Chemistry 460 Fall 2017 Dr. Jean M. Standard September 11, 2017 QUANTUM CHEMISTRY PROJECT 1 OUTLINE This project focuses on applications and solutions of quantum mechanical problems involving one-dimensional piece-wise constant potentials. In the first part of the project, you will study a system called a quantum corral, which has been realized in the laboratory by arranging atoms in a circle on a metal surface. Using methods similar to those discussed in class, you will solve the quantum corral problem and investigate its wavefunctions and energy eigenvalues. In the second part of the project, you will apply a simple numerical method in order to solve the onedimensional Schrödinger equation for the finite well problem. You also will be able to compare the energies and wavefunctions of the infinite well with those of the half-infinite well problem discussed in class. DUE DATE The project is due on MONDAY, SEPTEMBER 25, 2017. GRADING Project 1 is worth 50 points and consists of two parts. Part A is worth 25 points and Part B is worth 25 points.
PART A (25 POINTS) THE QUANTUM CORRAL 2 This problem illustrates the use of a quantum mechanical model to represent the behavior of electrons in a quantum corral. The figure below shows a plot of the electron density on a Cu metal surface on which individual Fe atoms (shown in orange) have been placed in a ring (or corral). The Fe atoms were positioned on the Cu surface by using the tip of a scanning tunneling microscope. Figure 1. Quantum corral. Reference for quantum corral figure and information: IBM Corporation, http://science.sciencemag.org/content/262/5131/218. See also http://nisenet.org/catalog/media/scientific_image_-_quantum_corral_top_view. Electrons inside the ring of Fe atoms are effectively trapped. This is like a two-dimensional particle in a circular box problem. The coordinates can be chosen to be r, the radius out from the origin at the center of the ring and the angle φ around the ring, such that the cartesian coordinates are x = r cosφ and y = r sinφ, and x 2 + y 2 = r 2. In our model of this behavior, the potential is defined to be V (r) = $ & 0, 0 r L '. (1) (, r > L ) Here, L is the radius of the corral. Note that the potential and the observed electron density in this case are independent of the angle φ and hence the problem reduces to a one-dimensional one that depends upon r only. The radial part of the kinetic energy operator in these coordinates can be defined as T ˆ =!2 2m # d 2 $ dr 2 + 1 r d & dr (. (2) ' The Schrödinger equation for the quantum corral is therefore d 2 2m + dr 2!2 1 r d dr ψ(r) + V(r) ψ(r) = E ψ(r). (3)
3 Since the potential is infinite (our assumption) outside the corral, the wavefunction is zero outside, ψ out (r) = 0. Thus, we only have to worry about the solution inside the corral, where the potential is zero. The Schrödinger equation for the solution inside the quantum corral is given by d 2 2m dr 2 + 1 d r dr ψ in (r) = E ψ in (r), (4)!2 where the potential V has been set equal to zero. In order to simplify the equation, we can scale the radial coordinate such that ρ = α r, with α = 2mE! 2. (5) The Schrödinger equation for the solution inside the quantum corral expressed in terms of the scaled radial coordinate ρ is d 2 dρ 2 + 1 d ρ dρ ψ in (ρ) = ψ in (ρ). (6) When this equation is rearranged so that all the terms are moved to the left side, it becomes d 2 dρ 2 + 1 d ρ dρ + 1 ψ in (ρ) = 0. (7) This is a well-known equation. The solutions are functions called Bessel functions. For this particular equation, the solution is a Bessel function of the first kind, denoted J 0 (ρ). This function looks something like a damped cosine function. Thus, we have that ψ in (ρ) = J 0 (ρ). (8) In this part of the project, you will explore the wavefunctions of the quantum corral and use techniques similar to those demonstrated in class for applying the boundary conditions to obtain the energy eigenvalues of the quantum corral. 1. First, you need to get a feel for what the Bessel function J 0 (ρ) looks like. Microsoft Excel includes the Bessel functions as built-in functions just like sine and cosine. In Excel, Bessel functions of the first kind are denoted BESSELJ(Z,N), where Z is the argument of the Bessel function (i.e., in our case it would represent the scaled radial coordinate ρ ) and N is the order of the Bessel function (in our case, N=0). Use Microsoft Excel to make a plot of the Bessel function J 0 (ρ) for ρ ranging from 0 to 20. Include this plot in your project. Select an appropriate step size in ρ such that your data generates a smooth plot. 2. Next, plot the probability density ψ in 2 (ρ) for ρ ranging from 0 to 20. How does the probability density compare with the quantum corral picture shown in Figure 1? Remember, you are plotting the radial part only since the solution is independent of the angular coordinate.
