Advanced Chemistry - Mono Mole

December 7, 2022September 26, 2024

Koopmans’ theorem

Koopmans’ theorem states that the energy to remove an electron from an orbital of an atom, whose state is described by a single Slater determinant, is approximately the negative value of the corresponding Hartree-Fock orbital energy. In other words, the theorem states that the negative value of the orbital energy $E_k$ of an atom is approximately equal to the ionisation energy IE for the k-th electron in the atom.

To prove the theorem, we begin with the ionisation process, which is generally expressed as

$X(g)+h\nu\rightarrow X^+(g)+e^-$

where $X$ is an atom, which is irradiated with a photon of energy $h\nu$ , and $X^+$ is the corresponding ion with an electron $e^-$ removed.

In terms of energies,

$h\nu=E_{X^+}-E_X+E_{e^-}$

where $E_{X^+}$ is the state of the ion, $E_X$ is the state of the atom and $E_{e^-}$ is the kinetic energy of the expelled electron.

As $E_{e^-}\rightarrow 0$ ,

$h\nu=IE=E_{X^+}-E_X\;\;\;\;\;\;\;\;(143)$

Substituting eq95 in eq143 gives

$IE=\sum_{i\neq k}^nH_i+\frac{1}{2}\sum_{j,i\neq k}^n(J_{ij}-K_{ij})-\sum_{i=1}^nH_i-\frac{1}{2}\sum_{j,i=1}^n(J_{ij}-K_{ij})$

$IE=-H_k+\frac{1}{2}\left[\sum_{j,i\neq k}^n(J_{ij}-K_{ij})-\sum_{j,i=1}^n(J_{ij}-K_{ij})\right]\;\;\;\;\;\;\;\;(144)$

Question

Show that $\sum_{j,i\neq k}^n(J_{ij}-K_{ij})-\sum_{j,i=1}^n(J_{ij}-K_{ij})=-2\sum_{j=1}^n(J_{kj}-K_{kj})$ .

Answer

Let $C=\sum_{j,i\neq k}^n(J_{ij}-K_{ij})-\sum_{j,i=1}^n(J_{ij}-K_{ij})$ . Expanding the equation and eliminating common terms,

$C=-\left[\sum_{i=1}^n(J_{ik}-K_{ik})+\sum_{j\neq k}^n(J_{kj}-K_{kj})\right]$

Since $J_{kk}-K_{kk}=0$ , we can add it to $\sum_{j\neq k}^n(J_{kj}-K_{kj})$ , giving

$C=-\left[\sum_{i=1}^n(J_{ik}-K_{ik})+\sum_{j=1}^n(J_{kj}-K_{kj})\right]$

Furthermore, $J_{ik}=J_{ki}$ and $K_{ik}=K_{ki}$ . So,

$C=-\left[\sum_{i=1}^n(J_{ki}-K_{ki})+\sum_{j=1}^n(J_{kj}-K_{kj})\right]$

Relabelling i to j for $\sum_{i=1}^n(J_{ki}-K_{ki})$

$C=-2\sum_{j=1}^n(J_{kj}-K_{kj})\;\;\;\;\;\;\;\;(145)$

Substituting eq145 in eq144 yields

$IE=-\left[H_k+\sum_{j=1}^n(J_{kj}-K_{kj})\right]\;\;\;\;\;\;\;\;(146)$

Relabelling t with j and $\lambda_r^{'}$ with $E_r$ in eq109, we have

$\left[\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)+\sum_{j\neq r}\int\phi_j^{'*}(x') \frac{(1-P_{rj})\phi_j^{'}(x')}{\vert\boldsymbol{\mathit{ r}}-\boldsymbol{\mathit{ r}}'\vert}dx'\right]\phi_r^{'}(x)=E_r\phi_r^{'}(x)\;\;\;\;(147)$

Multiplying the k-th equation in eq147 by $\phi_k^{'*}(x)$ and integrating over $dx$ , we have

$E_k=H_k+\sum_{j\neq k}^n(J_{kj}-K_{kj})$

Since $J_{kk}-K_{kk}=0$

$E_k=H_k+\sum_{j=1}^n(J_{kj}-K_{kj})\;\;\;\;\;\;\;\;(148)$

Substituting eq148 in eq146 results in

$IE=-E_k\;\;\;\;\;\;\;\;(149)$

Eq149 is known as Koopmans’ theorem.

