# Infinitesimal Translation Operator

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.

• 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. $\hat{x}$). – A. Bordg Sep 23 '20 at 13:17

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 $$\mathbf{\hat{x}} |\mathbf{x}' + \text{d}\mathbf{x}'\rangle = (\mathbf{x}' + \text{d}\mathbf{x}')|\mathbf{x}' + \text{d}\mathbf{x}'\rangle.$$