Koopmans' theorem says that the energy of the HOMO of the Hartree-Fock orbitals is equal to the first ionization energy of whatever system is being studied. This is only approximate because it assumes no changes in the orbitals in the ionic state, no relativistic effects, and no electron correlation. Ignoring those approximations, there is another feature of this which I don't understand.

The wavefunction as determined using HF can be unitarily transformed so that the total energy and wavefunction are preserved, but the "orbitals" themselves are not. Under many of these transformations, I assume that the orbital energies change. Is this true? If it is true, I assume Koopmans' theorem is still valid, but is now more complicated so that the ionization energy is some combination of orbital energies. Can anyone expand on this and let me know if my thinking is correct?

Also, what is it that is so special about the Hartree-Fock orbitals that Koopmans' Theorem has such a simple interpretation, while, if what I say above is true, it seems to be quite convoluted for other sets of orbitals?

As a note, I asked a question which more or less contained this question, but was a bit broader. The question I'm asking right now is the question I am really more interested in having answered.

You can find that other question here: Observability of Orbitals and Orbital Energies

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    $\begingroup$ HOMO has a well defined energy only when it corresponds to an eigenvalue from HF. After such a unitary transformation the orbitals have no well defined energies any more $\endgroup$
    – Greg
    Commented Mar 1, 2017 at 0:37
  • $\begingroup$ I don't think that the orbital energies ever correspond to the eigenvalues of the Fock operator. Rather the eigenvalues are the sum of these orbital energies, but I'm pretty sure these orbital energies are not unique because the orbitals themselves are not unique. The second half of that sentence depends on the answer to this question which I'm not completely sure of. $\endgroup$
    – jheindel
    Commented Mar 1, 2017 at 0:45
  • $\begingroup$ They do, the Fock operator is an effective one electron operator and this is how you find the orbitals and the correspond energies. $\endgroup$
    – Greg
    Commented Mar 1, 2017 at 0:52
  • $\begingroup$ Ah sorry. I was confusing the Fock matrix and the Fock operator. The eigenvalues of the Fock matrix are the energy of the whole system which corresponds to the sum of eigenvalues of the Fock operator which are indeed the orbital energies. Sorry to suggest otherwise. What I'm pointing out in the question is that the set of orbitals and their eigenvalues given by the Fock operator is not the only solution we could come up with. So, I'm basically asking what is so special about the HF orbitals that Koopmans' theorem should be so simple when that doesn't seem to be the case for other orbital sets. $\endgroup$
    – jheindel
    Commented Mar 1, 2017 at 3:26
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    $\begingroup$ No, again, the eigenvalues of the Fock matrix (which is the Fock operator in some basis) are the canonical orbital energies. Moreover, the sum of orbital energies is NOT the total system energy. Check out this paper for more info: pubs.acs.org/doi/abs/10.1021/ed200673w $\endgroup$
    – levineds
    Commented Jul 24, 2017 at 18:46

2 Answers 2


Disclaimer: I like Tyberius' answer, but I'd like to go a little further. $% \newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle} \newcommand{\lb}{\left|}\newcommand{\rb}{\right|} \newcommand{\op}[1]{\mathbf{#1}}$

The beauty of Koopmanns' theorem is its simplicity. Quoting from the gold book:

Koopmans' theorem

Directly relates experimental ionization potentials with energy levels of molecular orbitals. The theorem states that the ionization potential required to remove an electron from the orbital $\Psi_i$ is given by the negative value of the energy of the orbital, $−\varepsilon_i$, as calculated within the Hartree–Fock approximation. The theorem is not applied to localized molecular orbitals, which are not eigenfunctions of the effective hamiltonian.

While Koopmans' text (in German: Physica 1934, 1, 104-113.) is a bit hard to comprehend, due to the ancient language I never learned, it basically boils down to two important facts.

