Where Am I Going Wrong with the Secular Determinant in MO Method?

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Not a problem par se, as much as me failing somewhere in algebra, and i can't find my mistakes.

Trying to work through the secular determinant
<br /> <br /> E_\pm = det\left(<br /> \begin{array}{cc}<br /> \alpha - E &amp; \beta - ES\\<br /> \beta - ES &amp; \alpha - E<br /> \end{array}<br /> \right)<br /> = 0<br /> <br />

which gives the answer

<br /> <br /> E_\pm = \frac{\alpha \pm \beta} {1 \pm S}<br /> <br />

This multiplies out to give

<br /> <br /> (\alpha - E)^2 - (\beta - ES)^2 = 0<br /> <br />

which with rearrangement gives

<br /> <br /> (1 - S^2)E^2 + 2(\beta S - \alpha)E + (\alpha^2 - \beta^2) = 0<br /> <br />

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 can't 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
 
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Are you sure you have the correct determinant equation?

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


The Secular determinant is then

<br /> <br /> <br /> E_\pm = det\left(<br /> \begin{array}{cc}<br /> \alpha - E &amp; \beta - ES\\<br /> \beta - ES &amp; \alpha - E<br /> \end{array}<br /> \right)<br /> = 0<br /> <br /> <br />

This equation (a quadratic equation in E when the determinant is expanded) has the solutions

<br /> <br /> <br /> E_\pm = \frac{\alpha \pm \beta} {1 \pm S}<br /> <br /> <br />

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

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

<br /> <br /> E_\pm = \frac{-b^2 \pm \sqrt{b^2 - 4ac}} {2a}<br /> <br />

where, in the case of the equation i have:

<br /> a = (1 - S^2)<br />

<br /> b = 2(\beta S - \alpha)<br />

<br /> c = (\alpha^2 - \beta^2)<br />

working out each part separately (for clarity):

<br /> b^2 = 4(\beta S - \alpha) = 4(\beta^2 E^2 + \alpha^2 - 2\alpha \beta S)<br />

<br /> 4ac = 4(1 - S^2)(\alpha^2 - \beta^2)<br />
<br /> = 4(\alpha^2 - \beta^2 - S^2 \alpha^2 + S^2 \beta^2)<br />

<br /> b^2 - 4ac = 4(\beta^2 - 2\alpha \beta S + S^2 \alpha^2)<br />
<br /> = 4(\beta - S \alpha)^2<br />

<br /> \sqrt{4(\beta - S \alpha)^2} = 2(\beta - S \alpha)<br />

<br /> -b^2 \pm \sqrt{b^2 - 4ac} = 4(-\beta^2 + 2\alpha \beta S - S^2 \alpha^2) \pm 2(\beta - S \alpha)<br />

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.
 
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 E_\pm = \frac{-b \pm \sqrt{b^2 - 4ac}} {2a} not E_\pm = \frac{-b^2 \pm \sqrt{b^2 - 4ac}} {2a}...i.e. the -b isn't squared.:smile:
 
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! that's almost something i think

Cheers =>
 
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