Electronic properties of Graphene and 2-D materials. Part 2

Transcription

1 Electronic roerties of Grahene and -D materials z Part y The laws of Quantum Mechanics continued Time indeendent Schroedinger equation Eamles Particle in a well Harmonic oscillator Quantum Tunneling The scanning tunneling microscoe STM 3 Dimensional otentials 3D bo The atom Angular momentum Sin Pauli eclusion rincile The eriodic table Electrons in solids Energy bands Insulators, metals, semiconductors roblems adated from hysics 14 UIUC

2 Quantum Mechanics: : Law 1 Law1. The state of a quantum mechanical system is comletely secified by a function Y(r, t). Dirac notation ( ) et *( ) bra Princile of suerosition. If Y 1 and Y are ossible states of a system then any linear suerosition. Y is also an allowed state. Inner roduct a a 1 1 *( r ) ( r) dr A wave function () corresonds to a state in the ˆ reresentation ( ) A wave function corresonds to a state in the reresentation

3 Quantum Mechanics: Law 量子力学 Law. To every observable in classical mechanics there corresonds an oerator which is used to obtain hysical information about the observable from the wave function Observable Classical Quantum oerator Canonical momentum Mechanical momentum i i( ˆ yˆ zˆ qa y z ) ( r, t) i Aˆ ( rˆ, t) Kinetic energy (KE) Total energy KE+U m U ( r ) m m m y z m ˆ H U ( rˆ)

4 Quantum Mechanics: Law 量子力学 Oerators acting on the state of the system roduce another allowed state. ˆ A ' Every oerator has set of states that are not changed by the action of the oerator, ecet for being multilied by a constant. These are the eigenstates and the numbers are the eigenvalues of the oerator. Aˆ a a a

5 Eamle: the momentum oerator Solve for the Eigenstates and eigenvalues of the momentum oerator ˆ i 1D : i 1. Which of these wavefunctions reresents an eigenstate with momentum? (a) (b) (c) (d) (a) C sin Ce i Ce C. What is the value of (a) (b) (c) (d)

6 Solution: the momentum oerator Solve for the Eigenstates and eigenvalues of the momentum oerator ˆ i 1D : i 1. Which of these wavefunctions reresents an eigenstate with momentum? (a) (b) (c) (d) (a) C sin Ce i Ce C PLANE WAVE i Ce. What is the value of (a) (b) (c) (d)

7 Solution: the momentum oerator Solve for the Eigenstates and eigenvalues of the momentum oerator ˆ i 1D : i 1. Which of these wavefunctions reresents an eigenstate with momentum? (a) (a) (b) C sin Ce i Ce C PLANE WAVE (c) Ce i ; where (d). What is the value of (a) (b) (c) (d)

8 The momentum oerator The Eigenstates of the momentum oerator are PLANE WAVES ˆ i Ce i ; h where de Broglie wavelength! The Eigenvalues of the momentum oerator are = all real numbers Plane waves have a well defined momentum with no uncertainty D=0 Plane waves are infinite in sace they have no have well defined osition D=00 consistent with the uncertainty rincile! D D ħ

9 Quantum Mechanics: Law 3 量子力学 Law 3. The only ossible result of the measurement of an observable A is one of the eigenvalues of the corresonding oerator. The eigenvalues of oerators corresonding to observables are real. (we measure only real numbers) Oerators with real eigenvalues are called hermitian. Eigenstates of hermitian oerators form a basis that sans the set of allowed states : Orthogonal a j a ( r) ( r) dr Comlete. This means than an arbitrary state can be eanded as a linear combination of the eigenstates. 1 * a j a j 0 c a c n n n n a n j j

10 Eigenstates of hermitian oerators form a basis that sans the set of allowed states. If is a continuous variable then : Orthogonal Comlete. This means than an arbitrary state can be eanded as a linear combination of the eigenstates. Quantum Mechanics: Law 3 量子力学 Law 3. The only ossible result of the measurement of an observable A is one of the eigenvalues of the corresonding oerator. The eigenvalues of oerators corresonding to observables are real. (we measure only real numbers) Oerators with real eigenvalues are hermitian. ' 0 ' ') ( ') ( r ) ( ) ( ' * ' d d r r a a ; ) ( ) ( ; ) ( ) ( dr r a r d r a r

11 Eamle: eigenstates of momentum oerator j j d r r a a j a j j a 0 1 ) ( ) ( * r are numbers c a c n n n n, number.remember real is a ; 1 ) ( where e i If the wavefunction of a system is This function can be eressed in momentum basis 1 ) ( 1 1 ) ( ) ( a i a i i e C d e e C d e d e d ) ( a Ce Gaussian n a Also a Gaussian!

