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.


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.)

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.