1Lecture 42 (Monday)Reminders:Reading Chapter 8 (higher dimensions)Homework #14 due this Wednesday Reminder about filling out the online surveyCULearn – will be posted by the end of today. Please check all your scores. Any corrections need to be to me by this Friday!Today’s topic:Schrodinger Equation in3-dimensions and theHydrogen AtomOnline end-of-term survey:http://tinyurl.com/2170FA09POSTPlease fill out by Wednesday (extended). Worth an extra homework problem. Link on class web page.Exam #3 ResultsMean = 73 of out 100Stand. Dev = 1787-100 A Range70-86 B Range55-69 C Range<55 FailingGood to review this material for the final exam as well.Solutions will be posted on the class web page today.Schedule for this week….Problem Solving Session Monday 2:30 – 4:00 pmProblem Solving Session Tuesday 2:30 – 4:00 pmWednesday review in class and details about the final exam. * Note that the final exam is comprehensive (i.e. it includes everything).I will schedule some additional help sessions for Thursday and Friday (to be announced on Wednesday).The Θ(θ)Part222222222sin)sin(1)()(sin))()((sin21mrRrRrrVErme−=⎟⎠⎞⎜⎝⎛∂Θ∂∂∂−∂∂−−−=∂Φ∂ΦθθθθθθθφhNow can we separate the r and θ dependencies?2222222sin)sin(1)()(sin))()((sin2mrRrRrrVErme−=⎟⎠⎞⎜⎝⎛∂Θ∂∂∂−∂∂−−−θθθθθθθhWhat is the next step?A) Multiply everything by 1/r2B) Multiply everything by 1/sin2θC) Multiple by sinθ/θD) One needs to know V(r) first.E) Give up all hope.2222222sin)sin(1)()(sin))()((sin2mrRrRrrVErme−=⎟⎠⎞⎜⎝⎛∂Θ∂∂∂−∂∂−−−θθθθθθθh222222sin1sinsin1)())((2mrRrRrrVErmeθθθθθθ−=⎟⎠⎞⎜⎝⎛∂Θ∂∂∂−∂∂−−−h⎟⎠⎞⎜⎝⎛∂Θ∂∂∂−+=∂∂+−+θθθθθθsinsin1sin1)())((2222222mrRrRrrVErmehAll the r dependence on the leftAll the θ dependence on the rightκθθθθθθκ+=⎟⎠⎞⎜⎝⎛∂Θ∂∂∂−++=∂∂+−+sinsin1sin1)())((2222222mrRrRrrVErmehSome constant2κθθθθθθ+=⎟⎠⎞⎜⎝⎛∂Θ∂∂∂−+ sinsin1sin122m0sinsinsin122=⎥⎦⎤⎢⎣⎡−+⎟⎠⎞⎜⎝⎛∂Θ∂∂∂θθκθθθθmNot so easy to solve. However, it turns out that thereare solutions whereAnd l is an integer: ||)1(m≥+=lllκThe solutions are Legendre functions θl,mThe solutions are Legendre functions θl,mThese are then combined with the φ dependence into spherical harmonics Yl,mso (except for m=ℓ=0)and there is no integer between ℓ and ℓ+1Clicker QuestionTotal angular momentum is hll )1( +=Lℓ can be 0, 1, 2, 3, …The z-component of the angular momentum is hmLz=where m can be 0, ±1, ±2, … ±ℓJust like with any vector, the total angular momentum can be written222zyxLLLL ++=Q. Given the rules above, can Lx=Ly=0? That is, can L=Lz?A. Yes, in more than one caseB. Yes, but only in one caseC. NeverIt is possible for Lx=Ly=Lz=0 in which case L = 0 so ℓ=0 and m=0In general, if Lx=Ly=0 then simplifies to222zyxLLLL ++=zLL=which means or .hllh )1( +=m)1(2+= llm22)1()1( +<+< llll)1(2+≠llmBut Spherical Harmonics)()()( ),,(φθφθψΦΘ=rRrWe have determined the angular part of the wave function soφθφθψimmerRr )()( ),,(lΘ=has becomewith the quantum numbers ℓ and m specifying the total angular momentum and the z-component of angular momentum.This angular solution works for any central force problem.The combination are the spherical harmonics φθimme)(lΘ()φθ,mYlSpherical harmonics are 3-D and complex (real and imaginary terms), making it very difficult to display.A Few of the Spherical Harmonics100∝Y)cos(01θ∝YφθieY±±∝ )sin(11φθθieY±±∝ )cos()sin(12φθieY2222)(sin±±∝1)(cos3202−∝θY3=m1=m0=m 1=m 2=m3=m2=m0=l1=l2=l3=lhttp://oak.ucc.nau.edu/jws8/dpgraph/Yellm.htmlReal part onlyColors give phaseRed = +1Cyan = -1Purple = +iGreen = -iClicker QuestionφθφθψimmerRr )()( ),,(lΘ=For any central force potential we can write the wave function as ),()( ),,(φθφθψmYrRrl=Q. What are the boundary conditions on the radial part R(r)?A. R(r) must go to zero as r goes to 0B. R(r) must go to zero as r goes to infinityC. R(∞) must equal R(0)D. R(r) must equal R(r+2π).E. More than one of the above.In order for ψ(r,θ,φ) to be normalizable, it must go to zero as r goes to infinity. Therefore, R(r)→0 as r→∞.Physically makes sense as well. Probability of finding the electron very far away from the proton is very small.3The Radial Component of ψφθφθψimmerRr )()( ),,(lΘ=For any central force potential we can write the wave function as ),()( ),,(φθφθψmYrRrl=To solve this equation we need to know the potential V(r).For the hydrogen atomrkerV2)( −=The radial part of the time independent Schrödinger equation can be written as)()(2)1()()(22222rRErRrmrVdrrRdmee=⎥⎦⎤⎢⎣⎡+++−llhThis is how we are going to get the energy Eand the r dependence of the wave functionNote that m does not appear. This makes sense because it just contains information on the direction of the angular momentum.The total angular momentum isrelevant so ℓ shows up.There are a special set of functions that solve this:⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛=−BnBnarnarGaZrerRBlll ,/)(Special Polynomial FunctionsAnd for integer values n=1,2,3,4,…. > lOne finds the energies that are allowed are:224212 nekmEeh−=Bohr’s Energy Levels!The Three (3) Quantum NumbersApplying boundary conditions to the radial equation gives us yetanother quantum number which we have already used: nIn order to work, n must be an integer which is > ℓPutting it all together, our wave function isφθφθψimmnerRr )()( ),,(llΘ=),()( ),,(φθφθψmnYrRrll=orThe quantum numbers are:n = 1, 2, 3, … is the principal quantum numberℓ = 0, 1, 2, … n-1 is the angular momentum quantum numberm = 0, ±1, ±2, … ±ℓ is thez-component angular momentum quantum numberThe Three (3) Quantum NumbersFor hydrogenic atoms (one electron), energy levels only depend on n and we find the same formula as Bohr:22/ nEZERn−=For multielectron atoms the energy also depends on ℓ.There is a shorthand for giving the n and ℓ values.n = 2 ℓ = 1p2Different letters correspond to different values of ℓs p d f g h …=l012 345Hydrogen Ground StateThe hydrogen ground state has a principal