Lecture13Notes - Chem 120A 02/15/06 Spring 2006 READING:...

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Chem 120A Particle in a Box, Continued II and the Finite Square-Well 02/15/06 Spring 2006 Lecture 13 READING: Engel (QCS): Chapter 4 (you should finish both Ch. 4 and 5 before the end of next week) Properties of PIB solutions, continued 5. The kinetic energy is related to the curvature or wiggliness of the wavefunction. We can see this by examining the particle in a box wavefunctions. The wavefunctions, ψ ( x ) , shown in Figure5 become Figure 2. The wavefunctions and the square of the wavefunctions for a particle in a box of length L. increasingly wiggly as the energy for the wavefunction increases. For example, E 1 < E 2 < E 3 and we know that the potential energy is zero inside of the box. This implies that the kinetic energy, KE , is increasing as n increases since the total energy is the sum of the kinetic and potential terms. The reason why the curvature is directly related to kinetic energy is that is proportional to d 2 ( x ) / dx 2 . 6. Symmetry is very important in quantum mechanics. It helps us to simplify complicated problems. Parity is associated with the symmetry with respect to the origin in a coordinate system. For example, the finite square well potential in Figure 6 has V ( x )= V ( x ) . This symmetry is called ’inversion symmetry.’ The same symmetry applies for the infinite square well. Let’s consider the Schr¨odinger equation for the infinite square well potential, ¯ h 2 2 m d 2 ( x ) 2 + V ( x ) ( x E ( x ) . (1) Chem 120A, Spring 2006, Lecture 13 1
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-V 0 x V(x) Figure 1: Potential for a quantum particle in a 1D well of length a. V(x) = -V 0 , -a/2 < x < a/2 V(x) = 0, |x| r a/2 -a/2 a/2 II I III V=0 In order to investigate the symmetry of this equation with respect to the origin, we replace x with x , E ψ ( x )= ¯ h 2 2 m d 2 ( x ) d ( x ) 2 + V ( x ) ( x ) E ( x ¯ h 2 2 m d 2 ( x ) dx 2 + V ( x ) ( x ) ( x ) is also a solution to the Schr¨odinger equation with energy , E (2) In Lecture 12, we saw that the solutions to the Schr¨odinger equation for a particle in an infinite box in 1D are all non-degenerate, and that the eigenstates must also be real and normalized. The fact that ( x ) and ( x ) have the same energy implies that ( x A ( x ) , where A is a constant because the solutions must be non-degenerate. But, the normalization condition tells us that
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Lecture13Notes - Chem 120A 02/15/06 Spring 2006 READING:...

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