How to prove $\mathrm{Tr}[|\alpha\rangle\langle\alpha|\hat{A}]=\langle\alpha|\hat{A}|\alpha\rangle$ For a coherent state $$|\alpha\rangle=e^{-\frac{|\alpha|^2}{2}}\sum_n\frac{\alpha^n}{\sqrt{n!}}|n\rangle$$ please show me how to prove,
$$
\mathrm{Tr}\left[|\alpha\rangle\langle\alpha|\hat{A}\right]=\langle\alpha|\hat{A}|\alpha\rangle,
$$
where $\hat{A}$ is a quantum mechanics operator.
 A: Another way to see this is to observe that any state $|\psi⟩\in\mathcal H$ can be extended to an orthonormal basis of the Hilbert space, and in that basis the trace $\operatorname{Tr}\left(|\psi⟩⟨\psi|\hat A\right)$ is exactly $⟨\psi|\hat A|\psi⟩$.
More explicitly, for any $|\psi⟩\in\mathcal H$ there exists a sequence $\renewcommand{\phi}{\varphi}\left\{|\phi_n⟩\right\}_n$ such that $⟨\phi_n|\phi_m⟩=\delta_{nm}$, $⟨\phi_n|\psi⟩=0$, and
$$|\psi⟩⟨\psi|+\sum_n|\phi_n⟩⟨\phi_n|=1.$$
In this basis, then,
\begin{align}
\operatorname{Tr}\left(|\psi⟩⟨\psi|\hat A\right)
=⟨\psi|\psi⟩⟨\psi|\hat A|\psi⟩+\sum_n⟨\phi_n|\psi⟩⟨\psi|\hat A|\phi_n⟩
=⟨\psi|\hat A|\psi⟩.
\end{align}
For a coherent state $|\psi⟩=|\alpha⟩$, this can be made even more explicit by setting the basis as a displaced number state basis sitting on top of the coherent state, i.e. $|\phi_n⟩=\hat D(\alpha)|n⟩$ for $n=1,2,3,\ldots$ and $|n⟩$ a number state.
A: Let $|n'\rangle$ be a basis of the Hilbert space, then
$$
\textrm{tr}\Big[|\alpha\rangle\langle\alpha|A\Big]=\sum_{n'}\langle n'|\alpha\rangle\langle\alpha|A|n'\rangle=\sum_{n'}\langle\alpha|A|n'\rangle\langle n'|\alpha\rangle = \langle\alpha|A\left(\sum_{n'}|n'\rangle \langle n'|\right)|\alpha\rangle=\langle\alpha|A|\alpha\rangle
$$ 
