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Construct an ONB for a triatomic molecule

  1. Apr 5, 2015 #1
    Consider a triatomic molecule with three identical atoms that are bound together with each atom at its own corner of an equilateral triangle of edge length a. An added electron on the molecule can be put in an identical atomic orbital on any of the three atoms. Denote the atomic states in which the electron is on atom i as |i>, and assume that these three states {|1>, |2>, and |3>}, form an orthonormal set. Let the mean energy associated with such a state be the same value ε0=<i|H|i> for each state |i>. Suppose also that the electron on an atom can move to either of its neighbors, with an amplitude <i|H|j>=V0 i≠j

    a) Construct a 3X3 matrix [H] that represents the Hamiltonian within the subspace spanned by these 3 atomic states, using the states {\|i>} as an ONB for the subspace. Find the trace of this matrix.

    [H] =
    ε_0 & V_0 & V_0\\
    V_0 & ε_0 & V_0\\
    V_0 & V_0 & ε_0

    Tr[H] = 3ε0

    b) Find the energy eigenvalues and the degeneracies of the molecule with an added electron.

    | ε0-ε V0 V0 |
    | V0 ε0-ε V0 |
    | V0 V0 ε0-ε |

    = (ε0-ε) [(ε0-ε)(ε0-ε) - V0]
    - V0 [ V00-ε) - V02 ]
    + V0 [ V02 - V00-ε) ]

    = (ε0-ε)3 - 3V020-ε) +2V03=0


    c) Construct an ONB of eigenstates {|εn> |n=1,2,3} of the system, as linear combinations of the atomic states.

    I could not get Latex to work according to the directions on this site. I apologize for the crude look to the matrices, etc.

    I am stuck on part c as follows. Could someone please help me with this part? I get the following which make no sense to me:

    (H-ε1) |χ1>=(H-ε0+V0) |χ1>=

    V0 V0 V0
    V0 V0 V01>
    V0 V0 V0

    (H-ε3) |χ1>=(H-ε0+2V0) |χ1>=

    2V0 V0 V0
    V0 2V0 V01>
    V0 V0 2V0

    Attached Files:

    Last edited by a moderator: Apr 5, 2015
  2. jcsd
  3. Apr 5, 2015 #2


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    Your post had broken (unpaired) [sub]-tags, that made it hard to read. I fixed those.

    What did you do to get those equations for (c)?
  4. Apr 5, 2015 #3
    I took the solutions for the eigenvalues from part b (two of which are degenerate) and subtracted them from the Hamiltonian matrix.

    These are supposed to be the eigenvalue equations where the chi's should be the eigenvectors.
  5. Apr 5, 2015 #4


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    Okay, so ##(H-Iε_1) ~|\chi_1\!>~ = 0## right?
    Something that won't have a unique solution as your eigenspace has dimension 2.

    I'm a bit surprised by the signs in your eigenvalues, shouldn't their sum be equal to the trace of the Hamiltonian?
  6. Apr 5, 2015 #5
    You are correct. I wrote down the third eigenvalue incorrectly. It is correct on the attachment but the matrix is incorrect.

    The third eigenvalue should be ε3=ε0+2V0

    The third matrix should be

    -2V0 V0 V0
    V0 -2V0 V0 |χ1> = 0
    V0 V0 -2V0
  7. Apr 5, 2015 #6


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    I can see a non-trivial solution for this equation.
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