3. Recall the solution that we obtained in class for the particle in a one-dimensional box. The quantized energies came from applying the boundary conditions at x = 0 and x = L in order to require the wavefunction to be continuous. At x = L, the matching condition required that the wavefunction inside the box, which was proportional to sin kl, vanish. Therefore, we had to find the roots of the sine function; that is the locations where the sine function equals zero. This turned out to be the condition that kl = nπ, because the sine function has roots (or zeroes) at the values given by nπ. 4 Now, for the quantum corral, we need to do the same thing. Since the wavefunction outside the corral is zero, matching at r = L (or equivalently, ρ = αl) requires that the Bessel function vanish, J 0 (αl) = 0. So, we need to find the roots (or zeroes) of the Bessel function. Your plot from part #1 should give you some idea of the position of the roots; however, there is no simple formula for the roots like there is for the sine function. You will have to use the Solver in Microsoft Excel to find the roots. Using Solver, if you want the root of a function, you start with a guess. As an example, suppose you want to find the roots of the function f (x) = x 2 + x 2. This is a trivial example, since we can factor the quadratic function to obtain the roots analytically (they are x = 1 and x = 2). However, to find the roots numerically, a guess for the root is entered in one cell and the equation for the function is entered in another cell. For example, for this function, the following might be entered in the spreadsheet: 2 4 In the first cell, a guess for the root is entered this is the value "2". If there are multiple roots, then the guess will dictate which root is found. In the second cell, the value of the function evaluated for the guess is entered. In this case, if the guess is in cell A1, the equation entered in cell B1 would be "=A1^2+A1 2", so that the function f (x) = x 2 + x 2 is represented. The result that appears in cell B1 is the value f (2) = 4. You may need to add Solver to the Tools menu of Excel. To do this, select Tools Excel Add-ins. Check the box for Solver Add-in and click OK. Then, to use Solver to determine the roots, it must be selected from the Tools menu. Once selected, the Solver window appears:
5 Three settings must be entered in the Solver window to find the roots: (1) The Set Objective: box should be set to the cell location for the equation (in this case, $B$1 for cell B1). This may be typed in or the cell may be selected on the worksheet. (2) Since we want to find a root of the function, the To: setting should be set to Value of: 0. (3) And third, the By Changing Variable Cells: box should be set to the cell holding the guess (in this case, $A$1 for cell A1). Once these settings are entered, click Solve and the guess cell will be updated to hold the root x=1 (or something very close to 1), and the equation cell will be updated to hold value of the function (something very close to 0). For the example function, the results are shown below. 0.9999998-6.07E-07 For the other root, a guess of 1 might be entered, for which Solver returns the second root, x= 2. Using this method, employ Microsoft Excel to find the first five roots of the Bessel function, such that J 0 (ρ) = 0. Report these values of ρ. 4. Use the definition of the scaled coordinate ρ, along with the roots that you found in part (3), to calculate the energy of the first five levels of the quantum corral if the corral radius is 71.3 Å (this is the radius of the circle of Fe atoms shown in the figure). Report your energies in electron volts.
