Commutation Relations

Multidimensional problems entail the possibility of having rotation as a part of solution. Just like in classical mechanics where we can calculate the angular momentum using vector cross product, we have a very similar form of equation. However, just like any observable in quantum mechanics, this angular momentum is expressed by a Hermitian operator. Similar to classical mechanics we write the angular momentum operator $\mathbf {L} \!$ as:

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

Working in the spatial representation, we have $\mathbf {r}$ as our radial vector, while $\mathbf {p}$ is the momentum operator.

$\mathbf {p} =-i\hbar \nabla$

Using the cross product in Cartesian coordinate system, we get a component of ${\mathbf {L}}\!$ in each direction:

$L_{x}=yp_{z}-zp_{y}={\frac {\hbar }{i}}\left(y{\frac {\partial }{\partial z}}-z{\frac {\partial }{\partial y}}\right)\!$

Similarly, using cyclic permutation on the coordinates x, y, z, we get 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 (the Einstein summation convention of summing over repeated indices is understood here)

$L_{\mu }=\epsilon _{\mu \nu \lambda }r_{\nu }p_{\lambda }\!$

with

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

Or we simply say that the even permutation gives 1, odd permutation gives -1, otherwise, we get 0.

We can immediately verify the following commutation relations:

$[L_{\mu },r_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }r_{\lambda }$

$[L_{\mu },p_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }p_{\lambda }$

$[L_{\mu },L_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }L_{\lambda }$ (this relation tells us :$\mathbf {L} \times \mathbf {L} =i\hbar \mathbf {L}$)

and

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

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

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

For example,

${\begin{aligned}\left[L_{\mu },r_{\nu }\right]&=[\epsilon _{\mu \lambda \rho }r_{\lambda }p_{\rho },r_{\nu }]=\epsilon _{\mu \lambda \rho }[r_{\lambda }p_{\rho },r_{\nu }]=\epsilon _{\mu \lambda \rho }r_{\lambda }[p_{\rho },r_{\nu }]\\&=\epsilon _{\mu \lambda \rho }r_{\lambda }{\frac {\hbar }{i}}\delta _{\rho \nu }=\epsilon _{\mu \lambda \nu }r_{\lambda }{\frac {\hbar }{i}}\\&=i\hbar \epsilon _{\mu \nu \lambda }r_{\lambda }\end{aligned}}$

Also, note that for $L^{2}=L_{x}^{2}+L_{y}^{2}+L_{z}^{2}=L_{\mu }L_{\mu }$,

${\begin{aligned}\left[L_{\mu },L^{2}\right]&=\left[L_{\mu },L_{\nu }L_{\nu }\right]\\&=L_{\nu }\left[L_{\mu },L_{\nu }\right]+\left[L_{\mu },L_{\nu }\right]L_{\nu }\\&=L_{\nu }i\hbar \epsilon _{\mu \nu \lambda }L_{\lambda }+i\hbar \epsilon _{\mu \nu \lambda }L_{\lambda }L_{\nu }\\&=i\hbar \epsilon _{\mu \nu \lambda }L_{\nu }L_{\lambda }-i\hbar \epsilon _{\mu \lambda \nu }L_{\lambda }L_{\nu }\\&=i\hbar \epsilon _{\mu \nu \lambda }L_{\nu }L_{\lambda }-i\hbar \epsilon _{\mu \nu \lambda }L_{\nu }L_{\lambda }\\&=0.\end{aligned}}$