Lecture23.pptx

# Lecture23.pptx - PHYS 360 Quantum Mechanics Mon Mar...

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Unformatted text preview: 3/8/10 PHYS 360 Quantum Mechanics Mon Mar 7, 2011 Lecture 23: How do we get spherical harmonics from spherical potenEals? Discussion QuesEons: 2 ˆd 1. Consider the operator: Q = 2 , and f (φ + 2π ) = f (φ ). dφ What steps do you take to prove if this is HermiEan? a. HW 7 due on Monday Mar 2: Ch 4, #2, 11, 13, 14, 15. b. c. 1 2 2 ˆd 1. Consider the operator: Q = 2 , and f (φ + 2π ) = f (φ ). dφ What steps do you take to prove if this is HermiEan? ˆ f Qg = 2π 2. The ground state wave funcEon of the inﬁnite square well is: ⎧ 0, x<a ⎪ ⎪ ψ 1(x) = ⎨ ⎪ ⎪ ⎩ 2 πx sin( ), 0 ≤ x ≤ a a a 0, a<x ∫ 0 dg f * 2 dφ dφ dg dφ 2π 2π 2 =f* − 0 ∫ 0 df * dg dφ dφ dφ 2π 2π a.  Is this an eigenfuncEon of momentum? d2 f * gdφ dφ 2 =f* dg dφ 2π − 0 df * g+ dφ 0 ∫ 0 b.  If so, what is its momentum? c.  If not, why not? b a ˆ = Qf g QED N.b. ∫ f dx dx = fg a b dg −∫ a b df gdx integration by parts 3 dx 4 a.  Is this an eigenfuncEon of momentum? b.  If so, what is its momentum? c.  If not, why not? ⎧ ⎪ ⎪ ψ 1(x) = ⎨ ⎪ ⎪ ⎩ 0, x<a 2 πx sin( ), 0 ≤ x ≤ a a a 0, a<x ⎧ ⎪ ⎪ ψ 1(x) = ⎨ ⎪ ⎪ ⎩ 0, x<a 2 πx sin( ), 0 ≤ x ≤ a a a 0, a<x ˆ pψ 1 ( x ) = d 2 πx 2π π x ⎡ π πx ⎤ sin( ) = cos( ) = ⎢ −i cot( ) ⎥ ψ 1 ( x ) i dx a a i aa a a⎦ ⎣a Can this wave funcEon be constructed by adding the momentum eigenfuncEons for +p1 and –p1? p12 = E1 2m → p1 = 2 mE1 px px ⎤ ⎡ φ p ( x ) = Ae− ipx / = A ⎢ cos( ) − i sin( ) ⎥ ⎦ ⎣ 5 6 1 3/8/10 OK, back to QM in 3 dimensions…. SeparaEon of variables: If the potenEal is only a funcEon of distance from the origin, i.e. V (r ) = V ( r ) = V (r ), …then it is best to use spherical coordinates: ⎡1 ∂⎛ ∂⎞ ⎛ ∂2 ⎞ ⎤ 1 ∂⎛ ∂⎞ 1 ∇2 = ⎢ 2 ⎜ r 2 ⎟ + 2 ⎜ sin θ ∂θ ⎟ + r 2 sin 2 θ ⎜ ∂φ 2 ⎟ ⎥ ⎠ ⎝ ⎠⎦ ⎢ ⎥ ⎣ r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝ Hence: ⎛ ∂ 2ψ ⎞ ⎤ 2 ⎡ 1 ⎛ 2 ∂ψ ⎞ 1 ∂⎛ ∂ψ ⎞ 1 r + sin θ + ⎢ ⎥ + V ( r )ψ = Eψ 2 m ⎣ r 2 ∂r ⎜ ∂r ⎟ r 2 sin θ ∂θ ⎜ ∂θ ⎟ r 2 sin 2 θ ⎜ ∂φ 2 ⎟ ⎦ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥ ⎢ − 7 8 First, just separate the radial from the angular funcEons: Set each independent term equal to a constant (=l(l+1)): The Radial EquaEon: 1 d ⎛ 2 dR ⎞ 2 mr 2 r − 2 ⎡V ( r ) − E ⎤ = l l + 1 ⎦ R dr ⎜ dr ⎟ ⎝ ⎠ ⎣ ψ ( r ,θ , φ ) = R( r )Y (θ , φ ) Insert…. 