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Hydrogen Atom in Electric Field–The Variational Approach
APPENDIX V
Hydrogen Atom in Electric Field–The Variational Approach Polarization of an atom or molecule can be calculated by using the finite field (FF) method described on p. 746. Let us apply this method to the hydrogen atom. Its polarizability was already calculated by using a simple version of the perturbation theory (p. 743). 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 (with 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 it was in the perturbation theory (p. 744), the total Hamiltonian has the form Hˆ = Hˆ
0
+V
with V = z E , where z denotes the coordinate of the electron and the proton is in the origin (the derivation of the formula is given on p. 743, and the exchange of z to x does not matter). The variational wave function ψ is proposed in the form ψ = χ1 + cχ2 , where χ1 = normalized
1
√1 π
(V.1)
exp (−r ) is the 1s orbital of the hydrogen atom (ground state) and χ2 is the
p-type orbital χ2 = N z 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 that the only variational parameter is c. The wave function ψ is a linear combination of two expansion functions (“two-state model”): χ1 and χ2 . Therefore, the optimal [z exp(−ζ r )]2 d V = N 2 0∞ drr 4 −5 2 ζ5 2π exp(−2ζ r ) 0π dθ sin θ cos2 θ 02π dφ = N 2 4! 2ζ π . 3 2π = N ζ 5 . This gives N =
1 N can be easily calculated from the normalization condition 1 = N 2
Ideas of Quantum Chemistry, Second Edition. http://dx.doi.org/10.1016/B978-0-444-59436-5.00042-8 © 2014 Elsevier B.V. All rights reserved.
e159
e160 Appendix V energy follows from the Ritz method according to case III of see Appendix D available at booksite.elsevier.com/978-0-444-59436-5 on p. e63: E = E ar ± 2 + h 2 , (V.2) 22 22 where arithmetic mean energy E ar ≡ H11 +H , while ≡ H11 −H and h ≡ H12 = H21 , with 2 2
Hi j ≡ χi | Hˆ χ j = χi | Hˆ 0 χ j + χi |V χ j .
Let is compute all the ingredients given by Eq. (V.2). energy
of the First, let us note that H11 ≡ χ1 | Hˆ 0 χ1 = − 21 a.u., since χ1 is the ground state of the isolated hydrogen atom (p. 203), 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 1
5 ∞ The second integral is χ2 | r χ2 = N 2 0 drr 3 exp (−2ζ r ) 0 dθ sin θ cos2 θ 0 dφ = ζπ · −4 2 3! 2ζ · 3 · 2π = 21 ζ , where the dots separate the values of the corresponding integrals.2 In Appendix R available at booksite.elsevier.com/978-0-444-59436-5, the reader will find the (0) main ingredients needed to compute the first integral of H22 :
∂ 1 ∂ ∂ 1 ∂ 2 χ2 |e χ2 = N r cos θ exp (−ζ r )| 2 r 2 + 2 sin θ r ∂r ∂r r sin θ ∂θ ∂θ 2 ∂ 1 r cos θ exp (−ζ r ) + 2 2 r sin θ ∂φ 2 ⎡ ⎤ ∂ [r 2 exp −ζ r − ζ r 3 exp −ζ r ] r cos θ exp (−ζ r )| cos θ r12 ∂r ⎦ = N2 ⎣ −2 cos θ + r cos θ exp (−ζ r )| r 2 r exp (−ζ r ) + 0
2 2 2 =N − ζ − 3ζ + ζ r exp −ζ r r cos θ exp (−ζ r )| cos θ r −2 cos θ exp (−ζ r ) + r cos θ exp (−ζ r )| r −4 −3 ζ5 2 = · 2π 2 · 2 · 2ζ − 4ζ · 3! · 2ζ π 3 −5 −3 + ζ 2 · 4! · 2ζ − 2 · 2! · 2ζ = −ζ 2 . 2 Note that in the spherical coordinates, the volume element d V = r 2 sin θdr dθdφ. We have used the equality ∞ n −(n+1) . 0 drr exp (−αr ) = n!α
Hydrogen Atom in Electric Field–The Variational Approach e161 Thus, we obtain H22 = 21 ζ 2 − 21 ζ . This formula looks good, since for χ2 = 2 pz ; i.e., for ζ = 21 , we get (see p. 203) H22 = E 2 p = − 18 a.u., the energy of orbital 2 p. (0) Let us turn to the non-diagonal matrix element of the Hamiltonian: H12 = H12 + V12 . (0) of Hˆ (0) and χ1 |χ2 = 0. Thus, Note that H12 = 0 because χ1 is an eigenfunction
∞ π 1 h = N E r cos θ exp (−ζ r )|r cos θ √π exp (−r ) = N E √1π 0 drr 4 exp[−(ζ + 1)r ] 0 dθ √ √ 2π ζ5 ζ5 sin θ cos2 θ 0 dφ = E π · 4!(ζ + 1)−5 · 23 · 2π = 32 (ζ +1)5 E . Now we can write Eq. (V.2) as a function of ζ : 10 1 2 1 2 2 2 5 E = (ζ − ζ − 1) − (ζ − ζ + 1) + ζ E 2. (V.3) 4 16 ζ +1 We would like to expand this expression in a 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 atomic units) 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. (12.24) the polarizability (in atomic units) reads as 10 2 ζ5 αzz = 4 · 2 . |ζ − ζ + 1| ζ + 1
(V.4)
Several numerical values of αzz computed by using Eqs. (V.3) and (V.4) are given on p. 746. They are compared with the exact result αzz = 4.5 a.u.
