Physics 2D Lecture Slides Dec 1. Vivek Sharma UCSD Physics

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2 Physics D Lectue Sides Dec 1 Vivek Shama UCSD Physics

3 Lean to extend S. Eq and its soutions fom toy exampes in 1-Dimension (x) thee othogona dimensions ( x,y,z) = ix ˆ + ˆjy+ kz ˆ Then tansfom the systems Patice in 1D igid box 3D igid box 1D Hamonic Osciato 3D Hamonic Osciato Keep an eye on the numbe of diffeent integes needed to specify system 1 3 (coesponding to 3 avaiabe degees of feedom x,y,z) QM in 3 Dimensions x z y

4 Quantum Mechanics In 3D: Patice in 3D Box z z=l Extension of a Patice In a Box with igid was 1D 3D Box with Rigid Was (U= ) in X,Y,Z dimensions U()= fo (<x,y,z,<l) y= Ask same questions: Location of patice in 3d Box Momentum Kinetic Enegy, Tota Enegy Expectation vaues in 3D y x y=l To find the Wavefunction and vaious expectation vaues, we must fist set up the appopiate TDSE & TISE

5 The Schodinge Equation in 3 Dimensions: Catesian Coodinates Time Dependent Schodinge Eqn: x z y m Ψ(, xyzt,,) ( xy,, zt, ) Uxyz (,, ) ( xt, ) i...in 3D t = + + x y z Ψ + Ψ = So = + + = m m x m y m z = [K ] + [K ] + [K ] x so [ H] Ψ( x, t) = [ E] Ψ( xt, ) is sti the Enegy Consevation Eq x x [ K] Stationay states ae those fo which a pobabiities ae constant in time and ae given by the soution of the TDSE in sepeabe fom: ( xyzt,,,) (,) t = ()e -iωt Ψ =Ψ ψ ension of what we deive d in case of 1D This statement is simpy an ext time-independent potentia

6 Patice in 3D Rigid Box : Sepaation of Othogona Spatia (x,y,z) Vaiabes TISE in 3D: - ψ( xyz,, ) + Uxyz (,, ) ψ( xyz,, ) = Eψ ( xy,, z) m x,y,z independent of each othe, wite ψ( xyz,, ) = ψ ( x) ψ ( y) ψ ( z) 1 3 and substitute in the maste TISE, afte dividing thuout by ψ= ψ ( x) ψ ( y) ψ ( z) 1 3 and noting that U()= fo (<x,y,z,<l) 1 ψ1( x) 1 ψ ( y) 1 ψ 3( z) + + E Const = = mψ1( x) x m ψ ( y) y mψ 3( z) z This can ony be tue if each tem is constant fo a x,y,z m ψ ( x) x ψ ( y) E x = Eψ ( y); m y 1 = 1ψ 1( ); m ψ 3( z) z = Eψ ( z) 3 3 With E + E + E = E=Constant 1 3 (Tota Enegy of 3D system) Each tem ooks ike patice in 1D box (just a diffeent dimension) So wavefunctions must be ike ψ ( x) sink x, ψ ( y) s in k y, ψ ( z) sink z

7 Patice in 3D Rigid Box : Sepaation of Othogona Vaiabes Wavefunctions ae ike ψ ( x) sin k x, ψ ( y) sink y, ψ ( z) sin k z Continuity Conditions fo ψi and its fist spatia deivatives niπ = Leads to usua Quantization of Linea Momentum p= k...in 3D k i L p x π π π = n1 ; p y = n ; pz = n3 (n 1,n,n3 = 1,,3,.. ) L L L Note: by usua Uncetainty Pincipe agument neithe of n,n,n =! ( why?) 1 π + + = m ml Patice Enegy E = K+U = K + = ( px py pz) ( n 1 3 Enegy is again quantized and bought to you by integes n,n,n 1 3 (independent) and ψ ()=Asin k x sin k ysin k z (A = Ovea Nomaization Constant) E -i t Ψ(,t)= ψ () e = A[ sin k x sin k ysin k z] -i e E t + n + n )

