Information • Textbooks • Media • Resources edited by
Computer Bulletin Board
Steven D. Gammon
A Graphical Approach to the Angular Momentum Schrödinger Equation
University of Idaho Moscow, ID 83844
Danny G. Miles Jr.* Department of Science, Mount St. Mary’s College, Emmitsburg, MD 21727-7796;
[email protected] Theresa A. Francis Department of Mathematics and Computer Science, Shippensburg University, Shippensburg, PA 17257-2299
Finding the general form of the exact solution of the onedimensional Schrödinger equation can require mathematical techniques that are unfamiliar to the average undergraduate. An alternative approach is to use numerical methods. Commercially available software, such as Mathcad, Maple, and Mathematica, allows students to graph the numerical solution of the differential equation, thus helping them to visualize fundamental concepts behind the theory. The student inputs a guess for the eigenvalue, observes a graph of the wave function on the computer, and successively improves the guess until a function is obtained with correct endpoint behavior. This is the eigenfunction. Numerical methods have been used to solve quantum mechanical problems such as the harmonic oscillator, the Morse oscillator, and the radial part of the hydrogen-like atom (1–9), but their application to angular momentum has received little attention in the literature (3). The material presented here can help students gain insight into this difficult topic by allowing them to generate for themselves angular momentum eigenfunctions and quantum numbers. Background and Rationale The Schrödinger equation for angular momentum can be written in spherical coordinates as
∂ 2 + cot θ ∂ + 1 ∂ 2 ψ θ, φ + βψ θ, φ = 0 (1) 2 ∂θ sin 2 θ ∂φ 2 ∂θ where the angle θ ranges from 0 to π, the angle φ ranges from 0 to 2π, ψ (θ, φ) is a function of the two angles, and β is a constant. Solution of the partial differential equation proceeds by assuming that the wave function ψ (θ, φ) can be written as the product of two wave functions ψ (θ, φ) = Θ(θ)Φ(φ)
(2)
where each function depends on one of the angular coordinates. Substitution of eq 2 into eq 1 separates the variables θ and φ and gives two ordinary differential equations:
d2Φ dφ 2
d2Θ
+ m 2Φ = 0
m2 + cot θ dΘ + β – Θ=0 dθ dθ 2 sin2 θ
(3)
(4)
where m is a constant. Equations 3 and 4 are one-dimensional Schrödinger equations. Equation 3 can be solved exactly for the wave function Φ without difficulty (10), and the solution forces m to be an integer (m = 0, ±1, ±2, …); different values of the quantum number m result in different wave functions. The component of the angular momentum along the z axis depends on m and is given by mh/(2π), where h is Planck’s constant. It is possible to solve eq 4 exactly for the wave function Θ using series (11–17), but the procedure can be intimidating to undergraduates. From the finite series solution, β = l(l + 1), where l must be an integer greater than or equal to |m|. The range of values of m is thus determined by a particular value of l; varying l or m results in distinct wave functions. The magnitude of the angular momentum, which depends only on l, is l l +1
h 2π
Introductory physical chemistry textbooks usually contain tables and graphs of the eigenfunction Θ and focus on the form of the solutions rather than on how they were obtained (10, 18). The requirement of a finite series solution for eq 4 restricts the possible values of m for a given value of l. However, not all students taking physical chemistry have the mathematical background to fully understand series solutions. If eq 4 is solved numerically rather than exactly, students could derive (instead of simply being told) the relationship between l and m. Numerical methods provide approximate values of the solution function at several points. The general procedure is to start with a given initial value of the function and its derivative and then to calculate subsequent values of the function at other points. Since the value of the function at a point is calculated from previous values, the error term accumulates. The “best” method can vary from one problem to another and depends upon how the error term is measured. Numerov’s method (2–9, 19) is easily programmed but it applies only to equations of the form y′′ = f (x)y, which makes rewriting eq 4 necessary (3). The Runge–Kutta method (1, 19) does not require rewriting eq 4, and a version of this method is already part of many mathematics software packages. The Runge–Kutta method included with Mathcad PLUS 6 for Macintosh was used for all calculations and graphs in the following section. We will provide Mathcad work sheets with examples upon request.
