Here I give a part of derivation of Hartree-Fock equations in case where basis functions (wavefunctions) are orthonormal and real: $$ \langle \psi_i | \psi_j \rangle = \langle \psi_j | \psi_i \rangle = \delta_{ij} $$

Trial wavefunction is defined as: $$ |\Phi \rangle = \sum_{i=1}^n c_i |\psi_i \rangle $$

where $|\psi_i\rangle$ is basis function $i$.

Expectation value of energy is given by: $$ \langle \Phi | H |\Phi \rangle = \sum_{ij} c_i c_j \langle \psi_i |H|\psi_j \rangle $$

I don't quite understand, why is expectation value of energy for trial wavefunction equal sum of expectation values for every combination of two basis functions multiplied by their respective coefficients ($c_i$ and $c_j$)? What justifies this summation?

  • $\begingroup$ Your question lacks elementary research. Before you go to Hartree-Fock, read about the basics of QM, such as (linear, self-adjoint) operators, inner products, expectation values etc. $\endgroup$ Commented May 5, 2022 at 16:30
  • $\begingroup$ I’m voting to close this question because it lacks elementary research. $\endgroup$ Commented May 5, 2022 at 16:33
  • $\begingroup$ I understand concept of expectation value, inner products, but I didn't study self-adjoint. Maybe that is the problem. $\endgroup$ Commented May 5, 2022 at 16:43
  • $\begingroup$ I did, I wouldn't be able to understand anything without basics. This is one thing I am not sure about. Could you point to me towards what I am missing (what to check or study)? I think that vote for closing has no sense because it won't help me answer my question. Point of SE, is to learn. If I don't know what I am missing, saying that question should be closed because I don't know basics is unhelpful. Your answer should provide what exactly I need to check or study, not just saying that I miss basics. $\endgroup$ Commented May 5, 2022 at 16:45
  • 1
    $\begingroup$ It is of course no problem to ask a basic question, but IMO you can read these things in any book on e.g. quantum mechanics or Wikipedia etc. Further, your question is not really about Hartree-Fock, nor, in principle, quantum mechanics, but in the simplest case just about linear algebra... I think you would learn most if you'd solve this task on your own - it is not difficult if you have a little knowledge of the basics. Again: Read about linear operators and inner products. I voted to close because your question shows no effort of you to better understand the problem. $\endgroup$ Commented May 5, 2022 at 17:06

1 Answer 1


It's really linear algebra: if $$ \left| \Phi \right \rangle = \sum_{i} c_i \left| \Psi_i \right \rangle $$ taking Hermitian operator ("transpose conjugate"): $$ \left| \Phi \right \rangle^\dagger = \left\langle \Phi \right| = \sum_{i} c^*_i \left\langle \Psi_i \right|. $$ Now "sandwiching" $H$ you get: $$ \left\langle \Phi \right| H \left| \Phi \right \rangle = \left(\sum_{i} c^*_i \left\langle \Psi_i \right| \right) H \left(\sum_{i} c_i \left| \Psi_i \right \rangle \right) $$ now, before we expand, we should change one of the indices to $j$ to account for products of different terms, just like, say: $$ (a_1 + a_2) \times (b_1 + b_2) = a_1 b_1 + a_1 b_2 + a_2 b_1 + a_2 b_2 = \sum_{i,j = 1}^{2} a_i b_j $$ and not $\sum_i a_i b_i$. So, really: $$ \begin{align*} \left\langle \Phi \right| H \left| \Phi \right \rangle &= \left(\sum_{i} c^*_i \left\langle \Psi_i \right| \right) H \left(\sum_{i} c_i \left| \Psi_i \right \rangle \right) \\ & = \left(\sum_{j} c^*_j \left\langle \Psi_j \right| \right)\left(\sum_{i} c_i H\left| \Psi_i \right \rangle \right) \end{align*} $$ and apply the distribution rule of algebra: $$ \left(\sum_{j} c^*_j \left\langle \Psi_j \right| \right) \left(\sum_{i} c_i H \left| \Psi_i \right \rangle \right) = \sum_{i, j} c^*_j c_i (\left\langle \Psi_j \right|) (H\left| \Psi_i \right \rangle) = \sum_{i, j} c_i c^*_j \left\langle \Psi_i \right| H\left| \Psi_j \right \rangle $$

  • 1
    $\begingroup$ There are some indices on your $|\Psi\rangle$s missing. $\endgroup$ Commented May 5, 2022 at 17:04
  • $\begingroup$ I forgot all of them. Fixed. Thanks for pointing out! $\endgroup$
    – Petrini
    Commented May 5, 2022 at 17:08
  • $\begingroup$ Ohh, yes. Now I get it. Thanks, man 😀. $\endgroup$ Commented May 5, 2022 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.