Is the density matrix corresponding to a state $|\alpha\rangle$ simply $\rho =| \alpha \rangle \langle \alpha \mid$? The eigenstate of the annihilation operator $a$ is given by the state $a\mid \alpha \rangle = \alpha \mid \alpha \rangle$. In the Fock state basis, we can expand this state as
$$\mid \alpha \rangle = e^{-\frac{1}{2}\vert \alpha\vert^2}\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}\mid n \rangle$$
I am bit confused about density matrices, I am not sure if $\rho = \mid \alpha \rangle \langle \alpha \mid$ in this case, as I do not know if $\mid \alpha \rangle$ is a pure state (I am still working on understanding these definitions).
Is the density matrix corresponding to the state $\alpha$ simply $\rho = \mid \alpha \rangle \langle \alpha \mid$? If so, is this true in general?
If I am trying to calculate $Tr(\rho \hat{O})$ for some observable $\hat{O}$, then would it be valid (by the cyclic property of trace) to say that $$Tr(\rho \hat{O}) = Tr(\langle \alpha \mid \hat{O} \mid \alpha \rangle) = \langle \alpha \mid \hat{O} \mid \alpha \rangle~?$$
 A: The density matrix is indeed $|\alpha\rangle\langle\alpha|$. Remember that $\alpha \in \mathbb{C}$ so when you form $\langle\alpha|$ you must conjugate $\alpha$ (I always forget).
So, $\rho_\alpha := |\alpha\rangle\langle\alpha| = e^{-\left|\alpha\right|^2} \sum_{n, m}^\infty \frac{\alpha^n \alpha^{*m}}{\sqrt{n! m!}} |n\rangle\langle m |$.
For a pure state $|\phi\rangle$ the density matrix is always formed as $|\phi\rangle\langle\phi|$.
$|\alpha\rangle$ is a pure state. Indeed, all bra/ket vectors are pure, and this is a key motivation for including the density matrix in our mathematical toolbox: it allows us to use the same notation for both pure and mixed states.
Yes it is fine to use the cyclical property of the trace in this way. In my undergrad courses, $Tr\left(\rho\hat{O}\right)$ was referred to as the "trace", while $\langle\alpha|\hat{O}|\alpha\rangle$ was referred to as the "overlap" (in some contexts the "overlap integral"). It took me a while to work out that these are doing the same thing.
