VANDERBILT UNIVERSITY. MATH 3120 INTRO DO PDES The Schrödinger equation

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1 VANDERBILT UNIVERSITY MATH 31 INTRO DO PDES The Schrödinger equation 1. Introduction Our goal is to investigate solutions to the Schrödinger equation, i Ψ t = Ψ + V Ψ, 1.1 µ where i is the imaginary number i = 1; = erg s is Planck s constant; µ is a positive constant called the mass; V = V t, x : R R 3 R is called the potential function; and the unknown is the complex-valued function Ψ = Ψt, x : R R 3 C called the wave-function. The variables t and x represent, respectively, the time and space variables. The Schrödinger equation describes the dynamics of a particle of mass µ interacting with a potential V, according to the laws of Quantum Mechanics. The physical interpretation of Ψ is as follows. If U R 3, then Ψt, x dx U represents the probability of finding the particle in the region U at a time t. In particular, one must have R 3 Ψt, x dx = Notice that, upon multiplying Ψ by a suitable constant, condition 1. can always be fulfilled as long as R 3 Ψt, x dx <. 1.3 Our treatment will be based on [FY, T, W], to which the student is referred for more details.. Separation of variables for a time-independent potential We shall assume that V does not depend on time, i.e., V t, x = V x. We will have to divide several expressions by Ψ. In order to make this sensible, it will be assumed that Ψ does not vanish or, at least, does not vanish on an open set. Look for solutions of the form Plugging.1 into 1.1 gives Ψt, x = T tψx,.1 i T T = 1 ψ + V, µ ψ The left-hand side depends only on t, whereas the right-hand side depends only on x. Thus, both sides have to be equal to a constant, which we denote be E. Therefore i T = ET,. 1

2 VANDERBILT and Equation. is easily solved. Its solution is ψ + V ψ = Eψ..3 µ T t = e ie t,.4 where we ignored an arbitrary constant of integration such constants will be neglected throughout, as an overall constant of integration can be fixed at the very end via condition The time-independent Schrödinger equation. We now focus on.3, known as the timeindependent Schrödinger equation. To solve it, we assume further that V is radially symmetric, i.e., that V x = V x 1 + x + x 3 or, in spherical coordinates, that V = V r. This assumption suffices to treat many physical systems of interest. Recall the expression for the Laplacian in spherical coordinates, where = r + r r + 1 r S,.5 S = φ + cos φ sin φ φ + 1 sin φ θ.6 is the Laplacian on the unit sphere, r [,, φ [, π], and θ [, π. From now on we shall work in spherical coordinates. We suppose that ψr, φ, θ = RrY φ, θ..7 Plugging.7 into.3, and using.5, r R + r µ R R + V Er = 1 µ Y S Y. Since the left-hand side depends only on r and the right-hand side only on φ, θ, both sides must be equal to a constant, which we denote by a. Thus, R + r µ R + V + a r R = ER,.8 and µ SY = ay..9.. The angular equation. We first investigate.9, which, in light of.6, becomes Supposing one gets Θ φ Y + cos φ sin φ φy + 1 sin φ θ Y = aµ Y. Θ = sin φ Φ Y φ, θ = ΦφΘθ,.1 Φ + sin φ cos φ Φ + aµ sin φ Φ..11

3 VANDERBILT 3 Once more, both sides ought to be equal to a constant, which we denote by b. becomes One equation Θ = bθ..1 To solve.1, we need to analyze the cases b >, b =, and b <. Notice the following boundary condition: the points with coordinates θ and θ + π must be identified as they correspond to the same point in R 3. Thus, Θθ + π = Θθ..13 We immediately see that the case b < does not yield a solution satisfying.13; b = and.13 give Θ =constant; and b > along with.13 give that Θ is a linear combination of cos bθ and sin bθ. All these cases can be summarized by setting and writing b = m, m,.14 Θθ = e imθ..15 Next, we move to the Φ-equation. From.11 and.14, one has sin φ d sin φ dφ m = λ sin φ,.16 Φ dφ dφ where λ = µ a,.17 and we used the product rule to rewrite the terms involving derivatives. In order to solve.16, let us make the following change of variables, x = cos φ, φ π. Notice that this change of variables is well-defined since cos is one-to-one for φ π. The chain rule gives so that.16 becomes sin φ d d = sin φdx dφ dφ dx = sin φ d dx = cos φ 1 d dx = x 1 d dx, To solve.18, we seek for a solution of the form where P solves d 1 x dφ + λ m dx dx 1 x Φ =..18 Φx = 1 x m d m P x dx m,.19 1 x d P dx xdp + λp =.. dx To see that this works, differentiate. m times, obtaining 1 x d m + P dx m + m + 1xd m +1 P dx m +1 + λ m m + 1 d m P dx m =..1

