1. Szabo & Ostlund: 2.1, 2.2, 2.4, 2.5, 2.7. These problems are fairly straightforward and I will not discuss them here.

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Georgiana Joseph
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1 Solutio set III.. Szabo & Ostlud:.,.,.,.5,.7. These problems are fairly straightforward ad I will ot discuss them here.. N! N! i= k= N! N! N! N! p p i j pi+ pj i j i j i= j= i= j= AA ˆˆ= ( ) Pˆ ( ) Pˆ = ( ) ( PP ˆ ˆ ) pk = ( ) Pˆ = N! Aˆ k The crucial step is true because for fixed i, ( PP ˆˆ) rus over all permutatios, while the parity of ( PP ˆˆ) is determied by the umber of traspositio i this permutatio, which i j is the sum of pi + p j, so the sig is correct. i j, 3. Pˆ = ( TT... T ) = ( T... T T ) = ( T... T T ) = ( TT... T ) = P, which has the i k k k k i same parity as P ˆi, because the umber of traspositios is the same. We ca hece write N! N! N! N! pi ˆ pi ˆ pi ˆ k ( ) ( ) ( ) p ( ) ˆ i i i k i= i= i= k= Aˆ = P = P == P = P = Aˆ Summig over all iverse permutatios is the same as summig over all permutatios.

2 ˆ ˆ. You ca use the commutatio relatios bb, =, the defiitios ˆ ˆ b = ( qˆ ipˆ), b = ( qˆ+ ipˆ), where ˆp = i, as well as the Hermiticty of ˆp ad ˆq q i your proofs. a. ˆ φ b ψ = φ qˆ ipˆ ψ = ψ qˆ φ * i ψ pˆ φ * ˆ = ( ψ qˆ φ + i ψ pˆ φ ) = ψ qˆ+ ipˆ φ * = ψ b φ φ, ψ b. The statemet is true for = 0. If we assume that it is true for -, the we ca show it is also true for, so the proof follows by iductio: ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ b,( b ) = b, b ( b ) = b, b ( b ) + b b,( b ) ˆ ˆ ˆ ˆ = ( b ) + b ( )( b ) = ( b ) The iductio step for the other statemet follows alog similar lies. * * 5. The orthoormal eigefuctios of the harmoic oscillator are defied as ˆ = ( b ). Usig the relatios discussed i the previous questio show:! a. Let us show that is ideed ormalized to. Use a proof by iductio. For =0, it is true. Let us assume the statemet is true up to (-), the it is also true for. Cosider ˆ ˆ b b = bˆ bˆ bˆ!! 0( )( ) 0 0( ) ( ) 0 ˆ ˆ ˆ ˆ ˆ = 0( b) ( b,( b ) ( b ) b)0! + ˆ ˆ = b b = bˆ bˆ! ( )! 0( ) ( ) 0 0( ) ( ) 0 = Above you see a example of the use of the commutatio relatios from questio 5. The key is to brig the aihilatio operators to the left, as they give a magical 0 whe actig o the groud state.

3 b. ˆ ˆ + + ( ) 0 ( ˆ + b = b = b ) 0 = + +! ( + )! c. d. ˆ ˆ ˆ ˆ ˆ ˆ ˆ b = bb = b b b b bˆ!! + = =!! ( ) 0,( ) 0 ( ) 0 ( ) 0 m bˆ bˆ = bˆ bˆ bˆ + = m bˆ bˆ = m m m 0( )( ) 0 0( ),( ) 0 0 0( ) ( ) This process ca be repeated util either the power of ˆb or the power of ˆb equals 0. If > m the we are left with somethig proportioal to 0( ˆ m b) 0 = 0. I the other case we are left with bˆ = bˆ =. m m * 0( ) 0 0( ) Write the followig operators i terms of the ˆb ad ˆb, rememberig that these operators correspod to the dimesioless coordiates mometum operator ˆp = i. q q = mω x, ad the cojugate a. b. x = q ˆ ˆ ( ) mω = mω b+ b q mω mω i = i = i = mω p= i bˆ bˆ x q x q ˆ ( ) c. Kietic eergy operator m = ˆ ˆ = ˆ ˆ m x m ω ω ( b b) ( b b) d. Potetial eergy mω x = ˆ ˆ ω ˆ ˆ = mω ( b+ b ) = ( b+ b ) mω e. The Hamiltoia + mω x = ω[ bb ˆˆ+ bb ˆˆ] = ω+ ωbb ˆˆ m x