Lastly, let’s study the relation between the relation between the orbital energy $E_k$ of an atom and the total energy E of the atom. From eq148,

$\sum_{k=1}^nE_k=\sum_{k=1}^nH_k+\sum_{k,j=1}^n(J_{kj}-K_{kj})$

$\sum_{k=1}^nE_k=\sum_{k=1}^nH_k+\sum_{j\neq k}\sum_{k=1}^n(J_{kj}-K_{kj})$

Substituting eq94 in the above equation gives

$E=\sum_{k=1}^nE_k-\frac{1}{2}\sum_{j\neq k}\sum_{k=1}^n(J_{kj}-K_{kj})\;\;\;\;\;\;\;\;(150)$

Question

Show that the alternative to eq150 is $E=\frac{1}{2}\sum_{k=1}^n(E_k+H_k)$ .

Answer

Adding $H_k$ to both sides of eq148 and summing throughout from $k=1,\cdots,n$ ,

$\sum_{k=1}^n(E_k+H_k)=2\sum_{k=1}^nH_k+\sum_{k,j=1}^n(J_{kj}-K_{kj})$ $=2\sum_{k=1}^nH_k+\sum_{j\neq k}\sum_{k=1}^n(J_{kj}-K_{kj})=2\left[\sum_{k=1}^nH_k+\sum_{k >j}\sum_{j=1}^n(J_{kj}-K_{kj})\right]$

Since $E=\sum_{k=1}^nH_k+\sum_{k >j}\sum_{k=1}^n(J_{kj}-K_{kj})$ ,

$E=\frac{1}{2}\sum_{k=1}^n(E_k+H_k)\;\;\;\;\;\;\;\;(151)$

Next article: Atomic orbital energies

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December 7, 2022March 29, 2025

Atomic orbital energies

Atomic orbital energies are eigenvalues of the Hartree-Fock equations. These eigenvalues are expressed by eq148, and are based on a specific state of the atom. As illustrated in the article on the analytical solution of the Hartree-Fock method for the ground state of Li, we can theoretically compute $E_k$ for any occupied orbital of an atom. For example, $E_k$ of the 2p orbital of Li can be determined using the excited electron configuration state of 1s²2p¹. However, if we want to determine the energies of unoccupied orbitals of an atom in its ground state, we would have to use the Hartree-Fock-Roothaan method, where the basis set includes a linear combination of the 1s, 2s and 2p wavefunctions. In general, $E_k$ for atoms with $Z=1$ to $Z=100$ are depicted in the diagram below.

For atoms with $7\leq Z\leq 19$ , their 4s orbital energies lie below their respective 3d orbital energies (see diagram below). An example is the atom $K(Z=19)$ with the ground state electron configuration [Ar]4s¹. This implies that the theoretical ground state energy of $K$ is closer to the experimental value when we use a 4s Slater-type orbital rather than a 3d Slater-type orbital for computing $E_{19}$ .

With that in mind, one may conclude that the ground state electron configuration of Sc is [Ar]3d³. However, it is [Ar]3d¹4s². This is because the orbital energies $E_{3d}$ and $E_{4s}$ for [Ar]3d³are different from those for [Ar]3d¹4s² (refer to eq148), even though $E_{3d}<E_{4s}$ for both configurations. For example, $\sum_{k=1}^nE_k$ for the electron configurations [Ar]3d¹4s²and [Ar]3d²4s²are (in Hartree units)

$\sum_{k=1}^nE_k=E_{Ar1}+E_{3d}+2E_{4s}=E_{Ar1}+(-0.344)+2(-0.210)=E_{Ar1}-0.764$

and

$\sum_{k=1}^nE_k=E_{Ar2}+E_{3d}+2E_{4s}=E_{Ar2}+(-0.192)+2(-0.186)=E_{Ar2}-0.570$

respectively, where $E_{Ar1}\neq E_{Ar2}$ .

In short, the Hartree-Fock computation using the electron configuration of [Ar]3d¹4s²gives a ground state energy that is closest to experimental values. Qualitatively, we rationalise the above with the fact that 3d orbitals are smaller in size compared to the 4s orbitals. Electrons occupying 3d orbitals therefore experience greater repulsions than electrons residing in 4s orbitals, with the order of increasing repulsion being:

$V(e_{4s},e_{4s})<V(e_{4s},e_{3d})<V(e_{3d},e_{3d})$

where V is the potential energy due to repulsion.

Therefore, to determine the stability of an atom in the ground state, we need to consider the net effect of the relative energies of 4s/3d orbitals and the repulsion of electrons. Calculations for the overall energies of Sc show that

$E[Ar]3d^14s^2<E[Ar]3d^24s^1<E[Ar]3d^3$

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