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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?

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  • $\begingroup$ 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. $\endgroup$ May 5 at 16:30
  • $\begingroup$ I’m voting to close this question because it lacks elementary research. $\endgroup$ May 5 at 16:33
  • $\begingroup$ I understand concept of expectation value, inner products, but I didn't study self-adjoint. Maybe that is the problem. $\endgroup$ May 5 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. $\endgroup$ May 5 at 16:45
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    $\begingroup$ 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. $\endgroup$ May 5 at 17:06

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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: $$ \left(\sum_{j} c^*_j \left\langle \Psi_j \right| \right) \left(\sum_{i} c_i H \left| \Psi_i \right \rangle \right) = \sum_{i, j} c^*_j c_i (\left\langle \Psi_j \right|) (H\left| \Psi_i \right \rangle) = \sum_{i, j} c_i c^*_j \left\langle \Psi_i \right| H\left| \Psi_j \right \rangle $$

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

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