12 Eamle: eigenstates of momentum oerator j j d r r a a j a j j a 0 1 ) ( ) ( * r, number.remember real is a ; 1 ) ( where e i If the wavefunction of a system is This function can be eressed in momentum basis 1 a e C ) ( a Ce Simly a Fourier transform! Gaussian n a Gaussian! 1 a

13 The energy oerator When the total energy is conserved, the system is best described by the eigenvalues of the total energy oerator, the Hamiltonian: The Eigenstates and eigenvalues of the total energy oerator, the Hamiltonian: ˆ E ˆ U ( ) m E E E ˆ ˆ i m U ( r ) E ( r ) E E ( r ) This is nown as the time-indeendent Schroedinger equation Short notation: SEQ

14 The Schrödinger Equation (SEQ) The wave function Y(r, t) for a state with a well defined total energy E is a solution of the Schrödinger Equation (SEQ) We focus on the time-indeendent SEQ in 1-dimension: (,y,z) (). in 1D: ˆ i ˆ m d ( ) d U ( ) ( ) E ( ) h Time does not aear in this equation. Therefore, (,y,z) is a standing wave, because the robability density, (), is not a function of time. We call (,y,z) a stationary state. Note that the KE of the article deends on the curvature (d /d ) of the wave function. This is sometimes useful when analyzing a roblem

15 Quantum Mechanics Only 3 eactly Solved Problems! Particle in a bo Harmonic oscillator Hydrogen atom r q r U ) ( L L U 0; 0 0 ) ( 1 ) ( m r U

16 Particle Wavefunctions: Eamles What do the solutions to the SEQ loo lie for general U()? Eamles of () for a article in a given otential U() (but with different E): () We call these wavefunctions eigenstates of the article. () These are energy eigenstates. () The corresonding robability distributions () of these states are: Key oint: Particle cannot be associated with a secific location.

17 Eercise: Particle Wavefunction The three wavefunctions below reresent states of a article in the same otential U(), and over the same range of : () () (a) (b) (c) () 1. Which of these wavefunctions reresents the article with the lowest inetic energy? (Hint: Thin curvature.). Which corresonds to the highest inetic energy?

18 Solution The three wavefunctions below reresent states of a article in the same otential U(), and over the same range of : () () (a) (b) (c) Highest KE Lowest KE () 1. Which of these wavefunctions reresents the article with the lowest inetic energy? (Hint: Thin curvature.) The curvature of the wavefunction reresents inetic energy: d ( ) m d m Since (b) clearly has the least curvature, that article has lowest KE.. Which corresonds to the highest inetic energy? (a) has highest curvature highest KE

19 Solutions to the time-indeendent SEQ d m ( ) d U ( ) ( ) E ( ) Notice that if U() = constant, this equation has the simle form: where C m ( U E) d ( ) d C ( ) is a constant that might be ositive or negative. For ositive C (i.e., U > E), what is the form of the solution? a) sin b) cos c) e a d) e -a For negative C (U < E) what is the form of the solution?. a) sin b) cos c) e a d) e -a

20 Solution m d ( ) d U ( ) ( ) E ( ) Notice that if U() = constant, this equation has the simle form: where C m ( U E) d ( ) d C ( ) is a constant that might be ositive or negative. For ositive C (i.e., U > E), what is the form of the solution? a) sin b) cos c) e a d) e -a For negative C (U < E) what is the form of the solution?. a) sin b) cos c) e a d) e -a

