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Homework Help: Secular determinant in MO method.

  1. Nov 21, 2008 #1
    Not a problem par se, as much as me failing somewhere in algebra, and i cant find my mistakes.

    Trying to work through the secular determinant
    [tex]

    E_\pm = det\left(
    \begin{array}{cc}
    \alpha - E & \beta - ES\\
    \beta - ES & \alpha - E
    \end{array}
    \right)
    = 0

    [/tex]

    which gives the answer

    [tex]

    E_\pm = \frac{\alpha \pm \beta} {1 \pm S}

    [/tex]

    This multiplies out to give

    [tex]

    (\alpha - E)^2 - (\beta - ES)^2 = 0

    [/tex]

    which with rearrangement gives

    [tex]

    (1 - S^2)E^2 + 2(\beta S - \alpha)E + (\alpha^2 - \beta^2) = 0

    [/tex]

    this is quadratic in E, so solutions will be given by the usual formula for solving quadratic equations. This is where I seem to fail, rather badly really, and I cant find my error. If someone could point me in the direction of somewhere this has been solved fully, each step so i can find my error, id be very much obliged. If not I can write out my own calculations on here, but that is rather time consuming. But I will if needed.

    Thanks
    Steve
     
  2. jcsd
  3. Nov 21, 2008 #2

    gabbagabbahey

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    Gold Member

    Are you sure you have the correct determinant equation?

    ...After a little bit of work, you should find that your equation gives two solutions: [itex]E=0[/itex] and [tex]E=\frac{\alpha+\beta}{1+S}+\frac{\alpha-\beta}{1-S}[/tex]....which is somewhat different from what you claim the answer is supposed to be.
     
  4. Nov 21, 2008 #3
    From Atkins 'Molecular Quantum Mechanics' page 254:


    Perhaps my post also wasn't very clear, the third equation is an expansion of the determinant, just looking over it realised that might not be 100% clear at first glance.

    Wouldn't E = 0 only be a solution to the quadratic equation if [tex] \alpha [/tex] = [tex] \beta [/tex], which it does not ([tex] \alpha [/tex] is the coulomb intergral, [tex] \beta [/tex] the resonance integral, although this isn't really important here, I'm more bothered about why I suddenly can't do basic algebra.)
     
  5. Nov 21, 2008 #4

    gabbagabbahey

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    hmmm....yes E=0 isn't a solution...I made an error in my algebra too....after correcting it, I get the desired solution....If you show me your work (a few lines at a time), I can point out where you are going wrong.
     
  6. Nov 21, 2008 #5
    Okay i will set out my calculations, might be useful might be i spot my own mistake, or what i need to do.

    Solving the quadratic formula:

    [tex]

    E_\pm = \frac{-b^2 \pm \sqrt{b^2 - 4ac}} {2a}

    [/tex]

    where, in the case of the equation i have:

    [tex]
    a = (1 - S^2)
    [/tex]

    [tex]
    b = 2(\beta S - \alpha)
    [/tex]

    [tex]
    c = (\alpha^2 - \beta^2)
    [/tex]

    working out each part separately (for clarity):

    [tex]
    b^2 = 4(\beta S - \alpha) = 4(\beta^2 E^2 + \alpha^2 - 2\alpha \beta S)
    [/tex]

    [tex]
    4ac = 4(1 - S^2)(\alpha^2 - \beta^2)
    [/tex]
    [tex]
    = 4(\alpha^2 - \beta^2 - S^2 \alpha^2 + S^2 \beta^2)
    [/tex]

    [tex]
    b^2 - 4ac = 4(\beta^2 - 2\alpha \beta S + S^2 \alpha^2)
    [/tex]
    [tex]
    = 4(\beta - S \alpha)^2
    [/tex]

    [tex]
    \sqrt{4(\beta - S \alpha)^2} = 2(\beta - S \alpha)
    [/tex]

    [tex]
    -b^2 \pm \sqrt{b^2 - 4ac} = 4(-\beta^2 + 2\alpha \beta S - S^2 \alpha^2) \pm 2(\beta - S \alpha)
    [/tex]

    Which is where i seem to lose myself in not being able to simplify any further.
    I get the feeling I am missing something trivial, but its one of those things where the longer i stare at it the less i can see it.
     
  7. Nov 21, 2008 #6

    gabbagabbahey

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    Homework Helper
    Gold Member

    You're gunna kick urself when you realize your mistake....take a deep a breath and remember to smile....are you ready?......


    The quadratic formula is [tex]E_\pm = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a}[/tex] not [tex]E_\pm = \frac{-b^2 \pm \sqrt{b^2 - 4ac}} {2a}[/tex].....i.e. the -b isn't squared.:smile:
     
  8. Nov 21, 2008 #7
    Well, when i vote for the most embarasing moment of the year i think i know what i will pick..

    Goddamit!
    Well, at least i can do algebra, i just can't remeber formulae! thats almost something i think

    Cheers =>
     
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