Quantum Chemistry Exam 2 Solutions (Take-home portion)

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1 Chemistry 46 Spring 5 Name KEY Quantum Chemistry Exam Solutions Take-home portion 5. 5 points In this problem, the nonlinear variation method will be used to determine approximations to the ground state of an anharmonic oscillator. The Hamiltonian for the oscillator can be written H ˆ x! m d dx + kx + 4 γ x 4, where k is the harmonic oscillator force constant and γ is a parameter. A nonlinear variational calculation employing the following normalized function will be carried out. x λ 4 e λ x. The variational parameter is λ. It is to be varied to determine the best approximate energy and wavefunction. For this problem, use atomic units and employ the following values for the parameters:! a.u., m a.u., k a.u., and γ.5 a.u. a. Plot the potential energy as a function of x. Use the parameters given above. On the same graph, plot the harmonic oscillator potential set γ. Use a range for x of to + a.u. You may use Excel or another graphing program to do this. What does the quartic anharmonicity do to the potential? A plot of the harmonic+quartic and the harmonic potentials is shown on the next page. The quartic anharmonicity causes the potential to rise more rapidly, which would be expected to increase the energy eigenvalues relative to the harmonic energies alone.

2 5 a. Continued Harmonic+Quartic Harmonic 3. Vx a.u x a.u. b. Using the approximate wavefunction given above, determine the energy expectation value. Your result may depend on the constants!, k, m, γ, and the nonlinear variation parameter λ. Do not use the numerical values of the parameters; work this out analytically. The energy expectation value is E ˆ H. For a normalized wavefunction as was indicated, this simplifies to E ˆ H. In order to evaluate this expectation value, we must operate the Hamiltonian operator on the approximate wavefunction. ˆ H x &! d m dx + kx + 4 γ x 4 + x H ˆ r! d m dx x + kx x + 4 γ x 4 x. Substituting the expression for the approximate wavefunction, this becomes H ˆ r! m ˆ H x λ, π + d dx x kx x + 4 γ x 4 x! m d dx e λ x + kx e λ x + 4 γ x 4 e λ x.

3 3 5 b. Continued Next, we can work out the second derivative, Substituting the derivative yields d $ & e λ x dx d $ dx & e λ x λx e λ x λ + 4λ x e λ x. ˆ H x λ 4 -.! e λ x + kx e λ x + 4 γ x 4 e λ x m λ + 4λ x λ 4-4 γ x 4 +!λ m + k & x +! λ. m e λ x. Inserting this expression into the energy expectation value yields H ˆ x ˆ H λ + - π, ˆ λ + - π, H x dx 4 e λ x λ π, 4 γ x 4 +!λ m + k + - x +! λ 4, m 3 e λ x 4 γ x 4 +! λ m + k + - x +! λ 4, m 34 e λ x dx λ + - π, 4 γ x 4 e λ x dx + λ + - π, + λ +! - λ e λ x dx π, m. The integrals may be evaluated using integral tables, 4 e λ x dx! λ m + k + - x e λ x dx, e bx dx x e bx dx x 4 e bx dx + b + 4b b 3 8b +. b Note that the integrals listed above must be multiplied by a factor of to account for the integration interval being [, ] rather than [, ].

4 4 5 b. Continued Substituting, the energy expectation value becomes ˆ H ˆ H λ 4 γ x 4 e λ x dx + λ + λ! λ m λ 4 γ 3 4 λ π & e λ x dx + & λ + λ! λ m π & λ 64λ +! λ m + k & 4λ +! λ m λ 64λ! m + k 8λ +! λ m k 64λ + 8λ +! λ m.! λ m + k & x e λ x dx λ π! λ m + k & π λ & λ c. Minimize the expectation value from part b with respect to the variational parameter λ to determine the equation for the value of λ that gives the best approximation to the energy. You may not be able to easily solve this equation to get the best λ. At this point, it is best to substitute in the numerical values of the parameters!, k, m, and γ. In order to solve the equation for λ, you will need to find the roots of the equation numerically. You can do this with a calculator or with Excel. If you want to use Excel to find the roots, follow a similar procedure to what you used in Project for the quantum corral. Once you have found the root λ, report its value. For a minimum, d E dλ. Evaluating the derivative and using the result from part 5b, d E dλ d dλ 3λ 3 $ & 64λ + k 8λ +! λ m k 8λ +! m. Setting this equal to zero yields 3λ 3 k 8λ +! m.

5 5 5 c. Continued The equation may be simplified somewhat by multiplying both sides by 3λ 3, 4kλ + 6! m λ3. This is a cubic equation in λ. Using the parameters m a.u., k a.u., and γ.5 a.u. and! a.u., of course, the equation becomes 6λ 3 4λ 3. Using Microsoft Excel to find the root yields the result λ.633. d. Using the value of λ that you found in part c, substitute this value into your expression for the energy expectation value obtained in part b to determine the best approximate energy. Use the numerical values of the parameters!, k, m, and γ. How does your eigenvalue compare to the ground state of the harmonic oscillator which is what you would get if γ? Is the quartic oscillator energy higher or lower than the harmonic oscillator energy? Discuss whether this result agrees with your expectations based upon the shape of the quartic potential plotted in part a. Substituting the value λ.633 into the energy expectation value yields the best approximation. From part 5b, the energy expectation value is E ˆ H k 64λ + 8λ +! λ m. Using the parameters m a.u., k a.u., γ.5 a.u., and! a.u., the energy expectation value becomes E H ˆ 3 8λ + 8λ + λ. Substituting λ.633, the best approximation to the ground state energy of the quartic oscillator for this trial wavefunction is E 3 8λ + 8λ + λ E.575 a.u

6 6 5 d. Continued The ground state eigenvalue of the harmonic oscillator is E hν, where the harmonic frequency ν is defined as ν $ k &. Substituting, π m E h $ k &! $ k &. π m m Using the parameters m a.u., k a.u., and! a.u., the ground state energy of the harmonic oscillator is E.5 a.u. So we see that the approximate energy of the ground state of the quartic oscillator is larger than that of the harmonic oscillator as expected. The quartic potential rises more steeply than the harmonic one, pushing the eigenvalues up in energy. e. Make a plot of the approximate wavefunction x using the value of λ that you determined in part c. Use a range for x from to a.u. On the same graph, plot the ground state harmonic oscillator eigenfunction, ψ x α 4 & π e α x # mk &, where α $!, for comparison this is what you would get if γ. You will need to use the numerical values of the parameters!, k, and m. How does the quartic anharmonicity affect the wavefunction? For the quartic oscillator, the approximate wavefunction x has the form x λ 4 e λ x, where the constant λ.633 for the best approximation. Using the parameters m a.u., k a.u., and! a.u., the ground state harmonic oscillator wavefunction has the form & π ψ x $ 4 e x, since the constant α with the defined parameters. Plots of the two wavefunctions are shown on the graph on the next page.

7 7 5 e. Continued.9.8 Quartic Harmonic.7 Wavefunction x a.u. Note that the wavefunction of the quartic oscillator has a narrower width than that of the harmonic oscillator. This is because the quartic potential rises more quickly than the harmonic oscillator potential, as we saw in part 5a.

Time dependence in quantum mechanics Notes on Quantum Mechanics

Time dependence in quantum mechanics Notes on Quantum Mechanics http://quantum.bu.edu/notes/quantummechanics/timedependence.pdf Last updated Thursday, November 20, 2003 13:22:37-05:00 Copyright 2003 Dan