Eigenvalues and Eigenfunctions of Basis states

This chapter begins with a mathematical article. People who want to know more about benzene should click here.

If we measure n-energy state (for instance, of a particle in a potential box) we will have the following mathematical expression
 

Applying <m| (Eigenfunction) and correspondingly integrating  ò y_m*.....dV the initial formula one will have
 

We can write down quantities <m|H|n> and ò y_m* y_n dV o=in a shorter matrix form H_mn that includes only diagonal components in the case of eigenfunction. And other components are equal to 0:
 

In this case, the diagonal matrix components give energy eigenvalues of the system under consideration. However, the background is that |n> and correspondingly |m> are eigenfunctions. What does this system of equations look like when we know only improper functions? Let's analyse our system in the following way: let's expand our eigenfunction |n> into other basis states, moreover the corresponding basis vectors |i> will have their own eigenfunction H|i>.

Substituting it into our Schroedinger equation (with E_n = E):

Application of <k| and correspondingly òf_k*...dV     (<k| is orthogonal to |i>) will give us

This is the linear set of equations:
 

This a well-known set of equations:
 

	.		=  E .
Matrix¾­		­¾Vector¾­

There are particular algebraic ways to determine eigenvalues E and use transformation which gives us eigenfunction (here eigenvectors).

We will describe a simple system with only 2 states which, however, can give us basic knowledge of spectroscopy and chemistry. For instance, let's have a look at benzene that has the following electronic structure:
 

Benzene  ≡

|Benzene>  =  a1|1> + a2|2> H |Benzene>  =  E |Benzene>

Applying corresponding abbreviation <1|H|1> = H₁₁ and so on, we will have the following result:
 

There are two stationary energy states E_I and E_II. If H₁₂ = H₂₁ = 0 there are no interactions between states |1> and |2> and then E_I = H₁₁ ≡ E₁ and E_II = H₂₂ ≡ E₂. That' why E_I and E_II correspond to energyvalues of unperturbed states:

The formulas will be simpler if terms H₁₁ and H₂₂ are equal (as for benzene): H₁₁ = H₂₂ = E₀. And if we assign interaction H₁₂ = H₂₁  by ΔE  then the energy of stationary states |I> and |II>:
 

We have energy of stationary states |I> and |II>. And how does the wavefunction look? Now we can construct the wavefunction from known wavefunctions of states |1> and |2> in the example considered (H₁₁ = H₂₂ = E₀ and H₁₂ = H₂₁ = −ΔE). The solution is as follows:
 

Now let's have a look at normalization, orthogonalization and diagonal energy matrix:

a) Normalization:

b) Orthogonalization:

c) The energy matrix  must be for eigenvectors |I>, |II> diagonal components:

In general case H₁₁ ¹ H₂₂, H₁₂ ¹ H₂₁ are stationary states |I>, |II> having initial states |1> and |2> 

Complex constants have the following relationships:

Here one can find few simple chemical applications of above-mentioned for ammonia, H₂⁺ and H₂.

Bindings of different ions with electron

Two atomic asymmetric molecules often have weak bindings. Why?

If we consider, for instance, proton and positive Li ion. So now initially terms H₁₁ and H₂₂ are different and |H₁₁− H₂₂| >>DE. Then it will be (for H₁₂ = H₂₁ = −ΔE):

Since x = 4ΔE²/(H₁₁− H₂₂)² is smaller than 1 and in the first approximation is (1 + x)^½  ≈  1 + ^x/₂  one will have:
 

EII  =  H22 + ΔE²/(H22−H11)
EI  =  H11 -DE²/(H22−H11)

The additional energy fragmentation "exchange energy" ^ΔE²/(H₂₂−H₁₁) is smaller than that of H₂⁺  in this case.

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