PART B (25 POINTS) PARTICLE IN A FINITE WELL 6 In this problem, you will consider a particle in a finite potential well of width L and depth V 0 : V (x) = # $ & V 0 0 V 0 x < 0 0 x L x > L ' (. ) V=V 0 I II III x=0 x=l Suppose for this problem that L=5.0 bohr and V 0 = 4.5 hartrees. It will be helpful to employ atomic units in this problem, and you will consider only the case for which E < V 0. While this problem may be solved in a manner similar to that shown in class for the half-infinite well, in this case you will solve the Schrödinger equation numerically to obtain the quantized energies and wavefunctions. The approach that you will take to find the solution will be to construct a numerical approximation of the solution using Microsoft Excel. The numerical method that you will apply to this problem is called the Numerov method. It is based on Taylor series approximation of the derivatives that appear in the Schrödinger equation. For small increments, δ, the second derivative of the wavefunction may be approximated as ψ ""(x) = ( ) ψ " ( x δ /2) ψ " x + δ /2 δ. (9) Similarly, finite difference approximations of the first derivative of the wavefunction are given as ψ ʹ(x + δ / 2) = ψ ʹ(x δ / 2) = ( ) ψ ( x) ψ x + δ ψ x δ ( ) ψ ( x δ) δ, (10). (11) Substituting these equations into the expression for the second derivative gives ψ "" (x) = ψ x + δ ( ) 2ψ x The one-dimensional Schrödinger equation may be written as [ ( ) E] ψ "" (x) = 2m V x! 2 ψ x ( ) ( ) + ψ x δ δ 2. (12) ( ). (13)
7 Equating the second derivative on the left side of the Schrödinger equation to the numerical approximation of the second derivative given above yields ( ) 2ψ( x) + ψ( x δ) ψ x + δ δ 2 = 2m V x [ ( ) E ]! 2 ψ x ( ). (14) This equation may be solved for ψ( x +δ) to yield a form of the Numerov algorithm for the Schrödinger equation, ψ( x + δ) = 2ψ( x) ψ( x δ) + δ 2 2m [ V( x) E ]! 2 ψ x ( ). (15) The Numerov algorithm works because if the right side of the equation is known, then the wavefunction ψ( x +δ) can be calculated. The mass and potential are determined by the system that is being investigated. The parameter δ is selected to be a very small increment in x such as 0.01 or 0.02. ( ) and ψ( x 0 δ), where x 0 is an initial value of x The Numerov algorithm is initiated by guessing values for ψ x 0 chosen to be in well into the classically forbidden region. Since the wavefunction is expected to be close to zero far into the classically forbidden region, it is usually acceptable to set ψ x 0 δ ( ) = 0 and ψ( x 0 ) = 0.0001, for example. Once the initial values for the wavefunction are guessed, the right hand side of the Numerov algorithm may be evaluated if the energy E is known. Since the energy is not known, it must be guessed. Once the guess for the energy is substituted into the right side of the Numerov equation, it may be evaluated to yield the left side, the value of the wavefunction at the next point, ψ x +δ the complete range of x. The calculated value of ψ x +δ ( ) for the next iteration. ψ x δ ( ). This process then can be iterated to generate the wavefunction over ( ) becomes ψ( x) and the previous value of ψ( x) becomes If the guess for the energy E was correct, then when the numerical solution of the wavefunction reaches the classically forbidden region for large x, it will go to zero. If the energy guess was wrong, then the wavefunction will diverge. A new guess must then be entered. This process is repeated until the boundary conditions for large x are satisfied. The energy and wavefunction obtained are the correct solutions of the Schrödinger equation. 1.) To begin, you will use Microsoft Excel to construct a set of data points to represent the potential energy of the finite well as a function of x. Since the Numerov algorithm must begin its iterations far into the classically forbidden region, you will want to generate the potential about 3 bohr on either side of the well. So, for example, if the well is defined from x=0 to x=5.0 bohr, then you should generate the potential from about x= 3 to x=+8 bohr. The step size for the Numerov algorithm also must be selected. A good recommendation is for you to use a step size δ=0.02 bohr. With the range of the potential and the step size selected, you will first generate the array of x values and then the array of data points for the potential. Start with the x array. Enter the value 3 in one of the cells (this is the smallest value of x that you will use in the Numerov method). In the cell below, define the next value to be the previous value plus the step size (0.02). The second cell should contain the value 2.98. Now, use the cursor to copy that calculation into the cells below and continue all the way down the column until you reach the value of x of 8.00. There are about 550 points. You now have generated the array of x values.
