Midterm #1 - TakeHome - Answers

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Randolph Hensley
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1 Midterm #1 - TakeHome - Answers Engel 5.1 (p. 77-8) describes how to obtain total energy eigenfunctions and eigenvalues for an electron in a finite depth box of width 1.00 x 10-9 m with V o = 1. x J. Engel also lists the five lowest eigenvalues for this box. 1. What are the five lowest eigenvalues (in J) for an electron in an infinite depth box of the same width? Solution: E n = h n ê 8 ma h = Js me = kg a = 10 9 m n = 81,, 3, 4, 5< kg m s kg m ,, 3, 4, 5<

2 Exam1takehome_answers.nb En = h n 8mea kg m s, kg m s, kg m s, kg m s, kg m s =. Which system has lower eigenvalues, the finite depth box or the infinite depth box? Why? Solution: Engel p. 78 gives the five lowest energies of the finite depth box as: 4.61 μ 10-0 J, 1.84 μ J, 4.09 μ J, 7.13 μ J, 1.07 μ J Each energy state of the finite depth box is lower than the corresponding state of the infinite depth box. This can be explained by the boundary condition for the infinite depth box that requires the wave function to vanish at x = a. This forces the infinite depth box to curve faster over the interval 0 x a, and larger curvature corresponds to larger energy. 3. Use the formulas provided in Engel, and any graphing tools at your disposal, to determine the lowest eigenvalue (1% accuracy in E is adequate) for an electron confined to a finite depth box of the same width, but with V o =.4 x J. Solution: Engel, p. 77 shows two equations that must be satisfied by the energy of a stationary state (the energy must satisfy one of the equations). The top equation is used to generate the lowest energy state (see figure on p. 78), and it is written below in two pieces (LHStop = left-hand side, RHS = right-hand side).

3 Exam1takehome_answers.nb 3 Clear@EnergyD V0 = me = a = 10 9 hbar = LHStop = $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% me HV0 EnergyL %%%%%%%%%%%% a hbar RHStop = $%%%%%%%%%%%%%%%%%%%%%%%%%%%%% TanA $%%%%%%%%%%%%%%%%%%%%%%%%%%%%% E hbar hbar è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Energy è!!!!!!!!!!!!!!! Energy TanA è!!!!!!!!!!!!!!! Energy E Any E that satisfies this equation, i.e., makes LHStop = RHStop, is acceptable, but we want the lowest energy possible. Unfortunately, the equations cannot be solved algebraically, so we must 1) plot the functions on the left-hand side for a range of possible energies, and ) plot the functions on the right-hand side for the same energy range. The state's energy is given by the crossing point of these two graphs. Note #1: We already know from the previous problems that the ground state energy of the finite depth box must be less than the ground state energy of the infinite depth box ( J). Since V o for this problem is larger than V o for the previous problem, we also know that the answer must be greater than ground state energy given in the previous problem ( J). So we need to graph the energy range J. The following graph was arrived at by using trial-and-error to adjust the horizontal (Energy) and vertical ranges. The crossing point occurs at about J.

4 4 Exam1takehome_answers.nb RHStop<, 8Energy, , <, PlotRange 89.6, 10<D Graphics

5 Exam1takehome_answers.nb 5 FindRoot@LHStop == RHStop, 8Energy, <D 8Energy < Referring to Engel s original finite depth box (V o = 1. x J, a = 1 x 10-9 m) and the general formulas that he provides for the eigenfunctions, 4a. What are the normalized eigenfunctions for E = 4.61 x 10-0 J and 4.09 x J? Solution: The general formulas for the wave functions in regions I-III are given on Engel p. 77. Engel states that A' = B' = 0, which greatly simplifies the formulas for region I and region III. To take the next step, it is helpful to note the symmetry of the potential function. The box has a point of symmetry at x = 0, and all wave functions must have either even or odd symmetry with respect to this point. The ground state should be described by an even function (no nodes). The first and second excited states should be described by odd (one node) and even (two nodes) functions, respectively. The symmetry requirements introduce additional restrictions on A, B, C, and D. An even wave function has the same value at x = a/ and -a/. This means C = 0 and A = B. As it happens, E = 4.61 x 10-0 J is the ground state, and E = 4.09 x J is the second excited state, so both energies correspond to states described by even wave functions. Using the formulas in Engel, p. 77, and letting A = B, we can obtain the relative values of A, B, and D: A e -khaêl = D cosik a ÅÅÅÅ M A ÅÅÅÅÅ = D ekhaêl cosik ÅÅÅÅ a M and also A k e -khaêl = D k sinik a ÅÅÅÅ M A ÅÅÅÅÅ = ÅÅÅÅ k D k ekhaêl sinik ÅÅÅÅ a M

