Hydrogen Atom in Electric Field – Variational Approach

APPENDIX F

Hydrogen Atom in Electric Field – Variational Approach Polarization of an atom or molecule can be calculated by using the finite field method described on p. 284. Let us apply this method to the hydrogen atom. Its polarizability was already calculated by using a simple version of the perturbation theory (p. 282). This time we will use the variational method. The Hamiltonian for the isolated hydrogen atom (within the Born–Oppenheimer approximation) reads as 1 1 Hˆ (0) = − e − , 2 r where the first term is the electronic kinetic energy operator, and the second is its Coulomb interaction energy with the nucleus (proton–electron distance is denoted by r). The atom is in the uniform electric field E = (0, 0, E ) with E > 0 and, similarly as in perturbation theory (p. 282), the total Hamiltonian has the form Hˆ = Hˆ (0) + V with V = zE , where z denotes the coordinate of the electron and the proton is in the origin (the derivation of the formula is given on p. 282, the exchange of z to x does not matter). The variational wave function ψ is proposed in the form ψ = χ1 + cχ2 , where χ1 =

√1 π

(F.1)

exp (−r) is the 1s orbital of the hydrogen atom (ground state) and χ2 is the

normalized1 p-type orbital χ2 = Nz exp(−ζ r). There are two variational parameters, c and ζ . Let us assume for a while that we have fixed the value of ζ , so the only variational parameter is c. The wave function ψ is a linear combination of 1 N can be easily calculated from the normalization condition 1 = N 2

N



[z exp(−ζ r)]2 dV =   4 exp(−2ζ r) 2 θ 2π dφ = N 2 4! (2ζ )−5 2 2π = N 2 π . This gives N = ζ 5 . drr dθ sin θ cos 5 0 0 0 π 3 ζ

 2 ∞

π

609

610 Appendix F two expansion functions (“two-state model”), χ1 and χ2 . Therefore, the optimal energy follows from the Ritz method according to case III of Appendix V1-D on p. V1-655, i.e.,  (F.2) E = Ear ± 2 + h2 , 22 22 , while  ≡ H11 −H and h ≡ H12 = H21 with where the arithmetic mean energy Ear ≡ H11 +H 2 2       Hij ≡ χi |Hˆ χj = χi |Hˆ (0) χj + χi |V χj .

Let us compute all the ingredients of the energy given by (F.2).   First, let us note that H11 ≡ χ1 |Hˆ (0) χ1 = − 12 a.u., since χ1 is the ground state of the isolated hydrogen atom (p. V1-232), and χ1 |V χ1  = 0, because the integrand is antisymmetric with respect to z → −z. (0)

Now, let us compute H22 = H22 + V22 . Note that V22 = 0 for the same reason as V11 . We have

1 1 (0) H22 = − χ2 |e χ2  − χ2 | χ2 . 2 r   π  2π ∞ The second integral is χ2 | 1r χ2 = N 2 0 drr 3 exp(−2ζ r) 0 dθ sin θ cos2 θ 0 dφ =

ζ5 π

·

3! (2ζ )−4 · 23 · 2π = 12 ζ , where the dots separate the values of the corresponding integrals2 . In Appendix V1-S the reader will find the main ingredients needed to compute the first integral (0) : of H22 ⎡ ⎤   1 ∂ 2 ∂ 1 ∂ ∂ r + sin θ + 2 2 ∂θ ⎦ r cos θ exp(−ζ r) = χ2 |e χ2  = N 2 r cos θ exp(−ζ r)| ⎣ r ∂r ∂r r1 sin θ∂∂θ 2 ⎡ 

r 2 sin2 θ ∂φ 2

 ⎤ 1 ∂ 2 exp (−ζ r) − ζ r 3 exp (−ζ r)] + r cos θ exp(−ζ r)| cos θ [r r 2 ∂r ⎦=   N2 ⎣ θ) r exp(−ζ r) + 0 r cos θ exp(−ζ r)| (−2rcos 2 ⎡   ⎤ 2 2 r] exp (−ζ r) + − ζ − 3ζ + ζ r cos θ exp(−ζ r)| cos θ[ r ⎦=   N2 ⎣ θ) exp(−ζ r) r cos θ exp(−ζ r)| (−2 cos r   ζ5 2 ( · 2π) 2 · 2 · (2ζ )−3 − 4ζ · 3! · (2ζ )−4 + ζ 2 · 4! · (2ζ )−5 − 2 · 2! · (2ζ )−3 = −ζ 2 . π 3 2 Note that in the spherical coordinates the volume element dV = r 2 sin θdrdθdφ. In derivations of this appendix  (and not only) we often use the equality 0∞ dxx n exp(−αr) = n!α −(n+1) .

Hydrogen Atom in Electric Field – Variational Approach 611 Thus, we obtain H22 = 12 ζ 2 − 12 ζ . This formula looks good, since for χ2 = 2pz , i.e., for ζ = 12 , we get correctly (see p. V1-232) H22 = E2p = − 18 a.u., the energy of orbital 2p. (0)

Let us turn to the nondiagonal matrix element of the Hamiltonian, H12 = H12 + V12 . Note that (0) = 0, because χ1 is an eigenfunction of Hˆ (0) and χ1 |χ2  = 0. Thus, H12

1 h = N E r cos θ exp(−ζ r)|r cos θ √ exp (−r) = π  ∞  π  2π 1 4 2 NE √ drr exp[−(ζ + 1)r] dθ sin θ cos θ dφ = π 0 0 0   ζ5 ζ5 −5 2 · 4!(ζ + 1) · · 2π = 32 E E. π 3 (ζ + 1)5 Now we can write Eq. (F.2) as a function of ζ , i.e.,  10  1 2 2 1 2 2 5 (ζ − ζ + 1) + ζ E 2. E = (ζ − ζ − 1) − 4 16 ζ +1

(F.3)

We would like to expand this expression in power series of E in order to highlight the coefficient at E 2 , because this coefficient is related to the polarizability. The expansion gives (in a.u.) 1 1 1 1 1 E ≈ (ζ 2 − ζ − 1) − (ζ 2 − ζ + 1) − αzz E 2 + ... = − − αzz E 2 + ..., 4 4 2 2 2 where according to Eq. (4.24) the polarizability (in a.u.) reads as  10 ζ5 2 αzz = 4 · 2 . |ζ − ζ + 1| ζ + 1

(F.4)

Several numerical values of αzz computed by using (F.3) and (F.4) are given on p. 284. They are compared with the exact result αzz = 4.5 a.u.