21 Eamle: Particle in a Bo Consider a quantum article confined to a region, 0 < < L, by infinite otential walls. We call this a one-dimensional (1D) bo. U = 0 for 0 < < L U() U = everywhere else We already now the form of when U = 0. It s sin() or cos(). However, we can constrain more than this. 0 L The waves have eactly the same form as standing waves on a string, sound waves in a ie, etc. The wavelength is determined by the condition that it fits in the bo. On a string the wave is a dislacement y() and the square is the intensity, etc. The discrete set of allowed wavelengths results in a discrete set of tones that the string can roduce. In a quantum bo, the wave is the robability amlitude ψ() and ψ() is the robability of finding the electron near oint. The discrete set of allowed wavelengths results in a discrete set of allowed energies that the article can have.

22 Boundary conditions Standing waves A standing wave is the solution for a wave confined to a region Boundary condition: Constraints on a wave where the otential changes Dislacement = 0 for wave on string E = 0 at surface of a conductor E = 0 If both ends are constrained (e.g., for a cavity of length L), then only certain wavelengths are ossible: n f 1 L v/l L v/l 3 L/3 3v/L n = L n = 1,, 3 mode inde L 4 L/ v/l n L/n nv/l

23 Boundary conditions We can solve the SEQ wherever we now U(). However, in many roblems, including the 1D bo, U() has different functional forms in different regions. In our bo roblem, there are three regions: 1: < 0 : 0 < < L 3: > L () will have different functional forms in the different regions. We must mae sure that () satisfies the constraints (e.g., continuity) at the boundaries between these regions. The etra conditions that must satisfy are called boundary conditions. They aear in many roblems.

24 Particle in a Bo (1) Regions 1 and 3 are identical, so we really only need to deal with two distinct regions, (I) outside, and (II) inside the well Region I: When U =, what is ()? U() d ( ) d m ( E U ) ( ) I I II I I For U =, the SEQ can only be satisfied if: I () = 0 0 L U = 0 for 0 < < L U = everywhere else Otherwise, the energy would have to be infinite, to cancel U.

25 Region II: When U = 0, what is ()? d ( ) d d ( ) d Particle in a Bo () m ( E U ) ( ) m E ( ) U() II 0 L The general solution is a suerosition of sin and cos: ( ) B sin B cos where 1 Remember that and E are related: me E m m because U = 0 B 1 and B are coefficients to be determined by the boundary conditions.

26 Particle in a Bo (3) U() Now, let s worry about the boundary conditions. Match at the left boundary ( = 0). I II I Region I: ( ) 0 I II I Region II: ( ) B1 sin B cos 0 L Recall: The wave function () must be continuous at all boundaries. Therefore, at = 0: (0) (0) 0 1 B 1 sin(0) B cos(0) 0 B 0 because cos(0) = 1 and sin(0) = 0 This boundary condition requires that there be no cos() term!

27 Particle in a Bo (4) Now, match at the right boundary ( = L). At = L: I I U() II II I I This constraint requires to have secial values: 0 L This is the same condition we find for confined waves, e.g., waves on a string, EM waves in a laser cavity, etc.: n (= v/f) 4 L/ 3 L/3 L 1 L For matter waves, the wavelength is related to the article energy: E = h /m

28 Particle in a Bo (5) The discrete E n are nown as energy eigenvalues : n (= v/f) E 4 L/ 16E 1 3 L/3 9E 1 L 4E 1 1 L E 1 Imortant features: Discrete energy levels. E 1 0 Standing wave (± for a given E) n = 0 is not allowed. (why?) an eamle of the uncertainty rincile U = E n U = n=3 n= n=1 0 L

29 Particle in a Bo (6) We now now that n ( ) B1 sin( ). How can we determine B 1? n We need another constraint. It is the requirement that total robability equals 1. The robability density at is () : n=3 Integral under the curve = 1 B Therefore, the total robability is : 0 L P ( ) L n d B1 sin( ) d B L 1 L 0 Requiring that P tot = 1 gives : B L

