ch03 - CHAPTER 3. 1. The linear operators are (a), (b), (f)...

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Unformatted text preview: CHAPTER 3. 1. The linear operators are (a), (b), (f) 2. We have dx ' x ' ( x ') = ( x )- x To solve this, we differentiate both sides with respect to x , and thus get d ( x ) dx = x ( x ) A solution of this is obtained by writing d / = (1/ ) xdx from which we can immediately state that ( x ) = Ce x 2 / 2 The existence of the integral that defines O 6 (x) requires that < 0. 3 , (a) O 2 O 6 ( x )- O 6 O 2 ( x ) = x d dx dx ' x ' ( x ')-- x dx ' x ' 2 d ( x ') dx '- x = x 2 ( x )- dx ' d dx '- x x ' 2 ( x ') ( 29 + 2 dx ' x ' ( x ')- x = 2 O 6 ( x ) Since this is true for every ( x ) that vanishes rapidly enough at infinity, we conclude that [O 2 , O 6 ] = 2O 6 (b) O 1 O 2 ( x )- O 2 O 1 ( x ) = O 1 x d dx - O 2 x 3 ( 29 = x 4 d dx- x d dx x 3 ( 29 = - 3 x 3 ( x ) = - 3 O 1 ( x ) so that [O 1 , O 2 ] = -3O 1 4. We need to calculate x 2 = 2 a dxx 2 sin 2 n x a a With x/a = u we have x 2 = 2 a a 3 3 duu 2 sin 2 nu = a 2 3 duu 2 (1- cos2 nu ) The first integral is simple. For the second integral we use the fact that duu 2 cos u = - d d 2 du cos u = - d d 2 sin At the end we set = n . A little algebra leads to x 2 = a 2 3- a 2 2 2 n 2 For large n we therefore get x = a 3 . Since p 2 = h 2 n 2 2 a 2 , it follows that p = h n a , so that p x n h 3 The product of the uncertainties thus grows as n increases. 5. With E n = h 2 2 2 ma 2 n 2 we can calculate E 2- E 1 = 3 (1.05 10- 34 J . s ) 2 2(0.9 10- 30 kg )(10- 9 m ) 2 1 (1.6 10- 19 J / eV ) = 0.115 eV We have E = hc so that = 2 h c E = 2 (2.6 10- 7 ev . m ) 0.115 eV = 1.42 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 2 = 2(0.9 10- 30 kg )(2 10- 2 m ) 2 (1.5 eV )(1.6 10- 19 J / eV ) (1.05 10- 34 J . s ) 2 2 = 1.59 10 15 so that n = 4 x 10 7 . (b) We have E = h 2 2 2 ma 2 2 n n = (1.05 10- 34 J . s ) 2 2 2(0.9 10- 30 kg )(2 10- 2 m ) 2 2(4 10 7 ) = 1.2 10- 26 J = 7.6 10- 8 eV 7. The longest wavelength corresponds to the lowest frequency. Since E is proportional to ( n + 1) 2 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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ch03 - CHAPTER 3. 1. The linear operators are (a), (b), (f)...

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