Translation operator and position operator I am a little bit confused about the translation and position operator and hope for some clarification.
Let $\hat{x}$ be the position operator, which satisfies $\hat{x} \vert x \rangle = x \vert x \rangle $ and $\hat{T}_{a}$ the translation operator, which satisfies $\hat{T}_{a} \vert x \rangle = \vert x+a \rangle $.
With that being said, it is known, that $[\hat{x},\hat{T}_a] \neq 0$, since $\hat{T}_a \hat{x} \vert x \rangle = x \vert x+a \rangle$, whereas $\hat{x} \hat{T}_a \vert x \rangle = (x+a) \vert x+a \rangle$.
With that being said I am confused about the same operators in the position representation. Let $\psi(x)$ be any wavefunction, that solves the Schrödinger equation.
To reproduce the same results in the position representation

$\hat{T}_a \hat{x} \psi(x) = \hat{T}_a x \psi(x) = x \hat{T}_a \psi(x) = x\psi(x+a)$

and

$\hat{x} \hat{T}_a \psi(x) = \hat{x} \psi(x+a) = (x+a) \psi(x+a)$

must be true.
However I previously thought, that the postion operator is always "just x" in the position representation. In this case, we would get

$\hat{T}_a \hat{x} \psi(x) = \hat{T}_a x \psi(x) =  (x+a) \psi(x+a)$

and

$\hat{x} \hat{T}_a \psi(x) = x \psi(x+a) = x \psi(x+a)$

which does not fit to the generalized statements mentioned at the beginning.
To conclude everything: which - if any - result is the correct one and how exactly is the position operator defined for the position representation?
 A: First, a quick clarification.  If you define the action of $T_a$ on the position ket $|x\rangle$ as $T_a|x\rangle = |x+a\rangle$, then one would have that
$$T_a|\psi\rangle = \int \mathrm dx \ \psi(x) T_a |x\rangle = \int \mathrm dx \ \psi(x) |x+a\rangle = \int \mathrm dx \ \psi(x-a) |x\rangle$$
Similarly,
$$T_a X|\psi\rangle = \int \mathrm dx \ \psi(x) T_a X |x\rangle = \int \mathrm dx\ \psi(x) x T_a |x\rangle = \int \mathrm dx \ \psi(x) x |x+a\rangle$$
$$=\int \mathrm dx \ (x-a) \psi(x-a)|x\rangle$$
whereas
$$X T_a\psi\rangle = \int \mathrm dx \ \psi(x-a) X|x\rangle = \int \mathrm dx \ x \psi(x-a) |x\rangle$$
Avoiding bra-ket notation and defining the action of these operators directly at the level of the wavefunction would yield
$$\big[(T_a X)\psi\big](x) = (x-a) \psi(x-a) \qquad \big[(XT_a)\psi\big](x) = x \psi(x-a)$$
The first equality could be understood by letting $X\psi \equiv \phi$, so $\phi(x) = x \psi(x)$.  From there, application of $T_a$ would yield $\big(T_a \phi\big)(x)=\phi(x-a) = (x-a)\psi(x-a)$.
Altogether, I think the important thing is to remember that $\psi$ is a function while $\psi(x)$ is a number, and operators act on functions (in the position representation). Case in point,

$$\hat{T}_a \hat{x} \psi(x) = \hat{T}_a x \psi(x) = x \hat{T}_a \psi(x) = x\psi(x+a)$$

is incorrect. It should be understood as acting on $\psi$ with $\hat x$, then acting on the result with $T_a$, and only then evaluating the result at $x$.  In other words, as $\big(\hat T_a \hat x \psi\big)(x)$.  This makes it much easier to understand that $T_a$ shifts the argument of $\hat x \psi$:
$$\big(\hat T_a \hat x \psi\big)(x) = \big(\hat x \psi\big)(x-a) = (x-a)\psi(x-a)$$
because $\big(\hat x\psi\big)(u) = u \psi(u)$.
A: In the position representation, we need to (or can) evaluate the action of $T_a$ on the bra $\langle x|$. To this end, note that
$$\langle x|T_a = \int \mathrm dy \, \langle x|T_a|y\rangle \langle y| = \int \mathrm dy \, \langle x|y+a\rangle \langle y|  = \langle x-a| \quad ,$$
where we've used the completeness relation $\displaystyle \int \mathrm d x\, |x\rangle \langle x| = \mathbb I$ and $\langle x|y\rangle = \delta(x-y)$.
Now let us define
$$ \psi (x) \equiv \langle x|\psi\rangle $$
and
$$(O \psi)\,(x) \equiv \langle x|O|\psi\rangle$$
for some operator $O$ and state $|\psi\rangle$. Then
$$(T_ax \psi)\, (x) = \langle x|T_a x|\psi\rangle = \langle x-a |x|\psi\rangle = (x-a) \langle x-a|\psi\rangle  =(x-a)\, \psi(x-a) \quad , $$
while
$$(xT_a\psi)\,(x) = \langle x| x T_a|\psi\rangle =x \langle x|T_a|\psi\rangle = x \langle x-a|\psi\rangle = x\, \psi(x-a)\quad . $$
Eventually, this shows that
$$ \langle x| [T_a,x] |\psi\rangle  =(x-a)\psi(x-a) - x \psi(x-a) = -a \psi(x-a) = -a \langle x-a|\psi\rangle = -a \langle x|T_a|\psi\rangle \quad . $$
Since this should hold for all $|\psi\rangle$ (in a specified domain) and $\langle x|$, we conclude
$$ [T_a,x] = -a \,T_a  \quad .$$