Hydrogen Atom in Electric Field–The Variational Approach Polarization of an atom or molecule can be calculated by using the finite field (FF) method described on p. 746. Let us apply this method to the hydrogen atom. Its polarizability was already calculated by using a simple version of the perturbation theory (p. 743). 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 (with 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 it was in the perturbation theory (p. 744), the total Hamiltonian has the form Hˆ = Hˆ
0
+V
with V = z E , where z denotes the coordinate of the electron and the proton is in the origin (the derivation of the formula is given on p. 743, and the exchange of z to x does not matter). The variational wave function ψ is proposed in the form ψ = χ1 + cχ2 , where χ1 = normalized
1
√1 π
(V.1)
exp (−r ) is the 1s orbital of the hydrogen atom (ground state) and χ2 is the
p-type orbital χ2 = N z 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 that the only variational parameter is c. The wave function ψ is a linear combination of two expansion functions (“two-state model”): χ1 and χ2 . Therefore, the optimal [z exp(−ζ r )]2 d V = N 2 0∞ drr 4 −5 2 ζ5 2π exp(−2ζ r ) 0π dθ sin θ cos2 θ 02π dφ = N 2 4! 2ζ π . 3 2π = N ζ 5 . This gives N =
1 N can be easily calculated from the normalization condition 1 = N 2
Ideas of Quantum Chemistry, Second Edition. http://dx.doi.org/10.1016/B978-0-444-59436-5.00042-8 © 2014 Elsevier B.V. All rights reserved.
e159
e160 Appendix V energy follows from the Ritz method according to case III of see Appendix D available at booksite.elsevier.com/978-0-444-59436-5 on p. e63: E = E ar ± 2 + h 2 , (V.2) 22 22 where arithmetic mean energy E ar ≡ H11 +H , while ≡ H11 −H and h ≡ H12 = H21 , with 2 2
Hi j ≡ χi | Hˆ χ j = χi | Hˆ 0 χ j + χi |V χ j .
Let is compute all the ingredients given by Eq. (V.2). energy
of the First, let us note that H11 ≡ χ1 | Hˆ 0 χ1 = − 21 a.u., since χ1 is the ground state of the isolated hydrogen atom (p. 203), 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 1
5 ∞ The second integral is χ2 | r χ2 = N 2 0 drr 3 exp (−2ζ r ) 0 dθ sin θ cos2 θ 0 dφ = ζπ · −4 2 3! 2ζ · 3 · 2π = 21 ζ , where the dots separate the values of the corresponding integrals.2 In Appendix R available at booksite.elsevier.com/978-0-444-59436-5, the reader will find the (0) main ingredients needed to compute the first integral of H22 :
∂ 1 ∂ ∂ 1 ∂ 2 χ2 |e χ2 = N r cos θ exp (−ζ r )| 2 r 2 + 2 sin θ r ∂r ∂r r sin θ ∂θ ∂θ 2 ∂ 1 r cos θ exp (−ζ r ) + 2 2 r sin θ ∂φ 2 ⎡ ⎤ ∂ [r 2 exp −ζ r − ζ r 3 exp −ζ r ] r cos θ exp (−ζ r )| cos θ r12 ∂r ⎦ = N2 ⎣ −2 cos θ + r cos θ exp (−ζ r )| r 2 r exp (−ζ r ) + 0
2 2 2 =N − ζ − 3ζ + ζ r exp −ζ r r cos θ exp (−ζ r )| cos θ r −2 cos θ exp (−ζ r ) + r cos θ exp (−ζ r )| r −4 −3 ζ5 2 = · 2π 2 · 2 · 2ζ − 4ζ · 3! · 2ζ π 3 −5 −3 + ζ 2 · 4! · 2ζ − 2 · 2! · 2ζ = −ζ 2 . 2 Note that in the spherical coordinates, the volume element d V = r 2 sin θdr dθdφ. We have used the equality ∞ n −(n+1) . 0 drr exp (−αr ) = n!α
Hydrogen Atom in Electric Field–The Variational Approach e161 Thus, we obtain H22 = 21 ζ 2 − 21 ζ . This formula looks good, since for χ2 = 2 pz ; i.e., for ζ = 21 , we get (see p. 203) H22 = E 2 p = − 18 a.u., the energy of orbital 2 p. (0) Let us turn to the non-diagonal matrix element of the Hamiltonian: H12 = H12 + V12 . (0) of Hˆ (0) and χ1 |χ2 = 0. Thus, Note that H12 = 0 because χ1 is an eigenfunction
∞ π 1 h = N E r cos θ exp (−ζ r )|r cos θ √π exp (−r ) = N E √1π 0 drr 4 exp[−(ζ + 1)r ] 0 dθ √ √ 2π ζ5 ζ5 sin θ cos2 θ 0 dφ = E π · 4!(ζ + 1)−5 · 23 · 2π = 32 (ζ +1)5 E . Now we can write Eq. (V.2) as a function of ζ : 10 1 2 1 2 2 2 5 E = (ζ − ζ − 1) − (ζ − ζ + 1) + ζ E 2. (V.3) 4 16 ζ +1 We would like to expand this expression in a 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 atomic units) 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. (12.24) the polarizability (in atomic units) reads as 10 2 ζ5 αzz = 4 · 2 . |ζ − ζ + 1| ζ + 1
(V.4)
Several numerical values of αzz computed by using Eqs. (V.3) and (V.4) are given on p. 746. They are compared with the exact result αzz = 4.5 a.u.