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Modern Physics Quiz #3 NAME: ________________________ 1. The particle in a finite one-dimensional box. Somewhat more realistic than the particle in an infinitely-walled box problem is the particle in a finite one-dimensional box. If the potential energy well has length L and the sides of the box have height U, then 2 ( x) U ( x) ( x) E ( x) 2m x 2 0 if L2 x L2 where U ( x) L U if x 2 where U 0 2 Before we begin it is useful to identify reasonable units in which to work. Joules and meters don’t make sense here. However, the problem contains a natural unit of length (L) and a natural unit of energy. a. Show that by making the substitution x L z we can rewrite Schrodinger’s equation as 2 2 ( z ) U ( z ) ( z) E ( z) 2 L m z 2 0 if 12 z 12 where U ( z ) 1 U if z 2 2 b. That doesn’t look much better, but if we remember the particle in the infinite walled box of length L, the ground state of that wave function had energy Eunit 2 2 2 L2 m . So if E Eunit , show that Schrodinger's equation becomes 2 2 2 2 u ( z ) ( z ) ( z ) z where u ( z ) U ( z) Eunit Modern Physics Quiz 3 Page 2 This is much better. Simply by measuring position in units of L and energy in units of Eunit we have simplified our algebra considerably. This is what physicists refer to as using “natural units.” We now solve: u z where U ( x) u( z) Eunit . 2 The presence of 2 might look awkward, but it allows us to compare functions and energies to something that we know (a particle in an infinitely deep box). Inside of the finite one-dimensional box: if z 12 , then u( z) 0 so 2 The potential energy is never negative, so there can’t be a solution with a negative value of ( would curve away from the axis forever so such a solution would not have finite probability). Since is positive, the solution of this differential equation involves sines and cosines. c. Since the box is symmetric about the origin, must be symmetric about the origin as well. Explain why this means that possible wave functions include 2 odd ( z ) A sin z but not neither or even ( z ) A cos z ( z ) A sin z cos z Outside the box: if z 12 , then u( z) u so 2 u d. Show that the solution to this equation is exp Cz if z 12 ( z) BBexp , Cz if z 12 (a graph might help explain the signs!) where C u Modern Physics Quiz 3 Page 3 e. Now think of the edges the box (where the potential energy is discontinuous). What is continuous and what is discontinuous at an edge of the box? i. The wave function itself? (continuous or discontinuous?) ii. The first derivative of the wave function? (continuous or discontinuous?) iii. The second derivative of the wave function? (continuous or discontinuous?) f. If we look at the edge where z 12 or x L2 , show that if we apply your answers from part e to the “odd” wave functions odd we find: cos cos A sin A u , and B u exp u , u sin B exp 2 2 2 2 so 2 2 g. Using the same logic, show that if we look at even we find: sin 2 u cos 2 Modern Physics Quiz 3 Page 4 Now the fun part (or maybe just fun for your computer). We have absorbed the width of the potential well into our units. The height of the potential well is given (in units of Eunit ) by “u”. We now need to find the energy eigenvalues for this system. That is the same as finding the values of that satisfy: cos either sin 2 2 u sin or u cos 2 2 Which is the same as finding values of that satisfy: cos 2 u sin 2 sin 2 u cos 2 0 This cannot be solved analytically. You will have to solve it numerically. Different values of u will give you different solutions (and different numbers of solutions). You can use a graphing calculator, Maple, or a program such as Excel (look for functions with names like “Solver” or “Goal Seek”). Just a hint: if you copy things down by hand, you will have to keep a fair number of significant figures (five or six?) in your calculations. h. First set u = 6, so the height of the potential wall is six times higher than the ground state energy would be if the walls were infinitely high. i. Figure out how many eigenvalues of (or roots of the last equation) there are. A graph might help. How many are there? ii. Find all of them and record them in Table I on the following page. i. Now set u = 20. i. How many roots are there now? ii. Find all of them and record them in Table I on the following page. j. Now set u = 100. You don’t need to find all of the eigenvalues. Just find the first six. k. Now compare your results with the values we would get if u = . The first one has been written into the table. Fill in the next five. l. Now fill in Table II on the following page. For this table, instead of