Introduction to Quantum Mechanics · Problem 6.19

Symmetries and Conservation Laws — David J. Griffiths

Physics — Quantum Mechanics · vector_operator_raising_lowering

Problem

For any vector operator V^\hat{\mathbf{V}} one can define raising and lowering operators as V^±=V^x±i V^y.\hat{V}_{\pm} = \hat{V}_x \pm i\,\hat{V}_y. (a) Using Equation 6.33, show that [L^z, V^±]=±ℏ V^+.\left[\hat{L}_z,\,\hat{V}_{\pm}\right] = \pm\hbar\,\hat{V}_+. [L^2, V^±]=2ℏ2 V^±±2ℏ V^±L^z∓2ℏ V^zL^±.\left[\hat{L}^2,\,\hat{V}_{\pm}\right] = 2\hbar^2\,\hat{V}_{\pm} \pm 2\hbar\,\hat{V}_{\pm}\hat{L}_z \mp 2\hbar\,\hat{V}_z\hat{L}_{\pm}. (b) Show that, if ψ\psi is an eigenstate of L^2\hat{L}^2 and L^z\hat{L}_z with eigenvalues ℓ(ℓ+1)ℏ2\ell(\ell+1)\hbar^2 and ℓ ℏ\ell\,\hbar respectively, then either V^+ψ\hat{V}_+\psi is zero or V^+ψ\hat{V}_+\psi is also an eigenstate of L^2\hat{L}^2 and L^z\hat{L}_z with eigenvalues (ℓ+1)(ℓ+2)ℏ2(\ell+1)(\ell+2)\hbar^2 and (ℓ+1)ℏ(\ell+1)\hbar respectively. This means that, acting on a state with maximal mℓ=ℓm_\ell = \ell, the operator V^+\hat{V}_+ either "raises" both the ℓ\ell and mm values by 1 or destroys the state.

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