8 Patice in 3D Box :Wave function Nomaization Condition Ψ(,t)= ψ() e [ sin ysin ] e E E -i t -i t = A sin k1x k kz 3 Ψ Ψ E i t * * (,t)= ψ () e = A[ sin k1x s k sin k3z] * (,t) Ψ in y e (,t)= A [ sin k1x sin kysin k3z] Nomaization Condition : 1 = = A x,y, z P()dx dyd z L L L sin k1x dx k k3 x= y= z= E i t sin y dy sin z dz = A L L L 3 3 E -i t and Ψ(,t)= [ sin k1x sin kysin kz 3 ] e A= L L

9 Patice in 3D Box : Enegy Spectum & Degeneacy π E n 1,n,n = ( n n+ n3); n i= 1,,3..., ni ml 3π Gound State Enegy E111 = ml 6π Next eve 3 Excited states E 11= E11 = E11 = ml Diffeent configuations of ψ ()= ψ (x,y,z) have same enegy degeneacy z=l z x x=l y=l y

10 Gound State Degeneate States E = E = E = 6π ml E 111 ψ z ψ z E 11 E 11 E 11 y y x x

11 Pobabiity Density Functions fo Patice in 3D Box Same Enegy Degeneate States Cant te by measuing enegy if patice is in 11, 11, 11 quantum State

12 Souce of Degeneacy: How to Lift Degeneacy Degeneacy came fom the theefod symmety of a CUBICAL Box (L x = L y = L z =L) Enegy To Lift (emove) degeneacy change each dimension such that CUBICAL box Rectangua Box (L x L y L z ) Then E n1π nπ n3π = + + ml x ml y mlz

13 The Hydogen Atom In Its Fu Quantum Mechanica Goy kze U () = 1 1 U () = Moe compicated fom of U than box x + y + z By exampe of patice in 3D box, need to use sepeation of vaiabes(x,y,z) to deive 3 independent diffeentia. eqns. This appoach wi get vey ugy since we have a "conjoined tipet" To simpify the situation, use appopiate vaiabes Independent Catesian (x,y,z) Inde. Spheica Poa (, θφ, ) Instead of witing Lapacian 1 m x y = + 1 sin, wite 1 ψ(, θ, φ) 1 ψ(, θ, φ) sinθ + + sinθ θ θ sin + z 1 1 = + sin θ + sinθ θ θ TISE fo ψ(x,y,z)= ψ(, θφ, ) becomes ψ(, θ, φ) + θ φ (E-U()) ψ(, θ, φ) = θ φ!!!! fun!!!

14 Spheica Poa Coodinate System Voume Eement dv dv = ( sin θdφ) ( dθ)( d) = si nθddθdφ

15 Don t Panic: Its simpe than you think! 1 ψ 1 ψ 1 ψ m + (E-U()) (,, ) = + sin θ + ψ θ φ sin θ θ θ sin θ φ Ty to fee up ast tem fom a except φ This equies mutipying thuout by ψ ψ ψ m sin θ ke sin θ + sin θ sin θ + + (E+ ) ψ = θ θ φ Fo Sepeation of Vaiabes, Wite ψ (, θ, φ)=r(). Θ( θ ). Φ( φ) Pug it into the TISE above & divide thuout by ψ (, θ, φ)=r(). Θ ( θ ). Φ ( φ ) Note that : Ψ (, θ, φ ) Ψ (, θ, φ) θ Ψ (, θ, φ) θ sin θ R sinθ Θ R Θ θ θ Reaange by taking the φ tem on RHS sin + sinθ + θ R sinθ sin =Θ( θ ). Φ( φ) R() = R( ) Φ( φ ) Θ( θ ) θ = R( ) Θ( θ ) Φ( φ ) φ θ when substituted in TISE 1 Φ m sin θ ke + (E+ ) = Φ φ + sinθ + (E+ ) =- R Θ θ θ φ m sin θ ke 1 Φ Φ LHS is fn. of, θ & RHS is fn of φ ony, fo equaity to be tue fo a, θ, φ LHS= constant = RHS = m Θ