30 Particle in a Bo - summary 1,..., ) sin( ) ( n L n L n n n ) ( L m n m E n n

31 Probability Density In the infinite well:. (Units are m -1, in 1D) P( ) N sin( ) n Notation: The constant is tyically written as N, and is called the normalization constant. For the square well: N L One imortant difference with the classical result: For a classical article bouncing bac and forth in a well, the robability of finding the article is equally liely throughout the well. For a quantum article in a stationary state, the robability distribution is not uniform. There are nodes where the robability is zero! N n=3 0 L

32 Probabilities Often what we measure in an eeriment is the robability density, (). n ( ) N sin( ) n Wavefunction = Probability amlitude P ( ) n ( ) N sin( n ) Probability er unit length (in 1-dimension) U= U= n=1 0 L 0 L Probability density = 0 node 0 L n= 0 L 0 L n=3 0 L

33 Quantum Wire Eamle An electron is traed in a quantum wire that is L = 4 nm long. Assume that the otential seen by the electron is aroimately that of an infinite square well. 1: Calculate the ground (lowest) state energy of the electron. : What hoton energy is required to ecite the traed electron to the net available energy level (i.e., n = )? U= E n U= n=3 n= n=1 0 L The idea here is that the hoton is absorbed by the electron, which gains all of the hoton s energy (similar to the hotoelectric effect).

34 Solution An electron is traed in a quantum wire that is L = 4 nm long. Assume that the otential seen by the electron is aroimately that of an infinite square well. 1: Calculate the ground (lowest) state energy of the electron. : What hoton energy is required to ecite the traed electron to the net available energy level (i.e., n = )? U= E n U= n= n=1 0 L E n n E1 n=3 So, the energy difference between the n = and n = 1 levels is: DE = ( - 1 )E 1 = 3E 1 = ev

35 Eamle of a microscoic otential well -- a semiconductor quantum well Deosit different layers of atoms on a substrate crystal: AlGaAs GaAs AlGaAs U() Quantum wells lie these are used for light emitting diodes and laser diodes, such as the ones in your CD layer. The LED was develoed at UIUC by Nic Holonya. An electron has lower energy in GaAs than in AlGaAs. It may be traed in the well but it leas into the surrounding region to some etent

36 Particle in a Finite Well (1) What if the walls of our bo aren t infinitely high? We will consider finite U 0, with E < U 0, so the article is still traed. This situation introduces the very imortant concet of barrier enetration. As before, solve the SEQ in the three regions. Region II: U = 0, so the solution is the same as before: U() U 0 E We do not imose the infinite well boundary conditions, because they are not the same here. We will find that B is no longer zero. I II III 0 L Before we consider boundary conditions, we must first determine the solutions in regions I and III.

37 Particle in a Finite Well () Regions I and III: U() = U o, and E < U 0 Because E < U 0, these regions are forbidden in classical articles. The SEQ d ( ) d m ( E U ) ( ) 0 can be written: where: U 0 > E: K is real. U() The general solution to this equation is: Region I: Region III: U 0 I II III 0 L E C 1, C, D 1, and D, will be determined by the boundary conditions.

38 Particle in a Finite Well (3) Imortant new result! For quantum entities, there is a finite robability amlitude,, to find the article inside a classically-forbidden region, i.e., inside a barrier. U() U 0 I II III E 0 L

39 Eercise U() In region III, the wave function has the form U 0 1. As, the wave function must vanish. (why?) What does this imly for D 1 and D? a. D 1 = 0 b. D = 0 c. D 1 and D are both nonzero. I II III 0 L E. What can we say about the coefficients C 1 and C for the wave function in region I? a. C 1 = 0 b. C = 0 c. C 1 and C are both nonzero.

40 Solution U() In region III, the wave function has the form U 0 1. As, the wave function must vanish (why?). What does this imly for D 1 and D? a. D 1 = 0 b. D = 0 c. D 1 and D are both nonzero. Since e K as, D 1 must be 0. I II III 0 L. What can we say about the coefficients C 1 and C for the wave function in region I? E a. C 1 = 0 b. C = 0 c. C 1 and C are both nonzero.

