Lecture 12: Particle in 1D boxes, Simple Harmonic Oscillators

Lecture 1: Particle in 1D boes, Simple Harmonic Oscillators U U() ψ() U n= n=0 n=1 n=3 Lecture 1, p 1

This week and last week are critical for the course: Week 3, Lectures 7-9: Week 4, Lectures 10-1: Light as Particles Schrödinger quation Particles as waves Particles in infinite wells, finite wells Probability Uncertainty Principle Midterm am Monday, week 5 It will cover lectures 1-11 and some aspects of lecture 1 (not SHOs). Practice eams: Old eams are linked from the course web page. Review Sunday before Midterm Office hours: Sunday and Monday Net week: Homework 4 covers material in lecture 10 due on Thur. after midterm. We strongly encourage you to look at the homework before the midterm! Discussion: Covers material in lectures 10-1. There will be a quiz. Lab: Go to 57 Loomis (a computer room). You can save a lot of time by reading the lab ahead of time It s a tutorial on how to draw wave functions. Lecture 1, p

Properties of Bound States Several trends ehibited by the particle-in-bo states are generic to bound state wave functions in any 1D potential (even complicated ones). 1: The overall curvature of the wave function increases with increasing kinetic energy. ħ d ψ ( ) p = for a sine wave m d m ψ() : The lowest energy bound state always has finite kinetic energy -- called zero-point energy. ven the lowest energy bound state requires some wave function curvature (kinetic energy) to satisfy boundary conditions. n= n=1 n=3 0 L 3: The n th wave function (eigenstate) has (n-1) zero-crossings. Larger n means larger (and p), which means more wiggles. 4: If the potential U() has a center of symmetry (such as the center of the well above), the eigenstates will be, alternately, even and odd functions about that center of symmetry. Lecture 1, p 3

Act 1 The wave function below describes a quantum particle in a range : 1. In what energy level is the particle? n = (a) 7 (b) 8 (c) 9 ψ(). What is the approimate shape of the potential U() in which this particle is confined? (a) U() (b) U() (c) U() Lecture 1, p 4

Solution The wave function below describes a quantum particle in a range : 1. In what energy level is the particle? n = (a) 7 (b) 8 (c) 9 ψ() ight nodes. Don t count the boundary conditions.. What is the approimate shape of the potential U() in which this particle is confined? (a) U() (b) U() (c) U() Lecture 1, p 5

Bound State Properties: ample Let s reinforce your intuition about the properties of bound state wave functions with this eample: Through nano-engineering, one can create a step in the potential seen by an electron trapped in a 1D structure, as shown below. You d like to estimate the wave function for an electron in the 5th energy level of this potential, without solving the SQ. Qualitatively sketch the 5th wave function: ψ 0 L 5 U o Consider these features of ψ: 1: 5th wave function has zero-crossings. : Wave function must go to zero at and. 3: Kinetic energy is on right side of well, so the curvature of ψ is there. Lecture 1, p 7

Bound State Properties: Solution Let s reinforce your intuition about the properties of bound state wave functions with this eample: Through nano-engineering, one can create a step in the potential seen by an electron trapped in a 1D structure, as shown below. You d like to estimate the wave function for an electron in the 5th energy level of this potential, without solving the SQ. Qualitatively sketch the 5th wave function: ψ 0 L 5 U o Consider these features of ψ: 1: 5th wave function has 4 zero-crossings. : Wave function must go to zero at = 0 and. = L 3: Kinetic energy is lower on right side of well, so the curvature of ψ is smaller there. The wavelength is longer. ψ and dψ/d must be continuous here. Lecture 1, p 8

ample of a microscopic potential well -- a semiconductor quantum well Deposit different layers of atoms on a substrate crystal: AlGaAs GaAs AlGaAs U() Quantum wells like these are used for light emitting diodes and laser diodes, such as the ones in your CD player. The quantum-well laser was invented by Charles Henry, PhD UIUC 65. This and the visible LD were developed at UIUC by Nick Holonyak. An electron has lower energy in GaAs than in AlGaAs. It may be trapped in the well but it leaks into the surrounding region to some etent Lecture 1, p 9

Act 1. An electron is in a quantum dot. If we decrease the size of the dot, the ground state energy of the electron will n n=3 a) decrease b) increase c) stay the same n= n=1 0 L. If we decrease the size of the dot, the difference between two energy levels (e.g., between n = 7 and ) will a) decrease b) increase c) stay the same Lecture 1, p 10

