Lecture 16: 3D Potentials and the Hydrogen Atom

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ectue 6, p ectue 6: 3D Potentials and the Hydogen Atom 4.5 g( ) x 4 x P() 4a = a z x 8 z y x n n n n n n m h E z y x ) ( ) ( ) ( ),, ( z y x z y x o a / 3 o e a ) ( 3 6 n ev. E n

Oveview of the Couse Up to now: Geneal popeties and equations of quantum mechanics Time-independent Schodinge s Equation (SEQ) and eigenstates. Time-dependent SEQ, supeposition of eigenstates, time dependence. Collapse of the wave function, Schodinge s cat Tunneling This week: 3 dimensions, angula momentum, electon spin, H atom Next week: Exclusion pinciple, peiodic table of atoms Molecules and solids, consequences of Q.M. Metals, insulatos, semiconductos, supeconductos, lases,.. ectue 6, p

Today 3-Dimensional Potential Well: Poduct Wave Functions Degeneacy Schödinge s Equation fo the Hydogen Atom: Semi-quantitative pictue fom uncetainty pinciple Gound state solution* Spheically-symmetic excited states ( s-states )* *contains details beyond what we expect you to know on exams. ectue 6, p 3

ectue 6, p 4

Quantum Paticles in 3D Potentials A eal (D) quantum dot So fa, we have consideed quantum paticles bound in one-dimensional potentials. This situation can be applicable to cetain physical systems but it lacks some of the featues of most eal 3D quantum systems, such as atoms and atificial stuctues. http://pages.unibas.ch/phys-meso/pictues/pictues.html One consequence of confining a quantum paticle in two o thee dimensions is degeneacy -- the existence of seveal quantum states at the same enegy. To illustate this impotant point in a simple system, let s extend ou favoite potential - the infinite squae well - to thee dimensions. ectue 6, p 5

Paticle in a 3D Box () The extension of the Schödinge Equation (SEQ) to 3D is staightfowad in Catesian (x,y,z) coodinates: U( x, y, z) E m x y z m Kinetic enegy tem: px py pz whee et s solve this SEQ fo the paticle in a 3D cubical box: ( x, y, z) z x U(x,y,z) = outside box, x o y o z < inside box outside box, x o y o z > y This U(x,y,z) can be sepaated : U(x,y,z) = U(x) + U(y) + U(z) U = if any of the thee tems =. ectue 6, p 6

Paticle in a 3D Box () Wheneve U(x,y,z) can be witten as the sum of functions of the individual coodinates, we can wite some wave functions as poducts of functions of the individual coodinates: (see the supplementay slides) ( x, y, z) f ( x) g( y) h( z) Fo the 3D squae well, each function is simply the solution to the D squae well poblem: nx h nx nx fn ( x) N sin x x E m Similaly fo y and z. D wave functions: n sin x n x sin y y (n x,n y ) = (,) (n x,n y ) = (,) y y x x Each function contibutes to the enegy. The total enegy is the sum: Etotal = E x + E y + E z http://www.falstad.com/qmdbox/ y (n x,n y ) = (,) x y (n x,n y ) = (,) x ectue 6, p 7

ectue 6, p 8

Paticle in a 3D Box (3) The enegy eigenstates and enegy values in a 3D cubical box ae: nx ny nz Nsin x sin y sin z h E n n n x y z 8m n n n x y z whee n x,n y, and n z can each have values,,3,. y z x This poblem illustates two impotant points: Thee quantum numbes (n x,n y,n z ) ae needed to identify the state of this thee-dimensional system. That is tue fo evey 3D system. Moe than one state can have the same enegy: Degeneacy. Degeneacy eflects an undelying symmety in the poblem. 3 equivalent diections, because it s a cube, not a ectangle. ectue 6, p 9

Cubical Box Execise z Conside a 3D cubic box: Show enegies and label (n x,n y,n z ) fo the fist states of the paticle in the 3D box, and wite the degeneacy, D, fo each allowed enegy. Define E o = h /8m. y x E (n x,n y n z ) Degeneacy 6E o (,,) (,,) (,,) D=3 3E o (,,) D= ectue 6, p

Act Fo a cubical box, we just saw that the 5 th enegy level is at E, with a degeneacy of and quantum numbes (,,).. What is the enegy of the next enegy level? a. 3E b. 4E c. 5E. What is the degeneacy of this enegy level? a. b. 4 c. 6 ectue 6, p

ectue 6, p

Anothe 3D System: The Atom -electons confined in Coulomb field of a nucleus Ealy hints of the quantum natue of atoms: Discete Emission and Absoption specta When excited in an electical dischage, atoms emit adiation only at discete wavelengths Diffeent emission specta fo diffeent atoms Atomic hydogen (nm) Geige-Masden (Ruthefod) Expeiment (9): Measued angula dependence of a paticles (He ions) scatteed fom gold foil. Mostly scatteing at small angles suppoted the plum pudding model. But Occasional scatteings at lage angles Something massive in thee! v Au Conclusion: Most of atomic mass is concentated in a small egion of the atom a nucleus!

Atoms: Classical Planetay Model (An ealy model of the atom) Classical pictue: negatively chaged objects (electons) obit positively chaged nucleus due to Coulomb foce. F -e Thee is a BIG PROBEM with this: +Ze As the electon moves in its cicula obit, it is ACCEERATING. As you leaned in Physics, acceleating chages adiate electomagnetic enegy. Consequently, an electon would continuously lose enegy and spial into the nucleus in about -9 sec. The planetay model doesn t lead to stable atoms.