recording the energy levels, you just record the ratio of each energy level to the ground state energy level for that value of u. Thus for Table II, the entire first row is filled with the number 1.0000 (the ground state value divided by itself). In the column for u = 6, fill in the cell beneath the 1.0000 with the value of for the first excited state divided by the value of for the ground state. Fill in the rest of the table. Modern Physics Quiz 3 Page 5 Table I: Values of : Energy eigenvalues (in units of Eunit) for the particle in a finite walled box where the wall height is given by u (in units of Eunit) . Note: there are more cells than eigenvalues here, so some cells will be left blank. Value of u : 6 20 100 Ground energy: 0 1.0000 Next energy: 1 Next energy: 2 Next energy: 3 Next energy: 4 Next energy: 5 Table II: Ratios of to 0 : Energy eigenvalues of excited states divided by the ground state energy eigenvalue for different values of u. Again, there are more cells than eigenvalues here, so some cells will be left blank. Value of u : 0 / 0 1 / 0 2 / 0 3 / 0 4 / 0 5 / 0 6 20 100 1.0000 1.0000 1.0000 1.0000 Modern Physics Quiz 3 Page 6 2. The quantum mechanical harmonic oscillator. The solution of the harmonic oscillator problem in quantum mechanics has applications in all branches of atomic, solid state, and particle physics. To construct the Schrodinger equation for the harmonic oscillator, the potential energy function is U ( x) 12 kx 2 or 1 2 m 2 x 2 where k m , so 2 2 1 m 2 x 2 E 2 2m x 2 a. Finding the natural units: Step One. The only constants in the problem that we can use to find dimensions of length are , m, and . i. Show that the combination with dimensions of length is 1 Losc 2 m ii. Show that if we make the substitution x L z then Schrodinger’s equation becomes: 1 2 1 z 2 E 2 z 2 2 b. Now the easy part. Show that if we measure the energy in units of then 1 2 2 z 2 , where E . 2 z Modern Physics Quiz 3 Page 7 The differential equation on the previous page can be tough. It doesn’t immediately yield to any of the standard methods for solutions of differential equations. But it can be made a lot simpler if one notices that part of it at least looks quadratic: 2 2 2 z z z z . z z "looks sort of like" c. Check the similarity (see what “sort of like” means). Show that 1 1 2 z z 2 z 2 z z 1 2 z 2 z z z z Which means that if for some solution , n 1 2 2 z 2 n , then n 2 z n 1 z z n 12 n 2 z z n d. By the same logic show that 1 1 2 z z 2 z 2 z z 1 2 z 2 z z z z 1 so z 2 z 1 z n n 2 z n Modern Physics Quiz 3 Page 8 The last equation is crucial. Since the potential energy function is never negative, the eigenvalues of E (and thus ) must be positive. That means that we don’t expect to find a solution where = -1/2 , but we just might find a solution where = +1/2. e. Show that if we can find a solution to the first order equation z 0 0, then it will also be a solution to z 1 z z 0 0 12 2 z z 0 , where 0 1 2 Now to solve the first order equation. Let 0 ( z) A0 exp 0 ( z) where A0 is just a constant. f. Show that z 0 0, becomes z 0 z 0, or 0 ( z ) 12 z 2 z Modern Physics Quiz 3 Page 9 The previous solution turns out to be the ground state of the quantum harmonic oscillator. There is no state with lower energy. So we have found: m 2 x 2 0 A0 exp 12 z 2 A0 exp g. Go back to equations with x in them and show that this really is a solution to 2 2 1 m 2 x 2 0 ( x) E0 0 ( x), and find E0 2 2m x 2 h. Now find A0. Use the requirement that 2 0 dx 1 to find A0. - note #1: That is an integral with respect to x, not z. - note #2: You may want to know that exp y dy 2 Modern Physics Now define 1 A1 z 0 z Quiz 3 where Page 10 A1 is another constant. i. Find 1 ( z ) . (You don’t need to find A1 but you can if you want. It turns out to be easier to find A1 than it was to find A0.) j. Show that 1 2 2 z 2 1 1 1 , where 1 2 z 3 2 so E1 3 2 E0 . Modern Physics Quiz 3 Page 11 It turns out that 1 is the first excited state (okay, there is another “A” sitting out front that you might have to find eventually but you don’t need it now). There are no states with energies between 0 and 1 . As a matter of fact, you can find all of the solutions of the quantum harmonic oscillator using the “recurrence relation” n 1 An 1 z n z where An 1 is another constant. k. Extra Credit! Prove it! Show that for any n 1 2 if 2 z 2 n n n , and n 1 An 1 z n , 2 z z 1 2 then 2 z 2 n 1 n 1 n 1 , where n 1 n 1, 2 z so En 1 En E0 (n 1) n 1 12 . The tools you need are: 1 z 2 z 1 z 2 z 1 z n n 2 z 1 z n n 2 z n, n. and Apply those equations to the following to prove the result: 1 2 1 1 2 z 2 n1 z z n1 2 z z 2 2 z