Kinetic Energy Evaluation Integral Evaluation Program Ostlund

I'm reading Ostlund's Modern Quantum Chemistry. In Appendix B, the kinetic energy integral is evaluated using the Gaussian Basis functions as



```
T11=T11+T(A1(I),A1(J),0.0D0)*D1(I)*D1(J)
T12=T12+T(A1(1),A2(J),R2).D1(I).D2(J) 
T22=T22+T(A2(1),A2(J).0.0D0).D2(1).D2(J) 

```

I understand `T11` to be the kinetic energy integral using the first atom orbitals so it will give the kinetic energy of the electron of the first atom, and this is why it is passing zero as the bond length parameter. The same is `T22` but for the second atom.


What is `T12` used for, aren't `T11` and `T22` enough to calculate the total kinetic energy?




TAR86 gives the correct mathematical answer, but given how you express yourself I thought I would try and explain it at a slightly lower level.


Your problem is in the sentence "I understand T11 to be the kinetic energy integral using the first atom orbitals so it will give the kinetic energy of the electron of the first atom"


The problem is that in a molecule the electron will be partially associated with the first atom, and partially with the second - this is almost the definition of a covalent bond! Thus yes there will be a contribution to the kinetic energy due to the first electron being on the first atom, but there will also be a contribution due to the first electron being on the second atom, and it seems to me that this is what you are missing.


Let's try and make that a bit more concrete with a little maths. As we are using an orbital approximation the first electron's behaviour is described by it's wavefunction. We don't know exactly what it looks like, but we hope it will be similar to the atomic wavefunctions on the two atoms. So let's write the wavefunction for the first electron as


$$
 \psi (1)= c\_1 \phi\_1 + c\_2 \phi\_2
$$


where $\phi\_1$ is the first atomic orbital (actually contracted Gaussian basis functions) and $c\_1$ is the weight of this function as determined by solving the approximation to the Schrodinger equation that we are interested in. Then if $\hat T$ is the kinetic energy operator we can obtain the Kinetic Energy of the first electron from


$$
T=<\psi(1)|\hat T|\psi(1)>=c\_1^2<\phi\_1|\hat T|\phi\_1>+2c\_1c\_2<\phi\_1|\hat T|\phi\_2>+c\_2^2<\phi\_2|\hat T|\phi\_2>
$$


(using Hermiticity of $\hat T$ and the fact here the wavefunction is real)


In this we can see a cross term which involves electron 1 on both atom 1 and 2, and in fact even a term which is due to electron 1 being on atom 2. So using your notation we can identify


$$
T11=<\phi\_1|\hat T|\phi\_1> \\
T22=<\phi\_2|\hat T|\phi\_2> \\
T12=<\phi\_1|\hat T|\phi\_2>
$$


Note more generally we should talk about basis functions rather than atoms, but here as the basis functions are associated with the atomic sites (this need not be the case) and there is only 1 basis function per site (in real calculations there will be many) we can be slightly sloppy. But also understand that the general case is not really much more complicated than the above, we simply expand the orbital for the first electron using all the basis functions, and carry on similarly.


 2

The off-diagonal elements $T\_{\mu\nu}, (\mu \ne \nu)$ are just as necessary as the diagonal elements $T\_{\mu\mu}$ to represent the kinetic energy operator in the chosen AO basis set. Substituting a few of Szabo/Ostlund's equations into each other (in my edition, 3.147-3.154), one finds
$$
F\_{\mu\nu} = T\_{\mu\nu} + V\_{\mu\nu}^{\text{nucl}} + \sum\_{\lambda\kappa} P\_{\lambda\kappa}\left[ \left( \mu\nu \vert \kappa\lambda \right) - \frac{1}{2} \left( \mu\lambda \vert \kappa \nu\right) \right]
$$
where $F$ is the Fock matrix, $V^{\text{nucl}}$ represents the nuclear attraction term, $P$ is a density matrix and $\left( \mu\nu \vert \kappa\lambda \right)$ are two-electron repulsion integrals over the AO basis functions (denoted greek indices). Of course, your Fock matrix may be diagonal already, if you have a very special case and basis set, but in general and to perform the SCF calculation, the off-diagonal parts cannot be ignored.


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