41 Solution U() In region III, the wave function has the form U 0 1. As, the wave function must vanish (why?). What does this imly for D 1 and D? a. D 1 = 0 b. D = 0 c. D 1 and D are both nonzero. Since e K as, D 1 must be 0. I II III 0 L. What can we say about the coefficients C 1 and C for the wave function in region I? E a. C 1 = 0 b. C = 0 c. C 1 and C are both nonzero. K is negative for < 0. e -K as -. So, C must be 0.

42 Particle in a Finite Well (4) Summarizing the solutions in the 3 regions: Region I: Region II: Region III: U() U 0 I II III 0 L E As with the infinite square well, to determine arameters (K,, B 1, B, C 1, and D ) we must aly boundary conditions. Useful to now: In an allowed region, curves toward 0. In a forbidden region, curves away from 0.

43 Particle in a Finite Well (5) The boundary conditions are not the same as for the finite well. We no longer require that = 0 at = 0 and = L. Instead, we require that () and d/d be continuous across the boundaries: is continuous d/d is continuous U() U 0 I II III 0 L E At = 0: At = L: Unfortunately, this gives us a set of four transcendental equations. They can only be solved numerically (on a comuter). si

44 Particle in a Finite Well (6) What do the wave functions for a article in the finite square well otential loo lie? U() They loo very similar to those for the infinite well, ecet The article has a finite robability to lea out of the well!! n=4 U 0 n= n=1 n=3 0 L Some general features of finite wells: Due to leaage, the wavelength of n is longer for the finite well. Therefore E n is lower than for the infinite well. K deends on U 0 - E. For higher E states, e -K decreases more slowly. Therefore, their enetrates farther into the forbidden region. A finite well has only a finite number of bound states. If E > U 0, the article is no longer bound.

45 1D Harmonic oscillator Another very imortant otential is the harmonic oscillator: U U() U U() = ½ (/m) 1/ Why is this otential so imortant? It accurately describes the otential for many systems. E.g., sound waves. It aroimates the otential in almost every system for small deartures from equilibrium. E.g., chemical bonds. To a good aroimation, everything is a harmonic oscillator U (ev) r o Chemical bonding otential Taylor eansion of U near minimum. r (nm)

46 1D Harmonic oscillator 1 U ( ) ; / m E n Eigenvalues ( n 1/ ) n 0,1,,.. Eigenstates ( ) n n 1 n! H ( ) e a ; m H n () is the Hermite olynomial n a si

47 HO Wave Functions (1) The imortant features of the HO otential are: It s symmetrical about = 0. It does not have a hard wall (doesn t go to at finite ). U U() U E Consider the state with energy E. There are two forbidden regions and one allowed region. Alying our general rules, we can then say: I II III () curves toward zero in region II and away from zero in regions I and III. () is either an even or odd function of. U U() U Let s consider the ground state: () has no nodes. () is an even function of.

48 HO Wave Functions () For the ecited states, use these rules: Each successive ecited state has one more node. The wave functions alternate symmetry. Unlie the square well, the allowed region gets wider as the energy increases, so the higher energy wave functions oscillate over a larger range U U() () n= n=0 n=1 n=3 U

49 Probability distribution Difference between classical and quantum cases U() Classical (article with same energy as in qunatum case) U() Quantum (lowest energy state) E E P() In classical mechanics, the article is most liely to be found where its seed is slowest P() = In classical mechanics, the article moves bac and forth coming to rest at each turning oint In quantum mechanics, the article is most liely to be found at the center. In quantum mechanics, the article can also be found where it is forbidden in classical mechanics.

50 Quantum Tunneling Classical article o Cannot go through barrier o Comletely reflected Quantum article tunneling o Partially transmitted o Transmission decreases eonentially with barrier width. Lecture 1, 50

51 Tunneling: Key Points In quantum mechanics a article can enetrate into a barrier where it would be classically forbidden. The finite square well: In region III, E < U 0, and () has the eonential form D 1 e -K. We did not solve the equations too hard! The robability of finding the article in the barrier region decreases as e -K. U() U 0 I II III 0 L The finite-width barrier: We consider a related roblem a article aroaching a finite-width barrier and tunneling through to the other side. The robability of the article tunneling through a finite width barrier is aroimately roortional to e -KL where L is the width of the barrier.

