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Homework Help: Tight binding model (covalent bonding)

  1. Jul 20, 2016 #1
    1. The problem statement, all variables and given/known data
    (linear combination of atomic orbitals):
    Lets consider two atoms which are bond together with a covalent bond. Lets consider any sets of wavefunctions [itex]|n\rangle[/itex] for n=1,2,....,N. Lets call orbital [itex]|1\rangle[/itex] around nucleus 1 and orbital [itex]|2\rangle[/itex] around nucleus 2 and so on.
    for simplicity we assume this basis is orthonormal [itex]\langle n|m\rangle=\delta_{n,m}[/itex]
    Let us write a trial wavefunction for our ground state as [itex]|\psi\rangle=\sum_n\phi_n|n\rangle[/itex].
    This is known as linear combination of atomic orbitals, LCAO,or tight binding.

    We would like to find the lowest -energy wavefunction we can construct in this form.i.e, the best approximation to the actual ground state wavefunction. ( The more states we use in our basis, generally, the more accurate our results will be).

    We claim that the ground state is given by the solution of the effective Schroedinger equation [itex]H\phi=E\phi[/itex] where [itex]\phi[/itex] is the vector of N coefficients [itex]\phi_{n}[/itex] and [itex]H[/itex] is the N by N matrix [itex]H_{n,m}=\langle n|H|m \rangle[/itex] with H the full system we are considering.

    we want to prove this by constructing the energy [itex]E=\frac{\langle\psi|H|\psi\rangle}

    The question is : Show that minimizing this energy [itex]E[/itex] w.r.t to each [itex]\phi_{n}[/itex] gives the same eigenvalue equation above.(Caution:[itex]\phi_n[/itex] is generally complex! )

    2. Relevant equations

    3. The attempt at a solution

    after expanding
    I tried to minimize the energy [itex]E[/itex] w.r.t to [itex]\phi_{n}[/itex] by letting the derivative of it w.r.t to[itex]\phi_{n}[/itex] equals to zero, but this didn't lead me to the eigenvalue equation, if any body can give me more hints? thanks!
  2. jcsd
  3. Jul 21, 2016 #2
    3. The attempt at a solution

    [itex]\frac{\phi^*_1\phi_1\langle1|H|1\rangle+...+\phi^*_1\phi_n\langle1|H|1\rangle+\phi^*_2\phi_1\langle 2|H|1\rangle+...+\phi^*_2\phi_n\langle 2|H| n\rangle+..........+\phi^*_n\phi_1\langle n|H|1\rangle+.....+\phi^*_n\phi_n\langle n|H|n\rangle}{ \left| \phi_1 \right|^2+\left| \phi_2 \right|^2+....+\left| \phi_n \right|^2 }[/itex]

    Let A=[itex]\left| \phi_1 \right|^2+\left| \phi_2 \right|^2+....+\left| \phi_n \right|^2[/itex]

    Let's minimize E by differentiating it w.r.t to [itex]\phi_1[/itex]

    [itex]dE=\frac{\partial E}{\partial \phi_1}d\phi_1+.....+\frac{\partial E}{\partial \phi_n}d\phi_n[/itex]
    then[itex]\frac{dE}{d\phi_1}=\frac{\partial E}{\partial \phi_1}\frac{d\phi_1}{d\phi_1}+.....+\frac{\partial E}{\partial \phi_n}\frac{d\phi_n}{d\phi_1}[/itex]

    as all the [itex]\phi_i 's[/itex] are independent then [itex]\frac{dE}{d\phi_1}=\frac{\partial E}{\partial \phi_1}[/itex]

    [itex]\frac{\partial E}{\partial \phi_1}=\frac{\phi_1^*A\langle1|H|1\rangle-\phi^*_1\phi_1\phi^*_1\langle1|H|1\rangle}{A^2}+\frac{\phi_2^*A\langle2|H|1\rangle-\phi^*_2\phi_1\phi^*_1\langle2|H|1\rangle}{A^2}+............+\frac{\phi_n^*A\langle n|H|1\rangle-\phi^*_n\phi_1\phi^*_1\langle n|H|1\rangle}{A^2}=0[/itex]

    then[itex] A(\phi^*_1\langle 1|H|1\rangle+\phi^*_2\langle 2|H|1\rangle+.....+\phi^*_n\langle n|H|1\rangle)-\left| \phi_1\right|^2(\phi^*_1\langle 1|H|1\rangle+\phi^*_2\langle2|H|1\rangle+...+\phi^*_n\langle n|H|1\rangle)=0[/itex]

    this means that [itex]\phi^*_1\langle1|H|1\rangle+...........+\phi^*_n\langle n|H|1\rangle=0[/itex]
    lets complex conjugate we will get [itex]\phi_1\langle1|H|1\rangle+...........+\phi_n\langle 1|H|n\rangle=0[/itex]

    moving in this with every variable [itex]\phi_i[/itex] we will get [itex]H\phi=0[/itex] where H is the N by N Hamiltonian matrix in the |1> ,|2>,.....|n> basis and [itex]\phi[/itex] is the column vector containg [itex]\phi_1,\phi_2,...\phi_n[/itex]

    I reached this part and I don't know how to proceed.
  4. Jul 21, 2016 #3
    We have ##|\psi\rangle = \sum_n \phi_n |n\rangle## and define ##H|n\rangle = E_n|n\rangle## which gives
    $$\langle \psi| H | \psi \rangle = \langle m | \sum_m \phi_m^* \sum_n E_n \phi_n| n \rangle\\
    = \sum_{m,n}\phi_m^*\phi_n E_n \langle m| n \rangle\\
    = \sum_{m,n}\phi_m^*\phi_n E_n\delta_{m,n}\\
    = \sum_n E_n |\phi_n|^2$$
    The denominator gives
    $$ \langle \psi | \psi \rangle = \sum_n |\phi_n|^2$$
    I think from here on it is a bit easier.
  5. Jul 22, 2016 #4
    But I think this approach will not work since the orbitals|1>,|2>,.....|n> are not the eigenstates of the Hamiltonian H which is the Hamiltonian of the electron when the two atoms are combined.
  6. Jul 23, 2016 #5
    I found the solution. :)
  7. Jul 24, 2016 #6
    Great :). Do you mind to give the solution? I find this problem also quite interesting.
  8. Jul 26, 2016 #7
    It is problem 6.2. My approach was right but I differentiated the Energy E w.r.t to phi1 wrongly.

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