David Eigen. MA112 Final Paper. May 10, 2002

Transcription

1 David Eigen MA112 Fina Paper May 1, 22 The Schrodinger equation describes the position of an eectron as a wave. The wave function Ψ(t, x is interpreted as a probabiity density for the position of the eectron. That is, the probabiity of the eectron being in a region Ω in space at a time t is P (eectron in Ω at time t = Ψ(t = t 2 dx Ω Ψ(t = t R 3 2 dx The Schrodinger equation for the hydrogen atom describes how the probabiity distribution for the position of the eectron in the atom evoves. It is { i t Ψ = 1 Ψ 1Ψ 2 r Ψ 2 (1 dx finite where r = x. The first step in soving the equation is to seperate variabes: et Ψ(x, t = T (tv(x Pugging this into (1, we get it v = 1 2 T v 1 r T v Mutipying by 2 and dividing both sides by T v, 2i T T = v v 2 r The eft side of this equation depends ony on t, and the right side depends ony on x. Therefore, both sides must equa a constant, λ. The equation for T (t is T = λ 2i T (2 1

2 We can sove this simpy by seperating variabes: Therefore, the T (t eigenfunction is t dt dt dt T dt T = λ 2i T = λ 2i dt = t og T og T ( = λ 2i t λ 2i dt T = T (e iλt/2 T (t = e iλt/2 (3 The equation for v(x is v 2 v = λv (4 r Thus, we have an eigenvaue probem for v. To sove it, we express v in spherica coordinates, and ook for seperabe soutions of the form v(x = v(r, θ, φ = R(rY (θ, φ Here, θ is the ongitude coordinate and φ is the atitude coordinate. Therefore, we have θ 2π and φ π. We aso require that Y is periodic in θ: Y (θ, φ = Y (θ + 2π, φ for a θ. In spherica coordinates, the Lapacian operator is = rr + 2 r r + Using this and v = RY, (4 becomes 1 r 2 sin φ 1 φ(sin φ φ + r 2 sin 2 φ θθ λry }{{} + 2 r RY + R Y λv }{{} r R + r 2 sin φ (sin(φry 1 φ φ + r 2 sin 2 φ RY θθ = }{{} 2 r v v Mutipying both sides by r 2 /RY, λr 2 + 2r + r2 R R + 2rR R + 1 Y ( 1 sin φ (sin(φy φ φ + 1 sin 2 φ Y θθ = and so λr 2 + 2r + r2 R R + 2rR R = 1 Y ( 1 sin φ (sin(φy φ φ + 1 sin 2 φ Y θθ (5 2

3 The eft side of this equation depends ony on r, whie the right side depends ony on θ and φ. Therefore, both sides must equa a constant. Ca this constant γ. Setting the eft side of (5 equa to γ and mutipying both sides by R/r 2, we get (λ + 2r γr 2 R + R + 2 r R = (6 Setting the right side of (5 equa to γ and mutipying both sides by Y, we get 1 sin φ (sin(φy φ φ + 1 sin 2 φ Y θθ + γy = (7 (6 is an ODE for R, and (7 is an eigenvaue probem for Y. Since we do not yet know the eigenvaues γ, our approach wi be to find the eigenfunctions Y first, and then sove the equation for R. Whie soving the equation for R, we wi aso find the eigenvaues λ for v. Once we do a of this, we wi have the eigenfunctions v(r, θ, φ = R(rY (θ, φ for the entire probem. To sove (7, we ook for seperabe soutions Y (θ, φ = Θ(θΦ(φ. Pugging this into (7, we get 1 sin φ (sin(φθφ φ + 1 sin 2 φ Θ Φ + γθφ = Mutipying both sides by sin 2 (φ/θφ, and so sin(φ (sin(φφ Φ + Θ Θ + γ sin2 (φ = Θ Θ = sin(φ(sin(φφ + γ sin 2 (φ (8 Φ The eft side of (8 is dependent ony on θ, whie the right side is dependent ony on φ. Therefore, both sides must equa a constant. Ca this constant α. Then by setting each side equa to α, we have Θ = αθ (9 and sin(φ (sin(φφ + γ sin 2 (φφ = αφ (1 We first sove the Θ equation, (9. α = is a degenerate soution, so we are ony interested in the cases where α. First consider the case where α <. Let α = k, k >. Then Θ = kθ The genera soution to this is Θ = Ae kθ + Be kθ 3 (11

