1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper
    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    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 =>
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Secular determinant in MO method.
  1. Secular equilibrium (Replies: 1)

  2. D'alembert method (Replies: 1)

  3. D'alembert method (Replies: 0)

Loading...