Principles of Quantum Mechanics · Problem 12.5.6

Rotational Invariance and Angular Momentum — R. Shankar

Physics — Quantum Mechanics · finite-rotations

Problem

Consider the family of states ∣jj⟩,…,∣jm⟩,…,∣j,−j⟩|jj\rangle, \ldots, |jm\rangle, \ldots, |j, -j\rangle. One regards them as states of the same magnitude but different orientation of angular momentum. If one takes this remark literally, i.e., in the classical sense, one is led to believe that one may rotate these into each other, as is the case for classical states with these properties. Consider the family ∣1,1⟩,∣1,0⟩,∣1,−1⟩|1,1\rangle, |1,0\rangle, |1,-1\rangle. It may seem, for example, that the state with zero angular momentum along the zz axis, ∣1,0⟩|1,0\rangle, may be obtained by rotating ∣1,1⟩|1,1\rangle by some suitable (12π\frac{1}{2}\pi?) angle about the xx axis. Using D(1)[R(θxi)]D^{(1)}[R(\theta_x \mathbf{i})] from part (2) in the last exercise show that ∣1,0⟩≠D(1)[R(θxi)]∣1,1⟩for any θx|1,0\rangle \neq D^{(1)}[R(\theta_x \mathbf{i})]|1,1\rangle \quad \text{for any } \theta_x The error stems from the fact that classical reasoning should be applied to ⟨J⟩\langle \mathbf{J} \rangle, which responds to rotations like an ordinary vector, and not directly to ∣jm⟩|jm\rangle, which is a vector in Hilbert space. Verify that ⟨J⟩\langle \mathbf{J} \rangle responds to rotations like its classical counterpart, by showing that ⟨J⟩\langle \mathbf{J} \rangle in the state D(1)[R(θxi)]∣1,1⟩D^{(1)}[R(\theta_x \mathbf{i})]|1,1\rangle is ℏ[−sin⁡θxi+cos⁡θxk]\hbar[-\sin\theta_x \mathbf{i} + \cos\theta_x \mathbf{k}].

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