Quantum canonical transformation

Question

This post is very similar in content to this one. I'm looking for a quantum implementation of the transformations

$$ x_i \to x_i + f(p) p_i, $$

$$ p_i \to h(p) p_i. $$

In these, the subindex $i$ denotes components of $\mathbf{x}$ and $\mathbf{p}$; and $f(p)$ and $h(p)$ are scalar functions of $p\equiv|\mathbf{p}|$. So I'm looking for the operator $T$ implementing

$$ T x_i T^{-1} = x_i + f(p) p_i, $$

$$ T p_i T^{-1} = h(p) p_i. $$

I tried guessing an expression for $T$ inspired by the accepted answer to the post mentioned at the beginnig, but I couldn't work out the calculations to verify if I guessed it right. When trying to do so, I used the relation

$$ [A, e^B] = \int_0^1 ds e^{(1-s)B}[A,B]e^{sB} $$

discussed here, but got trapped in seemly never-ending nested, multiple integrals.

ADDENDUM

As pointed out in the comments, the transformations above are (classicaly) canonical only for the trivial case of $h=1$. To make my question more pertinent, I rephrase it removing the required transformation for $x_i$. Therefore, I ask for the quantum implementation of the transformation $p_i \to h(p) p_i$, letting the transformation of $x_i$ be determined after requiring canonicity.

Have you first tried the trivial case of one-dimension? What have you found for h(p)? — Cosmas Zachos, Jan 21 '22 at 17:04
I tried $T = \text{exp}(ih \mathbf{p}\cdot\mathbf{x})$, both for one- and three-dimensions. I end up with the same problem with multiple integrals. Considering the commutator formula above, I believe the trick is to find $B$ such that $[A,B]$ commutes with $\text{exp}(B)$, but had no luck so far. Actually, I can generate the transformation for $x_i$ setting $T=\text{exp}(iF(p))$ for some adequate $F(p)$, but it keeps $p_i$ unchanged nevertheless. — andrehgomes, Jan 21 '22 at 17:27
It's an xy problem. *Have you confirmed* for what h(p) your transformation is classically canonical, i.e. it preserves Poisson Brackets? — Cosmas Zachos, Jan 21 '22 at 17:29
That would be any $h(p)$ satisfying $d(ph)/dp=1$ along with $h\to1$ for $p\to0$. — andrehgomes, Jan 21 '22 at 17:45
Oh, sorry! (Lots of distractions here.) I get $h=1$, so my original question is ill-posed except for the trivial case indeed. Thanks for pointing out the way to see that! — andrehgomes, Jan 21 '22 at 18:11
I believe I can make my question more pertinent removing the required transformation for $x_i$. — andrehgomes, Jan 21 '22 at 18:18
General tip: Let's not have posts look like revision histories. — Qmechanic, Jan 23 '22 at 12:25

Cosmas Zachos · Accepted Answer · 2022-01-21T23:01:51.780

Hint: well, to break down and modularize your problem to several technical ones, if you stuck to one dimension and dismissed QM, so you worked in the p-representation where $\hat x$ is proportional to $d/dp$, and you sought the operator T which scales p by a function thereof, $$ T p T^{-1}= p ~ h(p), $$ then, simply consider Lagrange's celebrated operator, $e^{d/dk} f(k)e^{-d/dk}= f(k+1)$ : $$ \Large T=e^{\frac{1}{k'(p)}\frac{d}{dp}} = e^{ \frac{d}{dk(p)}} ~~~\leadsto \\ TpT^{-1}= k^{-1} \Bigl ( k(p) + 1 \Bigr )= p~ h(p), ~~~\implies \\ k(p) + 1= k(p~h(p)) ~, $$ a functional equation specifying k in terms of h.

For example, confirm that $h(p)=\sqrt{1+1/p^2}~\leadsto k(p)=p^2$, so $$ e^{ \frac{d}{d(p^2)}}~ p ~e^{ -\frac{d}{d(p^2)}}=e^{ \frac{d}{d(p^2)}}~\sqrt{ p^2 }~e^{ -\frac{d}{d(p^2)}}= \sqrt{1+p^2} =p ~h(p) . $$ (This illustration does not satisfy your h(0)=1 condition, of course, but I opted for a simple and direct one for intuitiveness, rather than a messier one that would satisfy it.)

Works wonderfully! Also, that's a very nice trick writing $p$ as $h^{-1}(h(p))$. — andrehgomes, Jan 23 '22 at 11:46

Quantum canonical transformation

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