Opening the book…
Q1
The Hamiltonian consists of the kinetic and potential energy.
\[ \hat{H} = \hat{T} + \hat{V} \nonumber \]
In atomic units all the constants are set to equal 1, \(h=1\)
Therefore the kinetic term in the Hamiltonian is transformed
\[\hat{T} = \dfrac{-\hbar^2}{2m} \bigtriangledown^2 \psi \Longrightarrow \dfrac{-1}{2} \bigtriangledown^2 \psi \nonumber \]
The potential terms in the \(H_2^+\) molecule are then given in atomic units
\[\hat{V} = \dfrac{-1}{r_A} + \dfrac{-1}{r_B} \nonumber \]
Q2
The potential term for the interaction between the two protons in atomic units is
\[ \hat{V} = \dfrac{1}{r_{AB}} \nonumber \]
To describe the Hamiltonian for the electron both the kinetic term for the nuclei can be ignored. This term would impart a constant energy to the electron that is independent of the electron's motion for each position of the nuclei.
\[E_{total} = E_{e^-} + T_{molecule} \nonumber \]
You calculate the electron energy using the Hamiltonian in question 1 for a nuclei position, then move the nuclei and calculate the electron energy again. And repeat for as many nuclei positions as necessary, in a computer.
For the variation method, I would start with a linear combination of the hydrogen atomic orbitals, like in the next question.
\[\Phi = c_1\psi +c_2\psi \nonumber \]
When we do this problem in the computer almost all programs use a basis set of Gaussian functions to create the orbitals.
Q3
The terms that arise from multiplying the linear combination of the wave function are
\[\begin{align*} \langle \phi|\phi\rangle &= \langle c_11s_A + c_2 1s_B |c_11s_A + c_2 1s_B \rangle \\[4pt] &= \langle c_11s_A|c_11s_A\rangle + \langle c_11s_A|c_21s_B\rangle + \langle c_21s_B|c_11s_A\rangle +\langle c_21s_B|c_21s_B\rangle \end{align*} \]
The first term
\[\langle c_11s_A|c_11s_A\rangle = c_1^2 \langle 1s_A|1s_A\rangle = c_1^2 \nonumber \]
The overlap integral \(\langle c_11s_A|c_21s_B\rangle \) is how much of the 1s orbital atom A occupies the same space as the 1s orbital on atom B. What is size of the area where \(1s_A\) and \(1s_B\) overlap?
Using the overlap integral the simple form for \(\langle \phi|\phi\rangle \) is
\[\langle c_11s_A|c_11s_A\rangle + \langle c_11s_A|c_21s_B\rangle + \langle c_21s_B|c_11s_A\rangle +\langle c_21s_B|c_21s_B\rangle = c_1^2 + 2c_1c_2S + c_2^2 \nonumber \]
According to the graph as the nuclei become farther apart the overlap approaches zero
In the energy equation the same four terms arise with the Hamiltonian.
\[\begin{align*} \langle \phi|\hat{H}|\phi\rangle &= \langle c_11s_A + c_2 1s_B |\hat{H}|c_11s_A + c_2 1s_B \rangle \\[4pt] &= \langle c_11s_A|\hat{H}|c_11s_A\rangle + \langle c_11s_A|\hat{H}|c_21s_B\rangle + \langle c_21s_B|\hat{H}|c_11s_A\rangle +\langle c_21s_B|\hat{H}|c_21s_B\rangle \end{align*}\nonumber \]
To simplify the first term we remember that the Hamiltonian is the molecular version:
\[ \hat{H} = \dfrac{-1}{2} \bigtriangledown^2 + \dfrac{-1}{r_A} + \dfrac{-1}{r_B} +\dfrac{1}{r_{AB}} \nonumber \]\[\langle c_11s_A|\hat{H}|c_11s_A\rangle = \langle c_11s_A|\dfrac{-1}{2} \bigtriangledown^2 + \dfrac{-1}{r_A} + \dfrac{-1}{r_B} + \dfrac{1}{r_{AB}}|c_11s_A\rangle \nonumber \]
The first two terms of the hamiltonian are the same as atomic hydrogen and can be separated
\[\langle c_11s_A|\dfrac{-1}{2} \bigtriangledown^2 + \dfrac{-1}{r_A} + \dfrac{-1}{r_B} + \dfrac{-1}{r_{AB}}|c_11s_A\rangle = \langle c_11s_A|\hat{H}_{atomic}|c_11s_A\rangle + \langle c_11s_A|\dfrac{-1}{r_B}|c_11s_A\rangle + \langle c_11s_A|\dfrac{1}{r_{AB}}|c_11s_A\rangle \nonumber \]
Then since we have hydrogen wave functions we have already solved the energy for the 1s orbital.
