Operator acting on bras I need some help. Suppose, $\hat{\textbf{A}}$ and $\hat{\textbf{B}}$ are operators and $|\psi\rangle$ is any state, so that
$$ \hat{\textbf{A}}|\psi\rangle=a|\psi\rangle. $$
And I wonder if this operation is correct
$$ \langle \psi| \hat{\textbf{A}}^\dagger = \langle \psi|a^*.$$
And, can I get into bra with my operator like
$$ \hat{\textbf{A}} \langle \psi|  = \langle \psi|a^*.$$
or
$$  \langle \hat{\textbf{A}}\psi|  = \langle \psi|a^*.$$
Can somebody explain? I'm so confused about acting operators on bra...
 A: First of all, the phrasing of the question might be a bit misleading, since you are obviously talking not about any state, but about any eigenstate of $\mathbf{\hat{A}}$, since it satisfies the property
\begin{equation}
\mathbf{\hat{A}}|\psi\rangle = a|\psi\rangle.
\end{equation}
Note, btw, that $\mathbf{\hat{A}}$ is not necessarily Hermitian (otherwise $a$ would be real, and it woudln't make much sense to talk about $a^*$.)
Let us note that the following are equivalent notations (they mean the same thing):
\begin{equation}
\langle\varphi|\mathbf{\hat{A}}|\psi\rangle = \langle\varphi|\mathbf{\hat{A}}\psi\rangle.
\end{equation}
The complex conjugation in the bra-ket notation is expressed as:
\begin{equation}
\langle\varphi|\psi\rangle^* = \langle\psi|\varphi\rangle.
\end{equation}
Now, for any states $|\varphi\rangle$ and $|\psi\rangle$ we have
\begin{equation}
\langle\varphi|\mathbf{\hat{A}}|\psi\rangle^* = 
\langle\varphi|\mathbf{\hat{A}}\psi\rangle^* =
\langle \mathbf{\hat{A}}\psi|\varphi\rangle = 
\langle\psi|\mathbf{\hat{A}}^\dagger\varphi\rangle =
\langle\psi|\mathbf{\hat{A}}^\dagger|\varphi\rangle.
\end{equation}
The equality $\langle \mathbf{\hat{A}}\psi|\varphi\rangle = 
\langle\psi|\mathbf{\hat{A}}^\dagger\varphi\rangle$ is actually the definition of the Hermitian adjoint for operator $\mathbf{\hat{A}}$.
Now, if $|\psi\rangle$ is an eigenstate of $\mathbf{\hat{A}}$, then for any $|\varphi\rangle$ we have 
\begin{equation}
\langle\psi|\mathbf{\hat{A}}^\dagger|\varphi\rangle =
\langle\varphi|\mathbf{\hat{A}}|\psi\rangle^* =
\left(\langle\varphi|\psi\rangle a\right)^* = 
a^*\langle\psi|\varphi\rangle.
\end{equation}
Since $|\varphi\rangle$ is an arbitrary state, this means that 
\begin{equation}
\langle\psi|\mathbf{\hat{A}}^\dagger =
a^*\langle\psi| = \langle\psi|a^*.
\end{equation}
The remaining point in your question is the expression
\begin{equation}
\mathbf{\hat{A}}\langle\psi| = \langle\psi|a^*.
\end{equation}
This is incorrect notation: the opertaor always acts from the side of $|$, not from the side of the angular bracket.