4 7. Let us cotiue the process. ˆ ˆ qˆ = ( b + b) ˆ ˆ ˆ ˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆˆ ˆˆ qˆ = ( b + b)( b + b) = ( b b + b b+ bb + bb) = ( b b + b b+ bb+ ) ˆ ˆ ˆˆ ˆˆ ˆˆ ˆ ˆ 3 qˆ = qˆ ( b + b) = ( bb + bb+ bb+ )( b + b) ˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆˆˆ ˆ ˆ = ( bbb + bbb+ bbb + bbb+ bbb + bbb+ b + b) ˆˆˆ ˆˆˆ ˆ ˆ ˆˆˆ ˆˆˆ =... = ( bbb + 3bbb+ 3b + 3b+ 3 bbb+ bbb) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 3 = (( bˆ ) + ( bˆ ) bˆ+ 3( bˆ ) bb ˆˆ + 3( bˆ ) bˆ + 3bˆ + 3bˆ bˆ+ 3bb ˆˆ + 3bˆ ˆ ˆ ˆ ˆ ˆ 3 ˆ 3 ˆ ˆ + 3 b ( b) b + 3 b ( b) + ( b) b + ( b) ) qˆ = qˆ ( b + b) = (( b ) + 3( b ) b+ 3b + 3b+ 3 b ( b) + ( b) )( b + b) ˆ ˆ 3ˆ = (( b ) + ( b ) b+ 3( bˆ ) + 3( bˆ ) bˆ+ 3( bˆ ) bˆ + 3bˆ + 3bb ˆˆ bb ˆˆ+ 3bˆ + 3( bˆ ) ( bˆ) ˆˆ ˆ ˆ 3 ˆ ˆ ˆ 3 ˆ + 6bb+ 3 b( b) + 3( b) + b( b) + ( b) ) 3 = ˆ + ˆ + ˆ ˆ+ ˆ ˆ + ˆ ˆ+ + ˆ ˆ + ˆ + ˆ 3 3 (( b ) 6( b ) ( b ) b 6( b ) b b b 3 b ( b) 6( b) ( b) ) ˆˆ λ H = + bb+ bˆ + bˆ + bˆ bˆ+ bˆ bˆ + bb ˆˆ+ + bˆ bˆ + bˆ + bˆ ˆ 3 3 (( ) 6( ) ( ) 6( ) 3 ( ) 6( ) ( ) ) The Hermiticity of the Hamiltoia is most easily show at this operator level. ˆ ˆˆ λ (( ˆ) 6( ˆ) ˆ( ˆ) 6( ˆ) ˆ ˆˆ 3 ( ˆ) ˆ 6( ˆ) ( ˆ) ) The terms are simply reshuffled i the Hamiltoia. 3 3 H = + bb+ b + b + b b + b b + bb+ + b b+ b + b = H ˆ

5 ˆ λ Hm = ( + ) δm, + [ ( m+ )( m+ )( m+ 3)( m+ ) δm, ( m+ )( m+ ) δ + m ( m+ )( m+ ) δ + 6 mm ( ) δ + mδ + 3δ + m, + m, + m, m, m, ( m ) m( m ) δ + 6 m( m ) δ m, m, + mm ( )( m )( m 3) δ ] m, The plots ad umerical results are geerated i the associated Maple file. There is a symmetry i the problem. The Hamiltoia is symmetric i poit q=0, ad this meas that eigefuctios are either eve or odd. If we add a basis state, this is also either eve or odd. If it is eve, oly the eve eigevalues (startig to cout from 0) ca improve. The eigevalues correspodig to odd eigefuctios stay the same. The diagoalizatio problem ca be split ito two. Oe for the eve fuctios ad oe for the odd. You will see zero for all matrix elemets that couple a odd ad a eve fuctio. This is why eigevalues oly improve i jumps of addig fuctios. Let us look ext at the double well potetial. The Hamiltoia matrix elemets are the same as above except for the additioal cotributios due to ˆˆ ˆˆ µ q = µ [ b b + b b+ bb ˆˆ+ ], which yields additioal cotributios: µ ( ( m+ )( m+ ) δm, + + mδm, + δm, + m ( m ) δm, ) If you choose µ large, such that the well becomes very deep, the eve ad odd solutios are almost degeerate, ad you will create a almost evely spaced eergy diagram, that cosists of harmoic oscillator fuctios i the local harmoic wells. Each level is early degeerate however, ad you ca iduce trasitios betwee these close lyig odd ad eve levels. The double well is potetial describes for example the umbrella mode i NH 3 (the trasitio state is the plaar D 3h geometry) ad the trasitios are oly a fractio of a wave umber. MASERS (the precursors of LASERS) are built o this pheomeo.

