Materials Genome Assessment

Transcription

1 Materials Genome Assessment Lecture 12 : Quantum Mechanics Theory: Differential equations with boundary conditions Schrodinger s cat Particle in a Box Programming: Modification of the Pendulum program Prof. Cedric Weber

2 Examples (just to give an idea..) * Energy materials : solar cells? * Thermo- electric materials? (fridge, cars, ) * Quantum dot? For quantum computing? * Graphene * Carbon nanotubes * Superconductors * Material for energy storage batteries? * Materials for conversion of methane liquid/gas (amonia) 2

3 What did you learn last time? 1. Using functions to solve differential equations ý 2. Reading an initial condition from the keyboard ý 3. Plotting the solution of a differential equation ý 4. Modifying a differential equation to introduce friction ý 3

4 4

5 Today s experiment! * [ do not repeat this experiment ] 5

6 Schrodinger s equation * Erwin Schrödinger, , born in Vienna * Developed the wave equation of quantum mechanics * Idea: every particle (electron, proton..) can be described by a function * This particle has a probability P(x) to be observed at a point x of space, this probability is given by the function: * Condition: the probability to observer Z the particle somewhere Z (anywhere) has to be P=1 : P (x)dx = (x) P (x) = (x) 2 (x) 2 dx =1 6

7 Differential equation * How can we know this function? To make the connection with the last lecture, we simply change the notation, let s call this function y(x) * Answer: the function is the solution of the Schrodinger equation (x)! y(x) d 2 y(x) dx 2 = 2m(V (x) E) ~ 2 y(x) * m : mass of the particle * E : energy of the particle * x : the coordinate, the position in space of the particle * V(x) : a potential which affects the particle and defines the problem to solve. If the particle is a rolling ball, the potential V(x) would be the potential energy ~ * : Plank constant

8 Particle in a box Let s be more specific: * Let s describe the problem of a particle resting in a box * The box has only one dimension (to simply our problem) * The box has infinitely hard walls on each side * The particle is not able to enter into the wall, so the probability to find the particle in x=a and x=b is zero: y(a)=y(b)=0 x Wall V (x) =0 Wall V (x) =1 V (x) =1 a =0 b =5 x = a x = b

9 Shake the box? * Shake the box : give an energy E to the particle * Where is the particle? Sitting at the center of the box? * To answer this question, we need to find y(x) in the interval a<x<b * In this interval V(x)=0 d 2 y(x) dx 2 = 2mE ~ y(x) * With boundary conditions: (the ball cannot enter the walls) y(a) =0 y(b) =0 9

10 Shake the box? What is the difference * Shake the box : give an energy E to the particle * Where is the particle? Sitting at the center of the box? with the pendulum? * To answer this question, we need to find y(x) in the interval a<x<b * In this interval V(x)=0 (last lecture) d 2 y(x) dx 2 = 2mE ~ y(x) * With boundary conditions: (the ball cannot enter the walls) y(a) =0 y(b) =0 10

11 Simplified equation * For simplicity, we set the mass of the particle such that : * This is the equation that we have to solve: 2m ~ =1 d 2 y(x) dx 2 = Ey(x) y(a) =y(b) =0 * Reminder: Pendulum was a similar problem, but with initial condition. * Here, we need to impose y(a) =y(b)=0. We have a free parameter E. * Idea: can I start from y(a)=0, and find E such that y(b)=0? y(a) =0&E y(b) =0 11

12 ² Strategy: aim and shoot Solving differential equation with boundary conditions ² Solve the differential equation with initial condition y(a)=0 y(x = a) =0 ² and with an energy E, see what you obtain for y(b) y(x = b) =0 ² Obtain y(b) as a function of E, when y(b)=0 we find the right Energy E. ² Find which energies E satisfy the condition y(b)=0 (finding zeros of a function, check where the plot is crossing the horizontal axis) 12

13 Reminder : coupled first order equations ² We discussed in the last lecture that this equation can be decomposed in two coupled first order equations: d 2 y(x) ² Simple idea: let s define a new variable: ² We get the equations : ² We are trying to obtain : y(b=5)=0 Initial conditions dx 2 = Ey(x) y(a) =0 y 0 (a) =1 dy dx = z dz dx =... z(0) = 1 y(0) = 0 ² Plot what you obtain for y(b=5) for energies E ranging from 0 to 20 13

14 Energy quantization d 2 y(x) dx 2 = Ey(x) y(0) = y(5) = 0 Each time y(5)=0, we solved our problem! Yay! 14

15 Energy quantization d 2 y(x) dx 2 = Ey(x) y(0) = y(5) = 0 Each time y(5)=0, we solved our problem! Yay! But Wait. The energy of the particle only can take SOME particular values right? Indeed, this is what we call the quantification of the energy (hence quantum physics) 15 The particle can only take some quantized energies

16 Quantum wave- function y(x) * For each obtained energy E, we can also obtain the solution to the differential equation y(x), 0<x<5 * What do we get? * For each allowed energy: n=1 n=2 * First energy (n=1) * Second energy (n=2) * Third energy (n=3) n=3 16

17 Where is my particle? * Probability P(x)=y(x) 2 * The particle with first energy has two maxima n=1 * The particle with second energy has three maxima * The particle with third energy has four maxima * When the particle has a high energy it is everywhere! n=3 n=2 17

18 18

19 Finding the energies 1. We use the code written for the pendulum (last lecture) 2. The differential equation needs to be solved for many different energies, we want to scan energies from 0 to 20 19

20 PROGRAM program quantumbox implicit none integer,parameter :: N= integer :: j,i, ie real(8) :: h,a,b,energy,x(0:n),y(0:n),z(0:n) a = 0.0 b = 5.0 h = (b - a) / dble(n) do ie=1,1000 energy = ie / 1000 * 20. x(0)=a; y(0)=0; z(0)=1. do j = 1, N y( j ) = y( j-1 ) + h * ( z( j-1 ) ) z( j ) = z( j-1 ) + h * ( -energy * y(j-1) ) x( j ) = j*h enddo write(100,*) energy, [FILL IN] end do end program 20 Solve the equ. diff. for every energy

21 Plotting the quantum functions * We modify bits in the previous code, now additionally we want to detect each time y(x) satisfies y(x=5)=0 * We add a counter counte, which is incremented each time we obtain : y(x=5)=0 * how can I detect if y(x=5) is equal to zero? 21

22 program quantumbox use library implicit none real(8) :: right_boundary( 0:1000 ) right_boundary=0.0 counte = 0 do ie = 1, 1000 energy = ie/1000*20.. counte counts how many times we obtain y(x=5)=0 as the energy is increased. It will hence count the energy levels for the wave- function x(0)=a; y(0)=0; z(0)=1. do j = 1, N y( j ) = y( j-1 ) + h * ( z( j-1 ) ) z( j ) = z( j-1 ) + h * ( -energy * y(j-1) ) x( j ) = j * h enddo y(n) is the last point obtained for the solution of the equation, so y(n) is y(x=5) right_boundary(i) = y(n) if( right_boundary(i)*right_boundary(i-1) < 0 )then [Fill in] Then the end if end do end program Here we want to check if y(x=5) is equal to zero. When it is, we want to write into a new file the array y which contains the solution y(x) for this given energy E 22

23 Problems Problem 1 : Particle in a box Problem 2 : Infinite box Problem 3 : Particle tunneling through a barrier 23

24 Particle in a box / Energy barrier Wall V (x) =0 V (x) =5 Wall V (x) =0 Wall V (x) =1 V (x) =1 a =0 b =5 x = a x = b x 24