2 ⎡ Y ⎛ 2 ∂R ⎞ R ∂⎛ ∂Y ⎞ R ⎛ ∂ 2Y ⎞ ⎤ − r + sin θ + ⎢ ⎥ + V ( r ) RY = ERY 2 m ⎣ r 2 ∂r ⎜ ∂r ⎟ r 2 sin θ ∂θ ⎜ ∂θ ⎟ r 2 sin 2 θ ⎜ ∂φ 2 ⎟ ⎦ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥ ⎢ ( ) The Angular EquaEon: 1⎧ 1 ∂ ⎛ ∂Y ⎞ 1 ∂ 2Y ⎫ ⎪ ⎪ sin θ ⎨ ⎬ = −l l + 1 ⎟+ Y ⎪ sin θ ∂θ ⎜ ∂θ ⎠ sin 2 θ ∂φ 2 ⎪ ⎝ ⎩ ⎭ …and rearrange terms: ⎧ 1 d ⎛ 2 dR 2 mr 2 ⎫ 1⎧ 1 ∂ ⎛ ∂Y ⎞ 1 ⎛ ∂ 2Y ⎞ ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ ⎣ ⎦ ⎜r ⎟ − 2 ⎡V ( r ) − E ⎤ ⎬ + Y ⎨ sin θ ∂θ ⎜ sin θ ∂θ ⎟ + sin 2 θ ⎜ ∂φ 2 ⎟ ⎬ = 0 ⎝ ⎠ ⎝ ⎠⎭ ⎪ ⎪ ⎪ ⎪ ⎩ R dr ⎝ dr ⎠ ⎭ ⎩ 9 ( ) 10 sin θ θ⎛ ∂Y ⎞ ∂ 2Y sin θ + = − l ( l + 1) sin 2 θY ∂θ ⎜ ∂θ ⎟ ∂φ 2 ⎝ ⎠ ⎧1 ⎡ ⎫ 1 d 2Φ d⎛ dΘ ⎞ ⎤ ⎪ ⎪ 2 sin θ =0 ⎨ ⎢sin θ ⎥ + l ( l + 1) sin Θ ⎬ + 2 dθ ⎜ dθ ⎟ ⎦ ⎝ ⎠ ⎪ ⎪ ⎩Θ ⎣ ⎭ Φ dφ Separate variables again…. Let the two independent parts equal the constant m2: Y θ ,φ = Θ θ Φ φ Hence: () () () Theta equaEon: 1⎡ d⎛ dΘ ⎞ ⎤ 2 2 ⎢sin θ ⎜ sin θ ⎥ + l l + 1 sin θ = m Φ⎣ θ⎝ dθ ⎟ ⎦ ⎠ ( ) Phi equaEon: 1 d 2Φ = − m2 Φ dφ 2 ⎧1 ⎡ ⎫ 1 d 2Φ d⎛ dΘ ⎞ ⎤ ⎪ ⎪ 2 sin θ =0 ⎨ ⎢sin θ ⎥ + l ( l + 1) sin Θ ⎬ + 2 dθ ⎜ dθ ⎟ ⎦ ⎝ ⎠ ⎪ ⎪ ⎩Θ ⎣ ⎭ Φ dφ 11 12 2 3/8/10 Let’s actually solve the easy one: Phi equaEon: 1 d 2Φ = − m2 Φ dφ 2 i ∂Ψ ( r , t ) 2 2 =− ∇ Ψ ( r , t ) + V ( r , t )Ψ ( r , t ) ∂t 2m Ψn r ,t () − iE t / = ψ n ( r )e n − 2 2 ∇ ψ + V ψ = Eψ 2m d 2Φ = − m2 Φ ⇒ Φ φ = eimφ dφ 2 () ψ ( r ,θ , φ ) = R( r )Y (θ , φ ) = R( r )Θ(θ )Φ(φ ) The Radial EquaEon: Theta equaEon: Phi equaEon: 13 If we consider changing the wave funcEon by adding 2π to the value of ϕ, we require that the wave funcEon is unchanged: Φ φ + 2π = Φ φ 1 d ⎛ 2 dR ⎞ 2 mr 2 r − 2 ⎡V ( r ) − E ⎤ = l l + 1 ⎦ R dr ⎜ dr ⎟ ⎝ ⎠ ⎣ ( ) ( ) () This means m must be an integer, so: Φ φ = eimφ , 1⎡ d⎛ dΘ ⎞ ⎤ 2 2 ⎢sin θ ⎜ sin θ ⎥ + l l + 1 sin θ = m Θ⎣ θ⎝ dθ ⎟ ⎦ ⎠ ( ) () m = 0, ± 1, ± 2, .... 1 d 2Φ = − m2 Φ dφ 2 Φ φ = eimφ , () m = 0, ± 1, ± 2, .... 