SINGLET AND TRIPLET STATESFOR TWO ELECTRONS An angular momentum is a vector, and this pertains also to spin angular momenta see Chapter 1.. These operators will be created using Pauli ma
Trang 1Q 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
ˆS2= ˆS2
x+ ˆS2
y+ ˆS2
z= ˆS2
z+ ˆS+ˆS−− ¯h ˆSz (Q.1) where ˆS2≡ ˆS2and ˆS+and ˆS−are the lowering and raising operators, respectively,
lowering and
raising
operators ˆS+= ˆSx+ i ˆSy (Q.2)
ˆS−= ˆSx− i ˆSy (Q.3) which satisfy the useful relations justifying their names:
ˆS+α= 0 ˆS+β= ¯hα
ˆS−α= ¯hβ ˆS−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 12 (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= 0 and the triplet 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: MS=
1 See Postulate VI in Chapter 1.
1006
Trang 2Q SINGLET AND TRIPLET STATES FOR TWO ELECTRONS 1007
−S −S + 1 +S, i.e MS= 0 for the singlet state and MS= −1 0 +1 for the
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:
S2= (s1+ s2)2= s2
1+ s2
2+ 2s1s2 Thus to create operator ˆS2we need operatorsˆs2
1andˆs2
2, which will be expressed
by the lowering and raising operators according to eq (Q.1), and the scalar product
ˆs1ˆs2expressed 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 ˆS2acts on α(1)β(2),
after five lines of derivation we obtain
ˆS2 α(1)β(2)
= ¯h2 α(1)β(2)+ α(2)β(1)
similarly
ˆS2 α(2)β(1)
= ¯h2 α(1)β(2)+ α(2)β(1) Now we will use this result to calculate
ˆS2
α(1)β(2)− α(2)β(1) and ˆS2
α(1)β(2)+ α(2)β(1)
We have
ˆS2 α(1)β(2)− α(2)β(1)
= 0 × α(1)β(2)− α(2)β(1)
≡ S(S + 1)¯h2
α(1)β(2)− α(2)β(1) where S= 0 (singlet) and
ˆS2
α(1)β(2)+ α(2)β(1) = 2
¯h2 α(1)β(2)+ α(2)β(1) ≡ S(S + 1)¯h2
α(1)β(2)+ α(2)β(1) where S= 1 (triplet)
If operator ˆSz= ˆs1z+ ˆs2zacts on[α(1)β(2) − α(2)β(1)], we obtain
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 ˆSz= ˆs1z+ ˆs2zacts on[α(1)β(2) + α(2)β(1)], we have
0× α(1)β(2)+ α(2)β(1)
Trang 31008 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 Sz = ¯h and Sz= −¯h, re-spectively Therefore, after normalization2finally
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)]:
σ1
σ2
1
√ 2
α(1)β(2) − α(2)β(1) )2
=
σ1
σ2
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
σ1
α(1) 2
σ2
β(2) 2 +
σ2
α(2) 2
σ1
β(1) 2 − 2
σ2
α(2)β(2)
σ1
α(1)β(1) )
=1
2 {1 · 1 + 1 · 1 − 2 · 0 · 0} = 1
Trang 4R 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): 1sa and 1sb 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±=Haa± Hab
1± S where the Hamiltonian (in a.u.)1 Hˆ = −1
2 − 1
ra − 1
rb + 1
R and S stands for the overlap integral of the two atomic orbitals Thus we have
E±= 1
R+Haa± Hab
1± S =
1
R+
−1
2 − 1
ra− 1
rb
aa±−1
2 − 1
ra − 1
rb
ab
1± S
= 1
R+EH+ Vaa b± EHS± Vab b
R+Vaa b± Vab b
1± S where EH means the energy of the hydrogen atom, while the nuclear attraction integrals are
Vaa b= −
a
r1ba Vab b= −
a
r1bb 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, Vaa band Vab b We use the elliptic coordinates (Fig R.1):
μ=ra+ rb
R ν=ra− rb
R φ= arctan
y x
The volume element in the elliptic coordinates is dV = R3/8(μ2− ν2) dμ dν dφ, where 1
1 See Fig R.1 for explaining symbols.
1009
Trang 51010 R THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET
electron
Fig R.1. The elliptic coordinates μ =
ra+rb
R , ν =ra −rb
R built using distances r a
and rbfrom 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:
An(σ α)=
∞
σ
μnexp(−αx) dx = exp(−ασ)
n
k=0
n! (n− k)!
