quantum mechanics - Position Operator Eigenvectors change with Space Displacement

I am working through Ch. 3 of Ballentine where he finds the commutator relationships between various operators.

He begins on p.78 with a space displacement

$$\mathbf{x'} = \mathbf{x} + \mathbf{a}$$

Which involves a corresponding displacement of position eigenvectors

$$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}|\mathbf{x}\rangle = |\mathbf{x} + \mathbf{a}\rangle .$$

However, I believe it should be instead

$$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}|\mathbf{x}\rangle = |\mathbf{x} - \mathbf{a}\rangle .$$

Because, he mentions that we are taking the "active" point of view, where we keep the coordinates the same but instead shift our vectors and operators. In this case, $|\mathbf{x'}\rangle$ is shifted $+\mathbf{a}$ which in turns means that it is equal to unprimed position eigenvector $-\mathbf{a}$.

Answer

After going through the text again, I wanted to give an answer more motivated by the exposition in the text.

From, p.63, since the laws of nature are invariant under certain space-time transformations, if $A | \phi_n \rangle = a_n | \phi_n$ with $A$ representing an observable, and $\phi_n$ an eigenvector, then after the transformation $A' | \phi_n' \rangle = a_n | \phi_n' \rangle$ because they represent the same observable.

And from p.64 and p.65, given a space-time transformation $U$

$$|\phi_n' \rangle = U|\phi_n \rangle$$

$$A' = UAU^{-1}$$

Now, for a space displacement of the position operator we have

$$|\mathbf{x'}\rangle =e^{-i\mathbf{a}\cdot \mathbf{P}}|\mathbf{x}\rangle = |\mathbf{x}+\mathbf{a}\rangle$$ $$\mathbf{Q'} = e^{-i\mathbf{a}\cdot \mathbf{P}}\mathbf{Q}e^{i\mathbf{a}\cdot \mathbf{P}}$$ And for a single direction of Q, $$Q_\alpha'|\mathbf{x'}\rangle = x_\alpha |\mathbf{x'}\rangle$$

Which suggests,

$$\mathbf{Q'} = \mathbf{Q} - \mathbf{a}I$$ because for a single direction $$Q_\alpha'|\mathbf{x'}\rangle = x_\alpha |\mathbf{x'}\rangle = (Q_\alpha -a_\alpha I)|\mathbf{x}+\mathbf{a}\rangle$$ $$= Q_\alpha |\mathbf{x}+\mathbf{a}\rangle - a_\alpha |\mathbf{x}+\mathbf{a}\rangle$$ $$= x_\alpha + a_\alpha |\mathbf{x}+\mathbf{a}\rangle - a_\alpha |\mathbf{x}+\mathbf{a}\rangle$$ $$= x_\alpha|\mathbf{x}+\mathbf{a}\rangle$$

So, thinking back to the active/passive view of transformations where I was confused, what is happening is that we are shifting the position operator $Q$ forward by $\mathbf{a}$ which corresponds to an active point of view.

This results in the position eigenvectors for a specific eigenvector to be shifted forward or in other words for the eigenvalues of a given eigenvector to be shifted backwards! All very confusing, but I believe this clarifies my original question more in the spirit of the text exposition.