Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
1  
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
2  
You should post this as an answer so it can be accepted. –  Aaron Dec 9 '11 at 0:51
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.