Thông tin tài liệu
Q. SINGLET AND TRIPLET STATES FOR TWO ELECTRONS An angular momentum is a vector, and this pertains also to spin angular momenta (see Chapter 1). The spin angular momentum of a certain number of elementary particles is the sum of their spin vectors. To obtain the total spin vector, we there- fore have to add the x components of the spins of the particles, similarly for the y and z components, and to construct the total vector from them. Then we might be interested in the corresponding spin operators. These operators will be created using Pauli matrices. 1 In this way we find immediately that, for a single particle, the following identity holds ˆ S 2 = ˆ S 2 x + ˆ S 2 y + ˆ S 2 z = ˆ S 2 z + ˆ S + ˆ S − − ¯ h ˆ S z (Q.1) where ˆ S 2 ≡ ˆ S 2 and ˆ S + and ˆ S − are the lowering and raising operators, respectively,lowering and raising operators ˆ S + = ˆ S x +i ˆ S y (Q.2) ˆ S − = ˆ S x −i ˆ S y (Q.3) which satisfy the useful relations justifying their names: ˆ S + α = 0 ˆ S + β = ¯ hα ˆ S − α = ¯ hβ ˆ S − β =0 For any stationary state the wave function is an eigenfunction of the square of the total spin operator and of the z-component of the total spin operator. The one and two-electron cases are the only ones for which the total wave function is the product of space and spin parts. The maximum projection of the electron spin on the z axis is equal to 1 2 (in a.u.). Hence, the maximum projection for the total spin of two electrons is equal to 1. This means that in this case only two spin states are possible: the singlet state corresponding to S = 0andthetriplet state with S = 1 (see Postulate V). In the singlet state the two electronic spins are opposite (“pairing of electrons”), while in the triplet state the spin vectors are “parallel” (cf. Fig. 1.11 in Chapter 1). As always, the possible projection of the total spin takes one of the values: M S = 1 See Postulate VI in Chapter 1. 1006 Q. SINGLET AND TRIPLET STATES FOR TWO ELECTRONS 1007 −S−S +1+S, i.e. M S = 0 for the singlet state and M S =−1 0 +1forthe triplet state. Now it will be shown that the two-electron spin function α(1)β(2) −α(2)β(1) ensures the singlet state. First, let us construct the square of the total spin of the two electrons: S 2 =(s 1 +s 2 ) 2 =s 2 1 +s 2 2 +2s 1 s 2 Thus to create operator ˆ S 2 we need operators ˆ s 2 1 and ˆ s 2 2 , which will be expressed by the lowering and raising operators according to eq. (Q.1), and the scalar product ˆ s 1 ˆ s 2 expressed as the sum of the products of the corresponding components x, y and z (we know how they act, see Postulate V in Chapter 1). If ˆ S 2 acts on α(1)β(2), after five lines of derivation we obtain ˆ S 2 α(1)β(2) = ¯ h 2 α(1)β(2) +α(2)β(1) similarly ˆ S 2 α(2)β(1) = ¯ h 2 α(1)β(2) +α(2)β(1) Now we will use this result to calculate ˆ S 2 α(1)β(2) −α(2)β(1) and ˆ S 2 α(1)β(2) +α(2)β(1) We have ˆ S 2 α(1)β(2) −α(2)β(1) =0 × α(1)β(2) −α(2)β(1) ≡S(S +1) ¯ h 2 α(1)β(2) −α(2)β(1) where S =0 (singlet) and ˆ S 2 α(1)β(2) +α(2)β(1) =2 ¯ h 2 α(1)β(2) +α(2)β(1) ≡S(S +1) ¯ h 2 α(1)β(2) +α(2)β(1) where S =1 (triplet). If operator ˆ S z = ˆ s 1z + ˆ s 2z acts on [α(1)β(2) −α(2)β(1)],weobtain 0 × α(1)β(2) −α(2)β(1) This means that, in the singlet state, the projection of the spin on the z axis is equal to 0. This is what we expect from a singlet state function. On the other hand, if ˆ S z = ˆ s 1z + ˆ s 2z acts on [α(1)β(2) +α(2)β(1)],wehave 0 × α(1)β(2) +α(2)β(1) 1008 Q. SINGLET AND TRIPLET STATES FOR TWO ELECTRONS i.e. the function [α(1)β(2)+α(2)β(1)]is such a triplet function which corresponds to the zero projection of the total spin. A similarly simple calculation for the spin functions α(1)α(2) and β(1)β(2) gives the eigenvalue S z = ¯ h and S z =− ¯ h,re- spectively. Therefore, after normalization 2 finally 1 √ 2 [α(1)β(2) − α(2)β(1)] is a singlet function, while: 1 √ 2 [α(1)β(2) + α(2)β(1)], α(1)α(2) and β(1)β(2) represent three triplet functions. 