1
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ 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
2
$\begingroup$

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 $$ e^{it\omega~ a^\dagger a} g(a^\dagger) = g(e^{it\omega} ~ a^\dagger). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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