Dealing with dirac notation with regards to different basis' So this should be a pretty simple question.
So we say that $\langle x | \psi \rangle = \psi(x)$. In other words $\psi(x)$ is the ket $|\psi\rangle$ expressed in terms of the $x$ basis. 
Now suppose I wanted to get $\psi(x+a)$. How would I act on $|\psi\rangle$ to get it? I think I could just do
$$ \langle x' | \psi \rangle = \psi(x+a)$$
Where $x' = x+a$. But I don't really know what this means in terms of how $\langle x' |$ and $\langle x |$ are related, or the formal justification for this.
Perhaps my trouble is I don't thoroughly understand what the $x$-basis is or what $\langle x | \psi \rangle$ is really doing.
 A: There is a unitary operator, called the spatial translation operator, that implements translations in precisely the way you want.  In fact, for any $a$, it is defined as
\begin{align}
  T_a = e^{-iaP/\hbar}
\end{align}
where $P$ is the momentum operator, and we are here using the operator exponential.  This operator translates position basis elements:
\begin{align}
  T_a|x\rangle = |x+a\rangle
\end{align}
I encourage you to attempt to prove this!  Moreover, it follows that
\begin{align}
  \langle x |T_{-a}|\psi\rangle = \langle x| T_a^\dagger|\psi\rangle = \langle x+a|\psi\rangle = \psi(x+a)
\end{align}
so that the state $T_{-a}|\psi\rangle$ is your desired, translated state.
A: I just wanted to add some precisions concerning a possible misunderstanding.
$"\psi"$ is to be understood as a (complex) vector, you could write $\vec \psi$  (a column vector) instead of the ket $|\psi\rangle$, and $(\vec {\psi^*})^t$  (a row vector)   instead of the bra $\langle \psi|$.
Above, $^t$ means transposition (of a matrix), that is inversion of rows and columns indices. 
Now, each vector may be decomposed under some basis, for instance one can take the basis $\vec e_x$ :
$\vec \psi = \sum\limits_x  \psi_x \vec e_x \tag{1}$
Here $\psi_x$ is the complex coordinate of the vector $\vec \psi$, on the basis, or "axis" $\vec e_x$
Now, rewriting $(1)$ with kets, one has : 
$|\psi\rangle = \sum\limits_x  \langle x|\psi\rangle  |x\rangle \tag{2}$
It is exactly the same equation as $(1)$, $\langle x|\psi\rangle$ is the coordinate of the vector/ket $|\psi\rangle $ in the vector/ket $|x\rangle$ basis. Here $|\psi\rangle$ and $|x\rangle$ are complex column vectors, and $\langle x|\psi\rangle$ is a complex quantity.
You may also write $(1)$ with bras : 
$\langle\psi| = \sum\limits_x  \langle \psi|x\rangle  \langle x| \tag{3}$.
Here $\langle\psi|$ and $\langle x|$ are complex row vectors, and you have $\langle \psi|x\rangle = \overline {\langle x|\psi\rangle}$
