Anyone who can contemplate quantum mechanics without getting dizzy hasn t understood it. --Niels Bohr. Lecture 17, p 1

Transcription

1 Anyone who can contemplate quantum mechanics without getting dizzy hasn t undestood it. --Niels Boh Lectue 17, p 1

2 Special (Optional) Lectue Quantum Infomation One of the most moden applications of QM quantum computing quantum communication cyptogaphy, telepotation quantum metology Pof. Kwiat will give a special 14-level lectue on this topic Sunday, Feb. 4 3 pm, 141 Loomis Attendance is optional, but encouaged. Lectue 17, p

3 Lectue 17: Atomic States, Angula Momentum & Selection Rules Lectue 17, p 3

4 Today Schödinge s Equation fo the Hydogen Atom Radial wave functions Angula Momentum Quantization of L z and L Lectue 17, p 4

5 Potential Enegy in the Hydogen Atom To solve this poblem, we must specify the potential enegy of the electon. In an atom, the Coulomb foce binds the electon to the nucleus. U() This poblem does not sepaate in Catesian coodinates, because we cannot wite U(x,y,z) = U x (x)+u y (y)+u z (z). Howeve, we can sepaate the potential in spheical coodinates (,θ,φ), because: U(,θ,φ) = U () + U θ (θ) + U φ (φ) e κ Theefoe, we will be able to wite: (,, ) = R ( ) Θ( ) Φ ( ) ψ θ φ θ φ U( ) κe = 1 κ = 9 1 Nm /C 4πε = 9 Question: How many quantum numbes will be needed to descibe the hydogen wave function? Lectue 17, p 5

6 Wave Function in Spheical Coodinates We saw that because U depends only on the adius, the poblem is sepaable. The hydogen SEQ can be solved analytically (but not by us). Hee we show the solutions and thei physical significance. We can wite: (,, ) = R ( ) Y (, ) ψ θ φ θ φ nlm nl lm x θ φ y z Thee ae thee quantum numbes: n pincipal (n 1) l obital ( l< n-1) m magnetic (-l m +l) What befoe we called Θ ( θ ) Φ ( φ ) The Y lm ae called spheical hamonics. Fist, we will only conside l = and m =. These ae called s-states. This simplifies the poblem, because Y (θ,φ) is a constant and the wave function has no angula dependence: ψ (, θ, φ ) = R ( ) n n These ae states in which the electon has no obital angula momentum. This is not possible in Newtonian physics. (Why?) Note: Some of this nomenclatue dates back to the 19 th centuy, and has no physical significance. Lectue 17, p 6

7 Radial Eigenstates of Hydogen Hee ae gaphs of the s-state wave functions, R no (), fo the electon in the Coulomb potential of the poton. The zeos in the subscipts ae a eminde that these ae states with l = (zeo angula momentum!). E -1.5 ev 5 R 1 R R ev R e ( ) / a 1, 4a / a,( ) 1 e a R 1a R 15a / 3a 3,( ) 3 + e a 3 a ħ a =.53 nm m κe e You can pove these ae solutions by plugging into the adial SEQ (Appendix). E n 13.6 ev n You will not need to memoize these functions ev Lectue 17, p 7

8 ACT 1: Optical Tansitions in Hydogen An electon, initially excited to the n = 3 enegy level of the hydogen atom, falls to the n = level, emitting a photon in the pocess. 1) What is the enegy of the emitted photon? a) 1.5 ev b) 1.9 ev c) 3.4 ev ) What is the wavelength of the emitted photon? a) 87 nm b) 656 nm c) 365 nm Lectue 17, p 8

9 An electon, initially excited to the n = 3 enegy level of the hydogen atom, falls to the n = level, emitting a photon in the pocess. 1) What is the enegy of the emitted photon? a) 1.5 ev b) 1.9 ev c) 3.4 ev 13.6 ev En = n 1 1 En 13.6 ev i n = f ni nf 1 1 Ephoton = E3 = 13.6 ev 1.9 ev 9 4 = ) What is the wavelength of the emitted photon? a) 87 nm b) 656 nm c) 365 nm Solution E -15 ev n = 3 n = n = 1 Lectue 17, p 9

10 Solution An electon, initially excited to the n = 3 enegy level of the hydogen atom, falls to the n = level, emitting a photon in the pocess. 1) What is the enegy of the emitted photon? a) 1.5 ev b) 1.9 ev c) 3.4 ev 13.6 ev En = n 1 1 En 13.6 ev i n = f ni nf 1 1 Ephoton = E3 = 13.6 ev 1.9 ev 9 4 = ) What is the wavelength of the emitted photon? a) 87 nm b) 656 nm c) 365 nm hc 14 ev nm λ = = = 656 nm E 1.9 ev photon You will measue seveal tansitions in Lab. We ll see next lectue that thee ae othe constaints on which tansitions can occu. E -15 ev Atomic hydogen λ (nm) n = 3 n = n = 1 Question: Which tansition is this? (λ = 486 nm) Lectue 17, p 1

