## Molecular orbital

Using 2 hydrogen atom, we started from 2 atomic orbital (AO),

$\phi_1(\vec{r}) = \phi(\vec{r} - \vec{R_1})$

$\phi_2(\vec{r}) = \phi(\vec{r} - \vec{R_2})$

The atomic orbital $\phi$ is the ground state of hydrogen atom. From these 2 AO, we can construct linear combination of atomic orbitals (LCAO),

$\psi_1(\vec{r}) = A_1( \phi_1(\vec{r}) + \phi_2(\vec{r}))$

$\psi_2(\vec{r}) = A_2( \phi_1(\vec{r}) - \phi_2(\vec{r}))$

Notice that an electron can be in either $\psi_1$ or $\psi_2$, that means the electron is not belong to any atom, but belong to the molecule.

The last step is adding the spin, there are 2 spins $\alpha, \beta$, The molecular orbital (MO) are,

$\chi_1(\vec{r}) = \psi_1(\vec{r}) \alpha$

$\chi_2(\vec{r}) = \psi_1(\vec{r}) \beta$

$\chi_3(\vec{r}) = \psi_2(\vec{r}) \alpha$

$\chi_4(\vec{r}) = \psi_2(\vec{r}) \beta$

The Slater determinant can be formed by any 2 of them.

$\Psi = \sqrt{2}^{-1} ( \chi_1(\vec{r_1})\chi_2(\vec{r_2}) - \chi_2(\vec{r_1})\chi_1(\vec{r_2}) ) = |\chi_1\chi_2 \rangle$

We started by 1 atomic orbital, and there are and 2 atoms,  then 2 LCAOs are formed, and then multiplied by 2 spin states, gives 4 molecular orbitals. From these 4 MO and 2 electrons, we can from 6 different Slater determinant.

If we have different number of AO for different atom, we have AO

$\vec{n} = (n_1, n_2,..., n_k)$

The LCAO are

$\displaystyle \psi_i = \sum_{r=1}^k C_{ir} \phi_r, r = 1,2,..., n_k$

The total number of LCAO are all permutation of the AOs. A simple construction is that only plus or minus sign are used. Then the number of LCAO ( not orthonormal ) is

$\displaystyle K = 2^{-1} \prod_{i=1}^{k} (2n_i) = 2^{k-1} \prod_{i=1}^{k} n_i$

The MO are

$\chi_{2i-1} = \psi_i \alpha \\ \chi_{2i} = \psi_i \beta , i = 1,2,..., K$

Thus the number of Slater determinant are

$\displaystyle N = {}_{2K}C_p = \frac{(2K)!}{(2K-p)!p!}$

For example, we have 3 atoms, there are $(1, 2, 2)$ AO, there are 16 LCAOs. Thus, we have 32 MO. With 2 electrons, there are 496 Slater determinant.

For C-H molecule, there are 6 electrons for carbon and 1 electron for hydrogen, use 6 AOs for carbon and 1 AO for hydrogen, the number of MO is 24 and 346,104 Slater determinant.

When using all Slater determinant, it is called full CI.

When only increase the AO, but limited number of Slater determinant, The energy is called Hartree limit.

A problem is to construct LCAOs for arbitrary AOs. For example, for $(1,1,1)$ AOs, The number of possible combination with pule or minus sign is 4. However. I cannot construct orthonormal 4 LCAOs, because the system are over constrained (16 constrains with 12 unknown of $C_{ij}$)

## Covalent and Ionic bonding

hm…. yes, I am now studying theoretical chemistry, so, it is a bit different from nuclear physics. But since theoretical chemistry developed a lot of techniques, which were also used in nuclear physics, it is worth to study them. anyway~

In theoretical chemistry, for homonuclear diatomic molecule (such as H2, O2, N2, etc.) , the system should be the same under parity transfrom. There are two kind of symmetry that satisfy the partiy transform, which is even or odd function. In chemistry, they called the even function be gerade state, and odd function be the ungerade state. to see more.

Lets use H2 molecule to see how covalent and ionic bonding defined.

In H2 molecule, the two protons are located at A and B. An electron relative position to proton A and B are $r_A$ and $r_B$.

The spatial orbit can be even or odd for an single electron in ground state:

$\displaystyle \phi_g = \frac{1}{\sqrt{2}}(\psi(r_A) + \psi(r_B) )$

$\displaystyle \phi_u = \frac{1}{\sqrt{2}}(\psi(r_A) - \psi(r_B) )$

In chemistry, they use $\alpha$ and $\beta$ for an electron spin state, which can only be spin-up or spin-down. So, for two electrons H2 molecule, assume the orbit can only be s-orbit, the total spin can only be $S=0$ or $S=1$.

for spin-singlet $S=0$, the two electron wave functions or molecular orbitals must be symmetric, which are:

$\Phi_A = \chi_{00}(1,2) \phi_g(1) \phi_g(2)$

$\Phi_B = \chi_{00}(1,2) \phi_u(1) \phi_u(2)$

$\Phi_C = \chi_{00}(1,2) \frac{1}{\sqrt{2}} (\phi_g(1) \phi_u(2) + \phi_u(1) \phi_g(2))$

for spin-triplet $S=1$,

$\Phi_D = \chi_{1m_S}(1,2) \frac{1}{\sqrt{2}} (\phi_g(1) \phi_u(2) - \phi_u(1) \phi_g(2))$

We expand the $\phi_{g/u}$

$\displaystyle 2 \phi_g(1) \phi_g(2) = \begin{matrix} \psi(r_{A1})\psi(r_{B2}) + \psi(r_{B1})\psi(r_{A2}) + \\ \psi(r_{A1})\psi(r_{A2}) + \psi(r_{B1})\psi(r_{B2}) \end{matrix}$

The upper row is the covalent bonding, and the bottom row is the ionic bonding.

Let take a careful look on the covalent bonding

$\Phi_{cov} = \psi(r_{A1})\psi(r_{B2}) + \psi(r_{B1})\psi(r_{A2})$

The first term means the electron-1 is orbiting A and the electron-2 is orbiting B, the 2nd term is similar.

The electrons are being share among the nuclei.

The ionic bonding

$\Phi_{ion} = \psi(r_{A1})\psi(r_{A2}) + \psi(r_{B1})\psi(r_{B2})$

Both electrons are at the same nucleus at a time.

These agrees with our learning from high school chemistry class. yeah~

Note that, when minimizing the energy of H2 system, a single function $\Psi_A$ is not a good choice, it is better to include more molecular orbitals, say, $\Psi = a_A\Psi_A + a_B\Psi_B$. notice that the covalent and ionic component in $\Psi_B$ has different sign. After minimization, the ionic component is greatly suppressed.  In fact, the maximum component of ionic bond is only 20% when $R = 0.75 \textrm{\AA}$