2 1 Hydrogen orbitals 1.2 Formulation of the Schrodinger¨ equation for the hydrogen atom In this initial treatment, we will make some practical approximations and simplifi- cations. Since we are for the moment only trying to establish the general form of the hydrogenic wave functions, this will suffice. To start with, we will assume that the nucleus has zero extension. We place the origin at its position, and we ignore the centre-of-mass motion. This reduces the two-body problem to a single particle, the electron, moving in a central-field potential. To take the finite mass of the nu- cleus into account, we replace the electron mass with the reduced mass, m, of the two-body problem. Moreover, we will in this chapter ignore the effect on the wave function of relativistic effects, which automatically implies that we ignore the spins of the electron and of the nucleus. This makes us ready to formulate the Hamilto- nian. The potential is the classical Coulomb interaction between two particles of op- posite charges. With spherical coordinates, and with r as the radial distance of the electron from the origin, this is: Ze2 V(r) = − ; (1.1) 4pe0 r with Z being the charge state of the nucleus. The Schrodinger¨ equation is: h¯ 2 − ∇2y(r) +V(r)y = Ey(r) ; (1.2) 2m where the Laplacian in spherical coordinates is: 1 ¶ ¶ 1 ¶ ¶ 1 ¶ 2 ∇2 = r2 + sinq + : (1.3) r2 ¶r ¶r r2 sinq ¶q ¶q r2 sin2 q ¶j2 Since the potential is purely central, the solution to (1.2) can be factorised into a radial and an angular part, y(r;q;j) = R(r)Y(q;j). Substitution this into (1.2), the Schrodinger¨ equation becomes: 2 2 1 ¶ 2 ¶R(r) 2mr Ze r − 2 − − E R(r) ¶r ¶r h¯ 4pe0 r 1 1 ¶ ¶ 1 ¶ 2 = − sinq + Y(q;j) : (1.4) Y(q;j) sinq ¶q ¶q sin2 q ¶j2 Before proceeding we will simplify the notation, by introducing atomic units, and the angular momentum operator. The motivation for using atomic units is that when performing long derivations, a large number of constants make the work cum- bersome. To circumvent that, a number of constants are set to unity: 1.3 Solution of the radial equation 3 1 e = me = h¯ = = 1 : (1.5) 4pe0 Then, units for involved physical quantities have to be adapted accordingly, when- ever quantified answers are sought. A brief introduction to, and a list of, atomic units are given in appendix A. For the continuation of this book, we will use atomic units, when we do not explicitly state otherwise. The expression within the square brackets in (1.4) is identical to the quantum mechanical operator for the square of the orbital angular momentum, L2. A more thorough discussion on angular momentum is presented in appendix C. In that ap- pendix, it is also shown that for an angular wave function that is eigenfunction to L2, we have: L2 Y(q;j) = l(l + 1)Y(q;j) ; (1.6) where the introduced quantum number l has to be a positive integer, or zero. In atomic units, and using (1.6) and (1.1), the Schrodinger¨ equation can now be written as: 1 ¶ ¶R(r) 1 r2 + 2Zr + 2Er2 = − L2 Y(q;j) = l(l + 1) : (1.7) R(r) ¶r ¶r Y(q;j) Here, we have set m ≈ me, utilising the fact that for a one-electron system, the nucleus is at least 1800 times heavier than the electron. Equation (1.7) has to be valid for all spatial parameters, so when the radial and angular parts have been separated, both sides of (1.7) have to be constant, for a given wave function. In (1.7), we have used (1.6) and set that constant to l(l + 1). We are now left with two uncoupled differential equations, which can be solved independently. The angular part of (1.7) is independent of the potential, as is the case for any kind of central potential, and the solution will be in the form of the standard spherical harmonics. The energy solely appears in the radial part of the equation, and therefor the energies will, at this level of approximation, be independent of the angular coordinates. In the following sections, we will treat the radial and angular solutions separately. 