Solution 1. An electron is in a quantum dot. If we decrease the size of the dot, the ground state energy of the electron will n n=3 a) decrease b) increase c) stay the same = h 8mL 1 The uncertainty principle, once again! n= n=1 0 L. If we decrease the size of the dot, the difference between two energy levels (e.g., between n = 7 and ) will a) decrease b) increase c) stay the same Lecture 1, p 11

Quantum Confinement size of material affects intrinsic properties M. Nayfeh (UIUC) : Discrete uniform Si nanoparticles Transition from bulk to molecule-like in Si A family of magic sizes of hydrogenated Si nanoparticles No magic sizes > 0 atoms for non-hydrogenated clusters Small clusters glow: color depends on size Used to create Si nanoparticle 1 nm Blue 1.67 nm Green.15 nm Yellow.9 nm Red microscopic laser:

Particle in Infinite Square Well Potential π nπ ψ n( ) sin ( kn) = sin = sin for 0 L λn L ψ() n= n=1 n=3 nλ = n L 0 L ħ d ψn( ) + U( ) ψ ( ) = ψ ( ) n n n m d The discrete n are known as energy eigenvalues : electron U = n U = n p h 1.505 ev nm = = = m mλ λ h n = n where 8mL n 1 1 n n=3 n= n=1 0 L Lecture 11, p 14

Quantum Wire ample An electron is trapped in a quantum wire that is L = 4 nm long. Assume that the potential seen by the electron is approimately that of an infinite square well. 1: Calculate the ground (lowest) state energy of the electron. : What photon energy is required to ecite the trapped electron to the net available energy level (i.e., n = )? n n=3 n= n=1 The idea here is that the photon is absorbed by the electron, which gains all of the photon s energy (similar to the photoelectric effect). 0 L Lecture 1, p 15

Solution An electron is trapped in a quantum wire that is L = 4 nm long. Assume that the potential seen by the electron is approimately that of an infinite square well. 1: Calculate the ground (lowest) state energy of the electron. h 1.505 ev nm n = n = = 8mL 4L 1 with 1 1.505 ev nm = = 0.035 ev 4(4 nm) 1 Using: h = = mλ where λ =L. 1.505 ev nm λ : What photon energy is required to ecite the trapped electron to the net available energy level (i.e., n = )? n n=3 n= n=1 0 L n = n 1 So, the energy difference between the n = and n = 1 levels is: = ( - 1 ) 1 = 3 1 = 0.071 ev Lecture 1, p 16

Probabilities Often what we measure in an eperiment is the probability density, ψ(). ( ) sin nπ Wavefunction = ψn = N L ψ Probability amplitude ( ) sin nπ ψn = N L ψ n=1 Probability per unit length (in 1-dimension) 0 L 0 L ψ ψ Probability density = 0 node 0 L n= 0 L ψ ψ 0 L n=3 0 L Lecture 1, p 17

Probability ample Consider an electron trapped in a 1D well with L = 5 nm. Suppose the electron is in the following state: ψ 0 5 nm a) What is the energy of the electron in this state (in ev)? N ( ) sin nπ ψn = N L b) What is the value of the normalization factor squared N? c) stimate the probability of finding the electron within ±0.1 nm of the center of the well? (No integral required. Do it graphically.) Lecture 1, p 18

Solution Consider an electron trapped in a 1D well with L = 5 nm. Suppose the electron is in the following state: ψ N 0 5 nm a) What is the energy of the electron in this state (in ev)? 1.505 ev nm n = 3, n = n 1 = 3 = 0.135 ev 4(5 nm) b) What is the value of the normalization factor squared N? L P = 1= N N = = 0.4 nm tot -1 L ( ) sin nπ ψn = N L c) stimate the probability of finding the electron within ±0.1 nm of the center of the well? (No integral required. Do it graphically.) Probability = ( ) ψ (0. nm) N = 0.08 [(sin(3π/l)) 1 for L/] middle This works because the entire interval is very close to the middle peak. Lecture 1, p 19

Probability ample Consider a particle in the n = state of a bo. a) Where is it most likely to be found? b) Where is it least likely to be found? c) What is the ratio of probabilities for the particle to be near = L/3 and = L/4? ψ() n= 0 L/4 L/3 L Lecture 1, p 0