11 Next week: Laboatoy 4 Eyepiece Coss hais Focusing lens Gating Light shield Collimating lens Slit Light souce Deflected beam θ Undeflected beam E n = 13.6eV n Lectue 17, p 11

12 Pobability Density of Electons ψ = Pobability density = Pobability pe unit volume The density of dots plotted below is popotional to. 1s state s state R n R n fo s-states. A node in the adial pobability distibution. 5 R 1 R 4a 1a Lectue 17, p 1

13 Radial Pobability Densities fo S-states Summay of wave functions and adial pobability densities fo some s-states. 5 R 1 5 P 1 The adial pobability density has an exta facto of because thee is moe volume at lage. That is, P n () R n. 4a R 4a P This means that: The most likely is not!!! Even though that s whee ψ() is lagest. 1a 1a This is always a confusing point. See the supplementay slide fo moe detail. R 3 P 3 15a a adial wave functions adial pobability densities, P() Lectue 17, p 13

14 Wave Function Nomalization What is the nomalization constant fo the hydogen atom gound state? ψ θ φ / ao 1(,, ) = NR1( ) = Ne 5 R 1 4a Lectue 17, p 14

15 Solution What is the nomalization constant fo the hydogen atom gound state? ψ θ φ / ao 1(,, ) = NR1( ) = Ne 5 R 1 The pobability density is ψ = N exp(-/a o ). In 3D, this means pobability pe unit volume. 4a We equie that the total pobability = 1: ψ dv = 1 dv = sinθ d dθ dφ With spheical symmety, the angula integals give 4π, so we ae left with: / 4πN e a d 1 ψ = 1 e / ao 1( ) 3 πao = N = 1 πa 3 o You can look it up! Nomalized gound-state wave function of hydogen Lectue 17, p 15

16 Pobability Calculation Estimate the pobability of finding the electon within a small sphee of adius s =. a o at the oigin. ψ() = R 1 () s 5 4a ψ = / o ( ) Ne a Lectue 17, p 16

17 Solution Estimate the pobability of finding the electon within a small sphee of adius s =. a o at the oigin. If it says estimate, don t integate. The wave function is nealy constant nea = : 1 1 o ψ () = e = πa / a 3 3 o πao s 5 ψ = ψ() = R 1 () / o ( ) Ne a 4a Simply multiply ψ by the volume V = (4/3)π s3 : 4 s Pobability = ψ () V =.1 3 ao 3 Lectue 17, p 17

18 Maximum Radial Pobability At what adius ae you most likely to find the electon? ψ = / o ( ) Ne a 5 max 4a Lectue 17, p 18

19 Solution At what adius ae you most likely to find the electon? Looks like a no-baine. =, of couse! 5 ψ = / o ( ) Ne a Well, that s not the answe. You must find the pobability P() that the electon is in a shell of thickness at adius. Fo a given the volume, V, of the shell inceases with adius. The adial pobability has an exta facto of : max max 4a - / a o P( ) = ψ ( ) V = C e Set dp/d = to find: max = a! 5 P() ψ Moe volume at lage. V = 4π 4a No volume at =. Lectue 17, p 19

20 Summay of S-states of H-atom The s-states (l=, m=) of the Coulomb potential have no angula dependence. In geneal: but: (,, ) = R ( ) Y (, ) ψ θ φ θ φ ψ nlm nl lm (, θ, φ ) R ( ) n n because Y (θ,φ) is a constant. S-state wave functions ae spheically symmetic. Some s-state wave functions (adial pat): ψ (,θ,φ) : 5 R 1 R R 3 4a 1a 15a Lectue 17, p

21 Total Wave Function of the H-atom We will now conside non-zeo values of the othe two quantum numbes: l and m. θ (,, ) = R ( ) Y (, ) ψ θ φ θ φ nlm nl lm z y n pincipal (n 1) l obital ( l< n-1) m magnetic (-l m +l) * x φ The Y lm (θ,φ) ae known as spheical hamonics. They ae elated to the angula momentum of the electon. * The constaints on l and m come fom the bounday conditions one must impose on the solutions to the Schodinge equation. We ll discuss them biefly. Lectue 17, p 1

