Confusion with rotation operator definition in Shankar

Question

In Shankar quantum mechanics on page 306-307 it has the following:

12.2. Rotations in Two Dimensions

Classically, the effect of a rotation $\phi_0\mathbf{k}$, i.e., by an angle $\phi_0$ about the $z$ axis (counterclockwise in the $x\ y$ plane) has the following effect on the state of a particle: $$\begin{align} \begin{bmatrix}x \\ y\end{bmatrix} \to \begin{bmatrix}\bar{x} \\ \bar{y}\end{bmatrix} &= \begin{bmatrix} \cos\phi_0 & -\sin\phi_0 \\ \sin\phi_0 & \cos\phi_0 \end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} \tag{12.2.1}\\ \begin{bmatrix}p_x \\ p_y\end{bmatrix} \to \begin{bmatrix}\bar{p}_x \\ \bar{p}_y\end{bmatrix} &= \begin{bmatrix} \cos\phi_0 & -\sin\phi_0 \\ \sin\phi_0 & \cos\phi_0 \end{bmatrix} \begin{bmatrix}p_x \\ p_y\end{bmatrix} \tag{12.2.2} \end{align}$$ Let us denote the operator that rotates these two-dimensional vectors by $R(\phi_0\mathbf{k})$. It is represented by the $2\times 2$ matrix in Eqs. (12.2.1) and (12.2.2). Just as $T(\mathbf{a})$ is the operator in Hilbert space associated with the translation $\mathbf{a}$, let $U[R(\phi_0\mathbf{k})]$ be the operator associated with the rotation $R(\phi_0\mathbf{k})$. In the active transformation picture $$\lvert\psi\rangle \underset{U[R]}{\longrightarrow} \lvert\psi_R\rangle = U[R]\lvert\psi\rangle\tag{12.2.3}$$ The rotated state $\lvert\psi_R\rangle$ must be such that $$\langle X\rangle_R = \langle X\rangle\cos\phi_0 - \langle Y\rangle\sin\phi_0\tag{12.2.4a}$$

Specifically I'm confused about how it describes the rotation operator $R(\phi_0\mathbf{k})$ and then the operator $U[R(\phi_0\mathbf{k})]$.

It says that $R(\phi_0\mathbf{k})$ is the operator associated with the rotation but the second to last sentence before equation 12.2.3 seems to imply that $R(\phi_0\mathbf{k})$ just denotes the rotation itself and $U[R(\phi_0\mathbf{k})]$ denotes the operator. Where am I going wrong?

Answer 1

Shankar is being a little bit sloppy with the term "state," although not that sloppy since he does use the qualifiers "quantum" and "classical" to achieve clarity.

In classical mechanics, the position of a particle is represented by a point $\mathbf r$ in three dimensions $\mathbb R^3$. To rotate the configuration of the classical particle, then, one would act on the position with a $3\times 3$ rotation matrix $R$.

In quantum mechanics, the state of the particle is represented by a vector in a Hilbert space $\mathcal H$. To "rotate" its state (which is what happens to the system when you, for example, rotate your measurement apparatus according to the spatial rotation $R$), one would act on it with a unitary operator $U(R)$. Why unitary you might ask? This is a consequence of a deep theorem on symmetries in quantum mechanics called Wigner's theorem.

Note that both $R$ and $U$ are operators, but they operate on different vector spaces, $\mathbb R^3$ and $\mathcal H$ respectively, and they have distinct physical interpretations as a result.

Answer 2

$R(\phi_0,k)$ is the operator that rotates your co-ordinate system. But it is not suitable to apply a 2x2 matrix, as in this case, to a vector in Hilbert space. Mind that Hilbert space is unlike an ordinary orthogonal position space.

Thus, using $U$, you map the operator into an equivalent operator which can operate on vectors in Hilbert space.