Next, since the potential is piecewise constant, in the cell next to x= 3.0, enter the value of V 0 (in this case, 4.5 hartrees). Then, copy this value in the cells below until you reach x=0. At this point, change the value that you are entering to 0.0 hartrees and copy that value into the cells until you reach x=l (in this case, x=5.0 bohr). Finally, in the remaining cells from x> 5.0 bohr to x=8.0 bohr, enter the value of V 0. You have now generated the array of data points for the potential energy function. 8 2.) Next you will need to enter a guess for the first energy eigenvalue in a cell somewhere on the spreadsheet. Start with a guess of 0.1 hartrees. At this point you are ready to enter the first two data points for the wavefunction into the spreadsheet and implement the Numerov algorithm. To start this, enter the value 0.00 into a cell next to the first point in the potential array. This is ψ x 0 δ ( ). Now, in the cell below that, enter the value 0.0001 this is ψ( x 0 ). These are the two wavefunction initial conditions required to start the Numerov algorithm. Recall that the Numerov algorithm is defined as ψ( x + δ) = 2ψ( x) ψ( x δ) + δ 2 2m [ V( x) E ] In atomic units, since m =1 and! = 1, the equation becomes! 2 ψ x ( ). ψ( x + δ) = 2ψ( x) ψ( x δ) + 2δ 2 [ V( x) E] ψ( x). In the third cell corresponding to the wavefunction array, enter this equation. Use the guess for the energy (0.1 hartrees) for E and for V x ( ), select the corresponding point from the data array that you generated in step 1. For the fourth and subsequent points of the wavefunction array, copy the formula that you entered in the third cell. If you have done this correctly, the third cell of the wavefunction array should hold the value 0.00020035 and the fourth cell should hold the value 0.00030141 for an energy guess of 0.1 hartrees. If you need assistance getting this to agree, please see Dr. Standard. 3.) In order to see whether or not the energy guess corresponds to the eigenvalue, you will need to plot the wavefunction. In the same worksheet, create a chart with the wavefunction on the y-axis and the x data on the x-axis. With E guess =0.1 hartrees, the plot of the wavefunction should look similar to the figure below. 100000 90000 80000 70000 60000 ψ(x) 50000 40000 30000 20000 10000 0-5.0-3.0-1.0 1.0 3.0 5.0 7.0 9.0 Trial wavefunction obtained for E guess =0.1 hartrees.
9 Notice that the trial wavefuntion blows up at large x. This means that the guessed energy did not correspond to an eigenvalue, and so the wavefunction does not satisfy the boundary conditions it is not a valid solution. You should be able to change the energy guess, and if you have done it correctly, the wavefunction and the graph should automatically be updated. If an energy guess of 0.2 hartrees now is entered, the wavefunction again blows up, but this time it gets very negative instead of very positive. When the trial wavefunction flips sign like this, it is an indication that the true energy eigenvalue lies between the two guesses (in this case, between E=0.1 and 0.2 hartrees). 5000 0-5.0-3.0-1.0 1.0 3.0 5.0 7.0 9.0-5000 -10000-15000 ψ(x) -20000-25000 -30000-35000 -40000-45000 Trial wavefunction obtained for E guess =0.2 hartrees. To complete the solution, all you need to do is systematically vary the energy guess until the trial wavefunction goes to zero for large values of x. When you are searching for energy eigenvalues using the Numerov method, you will need to be patient and carefully zoom in on the energy. You will need from about 3-6 decimal places in order to find a converged energy. A larger number of decimal places is required for the lower energy levels. The converged ground state wavefunction should look like the one in the figure below (note that this function is not necessarily normalized). 40 30 20 ψ(x) 10 0-5.0-3.0-1.0 1.0 3.0 5.0 7.0 9.0-10 Converged ground state wavefunction for finite well potential. 4.) To complete this portion of the project, use your Numerov algorithm to determine the energies and wavefunctions for all the bound states of the finite well with L=5.0 bohr and V 0 = 4.5 hartrees. Note that for a state to be bound, its energy must be lower than V 0. Tabulate the energy eigenvalues that you found using the Numerov method.