22 Quantized Angula Momentum Linea momentum depends on the wavelength (k=π/λ): p = ħk whee ψ ( x) e Angula momentum depends on the tangential component of the momentum. Theefoe L z depends on the wavelength as one moves aound a cicle in the x-y plane. Theefoe, a state with L z has a simila fom: im L = mħ whee ψ ( ) Y ( θ, φ) e φ L z Z An impotant bounday condition: An intege numbe of wavelengths must fit aound the cicle. Othewise, the wave function is not single-valued. ikx lm We e ignoing R() fo now. Re(ψ) Reminde: φ e imφ = cos(mφ) + i sin(mφ) This implies that and m =, ±1, ±, ±3, L z =, ±ħ, ±ħ, ±3ħ, Angula momentum is quantized!! Lectue 17, p

23 The l Quantum Numbe The quantum numbe m eflects the component of angula momentum about a given axis. L = mħ whee m =, ± 1, ±,... z In the angula wave function ψ lm (θ,φ) the quantum numbe l tells us the total angula momentum L. L = L x + L y + L z is also quantized. The possible values of L ae: L = ll + ħ l = ( 1) whee,1,,... Wave functions can be eigenstates of both L and L Z. Fo spheically symmetic potentials, like H-atom, they can also be eigenstates of E. Such states ae called obitals. Summay of quantum numbes fo the H-atom obitals: Pincipal quantum numbe: n = 1,, 3,. Obital quantum numbe: l =, 1,,, n-1 Obital magnetic quantum numbe: m = -l, -(l-1),, (l-1), l Lectue 17, p 3

24 Angula Momentum & Uncetainty Pinciple ( l ) L = ll ( + 1) ħ not ħ Note that. Also, we descibe angula momentum using only two numbes, l and m. Q: Why can t we specify all thee components (e.g., L =(,,l) so that L = l? A: The uncetainty pinciple doesn t allow us to know that both L x = and L y = unless L z = also. Poof by contadiction: Assume L =(,,l). L = p, so if L points along the z-axis, both and p lie in the x-y plane. This means that z = and p z =, violating the uncetainty pinciple. Thus, L must have a nonzeo L x o L y, making L somewhat lage. We can t specify all thee components of the angula momentum vecto. This logic only woks fo L. L = (,,) is allowed. It s the s-state. All physical quantities ae subject to uncetainty elations, not just position and momentum. Lectue 17, p 4

25 Classical Pictue of L-Quantization e.g., l = L = l(l + 1) ħ = ( + 1) ħ = 6ħ L z L = 6 ħ +ħ +ħ -ħ -ħ L = p Lectue 17, p 5

26 Act We ve just seen that we cannot have the total obital angula momentum vecto of an atom pointing definitely along the z-axis. If we measue L z = ħ, what can we say about L x, the x-component of the angula momentum? a) L x = b) L x = ħ c) L x = -ħ d) any of the above e) none of the above Lectue 17, p 6

27 Solution We ve just seen that we cannot have the total obital angula momentum vecto of an atom pointing definitely along the z-axis. If we measue L z = ħ, what can we say about L x, the x-component of the angula momentum? a) L x = b) L x = ħ c) L x = -ħ d) any of the above e) none of the above As soon as we detemine the angula momentum pojection along one axis, we lose definite knowledge of its value along any pependicula axis. We will again find L x = m ħ, but we don t know what m is until we do the measuement. And if we find m = 1, i.e., L x = m ħ, then we won t know anything about L z except that L z = m ħ, whee again we don t know m. Lectue 17, p 7

28 Next Time Electon obitals in atoms Electon spin and the Sten-Gelach expeiment Next Week Multi-electon atoms Covalent bonds in molecules Electon enegy bands in solids QM in eveyday life Lectue 17, p 8

29 Supplement: Why Radial Pobability Isn t the Same as Volume Pobability Let s look at the n=1, l= state (the 1s state): ψ(,θ,φ) R 1 () e -/a. So, P(,θ,φ) = ψ e -/a. This is the volume pobability density. 5 P(,θ,φ) If we want the adial pobability density, we must emembe that: dv = d sinθ dθ dφ We e not inteested in the angula distibution, so to calculate P() we must integate ove θ and φ. The s-state has no angula dependence, so the integal is just 4π. Theefoe, P() e -/a. The facto of is due to the fact that thee is moe volume at lage. A spheical shell at lage has moe volume than one at small : Compae the volume of the two shells of the same thickness, d. 4a 5 P() 4a Lectue 17, p 9

30 Appendix: Solving the Radial SEQ fo H --deiving a o and E 1 e ħ κ Substituting α R( ) = Ne into R( ) = ER( ), we get: m d ħ m 1 κe ( α α ) α α αe + α e e = Ee Fo this equation to hold fo all, we must have: ħ α = κ e m AND α ħ m = E mκe 1 ħ α = E = ħ a ma Evaluating the gound state enegy: ħ E = ma ħ = mc c a = (.51)(1 (197) 6 )(.53) = 13.6eV