1f5d745abd510a76cc199eef262af6e4.ppt

- Количество слайдов: 12

P 301 Lecture 19 “Fourier Decomposition/analysis” http: //demonstrations. wolfram. com/Wavepacket. For. AFree. Particle/ http: //www. jhu. edu/signals/fourier 2/index. html

Two-slit experiment with particles http: //en. wikipedia. org/wiki/Double-slit_experiment

Schrodinger’s Equation The above is taken from Wikipedia, and here the “Laplacian” operator in the first term on the right hand side is simply a short-hand for (s 2= d 2/dx 2 + d 2/dy 2 d 2/dz 2). We will concentrate (for the most part) upon the version that does not involve time and restrict ourselves to one dimension. The book goes through the details through the “separation of variables” technique, and the result is (for one dimension): This is the “time-independent Schrodinger equation” (the version we will deal with exclusively, in one dimension

P 301 Lecture 19 “JITT question” 1. Describe, as simply as you can, the “boundary conditions” that were imposed on the open and closed ends of organ pipes in P 221 (or whatever first semester intro Physics course you took). • 7 answers were confused or confusing • The boundaries were such that they did not cause destructive interference, that is, they were integer multiples of the wavelength of sound wave. (About 6 People concentrated on continuity of the function and wavelength. ) • [Never took P 221 or P 201, but I'll give it a shot. . . ] Nodes must occur at closed ends, antinodes at open ends, and the wavelengths of the resonant frequencies must occur in such a way to allow this restriction for the ends of the pipes. (10 people concentrated on this: This is exactly what “BOUNDARY CONDITIONS” refers to) • 18 didn’t answer.

Schrodinger’s Equation Conditions on the function y(x) 1. y(x) must be finite “everywhere” (except possibly for a set of measure zero). 2. y(x) must be single valued 3. For finite potentials both y(x) and dy(x)/dx must be continuous. 4. The integral of |y(x)|2 must be finite (and is normally chosen to be equal to one for a “properly normalized” wavefunction) 5. Behaviour at boundaries must be correct in order for all of these above conditions to be met, e. g. for a particle in a box, y(x) must be equal to zero at the box’s boundaries. There is a nice “live” graph showing the wavefunctions for the particle in a box (1 -D infinite square well) problem at: http: //bouman. chem. georgetown. edu/S 02/lect 13. htm Interesting “shooting algorithm example of ODE solving at: http: //www. cse. illinois. edu/iem/ode/shoot/

Schrodinger’s Equation Solutions Infinite Square Well http: //integrals. wolfram. com/index. jsp Useful web site for integrating

OPEN HOUSE 31 Oct 2009 • Volunteer via email to: [email protected] edu (Susie Rogers) • Or go to the front office.

Schrodinger’s Equation Solutions Finite Square Well Unlike the infinite case, in this case, there will only be a finite number of bound states (i. e. states for which there are classically forbidden regions where V>E.

Lecture 20 JITT question What is the most significant difference you see between the wave functions for a particle in an “infinite square well” and those for a particle in a “finite square well” of the same width? Which of well would you expect to have higher energy for its ground state (again, assuming both have the same width)? • Differences: • Particle can exist outside the well for the finite case (11 answers) • Larger l for the finite case (3 answers) • Other 5 • No answer (21 answers) • Higher Energy in which well? • Infinite case (15 answers) • Finite case (2 answers) • ? (2 answers)

Schrodinger’s Equation Solutions Finite Square Well For this case, you must solve a transcendental equation to find the solutions that obey the boundary conditions (in particular, continuity of the function and its derivative at the well boundary). For the geometry we considered in class, this takes on one of two forms: Even solutions k=k tan(ka/2) ; Odd solutions k= -k cot(ka/2) Recall from class that we can recast the transcendental equation into a dimensionless form, where the controlling parameter is the ratio of the potential well depth to the “confinement energy”. You can get the spreadsheet for the even solutions from the website: http: //physics. indiana. edu/~courses/p 301/F 09/Homework/ The fig. shows the marginal case for two even solutions (3 overall) a=p 2 9 7 5 LHS 3 RHS 1 -1 0 -3 -5 0. 2 0. 4 0. 6 0. 8 1 1. 2

Schrodinger’s Equation Solutions Harmonic Oscillator There an infinite number of possible states, since the potential is defined to keep going up (of course this is an idealization). Interestingly, the energy levels are evenly spaced in this case: E = hw(n + ½) n = 0, 1, 2, …

Lecture 20 JITT question How would you expect the light absorption spectrum for an electron trapped in a harmonic potential to differ from that for an electron trapped in an infinite square well potential? • Yes, because just like in the finite square well, harmonic oscillators don't need to have nodes at their edges unlike infinite square wells. Therefore, they work with different types of waves. (several, like this, were tangential to the question, e. g. focused on wavefunction not absorption) • I would expect it to take more energy to knock the infinite square well particles to a lower energy level, so they would emit more energetic photons. (about 4 people focused on absolute energy differences, but this depends on the parameters defining the wells, not just their shape). • Energy levels for the infinite well vary with the square of quantum #, whereas for SHM they vary with (n+0. 5). So spectrum bands for the SHM case should come at integral multiples of h-bar*omega whereas in the infinite well the bands will appear more irregular b/c of the exponential growth of E. (this is the key).