The Three-imensional Schöinger Equation R. L. Herman November 7, 016 Schröinger Equation in Spherical Coorinates We seek to solve the Schröinger equation with spherical symmetry using the metho of separation of variables. The time-epenent Schröinger equation is given by i h Ψ t = h m Ψ V r Ψ. 1 Here Ψr, t is the wave function, which etermines the quantum state of a particle of mass m subject to a time inepenent potential, V r. From Planck s constant, h, one efines h = π h. The probability of fining the particle in an infinitesimal volume, V, is given by Ψr, t V, assuming the wave function is normalize, all space Ψr, t V = 1. One can separate out the time epenence by assuming a special form, Ψr, t = ψre iet/ h, where E is the energy of the particular stationary state solution, or prouct solution. Inserting this form into the time-epenent equation, one fins that ψr satisfies the timeinepenent Schröinger equation, h m ψ V r ψ = Eψ. Since the potential epens only on the istance from the origin, V = Vρ, ρ = r, we can further separate out the raial part of this solution using spherical coorinates. Recall that the Laplacian in spherical coorinates is given by = 1 ρ ρ 1 ρ ρ ρ sin θ sin θ θ θ 1 ρ sin θ φ. 3 Then, the time-inepenent Schröinger equation can be written as [ h 1 m ρ ρ ψ 1 ρ ρ ρ sin θ ψ 1 ] ψ sin θ θ θ ρ sin θ φ = [E Vρ]ψ. 4 We will use the metho of separation of variables. We assume that the wave function takes the form ψρ, θ, φ = RρYθ, φ. Inserting
the three-imensional schöinger equation this form into Equation 4, we obtain [ h Y m ρ ρ R R ρ ρ ρ sin θ Y R ] Y sin θ θ θ ρ sin θ φ = [E Vρ]RY. 5 Diviing by ψ = RY, multiplying by mρ, an rearranging, we have h 1 ρ R mρ R ρ ρ h [Vρ E] = 1 LY, 6 Y where L = 1 sin θ sin θ 1 θ θ sin θ φ. In Equation 6 we have a function of ρ equal to a function of the angular variables. So, we set each sie of Equation 6 equal to a constant, λ. The resulting equations are then ρ R mρ ρ ρ h [Vρ E] R = λr, 7 1 sin θ sin θ Y θ θ 1 sin θ We solve these equations in the next two sections. Y = λy. 8 φ Angular Solutions We seek solutions of Equation 8 by seeking prouct solutions of the form Yθ, φ = ΘθΦφ. Inserting this form into the equation, we have 1 sin θθ sin θ Θ θ θ 1 sin θφ Φ = λ. 9 φ The final separation can be performe by multiplying this equation by sin θ, rearranging the terms, an introucing a secon separation constant: sin θ Θ sin θ Θ λ sin θ = 1 θ θ Φ Φ = µ. 10 φ From this expression we can etermine the ifferential equations satisfie by Θθ an Φφ: sin θ sin θ Θ λ sin θ µθ = 0, 11 θ θ Equation 9 is a key equation which occurs when stuying problems possessing spherical symmetry. It is an eigenvalue problem for Yθ, φ = ΘθΦφ, LY = λy, where L = 1 sin θ θ sin θ θ 1 sin θ φ. The eigenfunctions of this operator are referre to as spherical harmonics. an Φ µφ = 0. 1 φ
the three-imensional schöinger equation 3 The simplest of these ifferential equations is Equation 1 for Φφ. The general solution is a linear combination of sines an cosines. Furthermore, in this problem uρ, θ, φ is perioic in φ, uρ, θ, 0 = uρ, θ, π, u φ ρ, θ, 0 = u φ ρ, θ, π. Since these conitions hol for all ρ an θ, we must require that Φφ satisfy the perioic bounary conitions Φ0 = Φπ, Φ 0 = Φ π. The eigenfunctions an eigenvalues for Equation 1 are then foun as Φφ = {cos mφ, sin mφ}, µ = m, m = 0, 1,.... 13 Next we turn to solving equation, 11. We first transform this equation in orer to ientify the solutions. Let x = cos θ. Then the erivatives with respect to θ transform as θ = x θ x = sin θ x. Letting yx = Θθ an noting that sin θ = 1 x, Equation 11 becomes 1 x y λ m x x 1 x y = 0. 14 We further note that x [ 1, 1], as can be easily confirme by the reaer. This is a Sturm-Liouville eigenvalue problem. The solutions consist of a set of orthogonal eigenfunctions. For the special case that m = 0 Equation 14 becomes x 1 x y λy = 0. 15 x In a course in ifferential equations one learns to seek solutions of this equation in the form yx = a n x n. n=0 This leas to the recursion relation a n = nn 1 λ n n 1 a n. Setting n = 0 an seeking a series solution, one fins that the resulting series oes not converge for x = ±1. This is remeie by choosing λ = ll 1 for l = 0, 1,..., leaing to the ifferential equation x 1 x y x ll 1y = 0. 16
