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What does $\langle\phi|Q|\psi\rangle $ actually mean, where $|\psi\rangle$ and $|\phi\rangle$ are states and $Q$ is a linear operator?

I know that $\langle\psi|Q|\psi\rangle$ is the expected value of $Q$ in a given state and that $\langle\phi|\psi\rangle $ is the probability it is measured in state $|\phi\rangle$ given that it was in state $|\psi\rangle$.

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  • $\begingroup$ Is $Q$ assumed to be selfadjoint? $\endgroup$ – Qmechanic Jul 3 '17 at 19:03
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There are two important operators:

  1. Hermitian operator: these are the observables of the theory
  2. Unitary operators: these are transformations that we can make in the state

For a given state $|\psi\rangle$ we can extract the information of the expectation value of the observable $O$ by computing $\langle \psi|O|\psi\rangle$. Also, there is an hermitian operator $|\xi\rangle\langle\xi|$ (convince yourself that this is an hermitian operator) that is the following observable: "what is the probability of the $|\psi\rangle$ to behave in the same way as the state $|\xi\rangle$?" if we take the expectation value:

$$ \langle\psi|\xi\rangle\langle\xi|\psi\rangle=|\langle\xi|\psi\rangle|^2 $$

Now, the unitary operators $U$ describes a transformation of the state $|\psi\rangle\rightarrow U|\psi\rangle$, preserving the fact that probabilities add to one, i.e. preserving $\langle\psi|\psi\rangle=1$. They are related with the preparation of the system and time evolution. If you compute $\langle\phi|U|\psi\rangle$ you are going to obtain the amplitude of probability of the state $\psi$ behave as $\phi$ after the the $U$-transformation. This provides an interpretation for matrix elements of unitary operators.

Now if we allow the unitary operator to depend on a parameter $U(\theta)$ such that $U(0)=1$, we have that:

$$ U(\theta)=e^{i\theta G} $$

where $G$ is an hermitian operator. This means that $G$ is an observable of the theory, and for small $\theta$ we have:

$$ \langle\phi|U(\theta)|\psi\rangle \sim \langle\phi|\psi\rangle + i\theta\langle\phi|G|\psi\rangle $$

and this provide an interpretation for matrix elements of hermitian operators. In almost all situations $\theta$ is the time $t$ and $G$ is some interacting term of the hamiltonian $\Delta H$ and we are interesting in the rate of

$$ (\langle\phi|U(t)|\psi\rangle - \langle\phi|\psi\rangle)\sim it\langle\phi|\Delta H|\psi\rangle $$

if $\psi$ and $\phi$ are orthgonal we have just:

$$ \langle\phi|U(t)|\psi\rangle \sim it\langle\phi|\Delta H|\psi\rangle $$

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Suppose we were working with more familiar vectors and matrices. I might, for example, work in a basis for which $\left|\phi\right\rangle$ is like $\mathbf{i}$ and $\left|\psi\right\rangle$ like $\mathbf{j}$. Then $\left\langle\phi\right|\hat{Q}\left|\psi\right\rangle=\mathbf{i}^\dagger\hat{Q}\mathbf{j}=\mathbf{i}\cdot\hat{Q}\mathbf{j}=Q_{12}$. In other words, we're just computing one matrix element, and in general an off-diagonal one. In the above example the vectors are orthogonal; more generally, we have a linear combination of matrix elements. Either way, you can think of the product another way by noting $\mathbf{i}\cdot\left(\hat{Q}\mathbf{j}\right)=\left(\hat{Q}^\dagger\mathbf{i}\right)\cdot\mathbf{j}$.

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    $\begingroup$ I understand it mathematically however I am not sure of the meaning. $\endgroup$ – Toby Peterken Jul 3 '17 at 15:51
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You say:

$\langle\psi |Q|\psi\rangle$ is the expected value of Q in a given state

Small correction: I believe expectation value is a better term, or average value. "Expected" might imply that you will actually measure this value. If $\psi$ is not an eigenstate of $Q$, $\langle \psi|Q|\psi\rangle$ will probably not be an eigenvalue, but all your measurement values will be eigenvalues.

$\langle\phi |\psi\rangle$ is the probability it is measured in state $|\phi\rangle$ given that it was in state $|\psi\rangle$

No, $\langle\phi |\psi\rangle$ is the probability amplitude that tells how $|\psi\rangle$ is linearly decomposed into $|\phi\rangle$. The probability is the absolute square of the amplitude, $|\langle\phi |\psi\rangle|^2$.

See Feynmann's lecture on probability amplitudes and the Hamiltonian matrix. In fact, all of Volume 3 would be useful for you (and everyone else!)

Finally

What does ⟨ϕ|Q|ψ⟩ actually mean, where |ψ⟩ and |ϕ⟩ are states and Q is a linear operator?

It's probability amplitude for the measurement of Q on the state $|\psi\rangle$ and whether (or how much) the resultant state overlaps $|\phi\rangle$.

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