Ground state of electrons in diatomic molecule and probability to find electron near a shell at the ground state

Question

Question:

The effective Hamiltonian of an outer-shell electron in a diatomic molecule is given by

$$H = \left(\begin{array}{cc} E_1 & t \\ t & E_2 \\ \end{array} \right)$$ where $E_1$ and $E_2$ are the energies of the orbital states near atom $1$ and $2$, and $t$ is the overlap matrix element between these orbitals.

Determine the ground state energy of the electron, given that $E_2$ = $4E_1$, $t = 2E_1$ and $E_1 > 0$. For the ground state, calculate the probability to find the electron near atom $1$.

Attempt at solution:

Eigenvalues $\lambda$ are given by the characteristic equation: $$(E_1 - \lambda)(E_2 - \lambda) - t^2 = 0$$ which becomes after rearranging and substituting expression for $E_2$ and $t$: $\lambda^2 = 5\lambda E_1$.

This suggests two eigenvalues: $\lambda_1 = 0$ and $\lambda_2 = 5E_1$; so I guessed that the ground state energy is $0$.

I was just wondering how to calculate the probability, I'm a bit lost on that part.

Answer 1

You need to find the eigenvectors for $H$. The one for the excited state comes out as $$ \vert \psi_{5}\rangle =\frac{1}{\sqrt{5}}\left(\begin{array}{c} 1 \\ 2 \end{array}\right)\, . $$ You can work out $\vert \psi_0\rangle$ by orthogonality to $\vert\psi_5\rangle$, i.e by using $\langle\psi_0\vert\psi_5\rangle=0$. The game is then to expand the basis vector $$ \vert 1\rangle=\left(\begin{array}{c}1 \\ 0\end{array}\right)=\alpha \vert \psi_0\rangle +\beta \vert \psi_{5}\rangle\, . $$ The probability of finding the electron in the ground state near atom $1$ is then $\vert\langle 1\vert \psi_0\rangle\vert^2= \vert\alpha\vert^2$.

(sorry I didn't realize you had $t=2E_1$ which lead to an erroneous initial comment.)