Is there a second/many order form of the infinitessimal unitary operator in quantum mechanics?

We know that a unitarily transformed system must be invariant, i.e. $\langle\psi|\psi\rangle = (\langle\psi|\hat{U}^{\dagger}) (\hat{U}|\psi\rangle) = \langle\psi|\hat{U}^{\dagger} \hat{U}|\psi\rangle = \langle\psi|\psi\rangle \implies \hat{U}^{\dagger} \hat{U} = \hat{I}$

Assuming that $ \hat{T}(\delta x) = \hat{I} +i\hat{G}_T\delta x $ (which is often done), where $\hat{T}$ is our Unitary Operator and $\hat{G}_T$ is our Hermitian Generator, we have to enforce also that $ \hat{T}(\delta x)\hat{T}^{-1}(\delta x) =\hat{I}$. We can then say that $\hat{T}^{-1} = \hat{T}^\dagger \because \hat{U}^{\dagger} \hat{U} = \hat{I}$

Applying all of this leads to: $\hat{T}(\delta x)\hat{T}^{-1}(\delta x) = (\hat{I} +i\hat{G}_T\delta x)(\hat{I} -i\hat{G}_T^\dagger\delta x) = \hat{I}^2 + i\delta x(\hat{G}_T-\hat{G}_T) + (\delta x)^2\hat{G}_T^2 = \hat{I}+(\delta x)^2\hat{G}_T^2$.

Assuming that $(\delta x)^2 \approx 0 $, $\hat{T}(x) = \lim_{n\to \infty} (\hat{T}(\frac{x}{n}))^n = \lim_{n\to \infty}(\hat{I} +i\hat{G}_T\frac{x}{n})^n = e^{ix\hat{G}_T }$, which is the common way to write a finite unitary transformation. From this, I think one can show that the generator represents a conserved quantity.

For this to be coherent, we have to assume that the $(\delta x)^2$ term is negligible (i.e we are only taking this to first order). Why is this a mathematically justifiable statement?

If not, is there a second order or many order infintessimal unitary operator definition? Also, what would be the implied finite unitary operator from the second-order infinitessimal operator?

Or is it that the unitary operator is not defined from its infinitessimal, but from its finite exponential definition?

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    $\begingroup$ Is $$U(\delta x)=\exp(i\,G\,\delta x) = \sum_{n=0}^\infty \frac{(i\,G\,\delta x)^n}{n!}$$ what you're looking for? $\endgroup$ – Chiral Anomaly Sep 29 at 0:29
  • $\begingroup$ How can $\hat{T}(\delta x) = \sum_{n=0}^{\infty} \frac{(i\hat{G}_T \delta x)^n}{n!} and = I + i\hat{G}_T \delta x$ at the same time? $\endgroup$ – Aidan Birdi Sep 29 at 13:03
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    $\begingroup$ $I+i\,G\,\delta x$ is not unitary. It is only unitary to first order in $\delta x$. Working to first order in $\delta x$ is sufficient for some purposes, just like the approximation $f(x+\delta x)\approx f(x) + (df/dx)\delta x$ is sufficient for some purposes. You wrote "is it that the unitary operator is not defined from its infinitesimal, but from its finite exponential definition?" That is correct. $\endgroup$ – Chiral Anomaly Sep 29 at 13:40
  • $\begingroup$ Thank you Chiral Anomaly, I think I understand now. $\endgroup$ – Aidan Birdi Sep 29 at 16:00

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