1.3 Solution of the radial equation The radial part of (1.7) can be rewritten as: 1 ¶ ¶ 2Z l(l + 1) r2 R(r) + + 2E − R(r) = 0 : (1.8) r2 ¶r ¶r r r2 Equation (1.8) represents a one dimensional problem of a particle moving in an effective potential, consisting of the central Coulomb term and a centrifugal term, as shown in fig. 1.1 (for the case Z = 1). Positive energies will give diffusive solutions, which are relevant for scattering phenomena, but which will not be dealt with in this 4 1 Hydrogen orbitals Fig. 1.1 Effective potential for the radial part of the Schrodinger¨ equation for the hydrogen atom (1.8), with Z = �/�� 1, for three different values ��� of the angular momentum �/�� quantum number: l = 0 (blue), � �� �� �� l = 1 (green), and l = 2 (red). -��� The axes are in atomic units -��� and zero energy corresponds to an electron infinitely distant -��� from the nucleus. The three dashed horizontal lines shows -��� the energies of the three -��� lowest energy eigenstates (se section 1.5.1). -��� chapter. In the current treatment we explicitly look for bound states, i.e., solutions for E < 0. In the following, we will give a very brief outline the solution. For a more thorough treatment, see appendix B and general works in the suggested further reading. A first step is to introduce the substitution U(r) ≡ r R(r). This leaves us with the equation: ¶ 2 2Z l(l + 1) U(r) + + 2E − U(r) = 0 : (1.9) ¶r2 r r2 Note that since we have formulated the Schrodinger¨ equation in atomic units, the energy in (1.9) will be in Eh and r has to be given in a0 (cf. appendix A). The solution to (1.9) is periodical and discretized in two quantum numbers: the orbital angular momentum quantum number l (also called the azimuthal quantum number) introduced in section 1.2, and the principal quantum number n. From the solutions, we also get the following constraints for the integer quantum numbers n and l (cf. appendix B): 0 ≤ l < n : (1.10) 1.3.1 Eigenstates The eigenstates found from the solution of (1.9) are in the form of associated La- guerre polynomials (cf. appendix B ): s (n − l − 1)! l+1 −r=2 2l+1 Unl(r) = − r e L (r) : (1.11) n2 [(n + l)!]3 n+l Here, r is a rescaled radial parameter: 1.3 Solution of the radial equation 5 2Zr r = ; (1.12) n and equation (1.11) has been normalised such that: Z ¥ ∗ Unl(r)Unl(r)dr = 1 : (1.13) 0 Moreover, the functions Unl(r) are mutually orthogonal. Throughout this book, we will use the standard spectroscopic notation for the orbital angular momentum quantum number l, as presented in table (1.1): With this Table 1.1 Standard letter symbols used for different values of the angular momentum quantum number l l-quantum number 0 1 2 3 4 5 6 spectroscopic symbol s p d f g h i notation, some of the lowest order normalized radial functions, in the format Unl(r), are presented in (1.14), for the case Z = 1. −r U1s = 2r e 1 −r=2 r U2s = p r e 1 − 2 2 1 2 −r=2 U2p = p r e 2 6 2 2 −r=3 2r 2r U3s = p r e 1 − + 3 3 3 27 8 2 −r=3 r U3p = p r e 1 − 27 6 6 4 3 −r=3 U3d = p r e (1.14) 81 30 1 3r r2 r3 U = r e−r=4 1 − + − 4s 4 4 8 192 p 2 5 2 −r=4 r r U4p = p r e 1 − + 16 3 4 80 1 3 −r=4 r U4d = p r e 1 − 64 5 12 1 4 −r=4 U4f = p r e : 768 35 6 1 Hydrogen orbitals From (1.14), we can note that the wave function will be non-zero at r = 0 only for s-states (l = 0). Hence, only an s-electron, which lacks a centrifugal term, has a finit probability of being very close to the nucleus. 1.4 Solution of the angular equation Since both the radial and the angular sides of (1.7) are equal to l(l + 1), for a given wave function, the angular part of the equation is: 1 ¶ ¶ 1 ¶ 2 − sinq + Y(q;j) = l(l + 1) Y(q;j) : (1.15) sinq ¶q ¶q sin2 q ¶j2 As already stated, the operator on the left side of (1.15) is the operator for the square of the orbital angular momentum (cf. appendix C) divided by h¯ 2, which justifies the definition of the constant as l(l + 1), and the eigenvalue equation in (1.6). For the 2 projection of L along a quantisation axise ˆz, we will use the quantum number ml (eigenvalue to Lz). With (1.15) being in the form of the standard generator for the spherical harmon- ics, the solution to this differential equation is very general, and it is outlined in appendix D. The solutions are: s (2l + 1)(l − jm j)! (ml +jml j)=2 l jml j ij ml Yl;ml (q;j) = (−1) Pl (cosq)e ; (1.16) 4p (l + jmlj)! where Pl(cosq) is a l’th order associated Legendre function, cf.