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 $t$

 $\lambda^2 = 5\lambda E_1$

which 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.

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.