# How to find probability in different basis representations?

The matrix representation of the Hamiltonian is given in the position basis: $\{|+a\rangle,|-a\rangle\}$. $$H = \begin{pmatrix} E_{0} & E_{-A} \\ E_{-A} & E_{0} \end{pmatrix}$$

The initial state of a system is also given in terms of the position basis:

$\mid \psi \rangle = \frac{1}{5}[3|+a\rangle + 4|-a\rangle]$.

To find the probability of the a particular energy eigenvalue, why is it incorrect to choose the coefficient of $|-a\rangle$ so that it would be $\left(\frac{4}{5}\right)^2$?

Instead, we have to transform to the Hamiltonian basis, using the eigenvectors of the Hamiltonian. Why can't we use the position basis?

Now, in you case, the probability $|4/5|^2$ corresponds to the following: $$P(\text{-}a) = |\langle \text{-}a|\psi\rangle|^2.$$ This is the probability of projecting (by measurement) the state of the system into $|-a\rangle$. This corresponds to computing the probability of obtaining the eigenvalue $\text{-}a$ as a result of your measurement. If you want to know the probability of being in a give energy state, you have to compute the following probability $$P(E) = |\langle E|\psi\rangle|^2.$$ where $|E\rangle$ is the state with energy $E$ (eigenstate of the Hamiltonian). Therefore, in order to compute the braket $\langle E|\psi\rangle$ you need to write $|\psi\rangle$ in the basis of Hamiltonian's eigenstates: in this basis, the solution is trivial.