### Operator-valued vectors in quantum mechanics

Let $$\boldsymbol{e}_1$$, $$\boldsymbol{e}_2$$, $$\boldsymbol{e}_3$$ be a right-handed triplet of mutually orthogonal unit vectors in a three-dimensional Euclidean space. An operator-valued vector $$\boldsymbol{A}$$ is a "vector" whose components along $$\boldsymbol{e}_1$$, $$\boldsymbol{e}_2$$, $$\boldsymbol{e}_3$$ are operators $$A_1$$, $$A_2$$, $$A_3$$, respectively: $\boldsymbol{A} = \boldsymbol{e}_1 A_1 + \boldsymbol{e}_2 A_2 + \boldsymbol{e}_3 A_3 \,.$ Examples include the position and momentum operators in quantum mechanics: $\boldsymbol{x} = \boldsymbol{e}_1 x_1 + \boldsymbol{e}_2 x_2 + \boldsymbol{e}_3 x_3 \,,$ \begin{align*} \boldsymbol{p} &= \boldsymbol{e}_1 p_1 + \boldsymbol{e}_2 p_2 + \boldsymbol{e}_3 p_3 \\[0.2cm] &= \boldsymbol{e}_1 \frac{\hbar}{i} \frac{\partial}{\partial x_1} + \boldsymbol{e}_2 \frac{\hbar}{i} \frac{\partial}{\partial x_2} + \boldsymbol{e}_3 \frac{\hbar}{i} \frac{\partial}{\partial x_3} \,. \end{align*}

Let $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ be some operator-valued vectors. The operator $\boldsymbol{A} \cdot \boldsymbol{B} = A_i B_i$ is called the dot product of $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$. (Hereinafter, we use the Einstein summation convention: repeated indices are summed over their three possible values. Thus, $$A_i B_i = A_1 B_1 + A_2 B_2 + A_3 B_3$$.) The square of $$\boldsymbol{A}$$ is defined as $\boldsymbol{A}^2 = \boldsymbol{A} \cdot \boldsymbol{A} = A_i A_i \,.$

The cross product of $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ is an operator-valued vector $$\boldsymbol{A} \times \boldsymbol{B}$$ whose $$i$$-th component is given by the operator $\left( \boldsymbol{A} \times \boldsymbol{B} \right)_i = \epsilon_{ijk} A_j B_k \,.$ Here $$\epsilon_{ijk}$$ is the Levi-Civita symbol: $\epsilon_{ijk} = \left\{ \begin{array}{ll} +1 \quad &\text{if } (i,j,k) = (1, 2, 3), (2, 3, 1), \text{ or } (3, 1, 2) \\ -1 \quad &\text{if } (i,j,k) = (3, 2, 1), (2, 1, 3), \text{ or } (1, 3, 2) \\ 0 \quad &\text{if } i = j \text{ or } j = k \text{ or } k = i \\ \end{array} \right.$ For example, the orbital angular momentum operator is given by \begin{align*} \boldsymbol{L} &= \boldsymbol{x} \times \boldsymbol{p} \\[0.2cm] &= \boldsymbol{e}_1 L_1 + \boldsymbol{e}_2 L_2 + \boldsymbol{e}_3 L_3 \\[0.2cm] &= \boldsymbol{e}_1 (x_2 p_3 - x_3 p_2) + \boldsymbol{e}_2 (x_3 p_1 - x_1 p_3) + \boldsymbol{e}_3 (x_1 p_2 - x_2 p_1) \\[0.2cm] &= \boldsymbol{e}_1 \frac{\hbar}{i} \left( x_2 \frac{\partial}{\partial x_3} - x_3 \frac{\partial}{\partial x_2} \right) + \boldsymbol{e}_2 \frac{\hbar}{i} \left( x_3 \frac{\partial}{\partial x_1} - x_1 \frac{\partial}{\partial x_3} \right) \\ &\qquad\qquad\quad + \boldsymbol{e}_3 \frac{\hbar}{i} \left( x_1 \frac{\partial}{\partial x_2} - x_2 \frac{\partial}{\partial x_1} \right) \,. \end{align*}

### Problem

Show that $\boldsymbol{x} \cdot \boldsymbol{p} = \boldsymbol{p} \cdot \boldsymbol{x} + 3 i \hbar$

$\boldsymbol{x} \times \boldsymbol{p} = -\boldsymbol{p} \times \boldsymbol{x}$

$\boldsymbol{x} \cdot \boldsymbol{L} = 0$

$\boldsymbol{p} \cdot \boldsymbol{L} = 0$

$\boldsymbol{x} \times \boldsymbol{L} = \boldsymbol{x} (\boldsymbol{x} \cdot \boldsymbol{p}) - \boldsymbol{x}^2 \boldsymbol{p}$

$\boldsymbol{p} \times \boldsymbol{L} = \boldsymbol{p}^2 \boldsymbol{x} - \boldsymbol{p} (\boldsymbol{p} \cdot \boldsymbol{x})$

$\boldsymbol{L}^2 = \boldsymbol{x}^2 \boldsymbol{p}^2 - (\boldsymbol{x} \cdot \boldsymbol{p})^2 + i \hbar \boldsymbol{x} \cdot \boldsymbol{p}$

$\boldsymbol{L} \times \boldsymbol{L} = i \hbar \boldsymbol{L}$

### Solution

The solution is provided in Issue #6 - Operator-valued vectors in quantum mechanics of Quantum Newsletter.