2 For example let us check the normalization of the singlet function 1 √ 2 [α(1)β(2) −α(2)β(1)]: σ 1 σ 2 1 √ 2 α(1)β(2) −α(2)β(1) 2 = σ 1 σ 2 1 2 α(1) 2 β(2) 2 + α(2) 2 β(1) 2 −2 α(2)β(2) α(1)β(1) = 1 2 σ 1 α(1) 2 σ 2 β(2) 2 + σ 2 α(2) 2 σ 1 β(1) 2 −2 σ 2 α(2)β(2) σ 1 α(1)β(1) = 1 2 {1 ·1 +1 ·1 −2 ·0 ·0}=1 R. THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET Consider the quantum mechanical description of the hydrogen molecular ion in its simplest version. Let us use molecular orbital theory with the atomic basis set composed of only two Slater Type Orbitals (STO): 1s a and 1s b centred on the nuclei a and b. The mean value of the Hamiltonian calculated with the bonding (+) and antibonding (−) orbital (see Chapter 8 and Appendix D) reads as E ± = H aa ±H ab 1 ±S where the Hamiltonian (in a.u.) 1 ˆ H =− 1 2 − 1 r a − 1 r b + 1 R and S stands for the overlap integral of the two atomic orbitals. Thus we have E ± = 1 R + H aa ±H ab 1 ±S = 1 R + − 1 2 − 1 r a − 1 r b aa ± − 1 2 − 1 r a − 1 r b ab 1 ±S = 1 R + E H +V aab ±E H S ±V abb 1 ±S =E H + 1 R + V aab ±V abb 1 ±S where E H means the energy of the hydrogen atom, while the nuclear attraction integrals are V aab =− a 1 r b a V abb =− a 1 r b b The energy E ± is a function of the internuclear distance R, which is hidden in the dependence of the integrals on R. We want to have this function explicitly. To this end we have to compute the integrals S, V aab and V abb . We use the elliptic coordinates (Fig. R.1): μ = r a +r b R ν= r a −r b R φ=arctan y x The volume element in the elliptic coordinates is dV =R 3 /8(μ 2 −ν 2 ) dμ dν dφ, where 1 μ<∞, −1 ν 1 0 φ 2π. 1 See Fig. R.1 for explaining symbols. 1009 1010 R. THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET electron Fig. R.1. The elliptic coordinates μ = r a +r b R , ν = r a −r b R built using distances r a and r b from the two foci (where the nu- clei are, their distance is R) of the ellipse. The angle φ measures the rotation of the plane defined by ab and the correspond- ing electron about the ab axis. We will need two auxiliary integrals: A n (σα) = ∞ σ μ n exp(−αx) dx =exp(−ασ) n k=0 n! (n −k)! σ n−k α k+1 B n (α) = +1 −1 x n exp(−αx) dx =A n (−1α)−A n (1α) The integrals A n ( σ α ) satisfy the following recurrence relation: A n (σα) =σ n A 0 (σα) + n α A n−1 (σα) A 0 (σα) = 1 α exp(−σα) These are some simple auxiliary integrals (we will need them in a moment): A 1 (σα) =σ 1 α exp(−σα)+ 1 α 1 α exp(−σα) = 1 α σ + 1 α exp(−σα) A 2 (σα) =σ 2 1 α exp(−σα)+ 2 α 1 α σ + 1 α exp(−σα) = 1 α exp(−σα) σ 2 + 2 α σ + 1 α B 0 (α) = 1 α exp(α) − 1 α exp(−α) = 1 α exp(α) −exp(−α) B 1 (α) = 1 α −1 + 1 α exp(α) − 1 α 1 + 1 α exp(−α) = 1 α 1 α −1 exp(α) − 1 α +1 exp(−α) Thus, the overlap integral S is calculated in the following way S = R 3 8π ∞ 1 dμ exp(−Rμ) +1 −1 dν μ 2 −ν 2 2π 0 dφ R. THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET 1011 = R 3 2 ∞ 1 dμμ 2 exp(−Rμ) − 1 3 ∞ 1 dμ exp(−Rμ) = R 3 2 A 2 (1α)− 1 3 A 0 (1α) = R 3 2 1 R exp(−R) 1 + 2 R + 2 R 2 − 1 3 1 R