  1. There is a unitary transformation of the Lagrange multipliers which diagonalises them. These form a characteristic set of elementary wave functions, matrix, which in turn have a characteristic eigenvalue each. If all eigenvalues are different, then all elementary wave functions are specified through the Hamiltonian. (If there are equal eigenvalues, a freely chosen unitary transformation can be performed.)

    In other words: There is one set of canonical orbitals that diagonalises the Fock matrix; each canonical orbital has an eigenvalue. For non-degenerate systems those orbitals are specified through the Hamiltonian.

  2. This choice of functions has physical meaning, as the eigenvalue (disregarding a small correction) can be equated to the ionisation energies of the corresponding electron.

The second statement implies that there is no relaxation of the orbitals when removing an electron. This is oftentimes referred to as "frozen MO" approximation. Koopmanns himself appreciates that there should be a contraction of orbitals when removing an electron, however, he doesn't go into detail.

Nowadays we explain Koopmans' theorem in some simpler terms, where we remove the $k$th electron:

\begin{align} && E_N &= \sum_{i=1}^N H_{ii} + \frac12\sum_{i=1}^N\sum_{j=1}^N (J_{ij} - K_{ij}) + V_{\mathrm{nuc}} \tag1\\ && E_{N-1}^k &=\sum_{i=1}^{N-1} H_{ii} + \frac12\sum_{i=1}^{N-1}\sum_{j=1}^{N-1} (J_{ij} - K_{ij}) + V_{\mathrm{nuc}} \tag2\\\hline && E_N - E_{N-1}^k &= H_{kk} + \frac12\sum_{i=1}^N (J_{ik}-K_{ik}) + \frac12\sum_{i=1}^N (J_{kj}-K_{kj}) \tag{$1-2$}\\ \therefore&& E_N - E_{N-1}^k &= H_{kk} + \sum_{i=1}^N (J_{ki}-K_{ki})\\ \therefore&& E_N - E_{N-1}^k &= \varepsilon_k\\[2ex] \text{with}&& H_{ii} &= \langle \phi_i(\mathbf{x}_1)| \mathbf{H}^\mathrm{c} | \phi_i(\mathbf{x}_1)\rangle\\ && J_{ij} &= \langle \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) | r_{12}^{-1} | \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) \rangle\\ && K_{ij} &= \langle \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) | r_{12}^{-1} | \phi_j(\mathbf{x}_1) \phi_i(\mathbf{x}_2) \rangle\\ \end{align}

As you can see, Koopmans' theorem is not limited to the HOMO, but can be used for any occupied MO.

There is a very important point to be always be kept in mind when working with any Hartree-Fock based approaches:
Even though the equation $$\op{F}_i\phi_i = \varepsilon_i\phi_i \tag{3}\label{fock-pseudo}$$ suggests an eigenvalue problem, it is not. Remember the definition of the Fock operator and the operators contained \begin{align} && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\ \text{with}&& \op{J}_j\lb \phi_i\rr &= \ll \phi_j(\op{x}_1) \rb r_{12}^{-1} \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\ \text{and}&& \op{K}_j\lb \phi_i\rr &= \ll \phi_j(\op{x}_1) \rb r_{12}^{-1} \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr. \end{align} As you can see, the "one-electron" Fock operator depends on the solution of all "one-electron" Fock operators (ref. Szabó-Ostlund p. 115). The Hamilton operator is not the sum of all Fock operators, and the total HF energy is not the sum of all orbital energies. As a result of that, the canonical orbitals are actually unique solutions.
The Fock operator is associated with the whole wave function, and while unitary transformations will keep the wave function and its energy equivalent, everything else will fall apart.
Another important consideration is the fact that the Fock operator is only well defined for occupied MO. You sometimes find the statement $$E_{N+1}^{l>N} - E_N = \varepsilon_l,$$ which can not be applied in the same way; it is very basis set dependent.

From all that above it is obvious, that Koopmans' theorem only works for systems where a single determinant approximation is reasonable. It also explains, why it only works for HF; although there are generalisations for DFT.

Is Koopmans' theorem still valid after a unitary transformation?

