1

In Weinberg's Lectures on Quantum Mechanics (pg 31), he said that the commutator relation $$[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$$ is true for any vector operator $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$, where $\textbf{L}$ is the angular momentum operator given by $\textbf{L}=-i\hbar\textbf{x} \times \nabla$.

An example for vector $\textbf{v}$ is the angular momentum $\textbf{L}$ itself: $$[L_i,L_j] = i\hbar \sum_k \epsilon_{ijk} L_k.$$ Other examples are $\textbf{v}=\textbf{x}$ and $\textbf{v}=\nabla: $ $$[L_i,x_j] = i\hbar \sum_k \epsilon_{ijk} x_k,$$ $$[L_i,\frac{\partial}{\partial x_j}] = i\hbar \sum_k \epsilon_{ijk} \frac{\partial}{\partial x_k}.$$

How can it be shown that the commutator relation $[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$ is indeed true for any vector operator $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$?

TaeNyFan
  • 751
  • 2
    This is a great mathematics question! However, I have to recommend you will get a better response on https://physics.stackexchange.com/ – Taylor Rendon Jun 24 '20 at 04:56
  • 1
    @TaylorRendon Thanks for the recommendation. I will post it on physics stack too. – TaeNyFan Jun 24 '20 at 05:20
  • You could try a proof by induction on the number of $x$ and $\nabla$ factors in $v$, using $[A,,BC]=[A,,B]C+B[A,,C]$ and Levi-Civita identities. – J.G. Jun 24 '20 at 14:06

1 Answers1

2

Recall \begin{align} L_i = -i\hbar \sum_{j, k=1}^3\epsilon_{ijk}x_j\frac{\partial}{\partial x_k} = -i\hbar\epsilon_{ijk}x_j\frac{\partial}{\partial x_k} \end{align} then we see that \begin{align} [L_i, x_\ell]f = L_ix_\ell f-x_\ell L_if. \end{align} Here, we see that \begin{align} L_iv_\ell f =&\ -i\hbar \epsilon_{ijk}x_j\frac{\partial}{\partial x_k}(x_\ell f)\\ =&\ -i\hbar \epsilon_{ijk}x_j\frac{\partial x_\ell}{\partial x_k}f-i\hbar \epsilon_{ijk}x_jv_\ell\frac{\partial f}{\partial x_k} \end{align} and \begin{align} v_\ell L_i f = -i\hbar \epsilon_{ijk}x_jx_\ell\frac{\partial f}{\partial x_k}. \end{align} Finally, we have that \begin{align} [L_i, v_\ell]f = -i\hbar \epsilon_{ijk}x_j\frac{\partial x_\ell}{\partial x_k}f \end{align} or equivalently \begin{align} [L_i, v_\ell]= -i\hbar \epsilon_{ijk}x_j\frac{\partial x_\ell}{\partial x_k}=-i\hbar \epsilon_{ijk}x_j\delta_{\ell, k} = -i\hbar \epsilon_{ij\ell}x_j = i\hbar \epsilon_{i\ell j}x_j. \end{align}

The proof for $v_i = \partial_i$ is similar.

Jacky Chong
  • 25,739
  • Thanks for the answer, but is it possible to show that the commutator expression is true for any general operator $\textbf{v}$ that is constructed from $\textbf{x}$ and $\nabla$? I might not have not phrased my question very appropriately. – TaeNyFan Jun 24 '20 at 05:27
  • I think the author means including but not limited to linear combinations. For example, a dot product between $\vec{x}$ and $\nabla$. I'm not entirely sure though. – TaeNyFan Jun 24 '20 at 05:34