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I'm basing my attempt at an answer off of Modern Quantum Chemistry by Szabo and Ostlund, p.119-122.

The general development of Hartree-Fock is done via functional variation of the ground state energy $E_0=\left<\psi_0\right|\!\hat{H}\!\left|\psi_0\right>$. At the end of this, we obtain the matrix form of a differential equation

$$ f\left|\chi_a\right> = \sum_{b\,=\,1}^N \epsilon_{ba}\left|\chi_b\right> $$

where $f$ is the Fock operator and the sum is over all $N$ occupied spin orbitals. By appropriate unitary transformation (that which diagonalizes the matrix $\epsilon$), we obtain the equation in its canonical form

$$ f\left|\chi_a'\right> = \epsilon_{a}\left|\chi_a'\right> $$

where the $\chi_a'$ are the canonical Hartree-Fock orbitals.

My understanding of why it would be difficult to use Koopmans' theorem with other orbitals is that the canonical Hartree-Fock orbitals are unique in putting the above matrix equation in diagonal form. In the noncanonical forms, there isn't a clear value that could considered the orbital energy for a particular $\left|\chi_a''\right>$ (i.e., they don't return an eigenvalue when acted on by the Fock operator).

I'm basing my attempt at an answer off of Modern Quantum Chemistry by Szabo and Ostlund, p.119-122.

The general development of Hartree-Fock is done via functional variation of the ground state energy $E_0=\left<\psi_0\right|\!\hat{H}\!\left|\psi_0\right>$. At the end of this, we obtain the matrix form of a differential equation

$$ f\left|\chi_a\right> = \sum_{b\,=\,1}^N \epsilon_{ba}\left|\chi_b\right> $$

where $f$ is the Fock operator and the sum is over all $N$ occupied spin orbitals. By appropriate unitary transformation (that which diagonalizes the matrix $\epsilon$), we obtain the equation in its canonical form

$$ f\left|\chi_a'\right> = \epsilon_{a}\left|\chi_a'\right> $$

where the $\chi_a'$ are the canonical Hartree-Fock orbitals and the elements along the diagonal of $\epsilon$ are interpreted as molecular orbital energies.

My understanding of why it would be difficult to use Koopmans' theorem with other orbitals is that the canonical Hartree-Fock orbitals are unique in putting the above matrix equation in diagonal form. In the noncanonical forms, there isn't a clear value of the orbital energy for a particular $\left|\chi_a''\right>$ (i.e., they don't return an eigenvalue when acted on by the Fock operator).

I'm basing my attempt at an answer off of Modern Quantum Chemistry by Szabo and Ostlund, p.119-122.

The general development of Hartree-Fock is done via functional variation of the ground state energy $E_0=\left<\psi_0\right|\!\hat{H}\!\left|\psi_0\right>$. At the end of this, we obtain the matrix form of a differential equation

$$ f\left|\chi_a\right> = \sum_{b\,=\,1}^N \epsilon_{ba}\left|\chi_b\right> $$

where $f$ is the Fock operator and the sum is over all $N$ occupied spin orbitals. By appropriate unitary transformation (that which diagonalizes the matrix $\epsilon$), we obtain the equation in its canonical form

$$ f\left|\chi_a'\right> = \epsilon_{a}\left|\chi_a'\right> $$

where the $\chi_a'$ are the canonical Hartree-Fock orbitals.

My understanding of why it would be difficult to use Koopman's theorem with other orbitals is that the canonical Hartree-Fock orbitals are unique in putting the above matrix equation in diagonal form. In the noncanonical forms, there isn't a clear value that could considered the orbital energy for a particular $\left|\chi_a''\right>$ (i.e., they don't return an eigenvalue when acted on by the Fock operator).

I'm basing my attempt at an answer off of Modern Quantum Chemistry by Szabo and Ostlund, p.119-122.

The general development of Hartree-Fock is done via functional variation of the ground state energy $E_0=\left<\psi_0\right|\!\hat{H}\!\left|\psi_0\right>$. At the end of this, we obtain the matrix form of a differential equation

$$ f\left|\chi_a\right> = \sum_{b\,=\,1}^N \epsilon_{ba}\left|\chi_b\right> $$

where $f$ is the Fock operator and the sum is over all $N$ occupied spin orbitals. By appropriate unitary transformation (that which diagonalizes the matrix $\epsilon$), we obtain the equation in its canonical form

$$ f\left|\chi_a'\right> = \epsilon_{a}\left|\chi_a'\right> $$

where the $\chi_a'$ are the canonical Hartree-Fock orbitals.

My understanding of why it would be difficult to use Koopmans' theorem with other orbitals is that the canonical Hartree-Fock orbitals are unique in putting the above matrix equation in diagonal form. In the noncanonical forms, there isn't a clear value that could considered the orbital energy for a particular $\left|\chi_a''\right>$ (i.e., they don't return an eigenvalue when acted on by the Fock operator).

I'm basing my attempt at an answer off of Modern Quantum Chemistry by Szabo and Ostlund, p.119-122.

The general development of Hartree-Fock is done via functional variation of the ground state energy $E_0=\left<\psi_0|\hat{H}|\psi_0\right>$. At the end of this, we obtain a differential equation

$$ f\left|\chi_a\right> = \sum_{b=1}^N \epsilon_{ba}\left|\chi_b\right> $$

where $f$ is the Fock operator and the sum is over all $N$ occupied spin orbitals. By appropriate unitary transformation (that which diagonalizes the matrix $\epsilon$), we obtain the equation in its canonical form

$$ f\left|\chi_a'\right> = \epsilon_{a}\left|\chi_a'\right> $$

where the $\chi_a'$ are the Hartree-Fock orbitals.

My understanding of why it would be difficult to use Koopmans theorem with other orbitals is that the Hartree-Fock orbitals are unique in putting the equation in canonical form. In the noncanonical forms, there isn't a clear value that could considered the orbital energy for a particular $\left|\chi_a\right>$ (i.e. they don't return an eigenvalue when acted on by the Fock operator).

I'm basing my attempt at an answer off of Modern Quantum Chemistry by Szabo and Ostlund, p.119-122.

The general development of Hartree-Fock is done via functional variation of the ground state energy $E_0=\left<\psi_0\right|\!\hat{H}\!\left|\psi_0\right>$. At the end of this, we obtain the matrix form of a differential equation

$$ f\left|\chi_a\right> = \sum_{b\,=\,1}^N \epsilon_{ba}\left|\chi_b\right> $$

where $f$ is the Fock operator and the sum is over all $N$ occupied spin orbitals. By appropriate unitary transformation (that which diagonalizes the matrix $\epsilon$), we obtain the equation in its canonical form

$$ f\left|\chi_a'\right> = \epsilon_{a}\left|\chi_a'\right> $$

where the $\chi_a'$ are the canonical Hartree-Fock orbitals.

My understanding of why it would be difficult to use Koopman's theorem with other orbitals is that the canonical Hartree-Fock orbitals are unique in putting the above matrix equation in diagonal form. In the noncanonical forms, there isn't a clear value that could considered the orbital energy for a particular $\left|\chi_a''\right>$ (i.e., they don't return an eigenvalue when acted on by the Fock operator).