52 Tunneling Through a Barrier (1) What is the the robability that an incident article tunnels through the barrier? It s called the Transmission Coefficient, T. Consider a barrier (II) of height U 0. U = 0 everywhere else. Getting an eact result requires alying the boundary conditions at = 0 and = L, then solving si transcendental equations for si unnowns: U o 0 U() I II III 0 L A 1, A, B 1, B, C 1, and C are unnown. K and are nown functions of E. This is more comlicated than the infinitely wide barrier, because we can t require that B 1 = 0. (Why not?)

53 Tunneling Through a Barrier () G In many situations, the barrier width L U() is much larger than the decay length 1/K of the enetrating wave (KL >> 1). In this case B 1 0 (why?), and the result resembles the infinite barrier. The tunneling coefficient simlifies: U 0 0 L E The imortant result is e -KL E/U

54 Tunneling Through a Barrier (3)

55 Eercise Consider a article tunneling through a barrier. 1. Which of the following will increase the lielihood of tunneling? a. decrease the height of the barrier b. decrease the width of the barrier c. decrease the mass of the article U 0 0 L U() E. What is the energy of the emerging articles? a. < initial energy b. = initial energy c. > initial energy

56 Solution Consider a article tunneling through a barrier. 1. Which of the following will increase the lielihood of tunneling? a. decrease the height of the barrier b. decrease the width of the barrier c. decrease the mass of the article U 0 0 L U() E. What is the energy of the emerging articles? a. < initial energy b. = initial energy c. > initial energy

57 Solution Consider a article tunneling through a barrier. 1. Which of the following will increase the lielihood of tunneling? a. decrease the height of the barrier b. decrease the width of the barrier c. decrease the mass of the article U 0 0 L U() E. What is the energy of the emerging articles? a. < initial energy b. = initial energy c. > initial energy The barrier does not absorb energy from the article. The amlitude of the outgoing wave is smaller, but the wavelength is the same. E is the same everywhere. Probability Energy

58 Alication: Tunneling Microscoy One can use barrier enetration to measure the electron density on a surface. Metal ti Real STM ti Na atoms on metal: Scanning Tunneling Microscoe images DNA Double Heli: material STM ti ~ 1 nm material STM ti STM demo: htt:// Barrier enetration is a wave henomenon, not only QM. It is used in otical microscoes also. See: htt://en.wiiedia.org/wii/total_internal_reflection_fluorescence_microscoe

59 Alication: Tunneling Microscoy Quantum Tunneling V b

60 Scanning Tunneling Microscoe STM

61 Grahene Carbon 0.5 cm 0.5 nm Lecture 14, 61

62 Building materials atom by atom a ~ nm Lecture 14, 6

63 Solution The STM (scanning tunneling microscoe) ti is L = 0.18 nm from a metal surface. An electron with energy of E = 6 ev in the metal aroaches the surface. Assume the metal/ti ga is a otential barrier with a height of U o = 1 ev. What is the robability that the electron will tunnel through the barrier? metal U 0 0 L U() E STM ti ga T << 1, so our use of the KL >> 1 aroimation is justified. Q: What will T be if we double the width of the ga? Lecture 14, 63

64 Quantum Particles in 3D Potentials A real (D) quantum dot So far, we have considered quantum articles bound in one-dimensional otentials. This situation can be alicable to certain hysical systems but it lacs some of the features of most real 3D quantum systems, such as atoms and artificial structures. One consequence of confining a quantum article in two or three dimensions is degeneracy -- the eistence of several quantum states at the same energy. htt://ages.unibas.ch/hys-meso/pictures/ictures.html To illustrate this imortant oint in a simle system, let s etend our favorite otential - the infinite square well - to three dimensions.

