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# ch6 - Chapter 6 The Schrdinger Equation 6-1 Also The...

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123 Chapter 6 ! The Schrödinger Equation 6-1. Also, . The Schrödinger equation is then, with these substitutions, . Because the left side is real and the right side is a pure imaginary number, the proposed does not satisfy Schrödinger’s equation. 6-2. For the Schrödinger equation: Substituting these into the Schrödinger equation yields: , which is true, provided , i.e., if For the classical wave equation: (from Equation 6-1) From above: and also Substituting into Equation 6-1 (with replacing and v replacing c ) , which is true for . 6-3. (a) Substituting into the time-independent Schrödinger equation, Solving for V(x) ,

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Chapter 6 The Schrödinger Equation 124 (Problem 6-3 continued) where This is the equation of a parabola centered at x = 0. (b) The classical system with this dependence is the harmonic oscillator. 6-4. (a) (b) The classical turning points are the points where or . That occurs when , or when . (c) For a harmonic oscillator , so Thus, 6-5. (a)
Chapter 6 The Schrödinger Equation 125 (Problem 6-5 continued) (b) 6-6. (a) For a free electron V(x) = 0, so Substituting into the Schrödinger equation gives: and, since for a free particle, (b) (c)

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Chapter 6 The Schrödinger Equation 126 6-7. (a) And (b) 6-8. Normalization between and + is not possible because the value of the integral is infinite.
Chapter 6 The Schrödinger Equation 127 6-9. (a) The ground state of an infinite well is For m = m p , L = 0.1 nm : (b) For m = m p , L = 1 fm : 6-10. The ground state wave function is (n = 1) (Equation 6-32) The probability of finding the particle in x is approximately: (a) for (b) (c) for x = L, 6-11. The second excited state wave function is (n = 3) (Equation 6-32). The probability of finding the particle in x is approximately: (a) (b) (c) for x = L,

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Chapter 6 The Schrödinger Equation 128 6-12. (Equation 6-24) 6-13. (a) (b) 6-14. (a) This is an infinite square well of width L . V(x) = 0 and . From uncertainty principle: and (b) The solution to the Schrödinger equation for the ground state is: So, The result in (a) is about 1/10 of the computed value, but has the correct dependence on h , m , and L . 6-15. (a) For the ground state, . (b) Recall that state n has n half-wavelengths between x = 0 and x = L , so for n = 3, or .
Chapter 6 The Schrödinger Equation 129 (Problem 6-15 continued) (c) in the ground state.

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ch6 - Chapter 6 The Schrdinger Equation 6-1 Also The...

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