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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 $\psi_a$?

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  • $\begingroup$ At least, $\left| \left\langle \psi_a \right| H \left| \psi_b \right\rangle \right|^2$ is not the "probability to find state $\psi_a$" given some intricate condition, as $\sum_a \left| \left\langle \psi_a \right| H \left| \psi_b \right\rangle \right|^2 = \left| \left\langle \psi_b \right| H^2 \left| \psi_b \right\rangle \right|^2 \neq 1$. $\endgroup$ May 5, 2020 at 14:28
  • $\begingroup$ So, at the end, $<\psi_a|H|\psi_b>$ doesn't have a physical meaning? $\endgroup$
    – Salmone
    May 5, 2020 at 16:43

2 Answers 2

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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.

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  • $\begingroup$ Can we say that $|<\psi_a|H|\psi_b>|^2$ is equal to the probability that the measurement of $H$ over the state $\psi_b$ gives the state $\psi_a$? $\endgroup$
    – Salmone
    May 3, 2020 at 17:01
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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.

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  • $\begingroup$ Can we talk about $<\psi_a|H|\psi_b>$ in terms of probability from the state $\psi_b$ to the state $\psi_a$? $\endgroup$
    – Salmone
    May 3, 2020 at 14:14

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