16 Now go beak up LHS to sepeate the & θ tems..... sin θ R sinθ Θ m sin θ ke LHS: + + (E+ ) R Θ θ θ sinθ =m θ Divide Thuout by sin and aange a tems with aw ay fom θ 1 R m ke m 1 Θ + (E+ )= sin θ R sin θ Θsinθ θ θ Same agument : LHS is fn of, RHS is fn of θ, fo them to be equa fo a, θ LHS = const = RHS = ( + 1) What do we have afte shuffing! d Φ + Φ = dφ m....(1) 1 d dθ m sin θ + ( + 1) ( θ )...() Θ = sinθ dθ dθ sin θ d R + + (E+ )- R( ) =...( 3) d 1 m ke ( 1) These 3 "simpe" diff. eqn descibe the physics of the Hydogen atom. A we need to do now is guess the soutions of the diff. equations Each of them, ceay, has a diffeent functiona fom

17 Soutions of The S. Eq fo Hydogen Atom d Φ + Φ = dφ The Azimutha Diff. Equation : m imφ Soution : Φ( φ ) = A e but need to check "Good Wavefunction Condition" Wave Function must be Singe Vaued fo a φ Φ( φ)= Φ ( φ + π) Φ( φ) = A e im φ = = ± ± ± Magnetic Quantum # im ( φ+ π) A e m, 1,, 3...( ) Φ 1 d dθ m The Poa Diff. Eq: sin θ + ( 1) ( θ) sinθ dθ dθ + sin θ Θ = Soutions : go by the name of "Associated Legende Functions" ony exist when the integes and m ae eated as foows =, ± 1, ±, ± 3... ± ; = positive numbe : Obita Quantum Numbe 1 Fo =, m = Θ( θ ) = ; Fo m = 1, m =, ± 1 Thee Possibiities fo the Obita pat of wavefunction 6 3 [ = 1, m = ] Θ ( θ) = cos θ [ = 1, m =± 1] Θ( θ) = sin θ 1 4 [ =, m = ] Θ( θ) = (3cos θ 1)... and so on and so foth (see book)

18 Soutions of The S. Eq fo Hydogen Atom 1 d R m ke ( + 1) The Radia Diff. Eqn: (E+ )- ( ) + R d = Soutions : Associated Laguee Functions R(), Soutions exist ony if: 1. E> o has negtive vaues given by m =, ± 1, ±, ± 3,... ± E=- a ke 1 a n mke ; = = Boh Radius. And when n = intege such that =,1,,3,4,,,( n 1) n = pincipa Quantum # o the "big daddy" qunatum # To Summaize : The hydogen atom is bought to you by the ettes n = 1,,3,4,5,... =,1,,3,,4...( n 1) Quantum # appea ony in Tapped systems The Spatia Wave Function of the Hydogen Atom (,, ) ( ). ( ). ( ) Y m Ψ θφ = R Θ θ Φ φ = R (Spheica Hamonics) n m n m

19 n Radia Wave Functions & Radia Pob Distibutions m R()= 1 3 e a 3/ -/a 1 (- )e a a 3/ 3/ - a ( ) e 81 3a a a n=1 K she n= L She n=3 M she n=4 N She 3a = s(hap) sub she =1 p(incipa) sub she = d(iffuse) sub she =3 f(undamenta) ss =4 g sub she..

