Chem_370_Handout__5_02_25_08

Chem_370_Handout__5_02_25_08 - 1. Chem 370 Handout Five...

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Unformatted text preview: 1. Chem 370 Handout Five Spring 200.8 s _ The Properties of the Stationary Quantum States In this handout we will prove the foilowing three (3) results for, the stationary states. '1. -Ifthe initial system wavefimction ‘I’(x,'t = 0) is . _ m ‘P(x.r=0)=*1’f(x) ' (eq 1) where ‘I’f(x) is an eigenfunction of the system’s Hamiltonian operator H; then the system '. wavefimetion for times t 2 0 is stx,r)=wn(x,r)=exp[—§Enk]w:(x); . (eqz) where E,T is the eigenvalue of 1:1 associated with the eigenfimction ‘I’f ' II. If ‘I’(x,t) is given by equation (2), then the probabuizy distributions P(A,t) for all system observablesA are independent of time. I "I. If ‘I’(x,t) is given by equation (2), then for all times 1‘ the system’s energy has the precise value Chem 370 Handout Five Spring 2008 The Properties of the Stationary Quantum States Proof of Result 1: We must show that T(x,t) of equation (2) both: (a) Satisfies the initial condition of equation (1). (b) Satisfies the time-dependent Schrodinger equation (TDSE) ‘ inflgf—“Lfifim. (eq 3) -t- Proof of point (a): From equation (2) I ‘I’(x,r=0)=expl0]‘i’..(x)=(1)‘I‘..(x)=‘1’f(x)- (es 4) Equation (4), howeVer, agrees with equation (1). Thus we have proven point (a). -l- Proof of point (5): Competing-equations g2) and (3) we must show that 0'“ . g m " x _i E - Ih-é—exp[-EE"I]‘FR (x)—I-Ie p[ hEflt]‘l'fl (eq 5) However, evaluating the left-hand side (th) of equation (5) gives lbs = m[— iEnJexp[— i134}? (3") (at; 6a) = Efl‘fix, t) Similarly evaluating the right-hand side (rhs) gives the = Iii exp[~— %Ent:|‘l’f _ _ _ “ E ._ exp[ Ent]H‘I’n (1:) (eq 65) = exp[_%gnt]g,wf(x) = E31167, 1‘) But compan'ng equations (64:) and (65) show that lhs = rhs. Therefore equation (5) is valid. Thus we'have proven point (5). ' ' Chem 370 Handout Five I Spring. 2008 The Properties of the Stationary Quantum States Proof of Result I I: We must show that _ (Ah:F¥'(x,t)[5xT(x,t)1dx=(A)Fo (eq7a) and m _ (A2),=1‘P'(x,t)[fiz‘l’(x,t)]abc=(Az)l=o; (em) since then both (A)! and (AA)' = [(443)t —(A):]p are-independent of time, tendering P(A,t) independent of time if it is a “normal” distribution. We wiil prove equation (7a). The proof of equation (71)) is very similar. a. Proof of equation (7a); Using equation (2), equation (7a) my be women as . (A)! : ilegp[mi}gn1]wf(x)]° A[exp[_%g,t]wf(x)]oc = fleet exp[+%Ent)exP(-%Ent)‘3[‘1'f(x)]<& ' _ = goo Mme ' (A), .—. ?\P'(x,r = 0)A‘I’(x,t = o)osc = (11):“ Thus we haveshowu (A): = {Aha , and hence have proven equotion (7a). 7 Chem 370 Handout Five Spring 2008 ' - The Properties of the Stationary Quantum States Proof of Result 1 l I: We must show that I ' (E), = E" . (eq 3a) ' and 7 r I (A15)? = 0 (eq 8b) if ‘I’(x,t) is given by equation (2). To prove equatiozls (8) we will evaluate (E), and ( 2),. Note by equations (7), (E)r : M and (1:72)r m<Ez) 'M. Thus we may determine (E)r and ex _ (E),=(E)m{[Wflxiifiwflxfldx ' (eq9a) and . (E2),=(E1)M=1[w:(x)Ihzw:(x)]e We.) But since ‘1? (x) is an eigenfunction of £1 with eigenvalue Eu fI-‘I'f(x)=E:i’f(x)- (eq 10a) Also applying fl to equation (1041) gives - mm) = 1&2 we) = omega}: En new) = Ef‘l’flx). (eq 1015) Using equation (10a) in equation (9a) gives (E), = ilwnxteeewe Menage =5". («1 1n) Similarly comparing equations (10b) and (915) gives (E2)! = E3. (eq 11b) Equations (11a) and (lib) imply that (Ag);=kfiz),—<E>:l”=[E:—(E..rl“=0. e 12) Equation (11a) is identicai to equation (So) while equation (12) is identical to equation (Sb). - Thus we have proven equations (3a) and (85). _ ...
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Chem_370_Handout__5_02_25_08 - 1. Chem 370 Handout Five...

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