65 Particle in a 3D Bo (1) The etension of the Schrödinger Equation (SEQ) to 3D is straightforward in Cartesian (,y,z) coordinates: where Let s solve this SEQ for the article in a 3D cubical bo: z L U(,y,z) = outside bo, or y or z < 0 0 inside bo outside bo, or y or z > L y L L This U(,y,z) can be searated : U(,y,z) = U() + U(y) + U(z) U = if any of the three terms =. Lecture 16, 65

66 Particle in a 3D Bo () Whenever U(,y,z) can be written as the sum of functions of the individual coordinates, we can write some wave functions as roducts of functions of the individual coordinates: For the 3D square well, each function is simly the solution to the 1D square well roblem: Similarly for y and z. Each function contributes to the energy. The total energy is the sum: Etotal = E + E y + E z htt:// Lecture 16, 66

67 Particle in a 3D Bo (3) The energy eigenstates and energy values in a 3D cubical bo are: n ny nz Nsin sin y sin z L L L h E n n n y z 8mL ( ) n n n y z z L where n,n y, and n z can each have values 1,,3,. y L L This roblem illustrates two imortant oints: Three quantum numbers (n,n y,n z ) are needed to identify the state of this three-dimensional system. That is true for every 3D system. More than one state can have the same energy: Degeneracy. Degeneracy reflects an underlying symmetry in the roblem. 3 equivalent directions, because it s a cube, not a rectangle. Lecture 16, 67

68 Cubical Bo Eercise z Consider a 3D cubic bo: Show energies and label (n,n y,n z ) for the first 11 states of the article in the 3D bo, and write the degeneracy, D, for each allowed energy. Define E o = h /8mL. y L L L E (n,n y n z ) Degeneracy 6E o (,1,1) (1,,1) (1,1,) D=3 3E o (1,1,1) D=1 Lecture 18, 68

69 Solution z Consider a 3D cubic bo: Show energies and label (n,n y,n z ) for the first 11 states of the article in the 3D bo, and write the degeneracy, D, for each allowed energy. Define E o = h /8mL. y L L L E (n,n y n z ) Degeneracy 1E o D=1 (,,) 11E o (3,1,1) (1,3,1) (1,1,3) D=3 9E o (,,1) (,1,) (1,,) D=3 6E o (,1,1) (1,,1) (1,1,) D=3 3E o (1,1,1) D=1 Lecture 18, 69

70 Atoms: Classical Planetary Model (An early model of the atom) Classical icture: negatively charged objects (electrons) orbit ositively charged nucleus due to Coulomb force. F -e There is a BIG PROBLEM with this: +Ze As the electron moves in its circular orbit, it is ACCELERATING. As you learned in introductory Physics, accelerating charges radiate electromagnetic energy. Consequently, an electron would continuously lose energy and siral into the nucleus in about 10-9 sec. The lanetary model doesn t lead to stable atoms.

71 Hydrogen Atom - Qualitative Why doesn t the electron collase into the nucleus, where its otential energy is lowest? We must balance two effects: As the electron moves closer to the nucleus, its otential energy decreases (more negative): e U r However, as it becomes more and more confined, its inetic energy increases: r KE mr Therefore, the total energy is: e E KE PE mr r E has a minimum at: r a m e nm The Bohr radius of the H atom. At this radius, E 4 m e 13.6 ev The ground state energy of the hydrogen atom. Heisenberg s uncertainty rincile revents the atom s collase. One factor of e or e comes from the roton charge, and one from the electron. Lecture 16, 71

72 Potential Energy in the Hydrogen Atom To solve this roblem, we must secify the otential energy of the electron. In an atom, the Coulomb force binds the electron to the nucleus. This roblem does not searate in Cartesian coordinates, because we cannot write U(,y,z) = U ()+U y (y)+u z (z). However, we can searate the otential in sherical coordinates (r,q,), because: U(r,q,) = U r (r) + U q (q) + U () e 0 0 r Therefore, we will be able to write: ( r,, ) R ( r ) ( ) ( ) q q Question: How many quantum numbers will be needed to describe the hydrogen wave function? Lecture 16, 7

