4.2: Total orbital angular momentum
- Page ID
- 20884
In the hydrogen atom or any system with a spherically symmetric potential \(V(r)\), we have learned that angular momentum
is conserved. The Hamiltonian will be of the form
\(\displaystyle H\) | \(\textstyle =\) | \(\displaystyle -{\hbar^2 \over 2m}\nabla^2 + V(r)\) | |
\(\textstyle =\) |
and will satisfy
so that \({\bf L}\) is a constant of the motion. This is illustrated schematically below:
However, what happens when the ``source'' of the potential is not so heavy and can move on a time scale similar to that of the particle. An example would be hydrogen with the proton replaced by a particle with positive charge and the same mass of the electron, i.e., a positron. The system, shown below,
is known as positronium. It will be described by a Hamiltonian of the form
where
and
Although this is the specific form of the potential for this example, what we will show will be general for any potential that depends only on \(\vert{\bf r}_1-{\bf r}_2\vert\).
Now, the individual angular momenta
are no longer conserved, i.e.,
To see that this is true, consider the \(z\) components of the angular momentum operators:
It is straightforward to compute the commutators (left as an exercise for the reader) and it is found that
\(\displaystyle \left[L_{1z},H\right]\) | \(\textstyle =\) | ||
\(\textstyle =\) |
Similarly,
\(\displaystyle \left[L_{2z},H\right]\) | \(\textstyle =\) | ||
\(\textstyle =\) | |||
\(\textstyle =\) |
However, if we add these together, it can be see that
\(\displaystyle \left[L_{1z},H\right] + \left[L_{2z},H\right]\) | \(\textstyle =\) | \(\displaystyle \left[(L_{1z}+L_{2z}),H\right]\) | |
\(\textstyle =\) |
Thus, the quantity \(L_{1z}+L_{2z}\) is a constant of the motion. The same can be shown to be true for the \(x\) and \(y\) components. Thus, The quantity
is a constant of the motion. \({\bf L}={\bf L}_1+{\bf L}_2\) is known as the total orbital angular momentum. It is conserved because the potential only depends on the distance between the two particles.
If we have an \(N\)-particle system with a Hamiltonian of the form
then the total orbital angular momentum
will be a constant of the motion.