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

[SporadicSmile]

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

which gives the answer

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

This multiplies out to give

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

which with rearrangement gives

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

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

 

[gabbagabbahey]

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.

 

[SporadicSmile]

From Atkins 'Molecular Quantum Mechanics' page 254:

The Secular determinant is then

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

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

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

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.)

 

[gabbagabbahey]

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.

 

[SporadicSmile]

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:

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

where, in the case of the equation i have:

$$a = (1 - S^2)$$

$$b = 2(\beta S - \alpha)$$

$$c = (\alpha^2 - \beta^2)$$

working out each part separately (for clarity):

$$b^2 = 4(\beta S - \alpha) = 4(\beta^2 E^2 + \alpha^2 - 2\alpha \beta S)$$

$$4ac = 4(1 - S^2)(\alpha^2 - \beta^2)$$
$$= 4(\alpha^2 - \beta^2 - S^2 \alpha^2 + S^2 \beta^2)$$

$$b^2 - 4ac = 4(\beta^2 - 2\alpha \beta S + S^2 \alpha^2)$$
$$= 4(\beta - S \alpha)^2$$

$$\sqrt{4(\beta - S \alpha)^2} = 2(\beta - S \alpha)$$

$$-b^2 \pm \sqrt{b^2 - 4ac} = 4(-\beta^2 + 2\alpha \beta S - S^2 \alpha^2) \pm 2(\beta - S \alpha)$$

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.

 

[gabbagabbahey]

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.

 

[SporadicSmile]

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 =>