[sanderol]

So far I've come to the expression for energie

<E> : (c1²H₁₁ + c2²H11 + 2c1c2H12) / c1² + c2² + 2c1c2S12

Now derivatives towards c1 and c2 yields:

<E>2c1 + <E>2c2S12 = 2c1H11 + 2c2H12

<E>2c2 + <E>2c1S12 = 2c2H11 + 2c1H12

So far so good.

But now these 2 equations are rewritten into a determinant which yields:

| H11-<E>      H12 - <E>|
|H12 - <E>     H11 - <E>|    =0

I dont have any idea how they get this, so this is my first question

Then the determinant is solved to <E> which yields:

<Eb> = (H11+H12)/(1+S12)

and

<Ea> = (H11-H12)/(1-S12)

Which are the bonding and antibonding orbitals (but I have no idea how they do this, so this is my second question)
Can someone please give me a hand in this. Thanks in advance.

[JoopLenthe]

Like... this is first year Quantum Basic Chemistry... you should be able to figure this one out by heart and intuition

[sanderol]

...

[Hunt]

<E>2c1 + <E>2c2S12 = 2c1H11 + 2c2H12

<E>2c2 + <E>2c1S12 = 2c2H11 + 2c1H12

So far so good.

But now these 2 equations are rewritten into a determinant which yields:

| H11-<E>      H12 - <E>|
|H12 - <E>     H11 - <E>|    =0

You are solving a system of 2 eq's and 2 unknowns , C1 and C2. However the system is homogenous i.e. of the form :

a C1 + b C2 = 0
c C1 + d C2 = 0

Such a system has either the trivial soln C1=C2=0 which is physically meaningless, or a set of infinite solns which requires the determinant to be zero , which is what you choose in this case. If u dont know what a determinant is , then you should review the basics of elementary linear algebra.

Which are the bonding and antibonding orbitals (but I have no idea how they do this, so this is my second question)
Can someone please give me a hand in this. Thanks in advance.

The orbital with higher energy is the anti-bonding while the lower is bonding. How to actually determine that is not covered in basic quantum chem courses because you have to numerically compute the overlap and coulomb integrals but if u consider that H12 is usually less than zero while S12 > 0, then Eb < Ea.