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Phys Orgo Secular Determinant question

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve for the wavefuntions using the Secular Determinant, of cyclopropene anion/cation/whocares.

    2. Relevant equations

    Secular determinant:
    | x 1 1|
    | 1 x 1|
    |1 1 x |

    3. The attempt at a solution

    I'm literally understanding every aspect of this concept, but what I cannot understand is the math behind the solutions. It looks very un-math like, more of an intuitive answer but I want to know how to solve the math out, because quite frankly its making me angry that something so simple is stumping me.

    Anyway: Solving the matrix yields x = -2,1,1

    Psi 1:

    -2c1 + c2 + c3 = 0
    c1 + -2c2 + c3 = 0
    c1 + c2 + -2c3 = 0
    c1^2 + c2^2 + c3^2 = 1

    The solution for the secular polynomial of x=-2 is simple: you can solve it out for c1=c2=c3 and the only solution that satisfies that is c = (√3 /3)

    My problem now lies with solving for the psi 2 and psi 3 solutions. I know the answers are

    psi2: c1 = 0 c2 = +√2 / 2 and c3 = - √2 / 2

    psi3: c1 = 2/√6 c2 = c3 = -1/√6

    I have no idea how the math from this system can solve for those variables. I'm completely missing something here.

    c1 + c2 + c3 = 0
    c1 + c2 + c3 = 0
    c1 + c2 + c3 = 0
    c1^2 + c2^2 + c3^2 = 1

    Thank you.
  2. jcsd
  3. Sep 21, 2013 #2


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    Note that the x = 1 solution occurs twice; i.e., is 2-fold degenerate. This implies that there will be freedom in choosing the solutions for ψ2 and ψ3. For ψ2 you can choose any values for c1, c2, and c3 such that c1 + c2 + c3 = 0 and c12 + c22 + c32 = 1.

    Then think about how to choose c1, c2, and c3 for ψ3.

    The solutions you gave for ψ2 and ψ3 represent one possible set of solutions out of an infinite number of possible choices.
  4. Sep 21, 2013 #3
    Right. Also by the Frost model you can see that there should be two degenerate orbitals. I understand that part.

    Are you saying that it's essentially okay to solve it intuitively instead of mathematically? Because I have a system of two equations each with three variables, and obviously as you said there are infinite solutions.

    But I do know that no matter what -c1 = c2+c3, implying that in at least one of the MO's, c1 = 2c2 = 2c3. Is this something that I should just "know" or "see" from the equations? Because the confusing part is the exact derivation of the wave function equations. I know from my MO diagrams that there must be one where all aspects are in phase, one where there is a node and one where there are two out of phase interactions and one in phase interaction. My problem is that I can't go through the math to get the MO diagrams without intuitively thinking about the MO diagrams. That's where I'm stuck.

    Thanks for your answer.
  5. Sep 21, 2013 #4


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    I am not familiar with molecular orbital theory.

    But if all you have to work with mathematically is the secular determinant, then ψ2 and ψ3 may be chosen in many different ways. For example you could choose ##(c_1, c_2, c_3) = (\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{-3}{\sqrt{14}})## for ψ2 and ##(c_1, c_2, c_3) = (\frac{5}{\sqrt{42}}, \frac{-4}{\sqrt{42}}, \frac{-1}{\sqrt{42}})## for ψ3.

    If there are additional mathematical constraints that come from MO theory, then of course that could lead to a unique choice for ψ2 and ψ3.
  6. Sep 21, 2013 #5
    Yeah I see what you're saying. There actually are more laws for MO theory. They revolve around the requirement that the systems be symmetric for homoatomic constructs, of which this is one. The solutions you provided are mathematically valid but they don't abide by that.

    This was my main problem. The math and secular polynomial IS correct, I'm just missing something non-mathematical I think.

    Don't worry though I understand. People who don't have to understand this **** should really just stay away LOL.

    I'm open to any ideas.

  7. Sep 21, 2013 #6


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    Do you have a reason why ##c_1## for ##\psi_2## should be 0? Mathwise, you usually do that just to make it easier to write down a solution. The solution for ##\psi_3## also appears to be orthogonal to that for ##\psi_2##. That's another common constraint to place on solutions. It probably has something to do with the "in phase, out of phase" mumbo jumbo you mentioned above. (I don't know MO theory either.)
  8. Sep 21, 2013 #7
    Orthogonality has something to do with it for sure but I'm not positive about the relationships it has with the solutions.

    The final wave functions for the different bonding/antibonding molecular orbitals must all be symmetric in some fashion.

    There is a reason why c1 or c2 or c3= 0. I know from my diagrams (that I just happened to unfortunately memorize) that there is an orbital in which one of the coefficient of one of the wave functions is zero. That is simply a fact and I don't mathematically know why. Hence, my problem.
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