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Unformatted text preview: CHAPTER 3. 1. The linear operators are (a), (b), (f) 2. We have dx ' x ' Ïˆ ( x ') = Î»Ïˆ ( x )âˆž x âˆ« To solve this, we differentiate both sides with respect to x , and thus get Î» d Ïˆ ( x ) dx = x Ïˆ ( x ) A solution of this is obtained by writing d Ïˆ / Ïˆ = (1/ Î» ) xdx from which we can immediately state that Ïˆ ( x ) = Ce Î» x 2 / 2 The existence of the integral that defines O 6 Ïˆ (x) requires that Î» < 0. 3 , (a) O 2 O 6 Ïˆ ( x ) O 6 O 2 Ïˆ ( x ) = x d dx dx ' x ' Ïˆ ( x ')âˆž x âˆ« dx ' x ' 2 d Ïˆ ( x ') dx 'âˆž x âˆ« = x 2 Ïˆ ( x ) dx ' d dx 'âˆž x âˆ« x ' 2 Ïˆ ( x ') ( 29 + 2 dx ' x ' Ïˆ ( x ')âˆž x âˆ« = 2 O 6 Ïˆ ( x ) Since this is true for every Ïˆ ( x ) that vanishes rapidly enough at infinity, we conclude that [O 2 , O 6 ] = 2O 6 (b) O 1 O 2 Ïˆ ( x ) O 2 O 1 Ïˆ ( x ) = O 1 x d Ïˆ dx  O 2 x 3 Ïˆ ( 29 = x 4 d Ïˆ dx x d dx x 3 Ïˆ ( 29 =  3 x 3 Ïˆ ( x ) =  3 O 1 Ïˆ ( x ) so that [O 1 , O 2 ] = 3O 1 4. We need to calculate âŒ© x 2 âŒª = 2 a dxx 2 sin 2 n Ï€ x a a âˆ« With Ï€ x/a = u we have âŒ© x 2 âŒª = 2 a a 3 Ï€ 3 duu 2 sin 2 nu = a 2 Ï€ 3 Ï€ âˆ« duu 2 Ï€ âˆ« (1 cos2 nu ) The first integral is simple. For the second integral we use the fact that duu 2 cos Î± u =  d d Î± Ï€ âˆ« 2 du cos Î± u =  Ï€ âˆ« d d Î± 2 sin Î±Ï€ Î± At the end we set Î± = nÏ€ . A little algebra leads to âŒ© x 2 âŒª = a 2 3 a 2 2 Ï€ 2 n 2 For large n we therefore get âˆ† x = a 3 . Since âŒ© p 2 âŒª = h 2 n 2 Ï€ 2 a 2 , it follows that âˆ† p = h Ï€ n a , so that âˆ† p âˆ† x â‰ˆ n Ï€ h 3 The product of the uncertainties thus grows as n increases. 5. With E n = h 2 Ï€ 2 2 ma 2 n 2 we can calculate E 2 E 1 = 3 (1.05 Ã— 10 34 J . s ) 2 2(0.9 Ã— 10 30 kg )(10 9 m ) 2 1 (1.6 Ã— 10 19 J / eV ) = 0.115 eV We have âˆ† E = hc Î» so that Î» = 2 Ï€ h c âˆ† E = 2 Ï€ (2.6 Ã— 10 7 ev . m ) 0.115 eV = 1.42 Ã— 10 5 m where we have converted h c from J.m units to eV.m units. 6. (a) Here we write n 2 = 2 ma 2 E h 2 Ï€ 2 = 2(0.9 Ã— 10 30 kg )(2 Ã— 10 2 m ) 2 (1.5 eV )(1.6 Ã— 10 19 J / eV ) (1.05 Ã— 10 34 J . s ) 2 Ï€ 2 = 1.59 Ã— 10 15 so that n = 4 x 10 7 . (b) We have âˆ† E = h 2 Ï€ 2 2 ma 2 2 n âˆ† n = (1.05 Ã— 10 34 J . s ) 2 Ï€ 2 2(0.9 Ã— 10 30 kg )(2 Ã— 10 2 m ) 2 2(4 Ã— 10 7 ) = 1.2 Ã— 10 26 J = 7.6 Ã— 10 8 eV 7. The longest wavelength corresponds to the lowest frequency. Since âˆ† E is proportional to ( n + 1) 2 â€“ n 2 = 2 n + 1, the lowest value corresponds to n = 1 (a state with n = 0 does not exist). We therefore have= 0 does not exist)....
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 Spring '05
 mokhtari
 wave function, Even and odd functions, Parity

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