From Ticcati's textbook, he asks to show that from the axioms of position operator we get that:
$$
\text{e}^{-ia\cdot P} |x\rangle = |x+a\rangle
$$

where the axioms are:
$$
X=X^{\dagger}
$$

If $\Delta_a$ is a space translation, then $U(\Delta_a)^{\dagger} X U(\Delta_a) = X + a$, where $U$ is the representation of a unitary operator, we know that $\text{e}^{ia\cdot P} X \text{e}^{-ia\cdot P} = X + a$.

If $R$ is a space rotation then $U(R)^{\dagger} X U(R) = RX$.

Here's what I tried so far to do with this:

$$
\text{e}^{ia\cdot P} X \text{e}^{-ia\cdot P}|x+a\rangle =(X+a) |x+a\rangle = (x+a) |x+a\rangle
$$

$$
X|y\rangle:=X \text{e}^{-ia\cdot P}|x+a\rangle = (x+a)\text{e}^{-ia\cdot P}|x+a\rangle=(x+a) |y\rangle
$$

Now I want to show somehow that $|y\rangle=|x+2a\rangle$, but I don't see how, any hints?

Thanks in advance.

This post has been migrated from (A51.SE)