## Xem mẫu

1026 U. SECOND QUANTIZATION One-electron operators The operator F =Pi h(i) is the sum of the one-electron operators7 h(i) acting on functions of the coordinates of electron i. The I Slater–Condon rule says (see Appendix M), that for the Slater determi-nant ψ built of the spinorbitals φi, the matrix element hψ|Fψi = i hii, where hij =hφi|hφji. In the second quantization X F = hiji j: ij Interestingly, the summation extends to inﬁnity, and therefore the operator is independent of the number of electrons in the system. Let us check whether the formula is correct. Let us insert F = ij hijı†j into hψ|Fψi. We have ¿ ¯X À X X X ψ Fψ = ψ¯ hijı jψ = hij ψ ı jψ = hijδij = hii: ij ij ij i This is correct. WhatabouttheIISlater–Condonrule(theSlaterdeterminants ψ1 and ψ2 differ by a single spinorbital: the spinorbital i in ψ1 is replaced by the spinorbital i in ψ )? We have ­ψ1¯Fψ2®= hij­ψ1¯ı†jψ2®: ij The Slater determinants that differ by one spinorbital produce an overlap inte-gral equal to zero,8 therefore hψ1|Fψ2i = hii0. Thus, the operator in the form F = ij hijı†j ensures equivalence with all the Slater–Condon rules. Two-electron operators Similarly, we may use the creation and annihilation operators to represent the two-electron operators G= 1 ij g(i;j). In most cases g(i;j)= 1 and Ghas the form: G= 1X0 1 = 1 ∞ hij|klij†ˆ†kl: ij ij ijkl 7Most often this will be the kinetic energy operator, the nuclear attraction operator, the interaction with the external ﬁeld or the multipole moment. 8It is evident, that if in this situation the Slater determinants ψ1 and ψ2 differed by more than a single spinorbital, we would get zero (III and IV Slater–Condon rule). U. SECOND QUANTIZATION 1027 Here also the summation extends to inﬁnity and the operator is independent of the number of electrons in the system. The proof of the I Slater–Condon rule relies on the following chain of equalities X X ψ Gψ = hij|kli ψ j ˆ klψ = hij|kli ıjψ klψ ijkl ijkl X X = hij|kli(δikδjl −δilδjk)= hij|iji−hij|jii ; ijkl ij because the overlap integral hıjψ|klψi of the two Slater determinants ıjψ and klψ is non-zero in the two cases only: either if i = k, j = l or if i = l, j = k (then the sign has to change). This is what we get from the Slater–Condon rules. For the II Slater–Condon rule we have (instead of the spinorbital i in ψ we have the spinorbital i0 in ψ2): ­ψ1¯Gψ2®= 1 XhIj|kli­ψ1¯j†I†klψ2®= 1 XhIj|kli­Ijψ1¯klψ2®; (U.1) Ijkl Ijkl where the summation index I has been introduced in order not to mix with spinor-bital i. In the overlap integral hIjψ1|klψ2i the sets of the spinorbitals in the Slater determinant Ijψ1 and in the Slater determinant klψ2 have to be identical, other-wise the integral will equal zero. However, in ψ1 and ψ2 we already have a differ-ence of one spinorbital. Thus, ﬁrst we have to get rid of these spinorbitals (i and i ). For the integral to survive9 we have to have at least one of the following conditions satisﬁed: • I =i and k=i0 (and then j =l), • j =i and k=i0 (and then I =l), • I =i and l =i0 (and then j =k), • j =i and l =i0 (and then I =k). This means that, taking into account the above cases in eq. (U.1), we obtain ­ψ1¯Gψ2® = 1 X­ij¯i0j®­ıjψ1¯ı0jψ2®+ 1 X­li¯i0l®­lıψ1¯ı0lψ2® j l + 2 X­ij¯ji0®­ıjψ1¯jı0ψ2®+ 2 X­ki¯ki0®­kıψ1¯kı0ψ2® = 2 X­ij¯i0j®− 2 X­li¯i0l®− 2 X­ij¯ji0®+ 2 X­ki¯ki0® = 2 X­ij¯i0j®− 2 X­ji¯i0j®− 2 X­ij¯ji0®+ 2 X­ji¯ji0® 9This is a necessary, but not a sufﬁcient condition. 