6 8a. The excited state Hamiltoia ca be writte as b. ω q H = E + ω( q d) q = E ω + ωq ωq d + ωd ˆ ˆ = E+ ω+ ωbˆˆ b ωd( bˆ + bˆ) + ωd = E+ ω+ ω[( b )( b)] = E+ ω+ ωb b b b = bˆ bˆ bˆ bˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ = bb db ( + b) + d ( bb db ( + b) + d) ˆ = [ b, bˆ] = [, ] ( )( ) ( )( ) It is ow easy to proof, usig the commutatio relatio that if is a eigestate with E = E+ ω + ω, the b is a eigestate with eigevalue E = E + + ω ( ) ω + +. Aalogous to the reasoig for the regular harmoic ( ) 0 oscillator the ormalized states are give by = b, where 0 is the groud! state of the displaced oscillator, satisfyig b 0 = ( bˆ d) 0 = 0. c. The excitatio eergies of the vibratioal levels i the electroically excited state w.r.t. the groud state vibratioal level are give by E E0 = E+ ω + ω ω = E+ ω. The 0-0 eergy is obtaied for = 0, ad is hece give by the costat E i the excited state Hamiltoia. The vertical excitatio eergy is give by the differece i potetial eergy betwee groud ad excited state at the groud state equilibrium geometry, which is defied by q = 0. Hece Evert = Vx( q= 0) V0 ( q= 0) = E+ ω ( q d) q = E+ ωd. The differece q= 0

7 i eergy betwee the bottom of the potetial wells is also give by E. This is related to the fact that the vibratioal frequecies i groud ad excited states are equal i this problem. I geeral the 0-0 trasitio is ot quite equal to the differece i eergy of the bottom of the potetial eergy wells. d.. * * * 0 = 0 = 0 b 0 = 0 ( b d) 0 =!! * * ˆ ˆ ( ) ( ) 0 (( d) + ( d) b+ ( )( d) b = 0 0 = 0 0!!! A formally better proof is obtaied by iductio. The statemet is true for = 0. Let us assume it is true for, ad cosider 0 + = 0 = 0 = ˆ = + ( ) ( ) 0 = = + +! ( + )! b b d b The crucial step occurs whe realizig that ˆ 0 b = 0 e. d ( d ) d 0 0 = = 0 0 = 0 0 = 0 0 e =!! = 0,... Hece 00 = e /, where I have assumed we choose the phases of the wave fuctio such that the overlap is real ad positive. You ca i priciple multiply by ay phase factor. f. For the trasitios 0, the eergy is give by E+ ω ( ω+ ω ) = E ω. The correspodig Frack_Codo factor is ˆ ( + d) 0 = 0 ( b ) 0 = 0 ( b + d) 0 = 0 0!!!. The square of this Frack- ( d) Codo factor is precisely the same as before e! g. I the plots you should see that the progressios become more itese with icreasig displacemets (larger d). You see far more lies that have appreciable itesity. The

8 emissio spectrum ad the absorbtio spectrum are mirror images of each other. They coicide for the 0-0 trasio. You may also ote that the 0-0 trasitio is ot ecessarily the most itese. This depeds o the displacemet.

Chapter 5 Vibrational Motion

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