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In the quote shown below, they use the second approach, i.e the operator changes with time and not the state kets, making use of the unitary operator for infinitesimal translation. I understand that you can get the second to last step by ignoring the higher order terms of $d\mathbf x$ (if I'm not wrong), but I am unable to figure out how the commutation bracket leads to the final equation $(2.2.7)$.

In contrast, if we follow approach 2, we obtain

$$\begin{aligned}[b]|\alpha\rangle&\rightarrow|\alpha\rangle\\\mathbf x&\rightarrow\left(1+\frac{i\mathbf p\cdot d\mathbf x'}{\hbar}\right)\mathbf x\left(1-\frac{i\mathbf p\cdot d\mathbf x'}{\hbar}\right)\\&=\mathbf x+\left(\frac i{\hbar}\right)[\mathbf p\cdot d\mathbf x',\mathbf x]\\&=\mathbf x+d\mathbf x'\end{aligned}\tag{2.2.7}$$

We leave it as an exercise for the reader to show that both approaches lead to the same result for the expectation value of $\mathbf x$:

$$\langle\mathbf x\rangle\rightarrow\langle\mathbf x\rangle+\langle d\mathbf x'\rangle\tag{2.2.8}$$

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  • $\begingroup$ Please note that images of text and mathematics are very strongly discouraged on the site. Please use text and Mathjax. $\endgroup$ Commented Jul 15, 2021 at 10:09
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    $\begingroup$ My bad.Did not know that that was the case. Will keep it mind. Thanks! $\endgroup$ Commented Jul 15, 2021 at 10:31

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You're asking for a proof that $\left[\mathbf{p}\cdot d\mathbf{x}^\prime,\,\mathbf{x}\right]=-i\hbar d\mathbf{x}^\prime$, i.e. $\left[p_i dx^\prime_i,\,x_j\right]=-i\hbar dx^\prime_j$ in Einstein notation (for convenience I'm keeping indices downstairs). Since $dx^\prime_i$ is a c-number, this follows from the CCR $[p_i,\,x_j]=-i\hbar\delta_{ij}$.

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  • $\begingroup$ Thank you! I was too stuck up on the operators to have thought about it this way. Also, its safe to assume that the higher order terms have been ignored right? $\endgroup$ Commented Jul 15, 2021 at 7:08
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    $\begingroup$ @NiranjanHaridasMenon They have, yes. That's why we don't have $dx^\prime_jdx^\prime_k$ terms. $\endgroup$ Commented Jul 15, 2021 at 7:09
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It's because momentum is the generator for translation. In classical mechanics , infinitesimal canonical transformation $\eta\to \eta+\delta \eta$ generated by the function G, will satisfy $\delta \eta=\epsilon[\eta, G]$ with $|\epsilon|<<1$. In this case we have $\epsilon G=p.dx$

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Here is the way I maneged the problem: first rewrite $\mathbf{p \cdot dx'}$ in terms of the translation operator $\mathscr{T}(\mathbf{dx'})$ $$ \mathscr{T}(\mathbf{dx'}) = 1 - \frac{i \mathbf{p \cdot dx'}}{\hbar} \rightarrow \mathbf{p \cdot dx'} = i\hbar\mathscr{T}(\mathbf{dx'}) - i\hbar$$

Now open the commutator $[\mathbf{p \cdot dx' , x}] = \mathbf{p \cdot dx' x} - \mathbf{xp \cdot dx'}$ and substitute $\mathbf{p \cdot dx'}$ by the relation above. You'll find: $$ [\mathbf{p \cdot dx' , x}] = i\hbar \mathscr{T}(\mathbf{dx'}) \mathbf{x} - i\hbar \mathbf{x} \mathscr{T}(\mathbf{dx'}) = i\hbar[\mathscr{T}(\mathbf{dx'}) ,\mathbf{x}] = -i\hbar[\mathbf{x} , \mathscr{T}(\mathbf{dx'})]$$

Finally, the book has already proved that $[\mathbf{x} , \mathscr{T}(\mathbf{dx'})] = \mathbf{dx'}$, so $[\mathbf{p \cdot dx' , x}] = -i\hbar \mathbf{dx'}$

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