4 Reca that we require Θ to be 2π periodic. However, (11 is not periodic, and so α cannot be ess than. Now, consider the case where α >. Let α = ω 2. Then the genera soution to (9 is Θ = Ae iωθ + Be iωθ Using the condition that Θ has period 2π, we get that Ae iωθ + Be iωθ = Ae iωθ+2πiω + Be iωθ 2πiω This happens when ω = m, m Z. Therefore, we get that the eigenvaues are α = m 2, and the eigenfunctions are Θ(θ = Ae imθ + Be imθ Note, however, that m can be negative. Thus, we can et B = without osing any soutions. This is because at the end, we are going to ineary combine the eigenfunctions Ψ = T RΘΦ. Letting C m = A m + B m, the part of the inear combination for Θ is A m e imθ + B m e imθ = (A + B + (A m + B m e imθ + (A m + B m e imθ m m> = C + m> C m e imθ + C m e imθ = m C m e imθ Thus, we can say that the eigenfunctions are Θ(θ = C m e imθ (12 so ong as there are the same number of negative vaues for m as positive vaues for m. The corresponding eigenvaues are α = m 2. These eigenfunctions are orthogona with respect to the inner product (Θ m, Θ m = This is because for m m, 2π 2π e imθ e imθ dθ = Θ m (θ Θ m (θdθ = 2π 2π e imθ e imθ dθ e i(m m θ θ dθ = ei(m m i(m m 2π θ= = Now that we know α = m 2, the Φ equation (1 becomes sin(φ (sin(φφ + γ sin 2 (φφ = m 2 Φ Bringing a the terms to one side and dividing by sin 2 φ, (sin(φφ + (γ m2 sin φ sin 2 Φ = (13 φ 4

5 Note there coud be singuarities at φ = or φ = π. Since we require that the soution to the Schrodinger equation Ψ = T RΘΦ be finite everywhere, we have the condition that Φ is finite at φ = and φ = π. Now, make a change of vairabes s = cos φ. Then ds = sin φdφ, and dφ = ds. Aso, sin φ sin 2 φ = 1 cos 2 φ = 1 s 2. Therefore, (sin(φφ sin φ = d dφ ( sin(φ dφ dφ sin φ = d ds ( sin 2 (φ dφ = d ( (1 s 2 dφ ds ds ds Thus, (13 turns into ( d (1 s 2 dφ + (γ m2 Φ = (14 ds ds (1 s 2 In addition, the condition that Φ(φ is finite at φ = and φ = π turns into the condition that Φ(s is finite at s = ±1. Equation (14 is caed the associated Legendre equation, and its eigenfunctions are Φ(s = P m (s = P m (cos φ (15 where is an integer m. P k (s is defined for k = m as the function (s = (k (1 s 2 k/2 d+k 2! ds +k (s2 1 (16 P k The corresponding eigenvaues are γ = ( This soution is finite at s = ±1. Since k = m, (s 2 1 is a poynomia of degree 2 + k. Thus, (d +k /ds +k (s 2 1 is a poynomia of degree 2 ( + k = k. So, (d +k /ds +k (s 2 1 is finite at s = ±1. Furthermore, (1 s 2 k/2 is finite at s = ±1, and ( k /2! is a finite constant. Therefore, a the factors are finite, and so this soution satisfies the condition that Φ is finite at s = ±1. The eigenfunctions P m are orthogona with respect to the inner product (P k, P k = 1 P k (sp k (sds That is, (P k, P k = 1 P k (sp k (sds = for 1 I am actuay not sure where the ( k in equation (16 comes from. The section in Strauss on Legendre functions seems to not have it. I was unabe to derive this equation with the information in the section. 5

6 As an exampe, consider the case when m =, = 1, and = 3. We have The third derivative is Thus, the inner product is 1 P1 (s = 1 d 2 ds (s2 1 = s d 3 P3 (s = 1 48 ds 3 (s2 1 3 d 3 ds 3 (s2 1 3 = d3 ds 3 (s6 3s 4 + 3s 2 1 = d2 ds 2 (6s5 12s 3 + 6s = d ds (3s4 36s = 12s 3 72s P 1 (sp 3 (sds = s(12s 3 72sds = 1 12s 4 72s 2 ds 48 = 1 [ s5 72 ] 1 3 s3 = 1 [ 24s 5 24s 3] 1 48 = Reca that Y (θ, φ = Θ(θΦ(φ. We can now put together an expression for Y : Y m (θ, φ = Θ m (θφ m (φ = e imθ P m (cos φ (17 The reation m puts a constraint on and m. They must be in the range, m Aso note that there are the same number of positive and negative vaues for m. This is exacty condition we needed earier in order to use Θ(θ = e imθ instead of Θ(θ = e imθ +e imθ. The functions Y m are caed spherica harmonics. They are orthogona with respect to the inner product (Y m, Y m = (Θ m, Θ m (Φ m, Φ m To see why this is true, consider the possibe cases for the indices, m,, and m. We can spit the cases up as foows: Case 1 : m m Case 2 : m = m and Case 3 : m = m and = 6

