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im trying to understand the meaning of this inner product:
$⟨\psi_a|H|\psi_b⟩$.
$H$ can be a time-independent hamiltonian.
I know that $⟨\psi_a|H|\psi_a⟩$ is the expectation value, but I don't know the meaning of the first inner product.
Can we say that $|\langle\psi_a|H|\psi_b\rangle|^2$ is equal to the probability that the measurement of H over the state $\psi_b$ gives the state $\psi_a$?
As in any inner product you can take it as a projection of the state vector $|\psi_a\rangle$ in the direction of $H|\psi_b\rangle.$ The latter is just another state vector, for example $H|\psi_b \rangle = |\psi_b\prime\rangle,$ then $\langle \psi_a |\psi_b\prime\rangle$ is just a projection of one in the direction of the other. You can also take it as the Matrix element of the Hermitian operator $H_{ab},$ if you allow for $|\psi_i\rangle$ to be a basis of the State Space. In a more physical interpretation you could argue that you could obtain the probability of a state |$\psi_b\rangle$ transitioning to $|\psi_a\rangle$ in time if $$|\psi_b (x,t)\rangle = e^{tH/\hbar} |\psi_b(x,0)\rangle,$$ then the term $\langle \psi_a |\psi_b\prime\rangle$ would determine the probability amplitude of the transition from $b$ to $a.$ (Expand $|\psi_b(x,0)\rangle$ in a basis of eigenvectors of $H$ and the same for $|\psi_a(x,0)\rangle$ for this to work as a possible interpretation.
You should think of $\langle \psi_a |H|\psi_b\rangle$ as a "matrix element" of H (i.e. $\langle \psi_a |H|\psi_b\rangle = (H)_{ab}$), especially in the case that the indices $a,b$ run over an orthonormal basis of states.
While $\langle \psi_a |H|\psi_a\rangle$ is real, if $a \neq b$ the same cannot be said for $\langle\psi_a |H|\psi_b\rangle^* = \langle \psi_b |H|\psi_a\rangle$. However, in any calculation of a measurable quantity that involves these states, you'll end up taking the real or imaginary part of this mixed quantity.
Asking about the "meaning" is subjective; it depends on what the states $|\psi_a \rangle$ are meant to represent. If they are energy eigenstates, then $\langle \psi_a |H|\psi_b\rangle \propto \delta_{ab}$. Otherwise, it is difficult to say.
