# 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 $\hat{a}$.

• @Qmechanic I mistakenly put it as a Hermitian operator. But I really wanted to find the expectation value of the Hermitian conjugate operator $\hat{a}^\dagger$ given the expectation value of $\hat{a}$. – Coderzz Jun 29 '18 at 3:48

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.