Modern Quantum Chemistry

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Exercises: 1

This does not correspond to any chapter in the book, but rather are extra derivations that I found necessary during my coverage of the book, starting with chapter 3.

How are equations (3.219)-(3.221) in the book derived? Let’s start with the STO-1G case and try to maximize the overlap integral

S =drϕ1sSF(ζ,r)ϕ 1sCGF(ζ,STO 1G,r)

or, using the explicit forms for ϕ1sSF and ϕ1sCGF

ϕ1sSF(ζ,r) = ζ3 π eζr

ϕ1sCGF(α,r) = 2α π 34eαr2

find α for any ζ such that

max = ζ3 π 2α π 34eζreαr2 dr

or

max = 4πζ3 π 2α π 340r2eζreαr2 dr

when we eliminate the θ and ϕ spatial variables.

Using Mathematica, one can actually solve the integral to get:

max = ζ3 2αζ + eζ2 4α π 2α + ζ2  Erfc ζ2α (2π)14α74

Differentiating wrt α we get:

0 = ζ3 2αζ 7α + ζ2 eζ2 4α π 6α2 + 9αζ2 + ζ4  Erfc ζ2α 4(2π)14α154

Since Slater exponent ζ0 and α then we must have that

0 = 2αζ 7α + ζ2 eζ2 4α π 6α2 + 9αζ2 + ζ4  Erfc ζ 2α

This nonlinear equation in α can be solved for different values of ζ. A plot of the dependency is given below:

PIC

Now let’s consider the STO-2G case. In this case

ϕ1sCGF(α 12,α22,d12,d22,r) = d12ϕ1sGF(α 12,r) + d22ϕ1sGF(α 22,r)

with

ϕ1sGF(α,r) = 2α π 34eαr2

Therefore, the overlap integral writes:

S = d12ϕ1sSF(ζ,r)ϕ 1sGF(α 12,r)dr + d22ϕ1sSF(ζ,r)ϕ 1sGF(α 22,r)dr

The overall strategy is as follows. First, write the normalization condition for ϕ1sCGF and use the fact that ϕ1sGF(α12,r) and ϕ1sGF(α22,r) are normalized. Thus express d22 in terms of d12, α12, and α22. Second, replace d22 expression into the overlap integral. Third, differentiate the overlap integral with respect to d12, α12, and α22 to obtain a system of 3 equations with 3 unknowns which can be solved numerically.

The normalization constraint is:

dr d12ϕ1sGF(α 12,r) + d22ϕ1sGF(α 22,r)2 = 1

or

d122drϕ1sGF2 (α12,r)+d222drϕ1sGF2 (α22,r)+2d12d22drϕ1sGF(α 12,r)ϕ1sGF(α 22,r) = 1

Since ϕ1sGF(α,r) are normalized, we end up with:

d122 + d 222 + 2d 12d22drϕ1sGF(α 12,r)ϕ1sGF(α 22,r) = 1

Remember that the product of two Gaussians is another Gaussian. In this case the Gaussians are at the same center, hence:

ϕ1sGF(α 12,r)ϕ1sGF(α 22,r) = 2α12α22 (α12 + α22)π34 2(α12 + α22) π 34e(α12+α22)r2 = 4α12α22 π2 34e(α12+α22)r2

Replacing into the normalization condition we get:

d122 + d 222 + 2d 12d22 4α12α22 π2 34 4π 0drr2e(α12+α22)r2 = 1

After evaluating the integral, we are left with:

d122 + d 222 + 2d 12d22 4α12α22 (α12 + α22)2 34 = 1

We can formulate this as a 2nd degree equation and choose the solution (positive value for d22):

d22 = d12 4α12α22 (α12 + α22)2 34 + d 122 4α12α22 (α12 + α22)2 32 (d122 1)

Now, let’s replace this value into the overlap integral S:

S = d12ϕ1sSF(ζ,r)ϕ 1sGF(α 12,r)dr + + d12 2 4α12 α22 (α12 + α22)2 32 (d122 1) d12 4α12α22 (α12 + α22)2 34ϕ1sSF(ζ,r)ϕ 1sGF(α 22,r)dr

