Legendre Polynomials and Angular Momentum

Transcription

1 University of Connecticut Chemistry Education Materials Department of Chemistry August 006 Legendre Polynomials and Angular Momentum Carl W. David University of Connecticut, Follow this and additional works at: Recommended Citation David, Carl W., "Legendre Polynomials and Angular Momentum" 006). Chemistry Education Materials. 6.

2 Legendre Polynomials and Angular Momentum C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut Dated: August, 006) I. INTRODUCTION USING CARTESIAN COÖRDINATES AND GUESSWORK There are so many different ways to introduce Legendre Polynomials that one searchs for a path into this subject most suitable for chemists. Consider Laplace s Equation: χ = χ x + χ y + χ z = 0.) which is a partial differential equation for χx, y, z). For our purposes, the solutions can be obtained by guessing, starting with χ =, and proceeding to χ = x, χ = y and χ = z. It takes only a little more imagination to obtain χ = xy and its associates χ = yz and χ = xz. It takes just a little more imagination to guess χ = x y, but once done, one immediately guesses its two companions χ = x z and χ = y z. Reviewing, there was one simple order 0) solution, three not so simple, but not particularly difficult solutions of order ) and six slightly more complicated solutions of order. Note that the first third-order solution one might guess would be χ = xyz! If one looks at these solutions, knowing that they are quantum mechanically important, the, x, y, and z solutions, accompanied by the xy, xz, yz, and x y solutions suggest something having to do with wave functions, where the first function ) is associated in some way with s-orbitals, the next three x, y, and z) are associated with p-orbitals, and the next of order ) are associated with d-orbitals. If all this is true, the perceptive student will wonder where is the d z orbital, the fifth one, and how come there are six functions of order two listed, when, if memory serves correctly, there are 5 d-orbitals! Ah. can be rewritten as which is or as y z ) + x z ) y z ) + x z ) + z z ) x + y + z z r z which combines the last two linearly dependent) solutions into one, the one of choice, which has been employed for more than 50 years as the d z orbital. II. CONVERTING TO SPHERICAL POLAR COÖRDINATES In Spherical Polar Coördinates, these solutions become x r sin cos ϕ y r sin sin ϕ z r cos xy r sin cos ϕr sin sin ϕ = r sin cos ϕ sin ϕ xz r sin cos ϕr cos = r sin cos cos ϕ yz r sin sin ϕr cos = r sin cos sin ϕ x y r sin cos ϕ r sin sin = r sin cos ϕ sin ϕ ) r z r r cos = r cos )..) Now, Laplace s equation in spherical polar coördinates is χ = χ χ sin r + r sin sin + χ ϕ = 0.) Next, we notice that our earlier solutions were all of the form χ = r l Y l,ml r,, ϕ).)

3 where l specifies the order, 0,,, etc., and the function of angles is dependent on this order, and on another quantum number as yet unspecified. Memory suggests that this last quantum number, m l is the famous substate quantum number in elementary atomic structure discussions. Substituting this form into the spherical polar form of Laplace s equation results in a new equation for the angular parts alone, which we know from our previous results. We substitute Equation. into Equation. to see what happens, obtaining r l Y l,ml sin l Y l,ml + sin r sin = l Y l,ml r + r l Y l,ml ϕ = 0.4) which, by the nature of partial differentiation, becomes +r l r sin r l Y l,ml = Y l,ml r sin sin l,ml and the first term in Equation.5 is l r + Y l,ml ϕ l = 0.5) = r ll + )rl.6) so, substituting into Equation.5 we obtain r l Y l,ml = ll + )r l Y l,ml + r l sin sin sin l,ml + Y l,ml ϕ = 0.7) which, of course, upon cancelling common terms, gives ll)y l,ml + sin sin sin l,ml + Y l,ml ϕ = 0.8) Laplace s Equation in Spherical Polar Coördinates on the unit sphere, i.e. Legendre s Equation! l = ; cos l = ; sin cos ϕ sin ϕ l = ; sin cos cos ϕ l = ; sin cos sin ϕ l = ; sin cos ϕ sin ϕ ) l = ; cos )..) III. INTERPRETING ATOMIC ORBITAL DESIGNATORS What then are these angular solutions we found? l = 0; l = ; sin cos ϕ l = ; sin sin ϕ and if you check back concerning their origin, you will see where the Hydrogenic Orbitals naming pattern comes from for s, p and d orbitals. These functions have been written in real form, and they have complex, not complicated) forms which allow a different simplification: s l = 0; p x l = ; sin eıϕ + e ıϕ e ıϕ e ıϕ ) p y l = ; sin ı p z l = ; cos e d xy l = ; sin ıϕ + e ıϕ ) e ıϕ e ıϕ ) ı d xz l = ; sin cos eıϕ + e ıϕ