7 For Case 1, since m m, we have (Θ m, Θ m =, and so (Y m, Y m =. For Case 2, since m = m and, we have (Φ m, Φ m = (Φm, Φ m m =, and so (Y, Y m =. Case 3 is the case where Y m = Y m m, and so this case does not affect orthogonaity. Thus, (Y, Y m = if m m or, and so the functions Y m form an orthogona set. Note that the formua for the inner product is 2π (Y m, Y m = e imθ e imθ dθ 1 P m (sp m (sds = Since s = cos φ, we have ds = sin φdφ, and so (Y m, Y m = 2π = π 2π π e imθ e imθ P m e imθ e imθ P m 2π 1 e imθ e imθ P m (φp m (φ( sin φdφdθ (φp m (φ sin(φdφdθ (sp m (sdsdθ At this point, we have found the eigenfunctions Y m, and have a soution for the time part T (t. So the ony part of the soution to Ψ(t, r, θ, φ = T (tr(ry (θ, φ that we no not know yet is R(r. Reca that the equation for R is (λ + 2r γr R + R r R = Pugging in γ = ( + 1 and rearranging terms, we get R + 2 ( r R + λ + 2 ( + 1 R = (19 r r 2 Beacuse of the condition that Ψ 2 dx is finite, we require that R(r is finite for a r (in particuar, R( must be finite, and aso that R( is finite. We can sove the ODE (19 for λ < by doing a change of variabes and then ooking for soutions in the form of a power series. 2 Let w(r = e βr R(r, where β = λ for λ <. Then R = e βr w, and so R = βe βr w + e βr w = e βr ( βw + w R = βe βr ( βw + w + e βr ( βw + w = e βr (β 2 w 2βw + w Substituting these for R, R, and R in (19, as we as λ = β 2, we get e βr (β 2 w 2βw + w + 2 ( r e βr ( βw + w + β ( + 1 e βr w = r r 2 2 I tried to show that there are no soutions for λ satisfying the condition R( finite, R( finite, but can t yet I actuay haven t taken an ODE course and don t have much experience with ODEs. 7 (18

8 Since e βr is never, we can divide by it. Doing this and distributing, Combining ike terms, β 2 w 2βw + w 2 r βw + 2 r w β 2 w + 2 ( + 1 w w = r r 2 w + 2 ( ( 1 2(1 β r β w + r ( + 1 w = (2 r 2 Now, consider the power series expansion of w, w(r = k= a kr k. Any anaytic function can be expressed as a power series, so any (nice soution is a soution of this form. We want to find the coefficients a k. Pugging in the series into (2, we get k= ( 1 k(k 1a k r k r β ka k r k + k= ( 2(1 β r ( + 1 a r 2 k r k = k= Note that we can say the sums for the derivatives start at k =. This is because a the coefficients before the constant term are zero. Thus, terms before the constant term are zero anyway, except possiby when r =. As we wi soon see, however, we wi find the coefficients by requiring the expression above to be zero for a r, and so the case when r = wi not affect our argument. Distributing the ast two terms and bringing the factors of 1/r and 1/r 2 into the sums, k(ka k r k 2 +2 ka k r k 2 2β k= k= ka k r k +2(1 β a k r k (+1 a k r k 2 = k= Now, change the indices in the third and fourth sums such that a sums are expressed in terms of factors of r k 2. This gives us k(ka k r k 2 +2 ka k r k 2 2β k= k= k=1 Combining the sums with the same indices, [k(k 1 + 2k ( + 1] a k r k 2 + k= k= (ka k r k 2 +2(1 β a k r k 2 (+1 a k r k 2 = k=1 k= [ 2β(k 1 + 2(1 β] a k r k 2 = k=1 Now, the eft hand side must be identicay zero. Therefore, the a the coefficients of r k 2 must be zero. Thus, for k =, we get For a other k, we get k= ( + 1a = (21 [k(k 1 + 2k ( + 1] a k + [ 2β(k 1 + 2(1 β] a k = 8

