# Inner product of wavefunctions

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 $$\psi_a$$?

• 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 $\sum_a \left| \left\langle \psi_a \right| H \left| \psi_b \right\rangle \right|^2 = \left| \left\langle \psi_b \right| H^2 \left| \psi_b \right\rangle \right|^2 \neq 1$. Commented May 5, 2020 at 14:28
• So, at the end, $<\psi_a|H|\psi_b>$ doesn't have a physical meaning? Commented May 5, 2020 at 16:43

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.
• 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 $\psi_a$? Commented May 3, 2020 at 17:01
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.
• Can we talk about $<\psi_a|H|\psi_b>$ in terms of probability from the state $\psi_b$ to the state $\psi_a$? Commented May 3, 2020 at 14:14