exp(−R) = exp(−R) R 2 3 +R +1 Thus we have explicit dependence on R.TheformulaforS satisfies correctly the limiting cases: lim R→∞ S(R) =0andlim R→0 S(R) =1 (normalization of the 1s orbital). Besides dS dR =−exp(−R) R 2 3 +R +1 +exp(−R) 2 3 R +1 =−exp(−R) R 2 +R 3 < 0 i.e. the overlap integral of the 1s functions decreases from 1 to 0, if R →∞(see Fig. R.2.a). We see that for small R the function S decreases gently, while for larger R it decreases fast. 2 Using the elliptic coordinates and the formulae for the integrals A n (σα) and B n (α) we obtain −V aab = a 1 r b a = 1 π exp(−2r a ) 1 r b dτ = R 3 8π 2 R ∞ 1 dμ exp −R(μ +ν) +1 −1 dν (μ 2 −ν 2 ) μ −ν 2π 0 dφ = R 2 4π 2π ∞ 1 dμ +1 −1 dν exp(−Rμ) exp(−Rν)(μ +ν) = R 2 2 ∞ 1 dμμexp(−Rμ) +1 −1 dν exp(−Rν) + ∞ 1 dμ exp(−Rμ) +1 −1 dννexp(−Rν) = R 2 2 A 1 (1R)B 0 (R) +A 0 (1R)B 1 (R) = 1 R −exp(−2R) 1 + 1 R 2 Just to get an idea: at R =5 a.u. (quite typical for van der Waals complexes) the value of the overlap integral is of the order of 0.1. 1012 R. THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET Fig. R.2. The hydrogen molecule in the simplest basis set of two 1s Slater type orbitals (STO). (a) The overlap integral S as a function of the internuclear distance R. (b) The penetration energy represents the difference between the electron–proton interaction calculated assuming the electronic charge distribution and the same energy calculated with the point charges (the electron is located on nucleus a). (c) The ener- gies E + and E − of the bonding (lower curve) and of the antibonding (upper curve) orbitals. Energies and distances in a.u. This is an interesting result. The integral −V aab means (a| 1 r b |a), which at large R should give the Coulombic interaction of the two unit point charges, i.e. 1 R .This is what we have as the first term. The second term: E penetr =−exp(−2R)(1 + 1 R ) represents what is known as penetration energy resulting from the non-point-likepenetration energy character of one of the interacting charges. 3 From Fig. R.2.b we see that the penetration energy vanishes much faster that the overlap integral. No wonder it vanishes as exp(−2R), while the overlap integral vanishes only as exp(−R). It is seen that the diffuse charges interact more weakly. 3 The electron cloud of electronic density a 2 . R. THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET 1013 On the one hand diffuse charges offer the chance to be close in space (this in- creases the interaction), on the other hand some charges become more distant. The second effect prevails and therefore the penetration energy makes the Coulombic interaction weaker. What will happen if R →0? Let us expand the exponential function in the Taylor series. We obtain lim R→0 V aab (R) =−lim R→0 1 R − 1 −2R + 1 2 R 2 +··· 1 + 1 R =−lim R→0 1 R −1 +2R − 1 2 R 2 − 1 R +2 + 1 2 R +··· =−1 This is exactly what we get for the hydrogen atom when calculating: V aaa =− dv 1 r (1s) 2 =− 1 π exp(−2r) 1 r r 2 sinθ dr dθ dφ =−4 ∞ 0 r exp(−2r)dr =−4 ×2 −2 =−1 Thus everything is all right. Similarly we calculate −V abb = a 1 r b b = 1 π exp −(r a +r b ) 1 r b dv = 1 π 2 R exp(−Rμ) 1 (μ −ν) R 3 8 μ 2 −ν 2 dμ dν dφ = R 2 2 ∞ 1 +1 −1 dμ dν μ exp(−Rμ) +ν exp(−Rμ) = R 2 2 2A 1 (1R)+0 =(1 +R)exp(−R) If R →∞,then−V abb →0, which is the correct behaviour. Do we get V aaa = −1, if R →0? Again, let us expand the exponential function: V aaa =−lim R→0 (1 +R) exp(−R) =−lim R→0 (1 +R) 1 −R + R 2 2 +··· =−lim R→0 1 +R −R −R 2 + R 2 2 +··· =−1 