6 6 Exam1takehome_answers.nb V0 = me = a = 10 9 hbar = Energy = , < k = $%%%%%%%%%%%%%%%%%%%%%%%%%% me Energy hbar κ = $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% mehv0 EnergyL %%%%%%%%%%% hbar , < , < , < I obtained two formulas for A/D above. I evaluate both formulas below, but I will just use the first formula to give me "AoverD" for future work.

7 Exam1takehome_answers.nb 7 AoverD = Hκ alê CosA ka E i k j k y z Hκ alê SinA ka κ { E , 17.13< , < The small discrepancies between the two estimates of A/D ("AoverD") can be blamed on round-off errors in our energies. The wave function will be normalized if the following is true: B Ÿ -aê - k x x + D Ÿ aê -aê cos HkxL x + A Ÿ aê - k x x = 1 which can be rearranged to give: D A I ÅÅÅÅÅ A D M Ÿ aê - k x x + Ÿ aê -aê cos HkxL xe = 1 So once we know the values of the integrals, we can solve for D ("Dcoef") and then use this and A/D ("AoverD" from above) to obtain A ("Acoef"), which is also B. int1 = κ x x aê aê int = Cos@k xd x aê Integrate::gener : Unable to check convergence. More Integrate::gener : Unable to check convergence. More , < , <

8 8 Exam1takehome_answers.nb Dcoef = $%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 1 %%%%%%%%%%%%%%%%%%%% AoverD int1 + int , 415.< Acoef = AoverD Dcoef k κ , < , < , < Now that A and D have been calculated, it is possible to write the wave functions for all three regions (recall B = A). E = J Region I: Region II: Region III: y = μ 1010 x y = cosh μ 10 9 xl y = μ 1010 x E = J Region I: Region II: Region III: y = μ 1010 x y = 415 cosh8.18 μ10 9 xl y = μ 1010 x The following calculations double-check the coefficient values by testing whether the boundary condition is satisfied at x = a/.

9 Exam1takehome_answers.nb 9 Acoef κ HaêL Dcoef Cos@k Ha ê LD , < , < 4b. What are the probabilities of finding an electron in Regions I, II, and III, respectively, when the electron s energy E = 4.61 x 10-0 J and 4.09 x J, respectively? Solution: The probabilities are simply given by B Ÿ -aê - y I x, D Ÿ aê -aê y II x, and A Ÿ aê y III x ProbIII = Acoef int1 ProbII = Dcoef int ProbI = ProbIII ProbI + ProbII + ProbIII , < General::spell1 : Possible spelling error: new symbol name "ProbII" is similar to existing symbol "ProbIII". More , < General::spell1 : Possible spelling error: new symbol name "ProbI" is similar to existing symbol "ProbII". More , < 81., 1.< As expected, the probability of staying in the classical region (region II) is higher for the ground state (0.995) and lower for the second excited state (0.949). 4c. Use Mathematica to plot the normalized Region III eigenfunctions from problem 4a. Solution:

10 10 Exam1takehome_answers.nb Acoef κ , < , < PlotA x, x =, 8x, a ê, a<e Graphics Consistent with the answer to part 4b, the wave function in region III for the second excited state has a larger magnitude than the wave function for the ground state.

Appendix A. The Particle in a Box: A Demonstration of Quantum Mechanical Principles for a Simple, One-Dimensional, One-Electron Model System

Appendix A. The Particle in a Box: A Demonstration of Quantum Mechanical Principles for a Simple, One-Dimensional, One-Electron Model System Appendix A The Particle in a Box: A Demonstration of Quantum Mechanical Principles for a Simple, One-Dimensional, One-Electron Model System Real quantum mechanical systems have the tendency to become mathematically