# ch03 - CHAPTER 3 1 The linear operators are(a(b(f 2.We have...

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CHAPTER 3. 1. The linear operators are (a), (b), (f) 2. We have dx ' x ' ψ ( x ') = λψ ( x ) -∞ x To solve this, we differentiate both sides with respect to x , and thus get λ d ψ ( x ) dx = x ψ ( x ) A solution of this is obtained by writing d ψ / ψ = (1 / λ ) xdx from which we can immediately state that ψ ( x ) = C e λ x 2 /2 The existence of the integral that defines O 6 ψ (x) requires that λ < 0. 3 , (a) O 2 O 6 ψ ( x ) - O 6 O 2 ψ ( x ) = x d dx dx ' x ' ψ ( x ') - -∞ x dx ' x ' 2 d ψ ( x ') dx ' -∞ x = x 2 ψ ( x ) - dx ' d dx ' -∞ x x ' 2 ψ ( x ') ( 29 + 2 dx ' x ' ψ ( x ') -∞ x = 2 O 6 ψ ( x ) Since this is true for every ψ ( x ) that vanishes rapidly enough at infinity, we conclude that [O 2 , O 6 ] = 2O 6 (b) O 1 O 2 ψ ( x ) - O 2 O 1 ψ ( x ) = O 1 x d ψ dx - O 2 x 3 ψ ( 29 = x 4 d ψ dx - x d dx x 3 ψ ( 29 =- 3 x 3 ψ ( x ) =- 3 O 1 ψ ( x ) so that [O 1 , O 2 ] = -3O 1

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4. We need to calculate x 2 ⟩= 2 a dxx 2 s in 2 n π x a 0 a With π x/a = u we have x 2 ⟩= 2 a a 3 π 3 duu 2 s in 2 nu = a 2 π 3 0 π duu 2 0 π (1 - co s2 nu ) The first integral is simple. For the second integral we use the fact that duu 2 cos α u =- d d α 0 π 2 du cos α u =- 0 π d d α 2 sin απ α At the end we set α = . A little algebra leads to x 2 ⟩= a 2 3 - a 2 2 π 2 n 2 For large n we therefore get x = a 3 . Since p 2 ⟩= h 2 n 2 π 2 a 2 , it follows that p = h π n a , so that p x n π h 3 The product of the uncertainties thus grows as n increases. 5. With E n = h 2 π 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 λ so that λ = 2 π h c E = 2 π (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 m a 2 E h 2 π 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 π 2 = 1 .59 × 10 15 so that n = 4 x 10 7 . (b) We have E = h 2 π 2 2 m a 2 2 n n = (1 .05 × 10 - 34 J . s ) 2 π 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 h c λ = 3 h 2 π 2 2 ma 2 If we assume that we are dealing with electrons of mass m = 0.9 x 10 -30 kg, then a 2 = 3 h πλ 4 m c = 3 π (1 .05 × 10 - 34 J . s )(4 .5 × 10 - 7 m ) 4 (0 .9 × 10 - 30 kg )(3 × 10 8 m / s ) = 4 .1 × 10 - 19 m 2 so that a = 6.4 x 10 -10 m.

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• Spring '05
• mokhtari
• wave function, Even and odd functions, Parity

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