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I'm not a student of quantum physics and I do know some very basics of quantum mechanics, which any engineer like me would know of. I tried to understand the concept of bonding using quantum mechanical superposition of wavefunctions, simply using a thought experiment. Suppose I have two waves $\left|\psi_1\right>$ and $\left|\psi_2\right>$ of energy $E_1$ and $E_2$, then I can write $$\hat{\mathcal{H}}\left|\psi_1\right>=E_1\left|\psi_1\right>$$ $$\hat{\mathcal{H}}\left|\psi_2\right>=E_2\left|\psi_2\right>$$ Now suppose I have a resultant wave $\left|\psi_r\right>=a\left|\psi_1\right>+b\left|\psi_2\right>$ of energy $E_r$, then I can write $$\hat{\mathcal{H}}\left|\psi_r\right>=E_r\left|\psi_r\right>$$ Which means $$a\hat{\mathcal{H}}\left|\psi_1\right>+b\hat{\mathcal{H}}\left|\psi_2\right>=E_r(a\left|\psi_1\right>+b\left|\psi_2\right>)$$ $$\implies \frac{E_1(a\left|\psi_1\right>)+E_2(b\left|\psi_2\right>)}{a\left|\psi_1\right>+b\left|\psi_2\right>}=E_r$$ This means that $E_1<E_r<E_2$, since $E_r$ is a weighted mean of $E_1$ and $E_2$, hence the resultant energy is between the energy of the two wavefunctions, so is my model of bonding absolutely wrong(since my proff. told me that bonding happens as a result of linear combination of atomic orbitals) or if it isn't then I don't understand why wavefunction $\left|\psi_1\right>$ would participate in this bonding, since it is going from a lower energy state, $E_1$, to a higher energy state, $E_r$. Any help is readily appreciated

P.S. I have assumed $E_1<E_2$

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Your last formula is meaningless as the kets are not mere numbers, so you can't divide by them. You have shown (your next-to-last formula) that

$$\newcommand{\ket}[1]{\mid #1 \rangle} \newcommand{\bra}[1]{\langle #1 \mid} \newcommand{\bra}[1]{\langle #1 \mid} \newcommand{\braket}[2]{\langle #1 \mid #2 \rangle} aE_1\ket{\psi_1}+bE_2\ket{\psi_2}=aE_r\ket{\psi_1}+bE_r\ket{\psi_2}.$$

If $E_1\ne E_2$, then $\ket{\psi_1}$ and $\ket{\psi_2}$ are orthogonal, and therefore we can equate the coefficients on both sides: $E_1 = E_r = E_2$, which is a contradiction.

Therefore, $E_1 = E_2$, which is then also equal to $E_r$.

Where did you go wrong? If you consider the bonding between two atoms for example, the Hamiltonian for the system reads

$H=H_1 + H_2 + H_{12},$

where $H_1$ (resp. $H_2$) is the Hamiltonian of the first (resp. second) atom and $H_{12}$ is the interaction terms between the two atoms. You would have then with your notations

$$\begin{align} H_1\ket{\psi_1}&=E_1\ket{\psi_1},\\ H_2\ket{\psi_2}&=E_2\ket{\psi_2},\\ H\ket{\psi_r}&\approx E_r\ket{\psi_r}. \end{align}$$

The difference is obvious, isn't it?

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While the math above is correct, your interpretation is off. I think your confusion comes from assuming that $|\psi_i\rangle$ are orbital wavefunctions (i.e. S, P states). For a Hamiltonian describing a crystal, the orbital states are not eigenstates of the system. Therefore your statement $H|\psi_i\rangle = E_i|\psi_i\rangle$ does not apply to orbital wavefunctions.

For a crystal, the eigenstates are called Bloch functions which extend across the entire length of the crystal and are a superposition of all orbital states. Defining your $|\psi_i\rangle$ as such, then your above analysis is correct and the ground state energy would simply be the lowest available energy state. This ground state defines a feature called the Fermi level which, along with the energy state density, defines the conductive properties of a material.

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