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The Hydrogen Atom (Finite Difference Equation/Discrete Wave Vectors/Bohr-Rydberg Formula) FREDERICK T

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The Hydrogen Atom (Finite Difference Equation/Discrete Wave Vectors/Bohr-Rydberg Formula) FREDERICK T Proc. Natl. Acad. Sci. USA Vol. 83, pp. 5753-5755, August 1986 Chemistry Discrete wave mechanics: The hydrogen atom (finite difference equation/discrete wave vectors/Bohr-Rydberg formula) FREDERICK T. WALL Department of Chemistry, B-017, University of California at San Diego, La Jolla, CA 92093 Contributed by Frederick T. Wall, March 31, 1986 ABSTRACT The quantum mechanical problem of the hy- its wave vector energy counterpart.* If V is small compared drogen atom is treated by use of a finite difference equation in to mc2, then V and l9 are substantially the same, but for large place of Schrodinger's differential equation. The exact solu- negative values of V, the two can be significantly different. tion leads to a wave vector energy expression that is readily The next question to be answered is how the potential en- converted to the Bohr-Rydberg formula. (The calculations ergy is introduced into the finite difference equation. In our here reported are limited to spherically symmetric states.) The problem, we shall write that wave vectors reduce to the familiar solutions of Schrodinger's equation as c -- w. The internal consistency and limiting be- O(r - A) - 2(X + V/mc2)4(r) + /(r + A) = 0, [2] havior provide support for the view that the equations em- ployed could well constitute an approach to a relativistic for- with mulation of wave mechanics. X = = (1 - 26/E)"2 [3] In an earlier paper (1), I set forth a simple difference equa- EI/E tion for the wave mechanical description of a free particle. This was accompanied by a postulatory basis for interpreting where & is the wave vector energy, and certain parameters related to the energy, thus appearing to provide a possible connection to relativity. The initial formu- +(r) = rq,(r), [4] lation, however, was limited to free particles-or particles subject to constant potential energy-and hence did not pro- +i(r) representing the wave vector. As far as V is concerned, vide an exhaustive test. Additional tests of the theory should Eq. 2 looks like a discrete counterpart of Schrodinger's be forthcoming by tackling problems involving variable po- equation. On the other hand, if we were to use the wave tential energies. With this end in mind, I shall now turn to the vector form of potential energy, a different kind of equation problem of the hydrogen atom. would be required. The hydrogen atom has long been regarded as a basic problem in quantum mechanics, so one naturally thinks Solution to Difference Equation about it when endeavoring to introduce and develop a new theory on the general subject. To deal with the hydrogen To find a solution to Eq. 2 for the hydrogen-like atom, let us atom, however, the equation valid for a one-dimensional free first introduce some simplification of notation as follows: particle must be modified in two ways. First, we must intro- duce the potential energy and second, we must adjust to r= (A, [S] three dimensions. Since the potential energy should depend only on the radi- al coordinate, r, the treatment is much simplified if the wave where vector, tA, is likewise assumed to be a function of r only. Under such circumstances, the problem becomes essentially A = h/mc. [Sb] a one-dimensional problem and the solutions of the differ- ence equation should resemble those of Schrodinger's equa- Also, tion for states of zero angular momentum. So long as angular dependence can be ignored, the three-dimensional differen- V/mc2 = -Ze2/(r mc2) = _,ulk, [6a] tial equation problem can be reduced to that of one dimen- sion by the simple expedient of replacing the dependent vari- provided able, fi, by a new variable, 4 = rt. Happily, the same can be done effectively in our finite difference approach, although I shall not offer a full justification here. (A two-dimensional ,u = Ze2/(hc) = Za, [6b] problem does not lend itelf to such a simple device.) In any event, we shall replace q, by 4/r and then introduce the po- where a is the fine structure constant. Upon substitution tential energy. into Eq. 2 we obtain We shall assume that the potential energy of a single elec- tron in the neighborhood