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Worksheet 10B Solutions

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  • 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

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


    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.


    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

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

    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

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

    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

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

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