the three-imensional schöinger equation 4 One might recognize this equation in the form 1 x y xy ll 1y = 0. The solutions are Legenre polynomials, enote by P l x. For the more general case, m = 0, the ifferential equation 14 with λ = ll 1 becomes x 1 x y x ll 1 m 1 x y = 0. 17 The solutions of this equation are calle the associate Legenre functions. The two linearly inepenent solutions are enote by Pl mx an Q m l x. The latter functions are not well behave at x = ±1, corresponing to the north an south poles of the original problem. So, we can throw out these solutions in many physical cases, leaving Θθ = Pl m cos θ as the neee solutions. In Table 1 we list a few of these. Pn m x Pn m cos θ P0 0x 1 1 P1 0 x x cos θ P1 1x 1 x 1 sin θ P 0x 1 3x 1 1 3 cos θ 1 P 1x 3x1 x 1 3 cos θ sin θ P x 31 x 3 sin θ P3 0x 1 5x 3 3x 1 5 cos 3 θ 3 cos θ P3 1x 3 5x 11 x 1 3 5 cos θ 1 sin θ P3 x 15x1 x 15 cos θ sin θ P 3 3 x 151 x 3 15 sin 3 θ The associate Legenre functions are relate to the Legenre polynomials by 1 Pl mx = 1m 1 x m/ m x m P lx, 18 for l = 0, 1,,,... an m = 0, 1,..., l. We further note that Pl 0x = P l x, as one can see in the table. Since P l x is a polynomial of egree l, then for m > l, m x m P l x = 0 an Pl m x = 0. Furthermore, since the ifferential equation only epens on m, P m l x is proportional to Pl m x. One normalization is given by Pl m m l m! x = 1 l m! Pm l x. The associate Legenre functions also satisfy the orthogonality conition 1 Pl mxpm l x x = l m! l 1 l m! δ ll. 19 1 associate Legenre functions Table 1: Associate Legenre Functions, P m n x. 1 The factor of 1 m is known as the Conon-Shortley phase an is useful in quantum mechanics in the treatment of agular momentum. It is sometimes omitte by some Orthogonality relation.
the three-imensional schöinger equation 5 Spherical Harmonics The solutions of the angular parts of the problem are often combine into one function of two variables, as problems with spherical symmetry arise often, leaving the main ifferences between such problems confine to the raial equation. These functions are referre to as spherical harmonics, Y lm θ, φ, which are efine with a special normalization as Y lm θ, φ = 1 m l 1 l m! 4π l m! Pm l cos θeimφ. 0 Y lm θ, φ, are the spherical harmonics. Spherical harmonics are important in applications from atomic electron configurations to gravitational fiels, planetary magnetic fiels, an the cosmic microwave backgroun raiation. These satisfy the simple orthogonality relation π π 0 0 Y lm θ, φy l m θ, φ sin θ φ θ = δ ll δ mm. As seen in the last section, the spherical harmonics are eigenfunctions of the eigenvalue problem LY = ll 1Y, where L = 1 sin θ sin θ 1 θ θ sin θ φ. This operator appears in many problems in which there is spherical symmetry, such as obtaining the solution of Schröinger s equation for the hyrogen atom as we will see later. Therefore, it is customary to plot spherical harmonics. Because the Y lm s are complex functions, one typically plots either the real part or the moulus square. One renition of Y lm θ, φ is shown in Figure for l, m = 0, 1,, 3. We coul also look for the noal curves of the spherical harmonics like we ha for vibrating membranes. Such surface plots on a sphere are shown in Figure 3. The colors provie for the amplitue of the Y lm θ, φ. We can match these with the shapes in Figure by coloring the plots with some of the same colors as shown in Figure 3. However, by plotting just the sign of the spherical harmonics, as in Figure 4, we can pick out the noal curves much easier.
the three-imensional schöinger equation 6 l = 0 m = 0 m = 1 m = m = 3 Table : The first few spherical harmonics, Y lm θ, φ l = 1 l = l = 3 l = 0 m = 0 m = 1 m = m = 3 Table 3: Spherical harmonic contours for Y lm θ, φ. l = 1 l = l = 3 l = 0 l = 1 l = l = 3 m = 0 m = 1 m = m = 3 Table 4: In these figures we show the noal curves of Y lm θ, φ Along the first column m = 0 are the zonal harmonics seen as l horizontal circles. Along the top iagonal m = l are the sectional harmonics. These look like orange sections forme from m vertical circles. The remaining harmonics are tesseral harmonics. They look like a checkerboar pattern forme from intersections of l m horizontal circles an m vertical circles.