73 Wave Function in Sherical Coordinates We saw that because U deends only on the radius, the roblem is searable. The hydrogen SEQ can be solved analytically. We will show you the solutions and discuss their hysical significance. We can write: ( r,, ) R ( r ) Y (, ) q q nlm nl lm r q y z There are three quantum numbers: n rincial (n 1) l orbital (0 l < n-1) m magnetic (-l m +l) What before we called ( q) ( ) The Y lm are called sherical harmonics. Today, we will only consider l = 0 and m = 0. These are called s-states. This simlifies the roblem, because Y 00 (q,) is a constant and the wave function has no angular deendence: These are states in which the electron has no orbital angular momentum. This is not ossible in Newtonian hysics. (Why?) Note: Some of this nomenclature dates bac to the 19 th century, and has no hysical significance. Lecture 16, 73

74 Radial Eigenstates of Hydrogen Here are grahs of the s-state wave functions, R no (r), for the electron in the Coulomb otential of the roton. The zeros in the subscrits are a reminder that these are states with l = 0 (zero angular momentum!). 0 E -1.5 ev -3.4 ev You can rove these are solutions by lugging into the radial SEQ (Aendi) ev Lecture 16, 74

75 Summary of S-states of H-atom 1 1 The s-states (l=0, m=0) of the Coulomb otential have no angular deendence. In general: r, q, R r Y q, but: ( ) ( ) ( ) nlm nl lm ( r, q, ) R ( r ) n00 n0 s-state wave functions (radial art): because Y 00 (q,) is a constant. ) 0.5 R 10 h( ) 0.5 R 0 d4( ) R a 4 0 r a r 0 15a 0 0 r htt:// Lecture 17, 75

76 Total Wave Function of the H-atom We will now consider non-zero values of the other two quantum numbers: l and m. ( r,, ) R ( r ) Y (, ) q q nlm nl lm n rincial (n 1) l orbital (0 l < n-1) m magnetic (-l m +l) * The Y lm (q,) are nown as sherical harmonics. They are related to the angular momentum of the electron. * The constraints on l and m come from the boundary conditions one must imose on the solutions to the Schrodinger equation. We ll discuss them briefly. Lecture 17, 76

77 Quantized Angular Momentum z comonent of angular momentum: Lˆ z i ˆ ( r ) L z ( r ) Y lm ( q, ) e m ( r ) im m 0, 1,,... An imortant boundary condition: An integer number of wavelengths must fit around the circle. Otherwise, the wave function is not single-valued. This imlies that and m = 0, ±1, ±, ±3, L z = 0, ±ħ, ±ħ, ±3ħ, Angular momentum is quantized!! We re ignoring R(r) for now. Reminder: e im = cos(m) + i sin(m) htt:// Lecture 17, 77

78 The l Quantum Number The quantum number m reflects the comonent of angular momentum about a given ais. In the angular wave function lm (q,) the quantum number l tells us the total angular momentum L. L = L + L y + L z is also quantized. The ossible values of L are: Wave functions can be eigenstates of both L and L Z. For sherically symmetric otentials, lie H-atom, they can also be eigenstates of E. Such states are called orbitals. Summary of quantum numbers for the H-atom orbitals: Princial quantum number: n = 1,, 3,. Orbital quantum number: l = 0, 1,,, n-1 Orbital magnetic quantum number: m = -l, -(l-1), 0, (l-1), l Lecture 17, 78

79 The Angular Wave Function, Y lm (q,) The angular wave function may be written: Y lm (q,) = P(q)e im where P(q) are olynomial functions of cos(q) and sin(q). To get some feeling for these angular distributions, we mae olar lots of the q-deendent art of Y lm (q, ) (i.e., P(q)): 1. y( t) 1 0 Parametric Curve q 1. 10( t) 1 0 Parametric Curve q 1. 11( t) 1 0 Parametric Curve q Length of the dashed arrow is the magnitude of Y lm as a function of q ( t) z 1. z y10( t) y11( t) z 1. y y y Lecture 17, 79

80 Why are these distributions imortant? They govern the bonding and chemistry of atoms. In articular, they determine the angles at which different atoms bond: the structure of molecules & solids. Historical Labeling of Orbitals Notation from 19 th century Angular momentum quantum # sectroscoy l = 0 s shar l = 1 rincial l = d diffuse l = 3 f fundamental Lecture 17, 80

81 End of art Lecture 17, 81