Translation operator properties

Question

Can somebody can give me a hand with these properties demonstrations, I am missing something.

Given the translation operator $T(dx)=1-ikdx$, where: $|\alpha'\rangle=T(dx)\vert\alpha\rangle$,

1) show that:

$\langle\alpha'\vert\alpha'\rangle=\langle\alpha\vert\alpha\rangle$, I have that:

$\langle\alpha'\vert\alpha'\rangle=\langle\alpha\vert T^\dagger(dx)T(dx)\vert \alpha\rangle=\langle\alpha\vert (1+ikdx)(1-ikdx)\vert \alpha\rangle$

$=\langle\alpha\vert (1+k^2dx^2)\vert \alpha\rangle=\langle\alpha\vert\alpha\rangle + k^2\langle\alpha\vert dx^2\vert\alpha\rangle$

The second term must be zero because it is a differential of second order?

2) Show that:

$T(dx)T(dx')=T(dx+dx')$, I have that:

$T(dx)T(dx')= (1-ikdx)(1-ikdx')=1-k^2dxdx'$,

How can I rewrite the terms in order to show the demonstration? I can not see the trick.

Answer 1

1. We are looking at an infinitesimal transformation, and hence we can discard any $dx^2$ terms.
2. Your expansion is wrong: $$T(dx)T(dx')=(1-ikdx)(1-ikdx')=1-ikdx-ikdx'-k^2dxdx'$$ From here, neglect the $dxdx'$ term, and you should see what you want.

However, if your expansion was correct, all you'd have to do is to note: $$T(dx)T(dx')=1-k^2dxdx'=1$$ And this is independent of $dx$ and $dx'$ (Which is a sign that this expression is wrong), and thus the order can be switched freely.

Answer 2

The “full” translation operator is $T(x_0)=e^{-ikx_0}$. The expression you give is for an infinitesimal translation : $$ T(dx)=e^{-ik~dx}\approx 1-i k~dx $$ where the series has been truncated after the term linear in $dx$. Thus, in any of your manipulations, you should throw out terms that are $(dx)^2$ or higher. Your manipulations and the results you are asked to find are consistent with this.