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In the book "Modern Quantum Mechanics" (by J.J. Sakurai and Jim Napolitano) page 44, infinitesimal Translation operator is given: $\mathscr{J}\left(d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}\right\rangle=\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle$ after this, book mentioned properties of the operator as being unitary and adding: $\mathscr{J}\left(d \mathbf{x}^{\prime \prime}\right) \mathscr{J}\left(d \mathbf{x}^{\prime}\right)=\mathscr{J}\left(d \mathbf{x}^{\prime}+d \mathbf{x}^{\prime \prime}\right)$ where I have confused is that:

$\mathbf{x} \mathscr{J}\left(d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}\right\rangle=\mathbf{x}\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle=\left(\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle$

The left hand side of the equation is so weird to me because middle one is directly driven from the first equation but to find a meaning at the right hand side of the above equation, I thought it might be true because $d \mathbf{x}^{\prime}$ is infinitesimally small but where is $\mathbf{x}^{\prime}$came from because initially it was $\mathbf{x}$ at the right hand side of the last equation.

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    $\begingroup$ Simply note that, apparently, Sakurai chooses to denote operators with a bold font (here the operator position $\mathbf{x}$) where everyone I know use a hat instead (e.g. $\hat{x}$). $\endgroup$
    – A. Bordg
    Sep 23, 2020 at 13:17

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I'm not sure I completely understood your question, but perhaps it will become clear if I explain what the last equation means. The left hand side of the equation means:

  1. Take a state $|\mathbf{x}'\rangle$
  2. Translate it by some infinitesimal amount $\text{d}\mathbf{x}'$ using the translation operator
  3. Act on it with the position operator, $\mathbf{\hat{x}}$.

By definition, translating the state $|\mathbf{x}'\rangle$ by $\text{d}\mathbf{x}'$ gives you the state $|\mathbf{x}' + \text{d}\mathbf{x}'\rangle$, and also by definition (since it is now a state of definite position with eigenvalue $\mathbf{x}'+ \text{d}\mathbf{x}'$), when you act on this state with the position operator you get $$\mathbf{\hat{x}} |\mathbf{x}' + \text{d}\mathbf{x}'\rangle = (\mathbf{x}' + \text{d}\mathbf{x}')|\mathbf{x}' + \text{d}\mathbf{x}'\rangle.$$

Does this help?

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    $\begingroup$ Thanks for the answer. $\endgroup$
    – asd.123
    Sep 23, 2020 at 13:08

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