9 Bringing the a k term to the other side of the equation and expanding products, [ k 2 k + 2k ( + 1 ] a k = [ 2βk + 2β + 2 2β] a k and so [k(k + 1 ( + 1] a k = 2(βk 1a k (22 Thus, (21 and (22 gives us a recursion reation for a k. Now, reca that is an integer with. If =, then a is arbitrary. If not, then by (22 we get that a 1 = if > 1. If = 1 then a 1 can be anything. We can show by induction that a a k are unti a, which is arbitrary. In the case where k =, (21 tes us that a = if >, and that a can be anything if =. Now, assume a k = for < k <. By (22, we get [k(k + 1 ( + 1] a k = And since k <, k(k + 1 ( + 1 and so a k =. In the case where k =, (22 gives us and so a can be anything. a = 2(β 1a = Now, consider the case when β = 1/n, where n is an integer, n >. Then (22 gives us ( 1 [n(n + 1 ( + 1]a n = 2 n n 1 a k = 2(1 1a k = and so a n =. Furthermore, once a n =, (22 tes us that a a k = for k > n as we. Thus, the power series w(r = k a kr k is a poynomia of degree n 1. Since R(r = e βr w(r, we get that R(r is a decreasing exponentia times a poynomia. Thus, R(r as r, and so the condition that R( = is satisfied. If β is not of the form 1/n, then using (22 and the fact that a coefficients before a are, we get k 2(βj 1 a k = a j(j + 1 ( + 1 Thus, for very arge k, we can say a k a k j=+1 2βj 2 j(j + 1 = 2a k j=+1 j=+1 ( β j a j(j + 1 k j=+1 β j = 2a β k (k k where I used the notation (p q = p(p 1(p 2 (p (q 1 for integers p and q. Now, for arge r, the power series is dominated by the terms a k for arge k. Therefore, for arge r, we can use the above approximation to get w(r = k a k r k 2a 9 k β k (k k r k

10 So, R(r = e βr w(r has the imit w(r im R(r = im r r e βr Mutipying by 1 in the form β /!, we get that this is β /! im R(r = im r r ( 2a k β k /(k k r k = im r k βk r k /k! 2a β /! k ( β k /k! r k k βk r k /k! = 2a β /! Thus, the condition that R( = is not satisfied, and so β must equa 1/n for an integer n >. Therefore, we get from the β = 1/n case that the eigenfunctions are where the a k satisfy n R(r = e r/n a k r k (23 k= a = 1 a k = 2(βk 1 k(k + 1 ( + 1 a k, k > and n >. We can set a = 1 because a a k are some constant number times a due to the recursion reation thus, a is a constant factor. The corresponding eigenvaues are λ = β 2 = /n 2. Now, we know the eigenfunctions R and Y, as we as the soution for T. We can now put the whoe probem together using (3, (23 and (17. The eigenfunctions are n Ψ nm (t, r, θ, φ = T n (tr n (ry m (θ, φ = e it/2n2 e r/n e imθ P m (cos φ a k r k (24 k= Thus, using inearity we get the more genera soution [ ] n Ψ(t, r, θ, φ = A mn e it/2n2 e r/n e imθ P m (cos φ a k r k = m= n=+1 k= (25 Using an initia probabiity distribution Ψ (r, θ, φ = Ψ(t =, r, θ, φ we shoud be abe to compute the coefficients A mn. First, note that the pugging in t = gets rid of the T eigenfunction, since T (t = = 1. Thus, Ψ = A mn R n Y m m n 1

11 Now, reca that the functions Y m (θ, φ are orthogona with respect to the inner product (18. So for a particuar and m, taking the inner product with Y m makes a terms in the and m sums zero except the th and m th. Thus, taking the inner product gives us 2π π Y Ψ Y m dφdθ = m 2 n= +1 A m nr n (26 I wasn t abe to find an inner product for the R functions that makes R n an orthogona set (either through research or on my own I did try to get one for a ong time. I aso actuay haven t seen anything that suggests there is or isn t a such an inner product. However, since the degree of each of the poynomias in R n is different for each vaue of n, the functions R n seem ike they woud be orthogona under some inner product. Even if they aren t orthogona, though, they woud at east be ineary independent. So even in the worst case, there shoud be some way to compute the coefficients A m n, even if it invoves soving a inear system numericay. If we do have that inner product in which the R n or orthogona, we can use it to isoate the particuar coefficient. That is, for a particuar n, we can take the inner product of both sides with R n and get 2π π (R n, Ψ Y m dφdθ 2 = Y m A m n R n 2 This gives us A m n = ( Y m 2 R n 2 ( R n, 2π π Ψ Y m dφdθ (27 References French, A. P. and Tayor, Edwin F., An Introduction to Quantum Mecahnics, New York: W. W. Norton & Company, Inc., 1978 Morse, Phiip M. and Feshbach, Herman, Methods of Theoretica Physics, New York: McGraw- Hi Book Company, Inc., 1953 Strauss, Water A., Partia Differentia Equations: An Introduction, New York: John Wiey & Sons, Inc.,