4 4 VANDERBILT Students are encouraged to verify.1 compute the first few derivatives to see that a pattern as.1 emerges. On the other hand, let Φ be defined by Φx = 1 x m Φx. and plug this into.18. Computing the derivative terms, d 1 x d 1 x m Φ = d m dx dx dx x1 x m = 1 x m +1 d Φ dx + 1 x m dφ m dx + 1 x + m + m x m 1 x m 1 x 1 x m Φ = 1 x m +1 d Φ dx x1 x m d Φ m + 1 = 1 x m dx + m 1 x d Φ d Φ x m + 1 dx dx + m Φ + 1 x m x 1 x m m x 1 x 1 By.18, this has to equal λ m 1 x Φ = λ m 1 x 1 x m Φx, what gives, after canceling 1 x m, But 1 x d Φ d Φ x m + 1 dx dx + λ Φ + m x m m 1 x 1 x +1 d Φ dx m x 1 x Φ Φ. Φ =. m x m m 1 x 1 x = m m + 1x m m x = m m + 1, and therefore 1 x d Φ d Φ x m + 1 dx dx + λ m m + 1 Φ =..3 Comparing.3 with.1, we see that if P solves., then. solves.18, as claimed. Therefore, it suffices to solve.. We seek a power series solution of the form P x = a k x k..4 Plugging.4 into. gives 1 x kk 1a k x k x ka k x k 1 + λ a k x k =, or yet, after rearranging some terms, k + k + 1a k+ kk + 1 λa k x k =,

5 VANDERBILT 5 which implies the following recurrence relation, a k+ = kk + 1 λ k + 1k + a k, k =, 1,, Relation.5 determines all coefficients a k except for a and a 1, which remain arbitrary this is consistent with the fact that we are solving a second order ODE. Furthermore, a determines all even coefficients, giving rise to an even power series, while while a 1 determines all odd coefficients, giving rise to an odd power series. These two power series, even and odd, are two linearly independent solutions of.. Next, we investigate the convergence of.4. Since it suffices to investigate the convergence of the even and odd expansions separately, as these are two linearly independent solutions, the ratio between two consecutive terms in the expansion is obtained from.5, yielding lim a k+ x x+ k a k x k = x, and thus.4 converges for x < 1 by the ratio test. We need to investigate the case x = 1 which corresponds to φ = or φ = π. Plugging x = ±1 into.4 gives P ±1 = ± a k..6 From.5 we have a k+ = k + Ok k + Ok a k = k + Ok k + Ok k k+ +Ok k+1 k + Ok k + Ok a k k = = k+ +Ok k+1 a, k k+1 +Ok k k k+1 +Ok k a 1, It follows that lim a k, k k even, k odd. and therefore.6 diverges by the divergence test, unless.4 is in fact a finite sum; i.e., unless a k = for all k greater than a certain l. Hence, we must have, form some non-negative integer l, which implies a l+ = = ll + 1 λ l + 1l + a l, λ = ll + 1,.7 provided that a l. Relation.7 determines λ, and hence the separation constant a in view of.17. The conclusion is that there is a family {P l } of solutions to. parametrized by l =, 1,,.... After conveniently choosing a and a 1 to obtain integer coefficients, the first few P s are P x = 1, P 1 x = x, P x = 1 3x, P 3 x = 3x 5x 3. Since P l is a polynomial of degree l, from.19 it follows that Φ = for m > l. Thus, the values of m are restricted to m l, i.e., the allowed m-values depend on l and satisfy m { l, l + 1,..., 1,, 1,..., l 1, l }..8

