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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.

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Just act by $X$ on $e^{-iaP}|x\rangle$: $Xe^{-iaP}|x\rangle=e^{-iaP}(X+a)|x\rangle=e^{-iaP}(x+a)|x\rangle=(x+a)e^{-iaP}|‌​x\rangle$. I.e. $e^{-iaP}|x\rangle$ is an eigenvector with eigenvalue $x+a$. – Pavel Safronov Dec 6 '11 at 18:29
Ah, that's what I did, I wasn't sure it was right cause I thought to myself that this eigenvalue could belong to another eigenvector, but now I see that it doesn't matter as long as it belongs to $|x+a>$. Foolish me... – MathematicalPhysicist Dec 7 '11 at 8:30
You should post this as an answer so it can be accepted. – Aaron Dec 9 '11 at 0:51
up vote 5 down vote accepted

As requested, reposting the comment.

Assuming one knows what an eigenvector for a continuous spectrum is (I don't), the solution is straightforward. Act by $X$ on $e^{-iaP}|x\rangle$: $$Xe^{-iaP}|x\rangle=e^{-iaP}(X+a)|x\rangle=e^{-iaP}(x+a)|x\rangle=(x+a)e^{-iaP}|x\rangle,$$ i.e. $e^{-iaP}|x\rangle$ is an eigenvector of $X$ with eigenvalue $x+a$. One of the axioms you are missing should state that the spectrum of $X$ is simple, which then implies that $e^{-iaP}|x\rangle\sim|x+a\rangle$. Normalization of $|x\rangle$ gives an equality.

Perhaps, this can be made precise using the machinery of Gelfand triples which I am unfamiliar with.

However, stating the condition of the simplicity of the continuous spectrum does not require anything fancy. The spectral theorem (for unbounded self-adjoint operators) allows us to view $X$ as a multiplication operator by a function $f:M\rightarrow\mathbf{R}$ on $L^2(M,\mu)$. The requisite condition is that $f$ is injective. Note, that it is automatically surjective unless $M$ is empty.

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