10 5.) Compare the energy eigenvalues that you found for the finite well in part #4 with the energies of a particle in an infinite box with the same well width and with a particle in a half-infinite box with the same well width and depth. What are the effects of the finite depth on the energies? Hint: You should be able to calculate the energy eigenvalues of the particle in an infinite box analytically for comparison. The energy eigenvalues of the particle in a half-infinite box with α=15.0 are listed below for [ ] 1/ 2. convenience. These energies were obtained by solving the equation z cot z = α 2 z 2 Half-Infinite Well Eigenvalues Eigenvalue No. Energy (hartrees) 1 0.173 2 0.692 3 1.548 4 2.726 5 4.157 6.) Include plots of the wavefunctions of each of the states that you found with the Numerov method for the particle in a finite well. Compare the tunneling behavior of these wavefunctions to those of the particle in an infinite box and to those of the particle in a half-infinite box. You should be able to generate wavefunctions for the particle in an infinite box from the analytic solutions (or from the Excel spreadsheet used in class) if you want to compare them to the wavefunctions generated from your Numerov method for the particle in a finite well. For the particle in a half-infinite well, you can generate the wavefunctions using the Excel spreadsheet from class; however, they are shown below and on the next page for your convenience. 0.7 0.8 0.6 0.6 0.5 0.4 ψ 1 (x) 0.4 0.3 0.2 0.1 ψ 2 (x) 0.2 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0-0.2-0.4-0.6 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 (a) (b) (a) Ground and (b) first excited state wavefunctions for particle in a half-infinite well with L=5.0 bohr and V 0 =4.5 hartrees. -0.8
11 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 ψ 3 (x) 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 ψ 3 (x) 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0-0.2-0.2-0.4-0.4-0.6-0.6-0.8 (c) (d) (c) Second and (d) third excited state wavefunctions for particle in a half-infinite well with L=5.0 bohr and V 0 =4.5 hartrees. -0.8 0.8 0.6 0.4 0.2 ψ 3 (x) 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0-0.2-0.4-0.6-0.8 (e) (e) Fourth excited state wavefunction for particle in a half-infinite well with L=5.0 bohr and V 0 =4.5 hartrees. To compare the half-infinite well wavefunctions shown above to the wavefunctions generated from your Numerov method for the particle in a finite well, you might want to normalize the Numerov wavefunctions. To do this, in another column in your Excel spreadsheet, calculate ψ 2 at each value of x. Then, the integral over all space can be approximated by summing up these values and multiplying by the step size, ψ * ( x) ψ( x) dx ψ 2 ( x i ) δ = δ ψ 2 ( x i ), n i=1 where δ is the step size for the Numerov method and we have assumed that the wavefunction is real. If the wavefunction is normalized, the integral and therefore the approximate summation should give the result of 1. If the sum is not equal to 1, then the wavefunction must be normalized and the sum can be set equal to1/n 2, where N is the normalization constant. Then, the normalization constant N may be determined as N = 1/ SUM, n i=1 where SUM is the summation defined above. Note that SUM(A1:A551) will sum the values in the cells from A1 to A551 in Excel. Remember to include the factor δ when defining the SUM. Then, to scale the wavefunction, in a new column in your Excel spreadsheet multiply all the values of ψ by the normalization constant N.