20 Symboic Notation of Atomic States in Hydogen s ( = ) p ( = 1) d ( = ) f( = 3) g( = 4)... n 1 1s s p 3 3 s 3 p 3d 4 4 s 4 p 4 d 4 f 5 5 s 5 p 5 d 5 f 5g Note that: n =1 non-degeneate system n1>1 ae a degeneate in and m. A states have same enegy But diffeent spatia configuation E=- a ke 1 n

21 Facts About Gound State of H Atom 1 1 n = 1, =, m = R = Θ θ = Φ φ = a -/a () e ; ( ) ; () 3/ π 1 -/a 1(, θφ, ) e...ook at it caef a π Ψ = 1. Spheicay symmetic no θφ, dependence (stuctue). Pobabiity Pe Unit Voume : Ψ (, θ, φ) = 1 3 π a Likeihood of finding the eecton is same at a θφ, and depends ony on the adia sepeation () between eecton & the nuceus. ke 3 Enegy of Gound State =- = 13.6eV a Ovea The Gound state wavefunction of the hydogen atom is quite boing Not much chemisty o Bioogy coud deveop if thee was ony the gound state of the Hydogen Atom! We need stuctue, we need vaiety, we need some cuves! 1 e a uy

22 Intepeting Obita Quantum Numbe () 1 d dr m ke ( + 1) Radia pat of S.Eq n: + (E+ )- R( ) d d = ke Fo H Atom: E = K + U = KRADIAL + K ORBITAL ;substitute this fom fo E d dr - d d m 1 m ( + 1) K + RADIAL + KORBI TAL d + [ KRADIAL ] R = ORBITAL R () = ( + 1) Examine the equation, if we set KORBITAL = then get a diff. eq. in m 1 d dr m d Futhe, we aso know that K () which depends ony on adius of obit 1 L = mvobit ; L= p ; L =mv ob K = m ORBITAL ( + 1) L ORBITAL = Ang. Mom L = = + Putting it a togathe: K magnitude of m m L p ( 1) Since = positive intege=,1,,3...( n-1) angua mome ntum L = ( + 1) = discete vaues L = ( + 1) : QUANTIZATION OF Eecton's Angua Momentum

23 L = p (Right Hand Rue) Cassicay, diection & Magnitude of L aways we defined QM: Can/Does L have a definite diection? Poof by Negation: Suppose L was pecisey known/defined (L z) ˆ Since L = p Eecton MUST be in x-y obit pane p z = ; pz z p z ; E =!!! m So, in Hydogen atom, L can not have pecise measuabe vaue Uncetainty Pincipe & Angua Momentum : L Abitaaiy picking Z axis as a efeence diectio n : L vecto spins aound Z axis (pecesses). The Z component of L L = m ; m = ± 1, ±, ± 3... ± Z Note : L Z since m< ( + 1) < L (aways) It can neve be that L = m= ( + 1) (beaks Uncetainty Pincipe) So you see, the dance has begun! Z Magnetic Quantum Numbe m z φ

24 L=, m =,±1, ± : Pictoiay Sweeps Conica paths of diffeent ϑ: Cos ϑ= L Z /L and aveage <L X > = <L Y > =

25 What s So Magnetic? Pecessing eecton Cuent in oop Magnetic Dipoe moment µ Moe in this in Tomoow s ectue when we ook at Enegy States

26 n Pobabiity Density Function in 3D: Radia Pobabiity Densities Ψ (, θφ, ) = R ( ). Θ ( θ). Φ ( φ) = R m P(, θφ, ) = ΨΨ = Ψ(, θ, φ) = R. Y Note * m n : 3D Voume eement dv=.sin... Pob. of finding patice in a tiny voume dv is m P.dV = Rn. Y..sin θ. d. dθ. m θ d The Radia pat of Pob. distibution: P()d π π P()d= Rn. d Θm ( θ ) dθ m ( φ) Φ When Θ m P()d= R m n Y m dθ dφ ( θ) & Φ ( φ) ae auto-nomaized then n dφ n dφ d in othe wods R n.. ; P()= Nomaization Condition: 1 = R d Expectation Vaues <f( )>= f().p()d

27 Gound State: Radia Pobabiity Density Pd d () = ψ().4π Pd () = a e 3 4 a a e 3 Pobabiity of finding Eecton fo >a P > a a 4 a To sove, empoy change of vaiabe Define z= ; change imits of integation a P > a = = 1 z e z dz d (such integas caed Eo. Fn) 1 z =- [ z + z + ] e = 5e = %!!