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In the book of Sakurai & Napolitano, the Canonical Commutation relation is derived using the unitary translation operator:

I agree with the derivation of the book until i reach this step, and sakurai simply approximates the RHS' translation operator to identity (1.206):

$[\hat{x}, \hat{T}(\delta x)] = \delta x \hat{T}(\delta x) \approx \delta x \hat{I}$

I also understand the rest of the reasoning from there, but this approximation step does not sit right with me. As far as I get it, the approximation means setting the limit as $\delta x \rightarrow 0$ but then this should be taken over the whole equation and all displacements $\delta x$ should tend to zero right?

Any help would be much appreciated.

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2 Answers 2

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Looking back on this, while the derivation of Sakurai is not rigorous, I see it can be solved with a quick taylor expansion. Recalling:

$\hat{T}(\delta \vec{r})=e^{-i\hat{\vec{K}}\cdot\delta\vec{r}}$

so in this case where $\delta\vec{r}=\delta x$, we have $\hat{T}(\delta x) = \hat{I}-i\hat{K}_x\delta x + O(\delta x^2)$

hence yielding:

$[\hat{x}, \hat{T}(\delta x)] = [\hat{x}, \hat{I}-i\hat{K}_x\delta x + O(\delta x^2)] = -i\delta x[\hat{x}, \hat{K}_x] = \delta x \hat{T}(\delta x)$

The $\delta x$ cancel on both sides and we can now safely take a limiting process for $\delta x \rightarrow 0$ yielding:

$\lim_{\delta x \rightarrow 0} [\hat{x}, \hat{K}_x] =[\hat{x}, \hat{K}_x] = \lim_{\delta x \rightarrow 0} i(\hat{I}+ O(\delta x))= i\hat{I} .$

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  • $\begingroup$ Please see my comment to the other similar answer. $\endgroup$ Oct 29, 2023 at 9:51
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$[\hat{x}, \hat{T}(\delta x)] = \delta x \hat{T}(\delta x) \approx \delta x \hat{I}$

...this approximation step does not sit right with me. As far as I get it, the approximation means setting the limit as $\delta x \rightarrow 0$ but then this should be taken over the whole equation and all displacements $\delta x$ should tend to zero right?

Yes. If you prefer, the approximate expression can be written as an equality using "big-oh" notation: $$ [\hat x, \hat T(\delta x)] = \delta x \hat T(\delta x) = \delta x \hat I + O(\delta x^2) $$

Thus, in Sakurai's notation we have: $$ [\hat x, 1-i\hat K\delta x+O(\delta x^2)] = \delta x \hat I + O(\delta x^2) $$ and, using the linearity of the commutator in the last arguement, and re-arranging the $O(\delta x^2)$ terms, we can write: $$ -i\delta x[\hat x, \hat K] = \delta x \hat I + O(\delta x^2) $$ or $$ -i[\hat x, \hat K] = \hat I + O(\delta x) $$ or, now taking the limit $\delta x \to 0$, we have: $$ [\hat x, \hat K] = i\hat I $$

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  • $\begingroup$ What topology are you using when writing $O(\delta x)$? Usually $O(f(x))$ means $|f(x)|< C|x|$ for some $C\geq 0$ and $x$ around $0$. Here the objects are operators, not numbers… $\endgroup$ Oct 29, 2023 at 9:46
  • $\begingroup$ @ValterMoretti A vague topology wherein I ignore everything other than the real number $\delta x$ and I wave my hands ferociously. $\endgroup$
    – hft
    Oct 29, 2023 at 17:29
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    $\begingroup$ The relevant one is the strong operator topology, which also needs to specify a vector as an argument. My point is that, or a technical answer is more or less complete or it adds nothing to the formal argument,which seems unsatisfactory to the author of the post. $\endgroup$ Oct 29, 2023 at 17:40

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