# Worksheet 10B Solutions

## Q1

The Hamiltonian consists of the kinetic and potential energy.

$\hat{H} = \hat{T} + \hat{V} \label{1}$

In atomic units all the constants are set to equal 1, $$h=1$$

Therefore the kinetic term in the Hamiltonian is transformed

$\hat{T} = \dfrac{-\hbar^2}{2m} \bigtriangledown^2 \psi \Longrightarrow \dfrac{-1}{2} \bigtriangledown^2 \psi \label{2}$

The potential terms in the $$H_2^+$$ molecule are then given in atomic units

$\hat{V} = \dfrac{-1}{r_A} + \dfrac{-1}{r_B} \label{3}$

## Q2

The potential term for the interaction between the two protons in atomic units is

$\hat{V} = \dfrac{1}{r_{AB}} \label{4}$

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.

$E_{total} = E_{e^-} + T_{molecule} \label{5}$

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.

$\Phi = c_1\psi +c_2\psi \label{6}$

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

$<\phi|\phi> = <c_11s_A + c_2 1s_B |c_11s_A + c_2 1s_B >$

$= <c_11s_A|c_11s_A> + <c_11s_A|c_21s_B> + <c_21s_B|c_11s_A> +<c_21s_B|c_21s_B> \label{7}$

The first term

$<c_11s_A|c_11s_A> = c_1^2 <1s_A|1s_A> = c_1^2 \label{8}$

The overlap integral $$<c_11s_A|c_21s_B>$$ 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 $$<\phi|\phi>$$ is

$<c_11s_A|c_11s_A> + <c_11s_A|c_21s_B> + <c_21s_B|c_11s_A> +<c_21s_B|c_21s_B>$

$= c_1^2 + 2c_1c_2S + c_2^2 \label{9}$

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.

$<\phi|\hat{H}|\phi> = <c_11s_A + c_2 1s_B |\hat{H}|c_11s_A + c_2 1s_B >$

$= <c_11s_A|\hat{H}|c_11s_A> + <c_11s_A|\hat{H}|c_21s_B> + <c_21s_B|\hat{H}|c_11s_A> +<c_21s_B|\hat{H}|c_21s_B> \label{10}$

#### To simplify the first term we remember that the Hamiltonian is the molecular version:

$\hat{H} = \dfrac{-1}{2} \bigtriangledown^2 + \dfrac{-1}{r_A} + \dfrac{-1}{r_B} +\dfrac{1}{r_{AB}} \label{11}$

$<c_11s_A|\hat{H}|c_11s_A> = <c_11s_A|\dfrac{-1}{2} \bigtriangledown^2 + \dfrac{-1}{r_A} + \dfrac{-1}{r_B} + \dfrac{1}{r_{AB}}|c_11s_A> \label{12}$

#### The first two terms of the hamiltonian are the same as atomic hydrogen and can be seperated

$<c_11s_A|\dfrac{-1}{2} \bigtriangledown^2 + \dfrac{-1}{r_A} + \dfrac{-1}{r_B} + \dfrac{-1}{r_{AB}}|c_11s_A> = <c_11s_A|\hat{H}_{atomic}|c_11s_A> + <c_11s_A|\dfrac{-1}{r_B}|c_11s_A> + <c_11s_A|\dfrac{1}{r_{AB}}|c_11s_A> \label{13}$

#### Then since we have hydrogen wave functions we have already solved the energy for the 1s orbital.

$<c_11s_A|\hat{H}_{atomic}|c_11s_A> + <c_11s_A|\dfrac{-1}{r_B}|c_11s_A> +<c_11s_A|\dfrac{1}{r_{AB}}|c_11s_A> = c_1^2E_{1s_A} + c_1^2<1s_A|\dfrac{-1}{r_B}|1s_A> + <c_11s_A|\dfrac{1}{r_{AB}}|c_11s_A> \label{14}$

#### Using the exchange and Coulomb integrals the first term is

$= c_1^2E_{1s_A} - c_1^2J + c_1^2\dfrac{1}{r_{AB}} \label{15}$

#### Similarly for the second term we have

$<c_11s_A|\dfrac{-1}{2} \bigtriangledown^2 + \dfrac{-1}{r_A} + \dfrac{-1}{r_B} + \dfrac{1}{r_{AB}}|c_21s_B> = <c_11s_A|\hat{H}_{atomic}|c_21s_B> + <c_11s_A|\dfrac{-1}{r_A}|c_21s_B> + <c_11s_A|\dfrac{1}{r_{AB}}|c_21s_B> \label{16}$

$<c_11s_A|\hat{H}_{atomic}|c_21s_B> + <c_11s_A|\dfrac{-1}{r_A}|c_21s_B> + <c_11s_A|\dfrac{1}{r_{AB}}|c_21s_B>$

$= c_1c_2E_{1s_B}<1s_A|1s_B> + c_1c_2\dfrac{1}{r_{AB}}<1s_A|1s_B> + c_1c_2<1s_A|\dfrac{-1}{r_A}|1s_B>$

$= c_1c_2E_{1s_B}S + c_1c_2\dfrac{1}{r_{AB}}S - c_1c_2K \label{17}$

#### Following these same patterns we would have for the third term

$= c_1c_2E_{1s_A}S + c_1c_2\dfrac{1}{r_{AB}}S - c_1c_2K \label{18}$

#### And the fourth term

$= c_2^2E_{1s_B} - c_2^2J + c_2^2\dfrac{1}{r_{AB}} \label{19}$

#### Thus we have

$E_+ = 2c^2E_{1s}S + 2c^2\dfrac{1}{r_{AB}}S - 2c^2K + 2c^2E_{1s} - 2c^2J + 2c^2\dfrac{1}{r_{AB}} \label{20}$

And with only two functions we did last week that $$c^2 = \dfrac{1}{2}$$

$E_+ = E_{1s}S +\dfrac{1}{r_{AB}}S - K + E_{1s} - J + \dfrac{1}{r_{AB}} \label{21}$

#### Rearrange them so that it becomes clear

$E_+ = E_{1s}S + \dfrac{1}{r_{AB}}S + E_{1s} + \dfrac{1}{r_{AB}} - K - J = (1 +S)( E_{1s} + \dfrac{1}{r_{AB}} ) - K - J \label{22}$

$\dfrac{E_+}{1+S} = E_{1s} + \dfrac{1}{r_{AB}} - \dfrac{K + J}{1+S} \label{23}$