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Lecture14NotesPart2

Lecture14NotesPart2 - Chem 120A Spring 2006 READING...

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Chem 120A Probability Current and Free Particles 02/17/06 Spring 2006 Lecture 14 READING: Engel (QCS): Chapter 4 (you should finish both Ch. 4 and 5 before the end of next week) The finite square-well E < 0 solutions Last time we used parity considerations to find the solutions for the allowed energies of the particle in a finite square-well potential (Figure ) with E < 0. For solutions with even parity, we applied the boundary conditions to find the condition for the allowed energies, k tan ka 2 = κ . (1) The allowed energies can be found graphically by plotting both sides of Eq.1 as a function of E and finding the energies where they are equal. Similarly, for the odd parity solutions we find k cot ka 2 = κ . (2) -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 I II III V=0 This is the general procedure we have used to solve the Schr¨odinger equation for the infinite and finite square-well potentials: 1. Divide into regions (i.e. I, II, and III) and examine symmetry. 2. Write the Schr¨odinger equation for each region. Chem 120A, Spring 2006, Lecture 14 1
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3. Apply boundary conditions and get either implicit or explicit solutions. Comments about finite square-well solutions 1. We will always have at least one bound-state solution. 2. The deeper the well (larger magnitude of V 0 ) the more possible bound solutions there are. 3. This is truly a 1D problem, meaning that the dimensions are x . In 3D we use spherical coordinates and define r as a relative coordinate with 0 r . The potential in Figure3 is one example of a problem where we use the relative coordinate r . In this particular case, we do not always have a bound state. There is some minimum value of | U 0 | that is required to have a bound state. V(x) 8 -U 0 V=0 r r 0 V = 0, r>r 0 V = -U 0 , 0<r<r 0 V = infinity, r<0 Figure 2: Pseudo-1D potential used to treat the deuteron. 0 Example: the deuteron. This is the nucleus of deuterium, 2 H+. The exact nature of the interaction between the proton and the neutron is not known, but the potential in Figure3 models this system well. There is only one bound state in this model, with energy E 1 = 2 . 8 MeV, and radius r 0 = 2 × 10 15 m. For hydrogen, 1 H, the electron-proton radius is r 0 0 . 5 × 10 10 m, which is about 5 orders of magnitude larger than the neutron-proton distance (i.e. the nucleus is small compared to the
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