m

25: 34 Boundary Conditions • The solutions in the regions a, b and c • What we mean is that discontinuities… ψ ψ need to match at the boundaries x=o and ( x ) or d / dx Discontinuous x=L. Taking this requirement into account puts a restriction on the allowed values of constants A,B,C and D. We can use this (x) restrictions to find the constants. These restrictions are called Boundary Conditions are not allowed because then the (BC’s). Schrödinger equation will not be satisfied at the point of discontinuity. Same is true for + ik x Ae 1 the derivative of the wavefunction. The De γ + Ce γ − x x reason is that in each case there will be an − ik x Be 1 unmatched delta function (infinite height) in the Schrödinger equation where all other (b) (a) (c) entities are finite: ∂ ∂ � 2 2 ? x = 0 x = L = − + � i ? U ( x ) ? ∂ ∂ 2 t 2 m x • The first BC is that the wavefunction has to be continuous across the boundaries. Note: if ? is discontinuous, then d ? / dx is a • The second BC is that the derivative of delta function. If d ? / dx is discontinuous, wavefunction has to be continuous across then d^2 ? /dx^2 is discontinuous. the boundaries.

27: 58 Utilizing BC’s to Determine the Constants + ik x • We will not get into the details of this Ae 1 Ce γ − De γ + problem. Instead we’ll use a more x x − ik x convenient method of solving and that is Be 1 the numerical method. (b) (a) (c) x = 0 x = L •We can utilize BC’s as follows: • Continuity of ? across x=0 gives us one equation that relates A, B and D: = + D A B • Continuity of ? across x=L gives us one equation that relates A, B and C. Similarly, continuity if d ? /dx gives us two equations at the two boundaries. One gets a set of equations that could be used to eliminate the unknown constants.

30: 40 ? and Electron Density n • In electronic devices we are interested in • Notice that for 1 electron inn the box, the two things the most: electron density shape of electron density n may indicate inside a device and current flow. that there is a fraction of electron at each • If an electron has a wavefunction , the point. This is not right. The correct view is = Ψ Ψ * n associated electron density is: that to look at these fractional values as • For example take the potential well: probabilistic values of finding the electron at each point. 8 8 Infinite • Notice for adding electrons in different Square states we have two ways to come up with Well n: ( ) ( ) * Ψ Ψ + Ψ Ψ Ψ + Ψ Ψ + Ψ * * or 1 1 2 2 1 2 1 2 • The correct answer is the first one. With electrons what we have to add are the x = 0 x = L electron densities not the wavefunctions. •The associated electron density would come from squaring the sin wave inside the box. (solid: ? & dashed :n)

39: 40 “n” , “I” and Equation of Continuity ( ) Ψ = − / � ikx e iEt x , t Ae (1) • For an electron that has the wavefunction: = Ψ * Ψ n (2) • Electron density is: • Current can be written as (the reason comes later):  ψ ψ  � * ( ) i d d   = − ψ − ψ * I q (3)     n = 2 m dx dx 2 A • (1) in (2) => ( ) � � ( ) i k = − − − = − 2 2 2 I q ikA ikA q A • (1) in (3) => 2 m m • Where did equation (3) come from? If one accepts (2), then (3) is the only consistent expression for current “I” considering the fact that ? has to satisfy the Schrödinger equation. Accepting (2) and (3) will satisfy the continuity equation and since the equation of continuity is general true argument (3) must be the right choice for “I”. ∂ − ∂ ( qn ) I = − Continuity Equation: ∂ ∂ t x This is a general argument and states that the electron density in a region cannot change over time unless the gradient of current is non-0 in the region. Moreover, the amount of current leaving the region has to be equal to the rate of change of electron density.