Chapter 6 The Hydrogen Atom 6.1 The One-Particle Central-Force Problem Before studying the hydrogen atom, we shall consider the more general problem of a single particle moving under a central force. The results of this section will apply to any central-force problem. Examples are the hydrogen atom (Section 6.5) and the isotropic three-dimensional harmonic oscillator (Prob. 6.3). A central force is one derived from a potential-energy function that is spherically symmetric, which means that it is a function only of the distance of the particle from the origin: V = V r . The relation between force and potential energy is given by (5.31) as (1F2= - ٌV x, y, z = -i 0V 0x - j 0V 0y - k 0V 0z (6.1 The partial derivatives in 1(6.1) can2 be found1 > by2 the chain1 > rule2 [Eqs.1 (5.53)–(5.55)].> 2 Since V in this case is a function of r only, we have 0V 0u r,f = 0 and 0V 0f r,u = 0. Therefore, 1 > 2 1 > 2 0V dV 0r x dV = = (6.2) 0x y,z dr 0x y,z r dr a b a b 0V y dV 0V z dV = , = (6.3) 0y x,z r dr 0z x,y r dr where Eqs. (5.57) and (5.58)a haveb been used. Equationa b (6.1) becomes 1 dV dV r r F = - xi + yj + zk = - (6.4) r dr dr r 1 2 where (5.33) for r was used. The quantity1 r r in (6.4)2 is a unit vector in the radial direc- tion. A central force is radially directed. Now we consider the quantum mechanics> of a single particle subject to a central force. The Hamiltonian operator is n n n (H = T + V = - U2 2m ٌ2 + V r (6.5 2 2 2 2 2 2 2 ,where ٌ K 0 0x + 0 0y + 0 0z [Eq.1 (3.46)].> 2 Since1 V2 is spherically symmetric we shall work in spherical coordinates. Hence we want to transform the Laplacian opera- tor to these coordinates.> We> already >have the forms of the operators 0 0x, 0 0y, and 0 0z in these coordinates [Eqs. (5.62)–(5.64)], and by squaring each of these operators and > > > 118 M06_LEVI3450_07_SE_C06.indd 118 9/18/12 9:48 AM 6.1 The One-Particle Central-Force Problem | 119 then adding their squares, we get the Laplacian. This calculation is left as an exercise. The result is (Prob. 6.4) 02 2 0 1 02 1 0 1 02 (cot u + (6.6 + + + = 2ٌ 0r2 r 0r r2 0u2 r2 0u r2 s in2 u 0f2 Looking back to (5.68), which gives the operator Ln2 for the square of the magnitude of the orbital angular momentum of a single particle, we see that 2 0 2 0 1 n (L2 (6.7 - + = 2ٌ 0r2 r 0r r2U2 The Hamiltonian (6.5) becomes 2 2 n U 0 2 0 1 n H = - + + L2 + V r (6.8) 2m 0r2 r 0r 2mr2 a b 1 2 In classical mechanics a particle subject to a central force has its angular momentum conserved (Section 5.3). In quantum mechanics we might ask whether we can have states with definite values for both the energy and the angular momentum. To have the set of eigenfunctions of Hn also be eigenfunctions of Ln2, the commutator Hn, Ln 2 must vanish. We have n n n n n n 3 4 H, L2 = T, L2 + V, L2 2 2 n n U 0 2 0 1 n n T, L2 =3 - 4 3 + 4 3 + 4 L2, L2 2m 0r2 r 0r 2mr2 c 2 2a b d 3n n 4 U 0 2 0 n 1 1 n n T, L2 = - + , L2 + L2, L2 (6.9) 2m 0r2 r 0r 2m r2 Recall that Ln 2 involves3 only4 u and fc and not r [Eq. (5.68)].d Hencec it commutesd with every operator that involves only r. [To reach this conclusion, we must use relations like (5.47) with x and z replaced by r and u.] Thus the first commutator in (6.9) is zero. Moreover, since any operator commutes with itself, the second commutator in (6.9) is zero. There- n n n fore, T, L2 = 0. Also, since L2 does not involve r, and V is a function of r only, we have n n V, L2 = 0. Therefore, 3 4 n n 2 3 4 H, L = 0 if V = V r (6.10) Hn commutes with Ln 2 when the potential-energy function is independent of u and f. 3 n 4 1 2 n Now consider the operator Lz = -iU 0 0f [Eq. (5.67)]. Since Lz does not involve r and n 2 n since it commutes with L [Eq. (5.50)], it follows that Lz commutes with the Hamiltonian (6.8): n n > H, Lz = 0 if V = V r (6.11) n n 2 n We can therefore have a set3 of 4simultaneous eigenfunctions1 2 of H, L , and Lz for the central-force problem. Let c denote