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Information • Textbooks • Media • Resources
Procedure and Examples The equation to be solved numerically is
Θ′′ + cot θ Θ′ + l l + 1 –
m2 sin2 θ
Θ=0
(5)
which is eq 4 with the derivative notation altered and with β replaced by l(l + 1). The solution of a second-order differential equation requires the value of the function and its derivative at some initial point. In the case of eq 5, the midpoint π/2 is used as the initial point for the integration because there are singularities in the differential equation (though not in the solution) at the endpoints 0 and π. The numerical integration then propagates symmetrically outward toward 0 and π. The general procedure is to make a guess for l (for a fixed value of m) and graph the solution. (It is assumed that students have solved eq 3 and know that m must be an integer.) Then the student makes another guess for l, graphs that solution, and studies both graphs. By this iterative method, students are able to zero in on a correct value of l. To critically compare the graphs, students should be aware that an eigenfunction exhibits a wavelike appearance and is mathematically well behaved (it and its first two derivatives are single-valued, continuous, and finite). The following examples illustrate the procedure. Since the approach presented here is graphical and is only concerned with the general shapes of the curves, the functions are not normalized. There are two important points to note. First, for m = 0, there is no component of angular momentum in the z direction. The maximum magnitude of the eigenfunction Θ is along the z-axis, and so Θ must be nonzero at the endpoints 0 and π. For m ≠ 0, there are components of angular momentum in the z direction. The maximum magnitude of Θ shifts away from the z axis, and Θ is zero at 0 and π (20–23). Second, for every physically meaningful combination of m and l, either Θ(π/2) or Θ′(π/2) must be zero; the other must be nonzero (in the examples below, it is arbitrarily set to one). If the eigenfunction has a node at π/2, then Θ(π/2) = 0 and Θ′(π/2) = 1; otherwise, Θ(π/2) = 1 and Θ′(π/2) = 0.
Figure 1. Graphs of eq 7 from Example 1 with l = m = 0. (a) c1 ≠ 0 (b) c1 = 0. In Figures 1–5, Cartesian graphs are on the left, polar graphs are on the right. (In Mathcad polar plots, the angle θ is measured in a counterclockwise direction from the positive horizontal axis.)
Example 1: m = 0, l = 0 The quantum number l can be associated with the magnitude of angular momentum, so a natural first guess is l = 0. For m and l both zero, eq 5 reduces to Θ′′ + (cot θ) Θ′ = 0
(6)
which can be solved analytically1 to give Θ = c1 ln |csc θ + cot θ| + c 2
(7)
where c1 and c2 are constants. It is obvious that c1 must be zero, resulting in a constant eigenfunction (Fig. 1). The value of c2 is arbitrarily set to one. Example 2: m = 0, l ≠ 0 The eigenfunction in Example 1 has no nodes; that is, it does not cross the θ axis. Since the number of eigenfunction nodes increases as the value of the quantum number l increases (analogous to the one-dimensional particle in a box and har-
406
Figure 2. Graphs from Example 2 with m = 0. (a) l = 0.20. (b) l = 0.80. (c) l = 0.92. To avoid the discontinuities in the differential equation, Figures 2–5 were created by joining segments of length π/2. One segment starts at π/2 and goes to 0; the other goes from π/2 to π. The Cartesian graphs consist of these two segments only; the range of the polar graphs has been extended to 2π by including the segments from 3π/2 to π and from 3π/2 to 2π.
Journal of Chemical Education • Vol. 78 No. 3 March 2001 • JChemEd.chem.wisc.edu
Information • Textbooks • Media • Resources Table 1. Graphical Parameters for ⌰ Θ(π/2)
Θ′(π/2)
0
0 1 2 3
1 0 1 0
0 1 0 1
0 1 2 3
≠.0 ≠.0 ≠.0 ≠.0
± .1
1 2 3
1 0 1
0 1 0
0 1 2
.0 .0 .0
± .2
2 3
1 0
0 1
0 1
.0 .0
± .3
3
1
0
0
.0
.
Figure 3. Graphs from Example 2 with m = 0. (a) l = 0.98. (b) l = 1.00. (c) l = 1.02.
