What does it mean for a ket to be multiplied by a vector, or how does a vector become an eigenvalue of an operator? I am reading Sakurai's Modern Quantum Mechanics, and I am so confused by the notations as I read the section Translation. The book throws these relations to me ($\mathscr T(\mathrm d \boldsymbol x')$ denotes the translation along $\mathrm d \boldsymbol x'$, i.e. $\vert \boldsymbol x' \rangle \mapsto \vert \boldsymbol x' + \mathrm d \boldsymbol x'\rangle$)
$$
\boldsymbol x \mathscr T(\mathrm d \boldsymbol x') \vert \boldsymbol x' \rangle 
= \boldsymbol x \vert \boldsymbol x' + \mathrm d \boldsymbol x'\rangle 
= (\boldsymbol x' + \mathrm d \boldsymbol x') \vert \boldsymbol x' + \mathrm d \boldsymbol x'\rangle
\,,
$$
and
$$
 \mathscr T(\mathrm d \boldsymbol x') \boldsymbol x \vert \boldsymbol x' \rangle 
= \boldsymbol x' \mathscr T(\mathrm d \boldsymbol x') \vert \boldsymbol x' \rangle 
= \boldsymbol x' \vert \boldsymbol x' + \mathrm d \boldsymbol x'\rangle
\,.
$$
I notice that the relation $\boldsymbol x \vert \boldsymbol x' \rangle = \boldsymbol x' \vert \boldsymbol x' \rangle$ is used. Here are the questions:

*

*What does the notation $\boldsymbol x' \vert \boldsymbol x' \rangle$ means? If $x'_i$ ($i = 1, 2, 3$) are simultaneous eigenvalues of operators $x_i$, then is $\boldsymbol x'$ a vector (whose components are these eigenvalues)?
How is it possible for a vector to become a factor of the ket $\vert \boldsymbol x' \rangle$?

The Hilbert space is defined over a field (often $\mathbb C$), and the eigenvalues of an operator $\boldsymbol x$ must be a number in the field. Please help me understand the notation.
 A: $\pmb x$ is defined as an operator. $\pmb x ' $ as a (real) vector $ | \pmb x' \rangle \equiv |x', y', z' \rangle$ as a ket (i.e. an element of a Hilbert space not projected on a basis).
The bold notation is to accentuate one is working on 3 degrees of freedom in space. $\pmb x$ is the operator embedding operators $x, y$ and $z$. With their respective eigenvalues $x | \pmb x' \rangle = x' | \pmb x' \rangle ,\quad$ $y | \pmb x' \rangle = y' | \pmb x' \rangle ,\quad$ $z | \pmb x' \rangle = z' | \pmb x' \rangle$
The starting point is the eigenvalue equation for the operator $\pmb x$
$$\pmb x | \pmb x' \rangle = \pmb x' | \pmb x' \rangle$$
Then he first applies the translation operator followed by position operator. I guess he defined somwhere early in the book the translation operator as
$$T(\pmb{dx}') |\pmb x'\rangle = |\pmb x' + \pmb{dx}'\rangle$$
and then again uses the eigenvalue equation (the one two equation above) but now on ket $| \pmb x' + \pmb{dx}' \rangle$. This is
$$ \pmb x | \pmb x' + \pmb{dx}' \rangle = \left(\pmb x' + \pmb{dx}' \right) | \pmb x' + \pmb{dx}' \rangle.$$
Same for the second equation. So basically position and translation operators do not commute
$$ \left[T (\pmb{dx}'), \pmb x \right] \neq 0, $$ where 0 defines the null operator.
have a nice reading
A: The notation
$$ \mathbf x |\mathbf x'\rangle = \mathbf x' |\mathbf x'\rangle $$
should be understood as
$$ \hat x_i |x'_1, x'_2, x'_3\rangle = x'_i |x'_1, x'_2, x'_3\rangle \qquad \text{(for all $i$).} $$
The vector "inside" the ket means that these vectors are labeled by the three degrees of freedom $x'_1$, $x'_2$ and $x'_3$. For briefness, this is written as $|\mathbf x'\rangle$.
Having vectors "multiply" the kets is a compact way of saying that the equation holds true for every entry of the vectors. The entries of the vector $\mathbf x$ are the operators $\hat x_i$ and the entries of $\mathbf x'$ are the eigenvalues $x'_i$ which label the eigenvectors.
