# Homework Help: Secular determinant in MO method.

1. Nov 21, 2008

### SporadicSmile

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

which gives the answer

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

This multiplies out to give

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

which with rearrangement gives

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

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. Nov 21, 2008

### 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.

3. Nov 21, 2008

### SporadicSmile

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 $$\alpha$$ = $$\beta$$, 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.)

4. Nov 21, 2008

### 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.

5. Nov 21, 2008

### 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:

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

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.

6. Nov 21, 2008

### 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.

7. Nov 21, 2008

### 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! thats almost something i think

Cheers =>

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