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The following section is from Modern Quantum Mechanics by Sakurai; can any one help me finding out how this is done?

In contrast, if we follow approach 2, we obtain \begin{align} \vert\alpha\rangle&\to\vert\alpha\rangle \\ \mathbf{x}&\to\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)\left[\mathbf p\cdot d\mathbf x',\,\mathbf x\right]\tag{this line} \\ &=\mathbf x +d\mathbf x' \end{align} We leave it as an exercise for the reader to show that both approaces lead to the same result for the expectation value of $\mathbf x$: $$\langle\mathbf x\rangle\to\langle\mathbf x\rangle+\langle d\mathbf x'\rangle$$

Sakurai just mentions that if operator is changed but state kets are kept frozen but he left mathematics for students.

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    $\begingroup$ I don't know if that is your problem, but he is neglecting terms of order higher than one in $d\mathbf{x}$ to understand the effect of an infinitesimal transformation. $\endgroup$ Commented Apr 11, 2019 at 18:04
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    $\begingroup$ (1+a)b(1-a) = (b+ab)(1-a) = b+ab-ba-aba, neglecting the last term we get b+ab-ba which is equal to b+[a,b] $\endgroup$ Commented Apr 11, 2019 at 18:37
  • $\begingroup$ My question how it is mathematically done? $\endgroup$ Commented Apr 12, 2019 at 14:39

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Expanding, \begin{align} &\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) \\ &= \left(\mathbf x + \frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar}\mathbf x\right)\left(1-\frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar}\right) \\ &=\mathbf x + \frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar}\mathbf x - \mathbf x\frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar} - \left(\frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar}\right)^2 \\ \end{align} But the last term is of order $(d\mathbf x')^2$, so it vanishes. Then \begin{align} &\to \mathbf x + \frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar}\mathbf x - \mathbf x\frac{i\mathbf p\cdot d\mathbf{x}'}{\hbar} \\ &=\mathbf x + \left(\frac{i}{\hbar}\right)\left[\mathbf p\cdot d\mathbf x',\,\mathbf x\right] \\ &=\mathbf x +d\mathbf x'. \end{align}

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  • $\begingroup$ What about the last step where from dx' comes and how? $\endgroup$ Commented Apr 25, 2019 at 2:05
  • $\begingroup$ How last step is done? $\endgroup$ Commented May 1, 2019 at 18:05
  • $\begingroup$ @FaheemAalijah : Your question was from a long time ago, but I think you use $d\mathbf x'$ is a constant vector and then use the known commutation relations between $\mathbf p$ and $\mathbf x$. $\endgroup$ Commented Dec 11, 2024 at 18:45

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