6 6 VANDERBILT We write m = m l when we want to stress this dependence of m on l. One obtains a family of solutions {Φ lml } to.18 parametrized by l and m l, where l =, 1,,... and m l satisfies.8. The first few Φ s are Φ x = 1, Φ 1 x = x, Φ 1,±1 x = 1 x 1, Φ x = 1 3x, Φ ±1 x = 1 x 1 x, Φ± x = 1 x, Φ 3 x = 3x 5x 3, Φ 3±1 x = 1 x 3 1 5x, Φ 3± x = 1 x x, Φ 3±3 1 x 3. Finally, it is necessary to rewrite our solutions in terms of the φ variable. Denoting and using 1 x = sin φ, Combining.1,.15, and.9 gives F lml = Φ lml, Φ lml φ = sin m l F lml cos φ, l =, 1,,..., m l l..9 Y lml φ, θ = e im lθ sin m l φf lml cos φ, l =, 1,,..., m l l..3 Notice that, in view of.9 and.17, Y l,ml solves S Y lml = ll + 1Y lml. We finish this section with some terminology. Equation. is known as Legene equation, and its solutions P l are known as Legene polynomials. The functions F lml are known as associated Legene functions. The functions Y lml are called spherical harmonics. Legene functions and spherical harmonics have many important applications in Physics. The interested reader is referred to [B] for details..3. The radial equation. We now turn our attention to equation.8. Using.17 and.7, equation.8 can be written as 1 d r r dr + µ E V r R = ll + 1 R r..31 It is important to stress that the results of section. are general, i.e., they apply to separation of variables to any radially symmetric potential V = V r. To solve.31, on the other hand, we need to specify the function V r. We shall assume that V is the potential describing the electromagnetic interaction of an electron with a nucleus. This covers the important case when one is solving the Schrödinger equation describing the evolution of an electron on a hyogen atom. In this situation, V takes the form V r = e 4πε r,.3 where is the nuclear charge for example, = 1 for the hyogen and = for an ionized helium atom, e is the electron charge, where e = C, and ε is the vacuum permitivity whose values is ε = F/m farads per meters. In order to investigate solutions to.31 with V given by.3, one needs more information about the separation constant E. We claim that E must be real and negative. To see this, multiply equation.31 by r R, where R is the complex conjugate of R, and integrate from to : R d r dr µ V R r ll + 1 R = µ E R r,.33

7 VANDERBILT 7 where we used that R = R R. Integrating by parts the first term, R d r dr dr dr = r + R r dr = where it has been assumed that R and dr where R R and R C are real-valued, it comes dr and we conclude that dr dr vanish sufficiently fast at. Writing R = R R + ir C, dr = dr R idr C dr R + idr C = dr drr + drc, is real-valued. But from.33 and.34 we have dr r.34 dr dr E = r + µ V R r + ll + 1 R µ..35 R r Therefore, since all terms on the right-hand side are real, we conclude that E is real as well. Students should notice that.35 gives an explicit expression for E in terms of the integral of R and other data of the problem although we shall derive a much more explicit expression for E, see below. Now that we know that E is real, let us show that it is negative 1. Let us investigate the behavior of.31 for large values of r, i.e., r 1. Then we can neglect the terms that contain 1 r and.31 gives, after expanding the terms in d, d R µe R..36 But for r 1 we also have the approximation r d R + dr r d R, so that d rr = r d R + dr r d R..37 Hence, multiplying.36 by r and using.37, d rr µe rr. This approximate equation can be easily solved, producing rr e ± µe r. If E, then R is a complex function which satisfies Then the integral R 3 Ψt, x dx = π π diverges since Rr r 1 for large r. produce a physically sensible solution. rr 1 for r 1. Y φ, θ sin φ dφdθ Rr r Consequently, condition 1.3 fails, and this does not 1 It is possible to obtain E < by a more delicate analysis of.35, but here we employ a simpler argument.