Hydogen Atom - Qualitative Why doesn t the electon collapse into the nucleus, whee its potential enegy is lowest? We must balance two effects: As the electon moves close to the nucleus, its potential enegy deceases (moe negative): e U Howeve, as it becomes moe and moe confined, its kinetic enegy inceases: p KE m Theefoe, the total enegy is: e E KE PE m E has a minimum at: m e a.53 nm The Boh adius of the H atom. At this adius, E 4 m e 3.6 ev The gound state enegy of the hydogen atom. Heisenbeg s uncetainty pinciple pevents the atom s collapse. One facto of e o e comes fom the poton chage, and one fom the electon. ectue 6, p 5

ectue 6, p 6

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 (,,f), because: U ( ) e U(,,f) = U () + U () + U f (f) e Theefoe, we will be able to wite:,, R f f 9 Nm /C 4 9 Question: How many quantum numbes will be needed to descibe the hydogen wave function? ectue 6, p 7

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). We will show you the solutions and discuss thei physical significance. We can wite:,, R Y, f f nlm nl lm x f y z Thee ae thee quantum numbes: n pincipal (n ) l obital ( l < n-) m magnetic (-l m +l) What befoe we called f The Y lm ae called spheical hamonics. Today, we will only conside l = and m =. These ae called s-states. This simplifies the poblem, because Y (,f) is a constant and the wave function has no angula dependence: (,, f) 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 9 th centuy, and has no physical significance. ectue 6, p 8

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 3 -.5 ev x).5 R h( x).5 R d4( x) R 3-3.4 ev 4 x R e () / a, 4a 4. 5 R e x /a,( ) a a.5 5 5a x 5 R e /3a 3,( ) 3 a 3 a a.53 nm m e e You can pove these ae solutions by plugging into the adial SEQ (Appendix). E n 3.6 ev n You will not need to memoize these functions. -3.6 ev ectue 6, p 9

ectue 6, p

ACT : 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. ) What is the enegy of the emitted photon? a).5 ev b).9 ev c) 3.4 ev ) What is the wavelength of the emitted photon? a) 87 nm b) 656 nm c) 365 nm ectue 6, p

Next week: aboatoy 4 Eyepiece Coss hais Focusing lens Gating ight shield Collimating lens Slit ight souce Deflected beam Undeflected beam E n 3.6eV n ectue 6, p

Pobability Density of Electons = Pobability density = Pobability pe unit volume The density of dots plotted below is popotional to. s state s state R n R n fo s-states. A node in the adial pobability distibution. f( x).5 R h( x).5 R 4 x 4a 4. 5 x a ectue 6, p 3

ectue 6, p 4

Radial Pobability Densities fo S-states Summay of wave functions and adial pobability densities fo some s-states. f( x).5 R g( x).5 P The adial pobability density has an exta facto of because thee is moe volume at lage. That is, P n (). R n ( x). 3 4a 4.5 x R 5 a x 4 4 4a x.5 h( x).4. 4 5 x P a This means that: The most likely is not!!! Even though that s whee () is lagest. This is always a confusing point. See the supplementay slide fo moe detail. x) R 3 P 3.5 5 5a x 5 adial wave functions a adial pobability densities, P() http://www.falstad.com/qmatom/ ectue 6, p 5

Next ectues Angula momentum Spin Nuclea Magnetic Resonance ectue 6, p 6

Supplement: Sepaation of Vaiables () In the 3D box, the SEQ is: U( x) U( y) U( z) E m x y z et s see if sepaation of vaiables woks. Substitute this expession fo into the SEQ: ( x, y, z) f ( x) g( y) h( z) NOTE: Patial deivatives. d f d g d h gh fh fg U( x) U( y) U( z) fgh Efgh m dx dy dz NOTE: Total deivatives. Divide by fgh: d f d g d h U( x) U( y) U( z) E m f x g dy h dz ectue 6, p 7

ectue 6, p 8

Supplement: Sepaation of Vaiables () Regoup: d f d g d h ( ) ( ) ( ) U x U y U z E m f x m g dy m h dz A function of x A function of y A function of z We have thee functions, each depending on a diffeent vaiable, that must sum to a constant. Theefoe, each function must be a constant: df U( x) E m f x dg ( ) U y E m g dy dh ( ) U z E m h dz Ex Ey Ez E x y z Each function, f(x), g(y), and h(z) satisfies its own D SEQ. ectue 6, p 9

Supplement: Why Radial Pobability Isn t the Same as Volume Pobability et s look at the n=, l= state (the s state): (,,f) R () e -/a. So, P(,,f) = e -/a. This is the volume pobability density. f( x).5 P(,,f) If we want the adial pobability density, we must emembe that: 4 dv = x 4 d sin d df We e not inteested in the angula distibution, so to calculate P() we must integate ove and f. The s-state has no angula dependence, so the integal is just 4. 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. g( x) 4a.5 P() 4 4a x 4 ectue 6, p 3

Appendix: Solving the Radial SEQ fo H --deiving a o and E e Substituting R( ) Ne into R( ) ER( ), we get: m d m e e e e Ee Fo this equation to hold fo all, we must have: e m AND m E m E a ma e Evaluating the gound state enegy: E ma mc c a (.5)( (97) 6 )(.53) 3.6eV