# Energy (Hamiltonian) of Trial Wavefunction

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?

• 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. Commented May 5, 2022 at 16:30
• I’m voting to close this question because it lacks elementary research. Commented May 5, 2022 at 16:33
• I understand concept of expectation value, inner products, but I didn't study self-adjoint. Maybe that is the problem. Commented May 5, 2022 at 16:43
• 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. Commented May 5, 2022 at 16:45
• 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. Commented May 5, 2022 at 17:06

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

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