\[\langle c_11s_A|\hat{H}_{atomic}|c_11s_A\rangle + \langle c_11s_A|\dfrac{-1}{r_B}|c_11s_A\rangle +\langle c_11s_A|\dfrac{1}{r_{AB}}|c_11s_A\rangle = c_1^2E_{1s_A} + c_1^2\langle 1s_A|\dfrac{-1}{r_B}|1s_A\rangle + \langle c_11s_A|\dfrac{1}{r_{AB}}|c_11s_A\rangle \nonumber \]
Using the exchange and Coulomb integrals the first term is
\[ = c_1^2E_{1s_A} - c_1^2J + c_1^2\dfrac{1}{r_{AB}} \nonumber \]
Similarly for the second term we have
\[\begin{align*} \langle c_11s_A|\dfrac{-1}{2} \bigtriangledown^2 + \dfrac{-1}{r_A} + \dfrac{-1}{r_B} + \dfrac{1}{r_{AB}}|c_21s_B\rangle &= \langle c_11s_A|\hat{H}_{atomic}|c_21s_B\rangle + \langle c_11s_A|\dfrac{-1}{r_A}|c_21s_B\rangle + \langle c_11s_A|\dfrac{1}{r_{AB}}|c_21s_B\rangle \\[4pt]
&=\langle c_11s_A|\hat{H}_{atomic}|c_21s_B\rangle + \langle c_11s_A|\dfrac{-1}{r_A}|c_21s_B\rangle + \langle c_11s_A|\dfrac{1}{r_{AB}}|c_21s_B\rangle \\[4pt]
&= c_1c_2E_{1s_B}\langle 1s_A|1s_B\rangle + c_1c_2\dfrac{1}{r_{AB}}\langle 1s_A|1s_B\rangle + c_1c_2\langle 1s_A|\dfrac{-1}{r_A}|1s_B\rangle \\[4pt]
&= c_1c_2E_{1s_B}S + c_1c_2\dfrac{1}{r_{AB}}S - c_1c_2K \end{align*}\nonumber \]
Following these same patterns we would have for the third term
\[= c_1c_2E_{1s_A}S + c_1c_2\dfrac{1}{r_{AB}}S - c_1c_2K \nonumber \]
And the fourth term
\[ = c_2^2E_{1s_B} - c_2^2J + c_2^2\dfrac{1}{r_{AB}} \nonumber \]
Now we need to combine all these terms. First realize that the\(E_{1s_B} = E_{1s_A} \) and \(c_1 = c_2 \)
Thus we have
\[E_+ = 2c^2E_{1s}S + 2c^2\dfrac{1}{r_{AB}}S - 2c^2K + 2c^2E_{1s} - 2c^2J + 2c^2\dfrac{1}{r_{AB}} \nonumber \]
And with only two functions we did last week that \(c^2 = \dfrac{1}{2} \)
\[E_+ = E_{1s}S +\dfrac{1}{r_{AB}}S - K + E_{1s} - J + \dfrac{1}{r_{AB}} \nonumber \]
Rearrange them so that it becomes clear
\[\begin{align*} E_+ &= E_{1s}S + \dfrac{1}{r_{AB}}S + E_{1s} + \dfrac{1}{r_{AB}} - K - J \\[4pt] &= (1 +S) \left( E_{1s} + \dfrac{1}{r_{AB}} \right) - K - J \\[4pt] \dfrac{E_+}{1+S} &= E_{1s} + \dfrac{1}{r_{AB}} - \dfrac{K + J}{1+S} \end{align*}\nonumber \]
Worksheet 10B Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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