Proceeding as above, the integrals evaluate to:

ϕ1sSF(ζ,r)ϕ 1sGF(α 12,r)dr = ζ3 2α12ζ + e ζ2 4α12 π 2α12 + ζ2  Erfc ζ 2α12 (2π)14α1274 ϕ1sSF(ζ,r)ϕ 1sGF(α 22,r)dr = ζ3 2α22ζ + e ζ2 4α22 π 2α22 + ζ2  Erfc ζ 2α22 (2π)14α2274

The overlap integral S is then:

S = d12ζ3 2α12ζ + e ζ2 4α12 π 2α12 + ζ2  Erfc ζ 2α12 (2π)14α1274 + + d12 2 4α12 α22 (α12 + α22)2 32 (d122 1) d12 4α12α22 (α12 + α22)2 34 × ×ζ3 2α22ζ + e ζ2 4α22 π 2α22 + ζ2  Erfc ζ 2α22 (2π)14α2274

or, after pulling out constant factors:

2π ζ3 S = S = d 12 2α12ζ + e ζ2 4α12 π 2α12 + ζ2  Erfc ζ 2α12 α1274 + + d12 2 4α12 α22 (α12 + α22)2 32 (d122 1) d12 4α12α22 (α12 + α22)2 34 × ×2α22ζ + e ζ2 4α22 π 2α22 + ζ2  Erfc ζ 2α22 α2274

Differentiating wrt d12, α12 and α22 we get:

0 = S d12 = 2α12ζ + e ζ2 4α12 π 2α12 + ζ2  Erfc ζ 2α12 α1274 + 2α22ζ + e ζ2 4α22 π 2α22 + ζ2  Erfc ζ 2α22 α2274 × ×d12 4α12α22 (α12+α22)2 32 1 d12 2 4α12 α22 (α12+α22)2 32 (d122 1) 4α12α22 (α12 + α22)2 34 0 = S α12 = d122α12ζ 7α12 + ζ2 πe ζ2 4α12 6α122 + 9α12ζ2 + ζ4  erfc ζ 2α12 4α12154 + + 2α22ζ + e ζ2 4α22 π 2α22 + ζ2  Erfc ζ 2α22 α2274 × ×3 α12 α22 α22d12 28 α12 α22 α12+α22 2 32 1d122 + 1 4 α12α22 α12+α22 2 34d12 2 α12α22 α12+α22 2 4 α12 + α22 38 α12 α22 α12+α22 2 32 1d122 + 1

0 = S α22 = d12 2 4α12 α22 (α12 + α22)2 32 (d122 1) d12 4α12α22 (α12 + α22)2 34 × ×2α22ζ 7α22 + ζ2 πe ζ2 4α22 6α222 + 9α22ζ2 + ζ4  erfc ζ 2α22 4α22154 + + 2α22ζ + e ζ2 4α22 π 2α22 + ζ2  Erfc ζ 2α22 α2274 × ×3α12 α12 α22 d12 4 α12α22 α12+α22 2 34d12 28 α12 α22 α12+α22 2 32 1d122 + 1 2 α12α22 α12+α22 2 4 α12 + α22 38 α12 α22 α12+α22 2 32 1d122 + 1

Now let’s consider the STO-3G case. The normalization condition writes:

dr d13ϕ1sGF(α 13,r) + d23ϕ1sGF(α 23,r) + d33ϕ1sGF(α 33,r)+ = 1

or

d132drϕ1sGF2 (α13,r) + d232drϕ1sGF2 (α23,r) + d332drϕ1sGF2 (α33,r) + 2d13d23drϕ1sGF(α 13,r)ϕ1sGF(α 23,r) + 2d13d33drϕ1sGF(α 13,r)ϕ1sGF(α 33,r) + +2d23d33drϕ1sGF(α 23,r)ϕ1sGF(α 33,r) = 1