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I am reading chapter 3.6.1 in the 3rd edition of Sakurai's Modern Quantum Mechanics. On page 191 the connection between the $\mathbf{L^2}$ operator and the angular part of the Laplacian is shown. The line $(3.227)$ reads:

$$\langle\mathbf{x'}|\mathbf{x \cdot p}|\alpha\rangle = \mathbf{x'} \cdot \left(- i \hbar \nabla' \langle\mathbf{x'}|\alpha\rangle \right)$$ $$=- i \hbar r' \frac{\partial}{\partial r'} \langle\mathbf{x'}|\alpha\rangle\tag{3.227}$$

The whole further derivation is based on this relation.

I cannot understand why we take only the radial part of the gradient. Why are we allowed to neglect the angular part?

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  • $\begingroup$ Was there any claim of symmetries (e.g., angular) expressed before or around the relevant equation? $\endgroup$ Commented Jul 21, 2023 at 21:18
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    $\begingroup$ Do you understand the identity ${\mathbf x}\cdot \nabla = r\partial_r$? $\endgroup$ Commented Jul 21, 2023 at 21:37

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In spherical coordinates, the position vector $\mathbf{x}$ is simply $r\hat{r}$. So only the $r$-component of the gradient contributes to the dot product.

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    $\begingroup$ Thank You! Indeed, $\mathbf{x'} = r \hat{r}$, so the angular part of the gradient disappears due to the scalar product $\mathbf{x'} \cdot \nabla'$ (thank You @Cosmas Zachos for the hint, it really helped). $\endgroup$ Commented Jul 22, 2023 at 12:06

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