14 The Angular EquaEon – not so simple…. sin θ d⎛ dΘ ⎞ sin θ + ⎡ l l + 1 sin 2 θ − m2 ⎤ Θ = 0 ⎦ dθ ⎜ dθ ⎟ ⎣ ⎝ ⎠ ( ) The soluEon: Θ θ = APl m (cos θ ) Pl m ( x ) ≡ 1 − x 2 () …where: ( ) m /2 ⎛d⎞ ⎜⎟ ⎝ dx ⎠ m Pl ( x ) Pl ( x ) ≡ 15 …and: l 1 ⎛d⎞ Pl ( x ) ≡ l ⎜ ⎟ x 2 − 1 2 l ! ⎝ dx ⎠ l ( ) l 1 ⎛d⎞ 2 l ⎜ ⎟ x −1 ⎝ dx ⎠ 2 l! l ( ) 16 Note: l must be a nonnegaEve integer, so there are 2l+1 values of m for each l. l = 0, 1, 2, ...; m = − l , − l + 1, ..., −1, 0,1, ..., l − 1, l Volume element: NormalizaEon: d 3r = r 2 sin θ dr dθ dφ ∫ψ 2 r 2 sin θ dr dθ d φ = ∫ R r 2 dr ∫ 0 ∞ 2 2π 0 ∫ π 0 Y sin θ dθ dφ = 1 2 Normalize two parts separately: Pl m ( x ) ≡ 1 − x 2 ( ) m /2 ⎛d⎞ ⎜ dx ⎟ ⎝⎠ m Pl ( x ) 17 ∫ ∞ 0 R r 2 dr = 1 and 2 ∫∫ 0 2π π 0 Y sin θ dθ dφ = 1 18 2 3 3/8/10 Yl m θ , φ = ε () ( 2l + 1)( l − m ) ! imφ m e Pl cos θ 4π ( l + m ) ! ( ) The Radial EquaEon: d ⎛ 2 dR ⎞ 2 mr 2 ⎣ ⎦ ⎜r ⎟ − 2 ⎡V ( r ) − E ⎤ R = l ( l + 1) R dr ⎝ dr ⎠ SubsEtute: u( r ) ≡ rR( r ) 2 d 2u ⎡ 2 l l + 1 + ⎢V + 2 m dr 2 ⎢ 2m r 2 ⎣ 2 d 2ψ New radial equaEon: − ( ) ⎤ u = Eu ⎥ ⎥ ⎦ Looks like 1 ­D Schrödinger equaEon, − 2 m 2 + Veff ψ = Eψ dx 2 l ( l + 1) 2m r 2 Orthogonality: ∫ 0 2π π 0 [Yl (θ , φ )] [Y m * m' l' (θ ,φ )] sin θ dθ dφ = δ …with: ll ' mm ' 19 δ Veff = V + (centrifugal term) 20 To get anywhere, we have to choose a potenEal V(r). Consider the inﬁnite spherical well: Choose l=0: d 2u = − k 2u dr 2 → u (r ) = A sin( kr ) + B cos( kr ) cos( kr ) → ∞ as r → 0, so B = 0. r ⎧0, if r ≤ a ⎫ V (r ) = ⎨ ⎬ ⎩∞, if r > a ⎭ Outside the well, the wave funcEon is zero. Inside, ⎤ d 2u ⎡ l l + 1 =⎢ − k2 ⎥u dr 2 ⎢ r 2 ⎥ ⎣ ⎦ R( r ) = u ( r ) / r → ( ) where k ≡ 2 mE Require: sin( ka ) = 0, En 0 = n 2π 2 2 , (n = 1, 2, 3, ...) 2 ma 2 22 Hence: 21 Add angular part, normalize, and ground state wave funcEon is: ψ n 00 = 1 2π a sin( nπ r / a ) r En 0 = n 2π 2 2 , (n = 1, 2, 3, ...) 2 ma 2 NoEce the wave funcEon depends on m, but the energy doesn’t. What about other excited states? Wave funcEon has spherical Bessel funcEons, includes centrifugal potenEal…. ψ nlm ( r ,θ , φ ) = Anl jl ( β nl r / a )Yl m (θ , φ ) Enl = 2 β2 2 ma 2 nl 23 4 ...
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## This note was uploaded on 04/23/2011 for the course PHYS 360 taught by Professor Durbin,stephen during the Spring '11 term at Purdue.

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