σn−k
αk+1
Bn(α)=
+1
−1 x
nexp(−αx) dx = An(−1 α) − An(1 α)
The integrals An(σ α)satisfy the following recurrence relation:
An(σ α)= σnA0(σ α)+n
αAn−1(σ α)
A0(σ α)=1
αexp(−σα)
These are some simple auxiliary integrals (we will need them in a moment):
A1(σ α)= σ1
αexp(−σα) + 1
α
1
αexp(−σα) = 1
α
σ+1 α
exp(−σα)
A2(σ α)= σ21
αexp(−σα) + 2
α
1 α
σ+1 α
exp(−σα)
= 1
αexp(−σα)
σ2+2 α
σ+1 α
B0(α)= 1
αexp(α)−1
αexp(−α) = 1
α
exp(α)− exp(−α)
B1(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= R3 8π
∞
1
dμ exp(−Rμ)
+1
−1 dν
μ2− ν2 2π
0
dφ
Trang 6R THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET 1011
=R3
2
∞
1
dμ μ2exp(−Rμ) −1
3
∞
1
dμ exp(−Rμ)
=R3
2
A2(1 α)−1
3A0(1 α)
=R3
2
1
Rexp(−R)
1+ 2
R+ 2
R2
−1 3
1
Rexp(−R)
= exp(−R)
R2
3 + R + 1
Thus we have explicit dependence on R The formula for S satisfies correctly
the limiting cases: limR→∞S(R)= 0 and limR →0S(R)= 1 (normalization of the 1s
orbital) Besides
dS
dR = − exp(−R)
R2
3 + R + 1
+ exp(−R)
2
3R+ 1
= − exp(−R)
R2+ 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 An(σ α) and
Bn(α) we obtain
−Vaa b=
a
r1ba= 1
π
exp(−2ra)1
rbdτ
=R3
8π
2 R
∞
1
dμ exp
−R(μ + ν) +1
−1 dν
(μ2− ν2)
μ− ν
2π
0
dφ
=R2
4π2π
∞
1
dμ
+1
−1 dν exp(−Rμ) exp(−Rν)(μ + ν)
=R2
2
∞
1 dμ μ exp(−Rμ)
+1
−1 dν exp(−Rν) +
∞
1 dμ exp(−Rμ)
+1
−1 dν ν exp(−Rν)
=R2
2
A1(1 R)B0(R)+ A0(1 R)B1(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.
Trang 71012 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−Vaa bmeans (a|1
rb|a), which at large R should give the Coulombic interaction of the two unit point charges, i.e R1 This
is what we have as the first term The second term: Epenetr= − exp(−2R)(1 + 1
R)
represents what is known as penetration energy resulting from the non-point-like
penetration
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 a2.
Trang 8R 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
Vaa b(R)
= − lim
R→0
1
R−
1− 2R +1
2R
2+ · · ·
1+ 1 R
= − lim
R →0
1
R− 1 + 2R −1
2R
2− 1
R+ 2 +1
2R+ · · ·
= −1
This is exactly what we get for the hydrogen atom when calculating:
Vaa a= −
dv1
r(1s)2= −1
π
exp(−2r)1
rr
2sin θ dr dθ dφ
= −4
∞
0 r exp(−2r) dr = −4 × 2−2= −1
Thus everything is all right
Similarly we calculate
−Vab b=
a
r1bb= 1
π
exp
−(ra+ rb) 1
rbdv
= 1 π
2 R
exp(−Rμ) 1
(μ− ν)
R3
8
μ2− ν2
dμ dν dφ
=R2 2
∞
1
+1
−1 dμ dν
μ exp(−Rμ) + ν exp(−Rμ)
=R2
2 2A1(1 R)+ 0 = (1 + R) exp(−R)
If R→ ∞, then −Vab b→ 0, which is the correct behaviour Do we get Vaa a=
−1, if R → 0? Again, let us expand the exponential function:
Vaa a= − lim
R →0(1+ R) exp(−R) = − lim
R →0(1+ R)
1− R +R2
2 + · · ·
= − lim
R→0
1+ R − R − R2+R2
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:
Trang 91014 R THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET
E±= EH+ 1
R+Vaa b± Vab b
1± S
= EH+ 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 EH for all sufficiently large R (a single minimum) The energy of the antibonding orbital is higher than EH for all R (no minimum) This simple theory predicts the position of the energy minimum for the ground state as Re= 25 a.u., while the experimental value is equal4ca 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).
Trang 10S 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 ϕi, we have (see (11.5)):
ρ(r)= 2 ϕ1(r)2+ϕ2(r)2+ · · · +ϕN
2
(r)2
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, then ρ 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 ϕi=rcriχr, we have (Srs stands for the overlap integrals between the atomic orbitals r and s, and c are the corresponding LCAO coefficients)
N=
ρ(r) dV = 2
N/2
i =1
ϕ
i(r)2
dV =
i
rs
2cri∗csiSrs
rs
where P is called the charge and bond-order matrix
Psr=
i
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