No. As stated above, the Fock operator is associated with the $N$-electron wave function. After a unitary transformation and then removal of an electron, the energy of the $N-1$-electron wave function is not conserved.
As an illustrative example: \begin{align} && \sum_{i=1}^{N} H_{ii} &= \sum_{i=1}^{N} H_{ii}'\\[2ex] \text{with}&& H'_{ii} &= \langle \phi'_i(\mathbf{x}_1)| \mathbf{H}^\mathrm{c} | \phi'_i(\mathbf{x}_1)\rangle\\ \text{and}&& |\phi_i\rangle &\color{red}{\neq} |\phi'_i\rangle\\[2ex] && \sum_{i=1,i\neq k}^{N-1} H_{ii} + H_{kk}&= \sum_{i=1,i\neq k}^{N-1} H_{ii}' + H_{kk}'\\ \text{for } H_{kk} = H_{kk}': && \sum_{i=1,i\neq k}^{N-1} H_{ii} &= \sum_{i=1,i\neq k}^{N-1} H_{ii}'&& \implies|\phi_i\rangle \color{red}{=} |\phi'_i\rangle\\ \end{align}

I have indicated the contradiction in red. Therefore it follows: $$|\phi_i\rangle \neq |\phi'_i\rangle \implies H_{kk} \neq H_{kk}' \implies \sum_{i=1,i\neq k}^{N-1} H_{ii} \neq \sum_{i=1,i\neq k}^{N-1} H_{ii}' $$

You can follow that trough for the other terms and will see that the expectation value of energy of the $N-1$-electron wave function needs to be different from the expectation value of energy of the $N-1$-electron wave function after unitary transformation.


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).

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    $\begingroup$ Correct, I would expect the "energies" to be linear combinations of the canonical Fock matrix eigenvalues, with the linear coefficients determined by the linear combination of eigenvectors from the canonical basis that form each new unitary-transformed (?) eigenvector. Sorry that's worded so awkwardly... $\endgroup$ Commented Mar 19, 2017 at 4:48
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    $\begingroup$ @pentavalentcarbon Or, well, maybe I should take that back. I'm reading Weinhold's NBO book, and apparently GENNBO does attribute energies to the individual NBOs. So, apparently there is some method for rationally assigning such energies -- and the method you describe seems sensible. $\endgroup$
    – hBy2Py
    Commented Mar 19, 2017 at 4:58
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    $\begingroup$ @hBy2Py I put energies in quotes because it isn't an energy. Not all eigenvalues are energies. Transformed orbitals may look more interesting physically but are more ill-defined, even more so for their eigenvalues. NBO eigenvalues may be interesting for comparison within a set but there is probably no quantitative meaning behind them at all. $\endgroup$ Commented Mar 19, 2017 at 5:03
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    $\begingroup$ @Tyberius To my best understanding, the "density matrix" used in the course of a HF calculation, as referred to by your link, is very different from the first-order reduced density operator, $\Gamma(\mathbf r_1)\equiv\int{\Psi\Psi^*\,d\mathbf r_2d\mathbf r_3\ldots d\mathbf r_n}$, whose matrix representation is diagonalized to generate natural orbitals. Personally, I would be inclined to call the former "density matrix" an "overlap matrix," instead, as it's what is used to probe the orthonormality of the canonical MOs. $\endgroup$
    – hBy2Py
    Commented Mar 21, 2017 at 16:15
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    $\begingroup$ @hBy2Py I know you said you would make a question on this but: 1) yes, the orbitals that diagonalize the Fock matrix also diagonalize the density matrix (in HF), 2) as such the natural orbitals are the canonical orbitals for HF, 3) the 1RDM is different from the n-particle density matrix. The 1RDM is the density matrix integrated over all deg. of freedom except 1 electron. HF is the approximation that all higher density matrices are direct products of the 1RDM and so the HF density specifies the full n-particle DM. 4) the density matrix is NOT any kind of overlap matrix $\endgroup$
    – levineds
    Commented Jul 24, 2017 at 19:09

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