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
ˆ2 =S2 +S2 +S2 =S2 +S+S− −hSz; (Q.1)
lowering and raising operators
where S2 ≡S2 and S+ and S− are the lowering and raising operators, respectively,
ˆ+ = Sx +iSy; (Q.2)
ˆ− = Sx −iSy; (Q.3)
which satisfy the useful relations justifying their names:
ˆ+α = 0;
S−α = hβ;
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 (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 =
1See Postulate VI in Chapter 1.
Q. 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 +s2 +2s1s2:
Thus to create operator S2 we need operators s2 and s2, which will be expressed bythe loweringand raisingoperatorsaccordingto eq. (Q.1), and the scalarproduct s1ˆ2 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 S2 acts on α(1)β(2), after ﬁve lines of derivation we obtain
Now we will use this result to calculate
ˆ2£α(1)β(2)−α(2)β(1)¤ and S2£α(1)β(2)+α(2)β(1)¤:
where S =0 (singlet) and
where S =1 (triplet).
If operator Sz =s1z +s2z acts 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 Sz =s1z +s2z acts on [α(1)β(2)+α(2)β(1)], we have
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 Sz = h and Sz = −h, re-spectively. Therefore, after normalization2 ﬁnally
1 [α(1)β(2) − α(2)β(1)] is a singlet function, while: 1 [α(1)β(2) + α(2)β(1)], α(1)α(2) and β(1)β(2) represent three triplet functions.
2For example let us check the normalization of the singlet function √ [α(1)β(2)−α(2)β(1)]: XX½ 1 £α(1)β(2)−α(2)β(1)¤¾2
=XX 1©£α(1)¤2£β(2)¤2 +£α(2)¤2£β(1)¤2 −2£α(2)β(2)¤£α(1)β(1)¤ª
= 1½X£α(1)¤2 X£β(2)¤2 +X£α(2)¤2 X£β(1)¤2 −2X£α(2)β(2)¤X£α(1)β(1)¤¾
σ1 σ2 σ2 σ1 σ2 σ1
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): 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
Haa ±Hab 1±S
where the Hamiltonian (in a.u.)1 H = −11 − 1 − 1 + 1 and S stands for the overlap integral of the two atomic orbitals. Thus we have
1 Haa ±Hab 1 ¡− 11− 1 − 1 ¢aa ±¡−11− 1 − 1 ¢ab R 1±S R 1±S
1 EH +Vaa;b ±EHS ±Vab;b 1 Vaa;b ±Vab;b R 1±S H R 1±S
where EH means the energy of the hydrogen atom, while the nuclear attraction
µ ¶ µ ¶
Vaa;b =− a¯ ¯a ; Vab;b =− a¯ ¯b : 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, Vaa;b and Vab;b. We use the elliptic coordinates (Fig. R.1):
μ= ra +rb ; ν = ra −rb ;
φ=arctan x :
The volume element in the elliptic coordinates is dV =R3/8(μ2 −ν2)dμdνdφ, where 1 6μ<∞, −1 6ν 61;0 6φ62π.
1See Fig. R.1 for explaining symbols.
1010 R. THE HYDROGEN MOLECULAR ION IN THE SIMPLEST ATOMIC BASIS SET
Fig. R.1. The elliptic coordinates μ = ra+rb , ν = ra−rb built using distances ra and rb 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:
Z ∞ X n−k An(σ;α) = σ μ exp(−αx)dx=exp(−ασ)k=0 (n−k)! αk+1 ;
Bn(α) = x exp(−αx)dx=An(−1;α)−An(1;α): −1
The integrals An (σ;α) satisfy the following recurrence relation: An(σ;α) = σnA0(σ;α)+ nAn−1(σ;α);
A0(σ;α) = α exp(−σα):
These are some simple auxiliary integrals (we will need them in a moment): µ ¶
A1(σ;α) = σα exp(−σα)+ α α exp(−σα)= α σ + α exp(−σα);
A2(σ;α) = σ2 α exp(−σα)+ α α σ + α exp(−σα)
= α exp(−σα) σ2 + α σ + α ;
B0(α) = 1 exp(α)− 1 exp(−α)= 1£exp(α)−exp(−α)¤; µ ¶ µ ¶
B1(α) = α −1+ α exp(α)− α 1+ α exp(−α) ·µ ¶ µ ¶ ¸
= α α −1 exp(α)− α +1 exp(−α) :
Thus, the overlap integral S is calculated in the following way
3 Z ∞ Z +1 Z 2π
S = dμ exp(−Rμ) dν μ −ν dφ 1 −1 0
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