these common eigenfunctions: n Hc = Ec (6.12) n2 2 L c = l l + 1 U c, l = 0, 1, 2, c (6.13) n 1 2 Lzc = mUc, m = -l, -l + 1, c, l (6.14) where Eqs. (5.104) and (5.105) were used. M06_LEVI3450_07_SE_C06.indd 119 9/18/12 9:48 AM 120 Chapter 6 | The Hydrogen Atom Using (6.8) and (6.13), we have for the Schrödinger equation (6.12) 2 2 U 0 c 2 0c 1 n - + + L2c + V r c = Ec 2m 0r2 r 0r 2mr2 2 a2 b 2 1 2 U 0 c 2 0c l l + 1 U - + + c + V r c = Ec (6.15) 2m r2 r r 2mr2 0 0 1 2 a n 2 b 1m2 n 2 The eigenfunctions of L are the spherical harmonics Y l u, f , and since L does m not involve r, we can multiply Y l by an arbitrary function of r and still have eigenfunc- n 2 n 1 2 tions of L and Lz. Therefore, m c = R r Y l u, f (6.16) 1 2 1 2 m Using (6.16) in (6.15), we then divide both sides by Y l to get an ordinary differential equation for the unknown function R r : 2 U2 2 1 2l l + 1 U - RЉ + RЈ + R + V r R = ER r (6.17) 2m r 2mr2 1 2 We have shown that,a for anyb one-particle problem 1with2 a spherically1 2 symmetric m potential-energy function V r , the stationary-state wave functions are c = R r Y l u, f , where the radial factor R r satisfies (6.17). By using a specific form for V r in (6.17), we can solve it for a particular problem.1 2 1 2 1 2 1 2 1 2 6.2 Noninteracting Particles and Separation of Variables Up to this point, we have solved only one-particle quantum-mechanical problems. The hydrogen atom is a two-particle system, and as a preliminary to dealing with the H atom, we first consider a simpler case, that of two noninteracting particles. Suppose that a system is composed of the noninteracting particles 1 and 2. Let q1 and q2 symbolize the coordinates x1, y1, z1 and x2, y2, z2 of particles 1 and 2. Because the particles exert no forces on each other, the classical-mechanical energy of the system is 1 2 1 2 the sum of the energies of the two particles: E = E1 + E2 = T1 + V1 + T2 + V2, and the classical Hamiltonian is the sum of Hamiltonians for each particle: H = H1 + H2. Therefore, the Hamiltonian operator is n n n H = H1 + H2 n n where H1 involves only the coordinates q1 and the momentum operators p1 that corre- spond to q1. The Schrödinger equation for the system is n n H1 + H2 c q1, q2 = Ec q1, q2 (6.18) We try a solution of (6.18) 1by separation2 1 of variables,2 setting1 2 c q1, q2 = G1 q1 G2 q2 (6.19) We have 1 2 1 2 1 2 n n H1G1 q1 G2 q2 + H2G1 q1 G2 q2 = EG1 q1 G2 q2 (6.20) Since Hn involves only the coordinate and momentum operators of particle 1, we have n 1 1 2 1 2n 1 2 1 2 n 1 2 1 2 H1 G1 q1 G2 q2 = G2 q2 H1G1 q1 , since, as far as H1 is concerned, G2 is a con- n stant. Using this equation and a similar equation for H2, we find that (6.20) becomes 3 1 2 1 24 n 1 2 1 2 n G2 q2 H1G1 q1 + G1 q1 H2G2 q2 = EG1 q1 G2 q2 (6.21) 1 2 1 2 1 2 1 2 1 2 1 2 M06_LEVI3450_07_SE_C06.indd 120 9/18/12 9:48 AM 6.3 Reduction of the Two-Particle Problem to Two One-Particle Problems | 121 n n H1G1 q1 H2G2 q2 + = E (6.22) G q G q 1 11 2 2 12 2 Now, by the same arguments used1 in2 connection1 with2 Eq. (3.65), we conclude that each term on the left in (6.22) must be a constant. Using E1 and E2 to denote these constants, we have n n H1G1 q1 H2G2 q2 = E1, = E2 G q G q 1 11 2 2 12 2 1 2 E = E1 + E21 2 (6.23) Thus, when the system is composed of two noninteracting particles, we can reduce the two-particle problem to two separate one-particle problems by solving n n H1G1 q1 = E1G1 q1 , H2G2 q2 = E2G2 q2 (6.24) which are separate Schrödinger1 2 equations1 2 for each particle.1 2 1 2 Generalizing this result to n noninteracting particles, we have n n n n H = H1 + H2 + g+ Hn c q1, q2, c, qn = G1 q1 G2 q2 gGn qn (6.25) 1 2 1 2 1 2 1 2 E = E1 + E2 + g+ En (6.26) n HiGi = EiGi, i = 1, 2, c, n (6.27) For a system of noninteracting particles, the energy is the sum of the individual energies of each particle and the wave function is the product of wave functions for each particle.