the three-imensional schöinger equation 7 Spherical, or surface, harmonics can be further groupe into zonal, sectoral, an tesseral harmonics. Zonal harmonics correspon to the m = 0 moes. In this case, one seeks noal curves for which P l cos θ = 0. Solutions of this equation lea to constant θ values such that cos θ is a zero of the Legenre polynomial, P l x. The zonal harmonics correspon to the first column in Figure 4. Since P l x is a polynomial of egree l, the zonal harmonics consist of l latituinal circles. Sectoral, or meriional, harmonics result for the case that m = ±l. For this case, we note that P l ±l x 1 x m/. This function vanishes for x = ±1, or θ = 0, π. Therefore, the spherical harmonics can only prouce noal curves for e imφ = 0. Thus, one obtains the meriians satisfying the conition A cos mφ B sin mφ = 0. Solutions of this equation are of the form φ = constant. These moes can be seen in Figure 4 in the top iagonal an can be escribe as m circles passing through the poles, or longituinal circles. Tesseral harmonics consist of the rest of the moes, which typically look like a checker boar glue to the surface of a sphere. Examples can be seen in the pictures of noal curves, such as Figure 4. Looking in Figure 4 along the iagonals going ownwar from left to right, one can see the same number of latituinal circles. In fact, there are l m latituinal noal curves in these figures In summary, the spherical harmonics have several representations, as show in Figures 3-4. Note that there are l noal lines, m meriional curves, an l m horizontal curves in these figures. The plots in Figure are the typical plots shown in physics for iscussion of the wavefunctions of the hyrogen atom. Those in 3 are useful for escribing gravitational or electric potential functions, temperature istributions, or wave moes on a spherical surface. The relationships between these pictures an the noal curves can be better unerstoo by comparing respective plots. Several moes were separate out in Figures 1-?? to make this comparison easier. Figure 1: Zonal harmonics, l = 1, m = 0. Figure : Zonal harmonics, l =, m = 0. Figure 3: Sectoral harmonics, l =, m =. Figure 4: Tesseral harmonics, l = 3, m = 1. Raial Solutions So, any further analysis of the problem epens upon the choice of potential, Vρ, an the solution of the raial equation 7. For this, we turn to the etermination of the wave function for an electron in orbit about a proton. Example 1. The Hyrogen Atom - l = 0 States Historically, the first test of the Schröinger equation was the etermination of the energy levels in a hyrogen atom. This is moele by an electron
the three-imensional schöinger equation 8 orbiting a proton. The potential energy is provie by the Coulomb potential, e Vρ = 4πɛ 0 ρ. Thus, the raial equation becomes ρ R [ mρ e ] ρ ρ h 4πɛ 0 ρ E R = ll 1R. 1 Solution of the hyrogen problem. Before looking for solutions, we nee to simplify the equation by absorbing some of the constants. One way to o this is to make an appropriate change of variables. Let ρ = ar. Then, by the Chain Rule we have ρ = r ρ r = 1 a r. Uner this transformation, the raial equation becomes r u ma r [ e ] r r h 4πɛ 0 ar E u = ll 1u, where ur = Rρ. Expaning the secon term, ma r [ e ] [ mae h 4πɛ 0 ar E u = we see that we can efine πɛ 0 h r mea h r ] u, a = πɛ 0 h me, 3 ɛ = mea h = πɛ 0 h me 4 E. 4 Using these constants, the raial equation becomes r u ru ll 1u = ɛr u. 5 r r Expaning the erivative an iviing by r, u r u 1 ll 1 u r r u = ɛu. 6 The first two terms in this ifferential equation came from the Laplacian. The thir term came from the Coulomb potential. The fourth term can be thought to contribute to the potential an is attribute to angular momentum. Thus, l is calle the angular momentum quantum number. This is an eigenvalue problem for the raial eigenfunctions ur an energy eigenvalues ɛ.
