📂Quantum Mechanics

Commutation Relations of the Angular Momentum Operator

Formula

The commutation relation of the angular momentum operators is as follows.

$$\left[L_{j}, L_{k} \right] = \i \hbar \epsilon_{jk\ell}L_{\ell} \tag{1}$$

Here, $\epsilon_{jk\ell}$ is the Levi-Civita symbol. When written out in full,

$$\left[ L_{x}, L_{y} \right] = \i \hbar L_{z} \\ \left[ L_{y}, L_{z} \right] = \i \hbar L_{x} \\ \left[ L_{z}, L_{x} \right] = \i \hbar L_{y}$$

Additionally, $L^{2} = L_{x}^{2} + L_{y}^{2} + L_{z}^{2}$ commutes with each component.

$$[L^{2}, L_{x}] = [L^{2}, L_{y}] = [L^{2}, L_{z}] = 0 \tag{2}$$

Explanation

$x$ and $p_{x}$ are, respectively, position operators and momentum operators.

Proof

$(1)$

Since the subscripts are cyclic, it suffices to compute for $\left[ L_{x},\ L_{y} \right]$.

Properties of Commutators](../297)

$$\left[ A + B, C \right] = \left[ A, C \right] + \left[ B, C \right]$$

From the properties of commutators, we obtain the following.

$$\begin{align*} [L_{x},L_{y}] &= [yp_{z}-zp_{y},zp_{x}-xp_{z}] \\ &= [yp_{z},zp_{x}-xp_{z}]- [zp_{y},zp_{x}-xp_{z}] \\ &= [yp_{z},zp_{x}] - [yp_{z},xp_{z}] - [zp_{y},zp_{x}] + [zp_{y},xp_{z}] \tag{1} \end{align*}$$

Here, different components of position and momentum operators commute.

$$[x, p_{y}] = [x, p_{z}] = [y, p_{x}] = [y, p_{z}] = [z, p_{x}] = [z, p_{y}] = 0$$

Additionally, the commutator of identical operators is also $0$. Consequently, when $(1)$ is expanded, only three terms of $[x,p_{x}],\ [y,p_{y}],\ [z,p_{z}]$ have a value different from $0$. Therefore, we need to consider only the terms that are not $0$. When the first term is expanded, all terms except for $y[p_{z},z]p_{x}$ are $0$. When the second term is expanded, all terms are $0$. When the third term is expanded, all terms are $0$. When the fourth term is expanded, all terms except for $x[z,p_{z}]p_{y}$ are $0$. (If you don’t understand, try expanding it yourself.) Hence, we obtain the following.

$$[L_{x},L_{y}] = y[p_{z},z]p_{+}x[z,p_{z}]p_{y}$$

Using the position-momentum commutator $[x, p_{x}] = \i\hbar$, the calculation proceeds as follows.

$$\begin{align*} [L_{x},L_{y}] &= y[p_{z},z]p_{x} - x[z,p_{z}]p_{y} \\ &= -\i\hbar (yp_{x}) + \i\hbar (xp_{y}) \\ &= \i \hbar (xp_{y} - yp_{x}) \\ &= \i \hbar L_{z} \end{align*}$$

By the same logic, we obtain the following.

$$[L_{y},L_{z}] = \i \hbar L_{x}, \qquad [L_{z}, L_{x}] = \i \hbar L_{y}$$

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$(2)$

Using the properties of commutators and $(1)$, since $[L_{z}, L_{z}] = 0$, we obtain the following.

$$\begin{align*} [L^2, L_{z}] &= [{L_{x}}^2 + {L_{y}}^2 + {L_{z}}^2, L_{z}] \\ &= [{L_{x}}^2,L_{z}] + [{L_{y}}^2, L_{z}] +[{L_{z}^2}, L_{z}] \\ &= L_{x}[L_{x}, L_{z}] + [L_{x}, L_{z}]L_{x} + L_{y}[L_{y}, L_{z}] + [L_{y}, L_{z}]L_{y} \\ &= (-\i\hbar L_{x}L_{y}) + (-\i\hbar L_{y}L_{x}) + \i\hbar L_{y}L_{z} + \i\hbar L_{x}L_{y} \\ &= 0 \end{align*}$$

Similarly, the following equation holds.

$$[L^2, L_{x}] = [L^2, L_{y}] = 0$$

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