This is what we expected. Bonding and antibonding orbital energy If we insert the results into the formula for the energy of the bonding and anti- bonding orbitals, we obtain the most important formulae for the problem being considered: 1014 R. THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET E ± = E H + 1 R + V aab ±V abb 1 ±S = E H + 1 R + − 1 R +exp(−2R) 1 + 1 R ±(−1 −R)exp(−R) 1 ± exp(−R) R 2 3 +R +1 The plots of E ± are shown in Fig. R.2.c. It is seen that in the quite primitive LCAO MO approximation, the bonding energy is lower than the energy of the hydrogen atom E H for all sufficiently large R (a single minimum). The energy of the antibonding orbital is higher than E H for all R (no minimum). This simple theory predicts the position of the energy minimum for the ground state as R e = 25 a.u., while the experimental value is equal 4 ca. 2.0 a.u. 4 These two quantities are not directly comparable, because the experimental value does not corre- spond exactly to the position of the minimum (because of anharmonicity). S. POPULATION ANALYSIS On p. 569 the electronic density ρ is defined. If the wave function is a Slater de- terminant (p. 332) and assuming the double occupancy of orbitals ϕ i ,wehave (see (11.5)): ρ(r) =2 ϕ 1 (r) 2 + ϕ 2 (r) 2 +···+ ϕ N 2 (r) 2 (S.1) The density distribution ρ may be viewed as a cloud carrying a charge −Ne and eq. (S.1) says that the cloud is composed of individual clouds of molecular orbitals, each carrying two electrons. On the other hand in the LCAO approximation any molecular orbital is represented by the sum of atomic orbitals. If we insert the LCAO expansion into ρ,thenρ becomes the sum of the contributions, each being the product of two atomic orbitals. There is a temptation to go even further and to divide ρ somehow into the contributions of particular atoms, calculate the charge corresponding to such contributions and locate the (point) charge on the nucleus. 1 We might say therefore, what the “electron population” residing on the particular atoms is (hence the name: population analysis). Mulliken population analysis Such tricks are of course possible, and one of them is called Mulliken popula- tion analysis. From (S.1), after using the LCAO expansion ϕ i = r c ri χ r ,wehave (S rs stands for the overlap integrals between the atomic orbitals r and s,andc are the corresponding LCAO coefficients) N = ρ(r)dV =2 N/2 i=1 ϕ i (r) 2 dV = i rs 2c ∗ ri c si S rs = rs P rs S rs =Tr (PS) (S.2) where P is called the charge and bond-order matrix P sr = i 2c ∗ ri c si (S.3) The summation over r and s may be carried out, being careful from which atom A the particular atomic orbital comes (we assume that the AO’s are centred on the nuclei). We get an equivalent formula (A ad B denote atoms): 1 This number need not be an integer. 1015 . angular momentum of a certain number of elementary particles is the sum of their spin vectors. To obtain the total spin vector, we there- fore have to add the x components of the spins of the particles,. of the square of the total spin operator and of the z-component of the total spin operator. The one and two-electron cases are the only ones for which the total wave function is the product of. the value of the overlap integral is of the order of 0.1. 1012 R. THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET Fig. R.2. The hydrogen molecule in the simplest basis set of two 1s
Ngày đăng: 06/07/2014, 09:20
Xem thêm: Ideas of Quantum Chemistry P105 pot
