Expectation value of quantum annihilation operator I already know the expectation value of a quantum annihilation operator $\hat{a} $. I want to find the expectation value of quantum creation operator $\hat{a}^\dagger$. Is it equal to following?
$$\langle\hat{a}^\dagger\rangle = \langle\hat{a}\rangle^* $$
$\langle\hat{a}\rangle^*$ is complex conjugate of the expectation value of $\hat{a}$.
 A: It is helpful here to remember that the expectation value $\langle \hat{a} \rangle$ is taken with respect to a quantum state $|\psi \rangle$ (or a density operator, but let's keep it simple). In this case,
$$
\langle \hat{a} \rangle = \langle \psi | \hat{a} | \psi \rangle
$$Recall now that $\langle \phi | \hat{O} | \psi\rangle^* = \langle \psi | \hat{O}^\dagger|\phi\rangle$. An intuitive picture for this relation comes from the matrix element interpretation of these objects - ie., a linear operator $\hat{O}$ can be expressed as an $N \times N$ matrix where $N$ is the dimension of the Hilbert space. In this case, the 'dagger' operator is the conjugate transpose, so the '$\psi$ by $\phi$'-th element of $\hat{O}^\dagger$ is the complex conjugate of the '$\phi$ by $\psi$'-th element of $\hat{O}$.
Given this relation,
$$
\langle \hat{a} \rangle^* = \langle \psi|\hat{a}|\psi\rangle ^* = \langle \psi | \hat{a}^\dagger |\psi\rangle = \langle \hat{a}^\dagger \rangle.
$$
So indeed the expectation values of $\hat{a}$ and $\hat{a}^\dagger$ are complex conjugates.
