As a counterpart to the quantum mechanical translation operator (see for example this post) is there a unitary operator which describes the stretching of a line. That is consider I have a chain of particles on a line, spaced at equal distances, $d$, from one another. The chain is assumed to be 1D and symmetric with respect to the origin.

I want to describe the transformation of this chain into another such that the equal separation distance is instead $2d$. This will correspond to a stretching of the whole line about the origin. Can a unitary, similar to that used to describe the translation of individial particles, be used to describe this?

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    $\begingroup$ It's called the "dilation" operator. $\endgroup$ – ACuriousMind Feb 29 '16 at 17:59
  • $\begingroup$ @ACuriousMind, Thank you, do you know of an example of its use in a manner similar to the post shown in the question? $\endgroup$ – Sid Feb 29 '16 at 18:05

By way of exemplifying @ACuriousMind 's succinct comment, first recall Lagrange's translation operator, $$ e^{b \frac{\partial}{\partial x}} f(x)= f(x+b). $$

Changes of variable produce arbitrary advective flows. For instance, for your dilation, $$ y\equiv e^x, \qquad \Longrightarrow \qquad x=\ln y . $$ Defining $g(y)=g(e^x)\equiv f(x)$, evaluate $$ e^{by \frac{\partial}{\partial y}} g(y)= g(e^{b+x})= g( e^b ~ y). $$ You have stretched y by a factor of $e^b$, and in your case you wanted doubling, so $b=\ln 2$. In terms of QM operators, the dilation operator is $\exp(i\frac{b}{\hbar}\hat{y}\hat{p}_y )$, a rotation in phase space.

You may well have used this in quantum optics without taking stock of it, since for the QSH oscillator $$ [a,a^\dagger ]=1 $$ combinatorially isomorphic to $$ [\frac{\partial}{\partial y}, y]=1, $$
so, then $$ e^{it\omega~ a^\dagger a} g(a^\dagger) = g(e^{it\omega} ~ a^\dagger). $$


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