# Transformation of a position for an infinitesimal rotation

I'm trying to figure out on how to work out this passage from Shankars $\textit{Principles of Quantum Mechanics 2nd Edition}$

Pg 309 // We could have also derived Eq. (12.2.11) for $\hat{L}^{z}$ by starting with the passive transformation equations for an infinitesimal rotation:

$$U^{\dagger}[R]X U[R] = X - Y \epsilon_{z}$$

where $Eq. (12.2.11)$ is the commutator definition of $\hat{L}^{z}$.

The passage uses a Unitary operator of an infinitesimal rotation $\epsilon_z$ defined as

$$U(R(\epsilon_z)) = 1 - \frac{i \epsilon_{z} L_{z}}{\hbar}$$

For a generalized position $r^i$ and an angle $\theta$,

$$U(\theta)\hat{x}U(\theta)^{\dagger} = \left( 1 - \frac{i \theta \hat{L}^{z}}{\hbar} \right)\hat{x} \left(1 + \frac{i \theta \hat{L}^{z}}{\hbar} \right)$$

$$= \hat{x} - \frac{i \theta}{\hbar}[\hat{x}, \hat{L}^{z}] + \frac{\theta^{2}}{\hbar^{2}}\hat{L}^{z}\hat{x}\hat{L}^{z}$$

$$= \hat{x} - \frac{i \theta}{\hbar} \left( \hat{x}[\hat{x}, \hat{p}^{y}] - [\hat{x}, \hat{p}^{y}]\hat{x} \right) + \frac{\theta^{2}}{\hbar}\hat{L}^{z}\hat{x}\hat{L}^{z}$$

From here on out, I do not know how reduce it any further. Expanding the last term is a complete mess and pretty difficult to reduce.

For the second term, I can of course use the relation of $[\hat{r}^{i}, \hat{p}^{j}] = i \hbar \delta_{ij}$, but that is all I know of.

## 1 Answer

To get you on track:

1. By definition, an infinitesimal rotation refers to the first order approximation to the rotation with respect to the rotation parameter (e.g. $\theta$ in your notation), so the last term should be neglected since it's second order in $\theta$.

2. You seem to have written $\hat L_z$ incorrectly, it should be $\hat L_z = \hat x \hat p_y - \hat y \hat p_x$