Today: Particle-in-a-Box wavefunctions

Transcription

1 Today: Particle-in-a-Bo wavefunctions 1.Potential wells.solving the S.E. 3.Understanding the results. HWK11 due Wed. 5PM. Reading for Fri.: TZ&D Chap. 7.1

2 PE for electrons with most PE. On top work function of copper 4.7 ev as more electrons fill in, potential energy for later ones gets flatter and flatter. For top ones, is VERY flat. How could you find out how deep the pit is for the top electrons in copper wire? This is just the energy needed to remove them from the metal. That is the work function!!

3 m ψ ( ) + V ( ) ψ ( ) Eψ ( ) Energy 4.7 ev <, V() 4.7 ev >, V() 4.7 ev <<, V() ev How to solve? 1. Plug and Chug approach: find Ψ in each region, make solutions match at boundaries, normalize. Works, and we ll do some of this but bunch of math.

4 4.7 ev ev -ais V() 4.7 ev for < and > V() ev for ><. Clever physicist approach. Reasoning to simplify how to solve. Electron KE not much more than ~kt.5 ev. Where is electron likely to be? What is chance it will be outside of well? a. zero chance b. very small chance c. small d. likely b..5 ev<< 4.7eV. So very small chance (e -4.7/. ) an electron could have enough energy get out. What does that say about boundary condition on ψ()? a. ψ() must be same <, <<, >, b. ψ(<)~, ψ(>) c. ψ() ~ ecept for << ans c.

5 Energy <, V() ~ infinite >, V() ~ infinite <<, V() so clever physicist just has to solve m ψ ( ) Eψ ( ) with boundary conditions, ψ()ψ() solution a lot like microwave & guitar string NOTE: Book uses infinite square well

6 ev m ψ ( ) Eψ ( ) functional form of solution: ψ ( ) Acos( k) + Bsin( k) Note: this looks the same as free electron until... Apply boundary conditions? ψ () A then A : ψ ( ) Bsin( k)? knπ (n1,,3,4 ) k knπ/ π λ λ n 1

7 λ What is E? a. can be any value (not quantized). π n π b. c. nm m nπ d. + PE e. n π m m n E quantized by B. C. s E k n π m m Does this dependence make sense? What value of when E -E 1 kt? ( when motion of e s depends on wire size) you should check, I estimate > ~1 atoms

8 Results: Quantized: knπ/ Quantized: π E n m Real(Ψ) n1 n E 1 For n4 KE What is potential energy of electron in the lowest energy state (n1)? a. E 1 b. c. d. could be anything Correct answer is b! V between and (We set it!) So electron has KE E1.

9 Normalize wavefunction / ) sin( ) ( ) ( ), ( iet e n B t t Ψ π φ ψ 1 ) / ( sin ), ( * Ψ Ψ Ψ d n B d d t π B (See Homework 11) Probability of finding electron between - and must be 1. / ) sin( ) ( ) ( ), ( iet e n t t Ψ π φ ψ Solving completely- everything there is to know about electron in small metallic object (flat V() with high walls).

10 Ψ(, t) nπ sin( ) e Quantized: knπ/ Quantized: π m E n iet / n E 1 How does probability of finding electron close to / if in n 3 ecited state compared to probability for when n ecited state? a. much more likely for n3. Correct answer is a! b. equal prob. for both n and 3. For n, Ψ c. much more likely for n For n3, Ψ at peak

11 Careful about plotting representations. Energy V() Total energy E (n3) E (n) ψ() V ev Careful plotting 3 things on same graph: E (n1) Potential Energy V() Total Energy E Wave Function ψ()

12 Ψ(, t) nπ sin( ) e iet / ψ n Quantized: knπ/ π m E n Quantized: 1 n E What you epect classically: What you get quantum mech.: Electron can have any energy owest energy in the bo has zero KE. Particle at rest. Electron can only have specific energies. (quantized) owest energy in the bo still has KE! ZERO POINT MOTION Additional new physics appears when the bo sides are not infinitely high (all real cases!).

13 ecture ended here

14 4.7eV Energy V() wire Need to solve Schrodinger Eqn: d m E electron ψ ( ) + V ( ) ψ ( ) Eψ ( ) d Region I Region II Region III In Region III total energy E < potential energy V d ψ ( ) d m ( V E) ψ ( ) Positive α ψ ( ) α is real What functional forms of ψ() work? a. e iα b. sin(α) c. e α d. more than one of these

15 4.7eV V() wire Energy E electron Region I Region II Region III ψ III ( ) Ae α + Be α In Region III total energy E < potential energy V d ψ ( ) d m ( V E) ψ ( ) Positive Answer is C: e α could also be e -α. Eponential decay or growth α ψ ( ) α is real Why not e iα? HS RHS α ψ ( ) α ψ ( )

16 Back to case of wire with workfunction of 4.7 ev Energy 4.7 ev V E particle d ψ ( ) d ev Positive number m ( E V ) ψ ( ) α ψ ( ) Answer is c could also be e -α. Eponential decay or growth ψ ( ) Ae α + Be α ψ ( ) ψ ( ) > < d ψ > d d ψ < d (curves upward) (curves downward) ψ

17 V() Good Approimation: Electrons never got out of wire ψ(< or >). (OK when Energy << work function) Energy What happens if electron Energy bigger? Eact Potential Energy curve (V): small chance electrons get out of wire ψ(< or >)~, but not eactly! What if two wires very close to each other? E_total Then whether ψ leaks out a little or not, is very important. How much coupling to other wire?

18 Need to solve for eact Potential Energy curve: V(): small chance electrons get out of wire ψ(< or >)~, but not eactly! wire Energy V() Finite Square Well Work function Important for thinking about Quantum tunneling : Radioactive decay Scanning tunneling microscope to study surfaces

19 4.7eV Energy V() wire Need to solve Schrodinger Eqn: d m E electron ψ ( ) + V ( ) ψ ( ) Eψ ( ) d Region I Region II Region III In Region II total energy E > potential energy V d ψ ( ) d m ( V E) ψ ( ) k ψ ( ) Negative number When E>V: Solutions sin(k), cos(k), e ik. Always epect sinusoidal functions k is real

20 4.7eV V() wire Energy E electron ψ ( ) I Region I Region II Region III Ee α + Fe α ψ ( ) C sin( k) D cos( k) II + ψ III ( ) Ae α + Be What will wave function in Region III look like? What makes sense for constants A and B? a. A must be b. B must be c. A and B must be equal d. A and B e. A and B can be anything, need more info. α

21 What will wave function in Region III look like? What makes sense for constants A and B? Answer is a. A must be.. otherwise ψ blows up as gets bigger. This doesn t make sense! ψ and probability should go to at large! Need to be able to normalize ψ 4.7eV V() wire Energy E electron Region I Region II Region III ψ I ( ) Ee α + Fe α ψ ( ) C sin( k) D cos( k) II + ψ III ( ) Ae α + Be α