4 e ıϕ e ıϕ ) d yz l = ; sin cos ı e d x y sin l = ; ıϕ + e ıϕ ) e ıϕ e ıϕ ) ) ı d z l = ; cos )..) Once they have been written in imaginary form, we can create intelligent linear combinations of them which illuminate their underlying structure. Consider the linear combination of p x and p y, i.e., p x + ıp y which becomes sin e ıϕ aside from irrelevant constants. Next consider p x ıp y which becomes We have a trio of functions sin e ıϕ p ml = = sin e ıϕ p ml =0 = cos p ml = = sin e +ıϕ.).4) which correspond to the m l values expected of p-orbitals. You will see that the same trick can be applied to d xz and d yz. Finally, the same trick, almost, can be applied to d xy and d x y with the m l = 0 value reserved for d z, all by itself. IV. CONNECTION TO THE H-ATOM Perhaps the next best place to introduce Spherical Harmonics and Legendre Polynomials is the Hydrogen Atom, since its eigenfunctions have angular components which are known to be Legendre Polynomials. The Schrödinger Equation for the H-atom s electron is h n, l, m l > Ze m e r n, l, m l >= E n n, l, m l > 4.) We know that this equation is variable separable, in the sense that n, l, m l >= R n,l r)y l,ml, ϕ) Here, R n,l is the radial wave function, and Y l,ml is the angular wave function a function of two variables), Since = r r + r sin sin sin + ϕ n, l, m l >= R n,l r)y l,ml, ϕ) = Y r R which means that the Schrödinger Equation has the form { h Y R R sin m e r + r sin sin Dividing through by R n,l Y l,ml abreviated as RY) we have { h R sin m e Rr + r Y sin sin + R sin r sin sin } + Y ϕ Ze r RY = E nry } + Y ϕ Ze = E n r + Y ϕ Multiplying through by r shows that the expected variable separation has resulted in a proper segregation of the angles from the radius. R sin + R Y sin sin + Y ϕ + mze r h = m h E nr }{{}

5 4 We can see this by noting that the underbracketed part of this last equation is a pure function of angles, with no radius explicitly evident. It is now standard to obtain recover?) the equation.8) Y sin sin sin + Y ϕ = ll + ) 4.) where the minus sign which is essentially arbitrary) is demanded by convention. Cross multipilying by Y, one has sin sin sin + Y ϕ = ll + )Y 4.) and so = sin = cos = µ = ) which is written in traditional eigenfunction/eigenvalue form. which is, substituting into Equation 4. V. CHANGING THE ϑ VARIABLE A change of variables can through this equation into a special form, which is sometimes illuminating. We write = ) µ µ = cos µ ) µ µ µ µ + Y ϕ = ll + )Y 5.) where the ll) form is a different version of a constant, looking ahead to future results! where ) µ ) + Y ϕ = ll + )Y 5.) Y l,ml, ϕ) = e ±ım lϕ S l,ml ) For m l = 0 we have ) µ ) S l,0 = ll + )S l,0 5.) which is Legendre s equation. VI. SCHMIDT ORTHOGONALIZATION We assume polynomials in µ n and seek orthonormal combinations which can be generated starting with a constant n=0) term. Then the normalization integral implies that < 0 0 >= 0 >= ψ 0dµ = We now seek a function > orthogonal to 0 > which itself is normalizeable. We have < 0 >= ψ dµ = 0 which has solution ψ = N x. Normalizing, we have which yields < >= ψ dµ = = = N N µ dµ

6 5 i.e., >= x To proceed, we need to generate a function > which is not only normalizeable but orthogonal to 0 > and >. < 0 >= ψ dµ = 0 and the other integral vanishes automatically. i.e., i.e., a = b so a + b = 0 >= bµ + b = b µ + ) in unnormalized form. Normalizing gives and < >= ψ µdµ = 0 where > is a polynomial of order in µ, i.e., i.e., < >= bµ + b) dµ = Substituting, we have and < 0 >= < >= ψ = >= aµ + b aµ + b) dµ = 0 aµ + b) µdµ = 0 We obtain two equations in two unknowns, a and b, ) a + b = 0 which gives < >= b µ + ) dµ = b = µ + ) dµ where, of course, we recognize the sign ambiguity in this result. One can continue forever with this Schmidt Orthogonalization, but the idea is clear, and there are better ways, so why continue? The limits correspond to = 0 to = π