# Commutation Relations

In many multidimensional problems, we often deal with rotational motion of particles, and thus we are interested in treating angular momentum in the framework of quantum mechanics. The (orbital) angular momentum operator in quantum mechanics is given by the cross product of the position of the particle with its momentum:

$\hat{\mathbf{L}}=\hat{\mathbf{r}}\times\hat{\mathbf{p}}$

Working in the position representation, this becomes

$\hat{\mathbf{L}}=\mathbf{r}\times\frac{\hbar}{i}\nabla.$

Evaluating the cross product in the Cartesian coordinate system, we get a component of $\mathbf{L}\!$ in each direction; for example,

$\hat{L}_x=\hat{y}\hat{p}_z-\hat{z}\hat{p}_y=\frac{\hbar}{i}\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right),$

and similarly the other two components of the angular momentum operator. All of these can be written in a more compact form using the Levi-Civita symbol as

$\hat{L}_{\mu}=\epsilon_{\mu\nu\lambda}\hat{r}_\nu\hat{p}_\lambda,$

where

$\epsilon_{\mu\nu\lambda} = \begin{cases} +1, & \mbox{if } (\mu,\nu,\lambda) \mbox{ is } (1,2,3), (3,1,2) \mbox{ or } (2,3,1), \\ -1, & \mbox{if } (\mu,\nu,\lambda) \mbox{ is } (3,2,1), (1,3,2) \mbox{ or } (2,1,3), \\ 0, & \mbox{otherwise: }\mu=\nu \mbox{ or } \nu=\lambda \mbox{ or } \lambda=\mu \end{cases}$

and we use the Einstein summation convention, in which sums over repeated indices are omitted. The above definition of the Levi-Civita symbol gives the "sign" of a permutation of 123 (it is 1 for even permutations, or -1 for odd permutations).

We can immediately verify the following commutation relations:

$[\hat{L}_\mu,\hat{x}_\nu]=i\hbar\epsilon_{\mu\nu\lambda}\hat{x}_\lambda$

$[\hat{L}_\mu,\hat{p}_\nu]=i\hbar\epsilon_{\mu\nu\lambda}\hat{p}_\lambda$

$[\hat{L}_\mu,\hat{L}_\nu]=i\hbar\epsilon_{\mu\nu\lambda}\hat{L}_\lambda$

The last relation may also be written as

$\mathbf{L}\times\mathbf{L}=i\hbar\mathbf{L}.$

Furthermore,

$[\hat{\mathbf{n}}\cdot\hat{\mathbf{L}},\hat{\mathbf{r}}]=i\hbar(\hat{\mathbf{r}}\times\hat{\mathbf{n}})$
$[\hat{\mathbf{n}}\cdot\hat{\mathbf{L}},\hat{\mathbf{p}}]=i\hbar(\hat{\mathbf{p}}\times\hat{\mathbf{n}})$
$[\hat{\mathbf{n}}\cdot\hat{\mathbf{L}},\hat{\mathbf{L}}]=i\hbar(\hat{\mathbf{L}}\times\hat{\mathbf{n}})$

For example,

\begin{align} \left[\hat{L}_\mu,\hat{x}_\nu\right] &= [\epsilon_{\mu\lambda\rho}\hat{x}_\lambda \hat{p}_\rho,\hat{x}_\nu] = \epsilon_{\mu\lambda\rho}[\hat{x}_\lambda \hat{p}_\rho,\hat{x}_\nu] = \epsilon_{\mu\lambda\rho}\hat{x}_\lambda[\hat{p}_\rho,\hat{x}_\nu] \\ &= \epsilon_{\mu\lambda\rho}\hat{x}_\lambda\frac{\hbar}{i}\delta_{\rho\nu} = \epsilon_{\mu\lambda\nu}\hat{x}_\lambda\frac{\hbar}{i} \\ &= i\hbar\epsilon_{\mu\nu\lambda}\hat{x}_\lambda. \end{align}

Also, note that for $\hat{mathbf{L}}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2=\hat{L}_{\mu}\hat{L}_{\mu},$

\begin{align} \left[\hat{L}_{\mu},\hat{L}^2\right] &= \left[\hat{L}_{\mu},\hat{L}_{\nu}\hat{L}_{\nu}\right] \\ &= \hat{L}_{\nu}\left[\hat{L}_{\mu},\hat{L}_{\nu}\right]+\left[\hat{L}_{\mu},\hat{L}_{\nu}\right]\hat{L}_{\nu} \\ &= \hat{L}_{\nu} i\hbar \epsilon_{\mu\nu\lambda} \hat{L}_{\lambda} + i\hbar \epsilon_{\mu\nu\lambda} \hat{L}_{\lambda} \hat{L}_{\nu} \\ &= i\hbar \epsilon_{\mu\nu\lambda} \hat{L}_{\nu}\hat{L}_{\lambda} - i\hbar \epsilon_{\mu\lambda\nu}\hat{L}_{\lambda}\hat{L}_{\nu} \\ &= i\hbar \epsilon_{\mu\nu\lambda} \hat{L}_{\nu}\hat{L}_{\lambda} - i\hbar \epsilon_{\mu\nu\lambda}\hat{L}_{\nu}\hat{L}_{\lambda} \\ &= 0. \end{align}

Therefore, the magnitude of the angular momentum squared commutes with any one component of the angular momentum, and thus both may be specified exactly in a given measurement.