of a nucleus of atomic number Z L(tu-n1)o-w2(-as/u)e()a + + 1) = 0. [7] will be given by Let us now assume that V = -Ze2/r. [1] +(C) = (1 + y)-l"FQ) [8] It may seem strange to say that we "assume" the validity of Eq. 1; after all, what else could we do? We simply emphasize *In a system without kinetic energy the2relationship between the that Eq. 1 represents the actual potential energy and noted two would be V = E[1 - (1 - 219/E)' i. 5753 Downloaded by guest on September 29, 2021 5754 Chemistry: Wall Proc. Natl. Acad. Sci USA 83 (1986) where y is a parameter to be determined later. Then, etc. In general, from Eq. 14 we can assert that (1 + y)"'2F( - 1) - 2(X - gl4)F(4) Ck+1 = [(k + 1)y - 2gv(1 + V)1/2ICk. [16] + F( + 1)/(1 + y)1/2 = 0. [9] Quantization and Discrete Energy Levels Now let us suppose that F(Q) can be expanded in a discrete series of It should be clear from Eq. 16 how we can proceed to set up power ~-namely,t the infinite discrete power series solution to Eq. 9. To assure proper finiteness for all values of (, however, we shall stipu- Cl Q-1) + CA Q - 1)(t - 2) FQ;) = 0!1!0 + 1!2! + [10a] late that the series must terminate as a polynomial, in analo- 2!.31 gy to the treatment of certain series solutions to differential =>(-1)k'Ck(- O)k+lI equations. (At this time we shall not carry through a detailed ( justification.) If we choose to make Ck = 0 for all values of k k=O k!(k + 1)! [lOb] equal to or greater than some integer n, then from Eq. 16 we see that with CO = 1. By the expression (X)k, we mean ny = 2,v(1 + y)1/2. [17] (x)k =X(X + 1)(x + 2) ... (x + k - 1), [11] Eq. 17 is the desired quantization condition. Evidently, provided (x)o = 1. Upon letting 2tIn = (1 + y)1/2 - 1/(l + Y)1/2 [18] 2K = + + + [12] (1 y)1/2 1/(1 y)1/2, Moreover, in the light of Eq. 12 we conclude that and substituting F(4) from Eq. lOa into Eq. 9, we find after K2 = 1 + ul/n2. [19] rearrangement that But the wave vector energy, & is given by (1/1l + V)1/2 - (1 + )1/2) & = (mc2/2)(1 - K2) [20a] Mc22 X + C1(- 1) + C2(A - 1)(C - 2) + [20b] 2n2 Upon substituting ,u from Eq. 6b, we obtain + 2u1 + CA(- 1) + C2(A - 1)(; - 2) + 1!2! 2!3! 2i'r2mZ2e4 &n= n2h2 [21] (i+ Eq. 21 will be recognized as precisely the Bohr equation, so no comments are necessary concerning its validity. Howev- - 1) C3(t - 1)( - er, we should recognize that C,, was obtained from K by use x {Ci + C2(A + 2)+ ... .4 = O. 1!2! 2!3! [13] of Eq. 3. All this suggests that there is a relativistic implica- tion in the treatment, with Eq. 3 providing the connection. It 13 can be written more as should be emphasized that the solutions presented here are Eq. compactly exact and not just good approximations to some reasonable order of magnitude. (1/(1 + Y)"12 - (1 + V)1/2) > (- 1)k(~+ l)kCk ) k=O 1) k . k Wave Vectors for the Hydrogen Atom (_-1)k(- + 1)kCk ++ 2,uA- The solutions to Eq. 7 are products of polynomials and an = (k+ exponential term as indicated by Eq. 8. In accordance with Eq. 17, the exponential part, which is also 4i for the lowest 1 + energy state, is given by + E (-1) (-4 l)kCk+l [14] (1 + V)1'2 k=O k!(k + 1)! (1 + y)-f2 = (2g/ny) . [22] For Eqs. 13 and 14 to be valid for all values of (, the coeffi- For the polynomial part of the wave vector, we note from cients of (- + 1)k must add up to zero. From Eq. 13 we see Eqs. 16 and 17 that that, Ck+j = -(n - k - l)YCk. [23] Ci = y- 2gi(1 + Y)112 [15al Therefore, C2 = (2 - 2v(11 + V)1/2)CC [15b] F.(O) = _ (n - 12-) = - + (n - 1)(n - 2) y24(q - 1)(C - 2) C3 (3y 2,*(l )1/2 C2, [15c] 2!3! I n= (-1)k(-n + + 1)k tThe precise form assumed for the series was chosen in anticipation 1)kyk(-C [24] of the final result. k=O k!(k + 1)! Downloaded by guest on September 29, 2021 Chemistry: Wall Proc. Natl. Acad. Sci. USA 83 (1986) 5755 The first three polynomials are become Fi(t) = t [25a] j (-n + 1)kp [28] k=k=O k'(k + 1)! ' F2(t)= t- 'yt(;- 1)/2 [25b] with F3(t) = t - ytq - 1) + y2t( - 1)(Q - 2)/6. [25c] 8i2m0Ze2r [29] These polynomials are a variety of the Hahn polynomials (2). P = -nh2 The 4,, values constitute a complete set of orthogonal vec- tors, so that Of course, it is no surprise that the wave vector solutions reduce to the wave functon solutions of Schrodinger's equa- tion. After all, the basic difference equation does become the Z 0"nwo" w = 0, [26] differential equation in the limit as c -A oo.
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