Solution Consider a particle in the n = state of a bo. a) Where is it most likely to be found? b) Where is it least likely to be found? c) What is the ratio of probabilities for the particle to be near = L/3 and = L/4? ψ() n= Solution: a) = L/4 and = 3L/4. Maimum probability is at ma ψ. 0 L/4 L/3 L b) = 0, = L/, and = L. Minimum probability is at the nodes. The sine wave must have nodes at = 0, = L, and, because n =, at = L/ as well. c) ψ() = Nsin(π/L) Prob(L/3) / Prob(L/4) Prob() = ψ(l/3) / ψ(l/4) = sin (π/3) / sin (π/) = 0.866 = 0.75 0 L Lecture 1, p 1

Harmonic Oscillator Potential Another very important potential is the harmonic oscillator: U U() U U() = ½ κ ω = (κ/m) 1/ Why is this potential so important? It accurately describes the potential for many systems..g., sound waves. It approimates the potential in almost every system for small departures from equilibrium..g., chemical bonds. To a good approimation, everything is a harmonic oscillator. U (ev) 10 5 0-5 r o Chemical bonding potential 0 0.5 1 1.5 r (nm) Taylor epansion of U near minimum. Lecture 1, p

Harmonic Oscillator () The differential equation that describes the HO is too difficult for us to solve here. Here are the important features of the solution. The most important feature is that the energy levels are equally spaced: n = (n+1/)ħω. The ground state (n = 0) does not have = 0. Another eample of the uncertainty principle. ω is the classical oscillation frequency nergy 7 n=3 ħω 5 n= ħω 3 n=1 ħω n=0 1 ħω Beware!! The numbering convention is not the same as for the square well. Spacing between vibrational levels of molecules in atmospheric CO and H O are in the infrared frequency range. = ħω = hf ~ 0.01 ev Molecular vibration... r This is why they are important greenhouse gases. Lecture 1, p 3

Harmonic Oscillator Wave Functions To obtain the eact eigenstates and associated allowed energies for a particle in the HO potential, we would need to solve this SQ: U U() U ħ d ψ ( ) 1 + κ ψ ( ) = ψ ( ) m d This is solvable, but not here, not now However, we can get a good idea of what ψ n () looks like by applying our general rules. The important features of the HO potential are: It s symmetrical about = 0. It does not have a hard wall (doesn t go to at finite ). Lecture 1, p 4

HO Wave Functions () Consider the state with energy. There are two forbidden regions and one allowed region. Applying our general rules, we can then say: ψ() curves toward zero in region II and away from zero in regions I and III. ψ() is either an even or odd function of. U U() U I II III Let s consider the ground state: ψ() has no nodes. ψ() is an even function of. U U() U This wave function resembles the square well ground state. The eact functional form is different a Gaussian but we won t need to know it in this course: / a ħ ψ n= 0( ) e a = mκ Lecture 1, p 5

HO Wave Functions (3) For the ecited states, use these rules: ach successive ecited state has one more node. The wave functions alternate symmetry. Unlike the square well, the allowed region gets wider as the energy increases, so the higher energy wave functions oscillate over a larger range. (but that s a detail ) U U() ψ() n= n=0 n=1 n=3 U Lecture 1, p 6

Harmonic Oscillator ercise A particular laser emits at a wavelength λ =.7 µm. It operates by eciting hydrogen fluoride (HF) molecules between their ground and 1 st ecited vibrational levels. stimate the ground state energy of the HF molecular vibrations. Lecture 1, p 7

Recall: Solution A particular laser emits at a wavelength λ =.7 µm. It operates by eciting hydrogen fluoride (HF) molecules between their ground and 1 st ecited vibrational levels. stimate the ground state energy of the HF molecular vibrations. photon = hc/λ = (140 ev-nm)/.7mm = 0.46 ev and: (by energy conservation) photon = = 1-0 = ħω = 0 ħω ½ħω n=1 n=0 Therefore, 0 = ½ħω = 0.3 ev Lecture 1, p 8

Leaky Particles Due to barrier penetration, the electron density of a metal actually etends outside the surface of the metal! Work function Φ Occupied levels V o F = 0 Assume that the work function (i.e., the energy difference between the most energetic conduction electrons and the potential barrier at the surface) of a certain metal is Φ = 5 ev. stimate the distance outside the surface of the metal at which the electron probability density drops to 1/1000 of that just inside the metal. Lecture 1, p 9