Orbital-angular momentum

See also angular momentum in quantum mechanics

In quantum mechanics, orbital angular momentum is a conserved property of a system of one or more particles that are in a centrally symmetric potential. If the radius of particle k with respect to the center of symmetry is r_k = (x_k, y_k, z_k) and if the momentum of the same particle is p_k, then the orbital angular momentum of particle k is defined as the following vector operator,

${\boldsymbol {\ell }}_{k}=\mathbf {r} _{k}\times \mathbf {p} _{k},$

where the symbol × indicates the cross product of two vectors. The total angular momentum of a system of N particles is

$\mathbf {L} =\sum _{k=1}^{N}{\boldsymbol {\ell }}_{k}.$

In the so-called x-representation of quantum mechanics, the vector r_k is a multiplicative operator and

$\mathbf {p} _{k}=-i\hbar {\boldsymbol {\nabla }}_{k}\equiv -i\hbar \left({\frac {\partial }{\partial x_{k}}},\;{\frac {\partial }{\partial y_{k}}},\;{\frac {\partial }{\partial z_{k}}}\right),\qquad k=1,2,\ldots ,N.$

The components of the orbital angular momentum satisfy the following commutation relations,

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

The fact that L is a conserved quantity is expressed by the commutation with the Hamiltonian (energy operator)

$[H,L_{\alpha }]=0,\quad \alpha =x,y,z.$

It can be shown that this condition is necessary and sufficient that the potential energy part of H be centrally symmetric.