the three-imensional schöinger equation 9 The solutions of this equation are etermine in a quantum mechanics course. In orer to get a feeling for the solutions, we will consier the zero angular momentum case, l = 0 : u r u 1 u = ɛu. 7 r Even this equation is one we have not encountere in this book. Let s see if we can fin some of the solutions. First, we consier the behavior of the solutions for large r. For large r the secon an thir terms on the left han sie of the equation are negligible. So, we have the approximate equation u ɛu = 0. 8 Therefore, the solutions behave like ur = e ± ɛr for large r. For boune solutions, we choose the ecaying solution. This suggests that solutions take the form ur = vre ɛr for some unknown function, vr. Inserting this guess into Equation 7, gives an equation for vr : rv 1 ɛr v 1 ɛv = 0. 9 Next we seek a series solution to this equation. Let vr = c k r k. k=0 Inserting this series into Equation 9, we have k=1 [kk 1 k]c k r k 1 k=1 [1 ɛk 1]c k r k = 0. We can re-inex the ummy variable in each sum. Let k = m in the first sum an k = m 1 in the secon sum. We then fin that [ mm 1cm 1 m ] ɛc m 1 r m 1 = 0. k=1 Since this has to hol for all m 1, c m = m ɛ 1 mm 1 c m 1. Further analysis inicates that the resulting series leas to unboune solutions unless the series terminates. This is only possible if the numerator, m ɛ 1, vanishes for m = n, n = 1,.... Thus, ɛ = 1 4n.
the three-imensional schöinger equation 10 Since ɛ is relate to the energy eigenvalue, E, we have me E n = 4 4πɛ 0 h n. Inserting the values for the constants, this gives 13.6 ev E n = n. This is the well known set of energy levels for the hyrogen atom. The corresponing eigenfunctions are polynomials, since the infinite series was force to terminate. We coul obtain these polynomials by iterating the recursion equation for the c m s. However, we will instea rewrite the raial equation 9. Let x = ɛr an efine yx = vr. Then Energy levels for the hyrogen atom. This gives r = ɛ x. ɛxy x ɛy 1 ɛy = 0. Rearranging, we have xy xy 1 ɛ 1 ɛy = 0. Noting that ɛ = n 1, this equation becomes xy xy n 1y = 0. 30 The resulting equation is well known. It takes the form xy α 1 xy ny = 0. 31 Solutions of this equation are the associate Laguerre polynomials. The solutions are enote by L α nx. They can be efine in terms of the Laguerre polynomials, n L n x = e x e x x n. x The associate Laguerre polynomials are efine as m L m n mx = 1 m L n x. x Note: The Laguerre polynomials were first encountere in Problem?? in Chapter 5 as an example of a classical orthogonal polynomial efine on [0, with weight wx = e x. Some of these polynomials are liste in Table 5 an several Laguerre polynomials are shown in Figure 5. Comparing Equation 30 with Equation 31, we fin that yx = L 1 n 1 x. The associate Laguerre polynomials are name after the French mathematician Emon Laguerre 1834-1886.
the three-imensional schöinger equation 11 L m n x L 0 0 x 1 L 0 1 x 1 x L 0 x 1 x 4x L 0 3 x 1 6 x 3 9x 18x 6 L 1 0 x 1 L 1 1 x x L 1 x 1 x 6x 6 L 1 3 x 1 6 x 3 3x 36x 4 L 0 x 1 L 1 x 3 x L x 1 x 8x 1 L 3 x 1 1 x 3 30x 10x 10 Table 5: Associate Laguerre Functions, L m n x. Note that L 0 nx = L n x. Figure 5: Plots of the first few Laguerre polynomials. Figure 6: Plots of Rρ for a = 1 an n = 1,, 3, 4 for the l = 0 states.
the three-imensional schöinger equation 1 In summary, we have mae the following transformations: 1. Rρ = ur, ρ = ar.. ur = vre ɛr. 3. vr = yx = L 1 n 1 x, x = ɛr. Therefore, Rρ = e ɛρ/a L 1 n 1 ɛρ/a. However, we also foun that ɛ = 1/n. So, In most erivation in quantum mechanics a = a 0. where a 0 = 4πɛ 0 h me is the Bohr raius an a 0 = 5.917 10 11 m. Rρ = e ρ/na L 1 n 1 ρ/na. In Figure 6 we show a few of these solutions. Example. Fin the l 0 solutions of the raial equation. For the general case, for all l 0, we nee to solve the ifferential equation u r u 1 ll 1 u r r u = ɛu. 3 Instea of letting ur = vre ɛr, we let This lea to the ifferential equation ur = vrr l e ɛr. rv l 1 ɛrv 1 l 1 ɛv = 0. 33 as before, we let x = ɛr to obtain [ xy l 1 x ] [ ] v 1 ll 1 v = 0. ɛ Noting that ɛ = 1/n, we have xy [l 1 x] v n ll 1v = 0. We see that this is once again in the form of the associate Laguerre equation an the solutions are yx = L l1 n l 1 x. So, the solution to the raial equation for the hyrogen atom is given by Rρ = r l e ɛr L l1 n l 1 ɛr ρ l ρ = e ρ/na L l1 na n l 1. 34 na Interpretations of these solutions will be left for your quantum mechanics course.