8 8 VANDERBILT In light of the above arguments, we assume, once and for all, that E <. In this case, we can define the real constants β = µe,.38 and and make the real change of variables γ = µe 4πε β,.39 ϱ = βr. With these definitions, equation.31, with V given by.3, becomes 1 d ϱ ϱ dr + 1 ll + 1 dϱ dϱ 4 ϱ + γ R =..4 ϱ Equation.4 will be solved using a power series expansion, but direct application of the method does not work. To see this, try plugging Rϱ = a k ϱ k into.4, obtaining kk + 1a k ϱ k 1 4 This can be rewritten as + a k ϱ k ll + 1 a k ϱ k + γ a k ϱ k 1 =. ll + 1a ϱ + ll + 1a 1 + γa ϱ 1 k + 3k + ll + 1 a k+ + γa k a k ϱ k =..41 Vanishing of each term order by order implies that a =, then a 1 =, and subsequently a k = for any k, so R =. We need, therefore, to try a different approach. We shall focus on the behavior of.4 when ϱ 1, in which case the equation simplifies to 1 d ϱ dϱ ϱ dr dϱ R 4..4 This approximate equation can be solved as follows. Look for a solution of the form e Aϱ. Plugging into the equation we find A = 1, i.e., e ϱ is a approximate solution of.4. This suggests looking for solutions of.4 in the form Plugging.43 into.4 gives an equation for G, d G G γ 1 dϱ + ϱ 1 ϱ + ϱ Rϱ = e ϱ Gϱ..43 ll + 1 ϱ G =..44 The reader may remember that when one solves second order ODEs with constant coefficients, sometimes we have to multiply a solution by a suitable power of the variable in order to produce a particular solution or a second linearly independent solution. What it is being done here resembles that: we have some information about solutions, i.e., that e ϱ solves the equation in an approximate sense for large values of ϱ. Thus, we try multiplying by e ϱ to construct the full, exact solution.

9 VANDERBILT 9 We seek a solution of the form Gϱ = ϱ s a k ϱ k = a k ϱ k+s,.45 where s is to be determined. The term ϱ s has been included due to the 1 ϱ terms in the equation, as these may lead to singular terms that do not fit into a general recurrence relation, as it occurred in.41. Notice that the traditional procedure is included in this approach by simply setting s =. Plugging.45 into.44 gives, after some algebra, ss + 1 ll + 1 a ϱ s + s + k + 1s + k + ll + 1a k+1 s + k + 1 γa k ϱ s+k 1 =. The vanishing of the term in ϱ s requires ss + 1 ll + 1 =,.46 which has roots s = l and s = l + 1. This latter root is rejected on the basis that it does not yield a finite solution when ϱ +, i.e., Gϱ blows up at the origin when s = l + 1 recall that l is non-negative. Using s = l, one finds from.46 the following recurrence relation, a k+1 = k + l + 1 γ k + l + 1k + l + ll + 1 a k..47 From.47 and the ratio test, we see at once that.45, with s = l, converges for all values of ϱ. In order for.45 to be an acceptable solution, we also must verify 1.3. From.47, it follows that a k+1 = k + k = ak k + ak, and so that Continuing this way, Remembering that a k = k 1 + k 1 + ak 1 = 1 + k 1 + ak 1, a k+1 = 1 + = k + ak k + k 1 + ak = kk 1 + ak 1. a k+1 = 1 + kk 1k k j + ak j. e ϱ = 1 k! ϱk, we see that Gϱ is asymptotic to ϱ s e ϱ, i.e., its series expansion behaves very much like the series of ϱ l e ϱ recall that s = l: Gϱ ϱ l e ϱ,

10 1 VANDERBILT which implies, upon recalling.43, Rϱ = e ϱ Gϱ e ϱ ϱ l e ϱ = ϱ l e ϱ, which diverges when ϱ. As a consequence, 1.3 is not satisfied. This will be the case, unless the series.45 terminates, i.e., unless a k = for all k greater than a certain n. From.47, this means i.e., In particular, γ has to be an integer, k + l + 1 γ =, γ = k + l + 1. γ = n, n = l + 1, l +,.... With this, the series terminates at the n l + 1 th term, and G is a polynomial of degree n 1. Recalling.38 and.39, we have found the possible values for the separation constant E = E n, namely, µ e 4 E n = 4πε, n = 1,, 3, n We write R nl to indicate that R is parametrized by the integers n and l, with n = l, l + 1,.... We can now write, for each n, the corresponding R nl by using.47 to find the polynomial G = G nl, and then R nl via.43. Unwrapping all our definitions, where R nl r = nα r nα l G nl α = 4πε µe. r In light of.7, we see that ψ is also parametrized by n, l, and m l. Instead of thinking of n varying according to n = l, l + 1,..., we can equivalently think of l as constrained by l =, 1,..., n 1, for each given n = 1,,..., what is more convenient in order to organize the parameters n, l, m l. We obtain, therefore, a family of solutions to.3, where n = 1,, 3,..., nα, ψ nlml = R nl Y lml,.49 l =, 1,,..., n 1, m l = l, l + 1,...,,..., l 1, l. Our final solution is then given, in view of.1 and.4, by Ψt, x = A nlml e ien t ψ nlml,.5 where n, l, and m l satisfy.5, E n and ψ nlml are given by.48 and.49, respectively, and A nlml is a constant depending on n, l, and m l that ensures 1., i.e., A nlml is given by 1 A nlml = ψ nlml. R 3