1028 U. SECOND QUANTIZATION = 2 X­ij¯i0j®− 2 X­ij¯ji0®− 2 X­ij¯ji0®+ 2 X­ij¯i0j® = X­ij¯i0j®−X­ij¯ji0®; j j where the coordinates of electrons 1 and 2 have been exchanged in the two sums. Notice that the overlap integrals ­ıjψ1¯ı0jψ2®=­kıψ1¯kı0ψ2®=1; because the Slater determinants iψ1 and i0ψ2 are identical. Also, from the anti-commutation rules lıψ1¯ı0lψ2 = ıjψ1¯jı0ψ2 =−1: Thus the II Slater–Condon rule has been correctly reproduced: ­ψ1¯Gψ2®=X£­ij¯i0j®−­ij¯ji0®¤: j We may conclude that the deﬁnition of the creation and annihilation operators and the simple anticommutation relations are equivalent to the Slater–Condon rules. This opens up the space spanned by the Slater determinants for us, i.e. all the integrals involving Slater determinants can be easily transformed into one- and two-electron integrals involving spinorbitals. V. THE HYDROGEN ATOM IN THE ELECTRIC FIELD – VARIATIONAL APPROACH Polarization of an atom or molecule can be calculated by using the ﬁnite ﬁeld method described on p. 639. Let us apply this method to the hydrogen atom. Its polarizability was already calculated using a simple version of perturbation theory (p. 636). This time we will use the variational method. The Hamiltonian for the isolated hydrogen atom (within the Born–Oppenhei-mer approximation) reads as H(0) =−11e − 1; where the ﬁrst term is the electronic kinetic energy operator, and the second its Coulomb interaction energy with the nucleus (proton–electron distance is denoted by r). The atom is in a homogeneous electric ﬁeld E = (0;0;E) with E >0 and as in perturbation theory (p. 636), the total Hamiltonian has the form H =H(0) +V with V = zE, where z denotes the coordinate of the electron and the proton is at the origin (the derivation of the formula is given on p. 636, the exchange of z to x does not matter). The variational wave function ψ is proposed in the form ψ=χ1 +cχ2; (V.1) where χ1 = 1 exp(−r) is the 1s orbital of the hydrogen atom (ground state) and χ2 is the normalized1 p-type orbital χ2 =Nzexp(−ζr): 1N can be easily calculated from the normalization condition 1 = N2 Z £zexp(−ζr)¤2 dV =N2 Z ∞ drr4 exp(−2ζr)Z π dθ sinθcos2 θZ 2π dφ 0 0 0 = N24!(2ζ)−5 22π =N2 π : q This gives N = π . 1029 1030 V. THE HYDROGEN ATOM IN THE ELECTRIC FIELD – VARIATIONAL APPROACH 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 ψis a linear combination of two expansion functions (“two-state model”): χ1 and χ2. Therefore, optimal energy follows from the Ritz method, according to case III of Appendix D on p. 948: p E =Ear ± 12 +h2; (V.2) where arithmetic mean energy Ear ≡ H11+H22 , while 1 ≡ H11−H22 and h ≡ H12 = H21 with Hij ≡ χi¯Hχj = χi¯H(0)χj +hχi|V χji: Let us calculate all the ingredients of the energy given by (V.2). First, let us note that H11 ≡hχ1|H(0)χ1i=−1 a.u., since χ1 is the ground state of the isolated hydrogen atom (p. 178), and V11 =hχ1|V χ1i=0, because the inte-grand is antisymmetric with respect to z →−z. Now let us calculate H22 = H(0) +V22. Note that V22 = 0, for the same reason as V11. We have ¿ ¯ À H(0) =−2hχ2|1eχ2i− χ2¯rχ2 : The second integral is ¿ ¯ À Z ∞ Z π Z 2π χ2 χ2 = N drr exp(−2ζr) dθ sinθcos θ dφ 0 0 0 = π ·3!(2ζ)−4 · 3 ·2π = 2ζ; where the dots separate the values of the corresponding integrals.2 In Appendix R, the reader will ﬁnd the main ingredients needed to calculate the ﬁrst integral of H(0): ¿ ¯· hχ2|1eχ2i = N2 rcosθexp(−ζr)¯ r2 ∂rr2 ∂r + r2 sinθ ∂θ sinθ∂θ 2 ¸ À + r2 sin2 θ ∂φ2 rcosθexp(−ζr) 2 ­rcosθexp(−ζr)¯cosθ 1 ∂ £r2 exp(−ζr)−ζr3 exp(−ζr)®+ rcosθexp(−ζr)¯(−2cosθ)rexp(−ζr) +0 ·¿ ¯ · ¸ À = N2 rcosθexp(−ζr)¯cosθ r −ζ −3ζ +ζ2r exp(−ζr) 2Note that, in spherical coordinates, the volume element dV =r2 sinθdrdθdφ. In derivations of this Appendix (and not only) we often use the equality 0 dxxn exp(−αr)=n!α−(n+1). ... - tailieumienphi.vn
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