No. of Endpoint Nodes Value
l
m
monic oscillator systems), the eigenfunction for the next value of l would then have one node, at π/2 by symmetry. The existence of a node at π/2 forces the initial conditions Θ(π/2) = 0, and Θ′(π/2) ≠ 0. Recall that this nonzero value is arbitrarily set to one and that, since m = 0, Θ is nonzero at the endpoints 0 and π. The student then chooses two fairly well-separated nonzero values for l and studies the Cartesian and polar graphs obtained (Fig. 2). As l gets closer to one, the behavior of each graph approaches that of an eigenfunction (Fig. 3). Note that when l is exactly one, the Cartesian graph of the function has no vertical asymptotes. Students should recognize the general shape of the polar graph, which can be more sensitive to slight changes in l as l gets closer to an acceptable value (Fig. 3). In a similar manner, the student can build up (l,m) pairs of (2,0) and (3,0). Table 1 contains Θ-eigenfunction initial values, number of nodes, and endpoint values for some physically meaningful l and m pairs. Example 3: |m| = 1, l = 1 For m ≠ 0, Θ(0) and Θ(π) are both zero. The eigenfunction for the lowest acceptable value of l has no nodes, so Θ(π/2) = 1 and Θ′(π/2) = 0. This leads to l = 1 exactly (Fig. 4), and then to other (l,m) pairs of (2, ±1), (2, ±2), (3, ±1), (3, ±2), and (3, ±3). Example 4: |m|= 2, l = 1 With the (l,m) pairs obtained so far, students determine that l is an integer. If they think any pair of integers is acceptable, they should try l < |m|; one such pair is (1,2). Whether students use Θ(π/2) = 0 and Θ′(π/2) = 1 as the initial conditions or vice versa, the resulting graph will not correspond to an eigenfunction (Fig. 5). Summary The approach to quantum mechanical angular momentum outlined here allows students in undergraduate physical chemistry to determine for themselves that 1. graphs of the Θ eigenfunctions have wavelike shapes, 2. l must be an integer,
Figure 4. Graphs from Example 3 with |m| = 1. (a) l = 0.98. (b) l = 1.00. (c) l = 1.02.
3. the value of l sets limits for values of m, and 4. there are 2l + 1 values of m for each value of l.
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Information • Textbooks • Media • Resources Next integrate both sides:
dχ χ =
cot θ dθ
so
ln |χ | = ln |sin θ | + ln c 1 = ln
c1 |sin θ |
Then
χ=
c1 sin θ
= c 1csc θ, dΘ = c 1csc θ, and dθ Θ = c1
csc θ dθ = c 1ln csc θ + cot θ + c 2
This is for any constants c1 and c2, so choosing c1 = 0 forces Θ = c2; normalizing then gives Θ = 1.
Literature Cited
Figure 5. Graphs from Example 4 with |m| = 2, l = 1. (a) Initial conditions are Θ(π/2) = 0 and Θ′(π/2) = 1. (b) Initial conditions are Θ(π/2) =1 and Θ′(π/2) = 0.
A worthwhile exercise is to have students compare and contrast features of the angular momentum problem with those of the one-dimensional particle in a box and harmonic oscillator. Quantum mechanical angular momentum is an extremely rich topic and offers many opportunities for further study. Applications include the angular dependence of the one-electron atom wave functions and probability densities, energy levels of the rigid rotor (in which the quantum number l is replaced by J ), and the directional properties of spherical harmonics (12, 23–25). Acknowledgments D.G.M. thanks Mount St. Mary’s College, the University of Wisconsin–Madison Department of Chemistry, and the New Traditions systemic chemistry initiative (NSF DUE 9455928) for support during sabbatical leave. Note 1. In the case l = m = 0, the differential equation can be solved exactly. In that case: Θ′′ + (cot θ)Θ′ = 0
Substitute χ = Θ′ (so χ′ = Θ′′) to get:
χ′ + cot θ χ = 0
or
dχ + cot θ χ = 0 dθ
Separate the variables by bringing cotangent to the other side and multiplying by dθ/χ:
dχ χ = cot θ dθ
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Journal of Chemical Education • Vol. 78 No. 3 March 2001 • JChemEd.chem.wisc.edu
