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# chap35 - QUANTUM MECHANICS 35 2 EXERCISES Section 35.2 The...

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35.1 QUANTUM MECHANICS E XERCISES Section 35.2 The Schrödinger Equation 10. The one-dimensional wave function is related to the probability by Equation 35.2, 2 ( ) . dP x dx ψ = Since probability (a pure number) is dimensionless, the units of ψ must be the square root of the inverse of the units of x (a length), or 1/ 2 (meters) . 11. I NTERPRET We are given the wave function of a particle and asked about its probability distribution. D EVELOP Since the quantity 2 ( ) x ψ represents the probability density of finding the particle, the particle is most likely to be found at the position where the probability density 2 ( ) x ψ is a maximum. E VALUATE (a) The maximum of 2 2 2 2 2 / ( ) x a x A e ψ = is at 0 x = (calculate 2 ( )/ 0 d dx ψ = for corroboration). (b) The probability density 2 ( ) x ψ falls to half its maximum value 2 2 ( (0) ) A ψ = when 2 2 2 / 1/2 , x a e = or 1 2 ln 2 0.589 . x a a = ± = ± A SSESS The probability distribution is shown below. Note that 2 2 2 2 2 / ( )/ x a x A e ψ = peaks at 0 x = and is halved at / 0.589. xa = ± 12. I NTERPRET We normalize the wave function that is given graphically in Figure 35.19. The particle must be somewhere , so the normalization constant is chosen such that the integral of the wave function over all space is one. D EVELOP The probability of finding the particle within dx of x is 2 ( ) ( ). P x x ψ = We will write an equation for the wave function, square it, and integrate. Then we’ll set the value of A such that the integral over the entire region is one. E VALUATE For 2 0 , L x 2 ( ) . A L x x ψ = Since the wave function is symmetric, we’ll just integrate from 0 x = to 2 L x = and multiply by 2 to obtain the integral over all space. 3 2 2 2 2 / 2 2 2 0 4 3 1 2 3 3 L A x A L A L dx A L L L = = = = A SSESS The actual wave function for the region 2 0 L x is then 3/ 2 2 3 ( ) . L x x ψ = 13. I NTERPRET We use the normalization constant and the wave function from Exercise 12 to find the probability of finding the particle in the region 4 0 . L x The probability of finding a particle in a region dx is 2 . dx Ψ D EVELOP We integrate the square of the normalized wave function 3/ 2 2 3 ( ) L x x ψ = from 0 x = to 4 L x = to find the probability of finding the particle in the region 4 0 . L x 35

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35.2 Chapter 35 E VALUATE 3 / 4 2 3 3 3 0 12 12 1 1 3 16 4 L L P x dx L L = = = A SSESS There is a one in sixteen chance of finding the particle in the region described. The region is a quarter of the entire region where the particle could be, but the wave function is smaller in this region than elsewhere so the probability is less than one in four. 14. The wave function is like the first excited state of an infinite square well found in Section 35.3, except that a x a rather than 0 x L as in Fig. 35.5. The correspondence is explicit if one takes /2 a L = and /2. x x L = Then sin( / ) sin[(2 / ) ] sin(2 / ), xa xL xL π π π π = = − which is the wave function in the equation ( ) sin( ) n x L x A π ψ = for 2, n = except for an overall phase. The normalization constant for this wave function is 2/ 1/ A L a = = (see Equation 35.6). (Of course,
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