11 t does not contribute to Ψ since e ien The term e ien is customary to absorb the constant A nlml VANDERBILT 11 t e ien into ψ nlml, in which case R 3 ψ nlml = 1. Or course, 1. is automatically satisfied in this case. t = e + ien t e ien t = 1. It 3. Final comments We close with some remarks about the physical meaning of the problem we just described. Readers are referred to [W] for a more thorough physical discussion. Below, we list some the first few ψ nlml. n l m l ψ nlml 3 1 ψ 1 = 1 π a ψ = 1 4 π 1 ψ 1 = 1 1 ± 1 ψ 1±1 = 1 3 ψ 3 = ψ 31 = 3 1 ± 1 ψ 31±1 = 81 3π 81 π 81 π 3 ψ 3 = π 3 ± 1 ψ 3±1 = 1 81 π 4 π 8 π 3 ± ψ 3± = 1 6 π a a a 3 r a 3 r a 3 r a a a r a 3 6 r a 3 6 r a 3 a 3 r a r a a 3 r a a a cos φ a sin φe ±iθ a + r z r a a r a a 3a 3a cos φ 3a cos φe ±iθ 3a 3 cos φ 1 3a sin φ cos φe ±iθ 3a sin φe ±iθ It is possible to show that the constants E n, ll+1, and m l have important physical interpretation: E n corresponds to the electron energy, ll + 1 to the magnitude of its orbital angular momentum, and m l to the projection of the orbital angular momentum onto the z-axis. The reader should notice that these quantities cannot be arbitrary, being allowed to take values only on a countable set of multiples of integers. This is a distinctive feature of Quantum Mechanics we say that the energy and orbital angular momentum are quantized. The indices n, l, and m l are called quantum numbers. One-electron atoms with l =, 1,, 3 are labeled s, p, d, f. In hyogen and hyogen-like atoms, this letter is preceded by a number giving the energy level n. Thus, the lowest energy state of the hyogen atom is 1s; the next to the lowest are s and p; the next 3s, 3p, 3d and so on. These are the so-called atomic orbitals that the student is likely to have learned in Chemistry. Remembering that Ψ is a probability density, what these orbitals represent are clouds of probability, highlighting the regions of three-dimensional space where it is more likely to find the electron. A few illustrations of the atomic orbitals are given in Figures 1,, and 3. These figures were generated with the Mathematica package Visualizing Atomic Orbitals that can be found at

12 1 VANDERBILT Figure 1. An illustration of the orbital 1s n = 1, l =, m l =. Figure. An illustration of the orbital p n =, l = 1, m l =. Figure 3. An illustration of the orbital 3d n = 3, l =, m l =. We finish mentioning that in a more detailed treatment of the problem, µ is not exactly the mass of the particle being described, but rather the reduced mass of the system. This is because, strictly speaking, the electron does not orbit the nucleus, but both orbit the center of mass of the system electron nucleus. This is very much like the situation of the Earth orbiting the Sun: both bodies move due to their reciprocal gravitational attraction, although the Sun, begin much more massive, barely feels the pull caused by Earth s gravitational field, and that is why one usually thinks of the Earth orbiting an standing-still Sun. A similar situation occurs for the nucleus and the electron. We remark, however, that the calculations we presented apply, with no change, to this more accurate situation: we only have to change the value of µ to be the reduced mass. References [B] Butkov, E. Mathematical Physics. Addison-Wesley Educational Publishers Inc; New edition [FY] Faddeev, L. D., and Yakubovskii, O. A. Lectures on Quantum Mechanics for Mathematics Students. American Mathematical Society 9. [T] Takhtajan, L. A. Quantum Mechanics for Mathematicians. American Mathematical Society 8. [W] Weinberg, S. Lectures on Quantum Mechanics. Cambridge University Press 1.