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4.2: Total orbital angular momentum

  • Page ID
    20884
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    In the hydrogen atom or any system with a spherically symmetric potential \(V(r)\), we have learned that angular momentum

    \begin{displaymath}
{\bf L}= {\bf r}\times {\bf p}
\end{displaymath}

    is conserved. The Hamiltonian will be of the form

    \(\displaystyle H\) \(\textstyle =\) \(\displaystyle -{\hbar^2 \over 2m}\nabla^2 + V(r)\)
    \(\textstyle =\) $\displaystyle -{\hbar^2 \over 2m}{1 \over r}{\partial^2 \over \partial r^2}r
+ {{\bf L}^2 \over 2m\hbar^2 r^2} + V(r)$

    and will satisfy

    \begin{displaymath}
\left[{\bf L},H\right]=0
\end{displaymath}

    so that \({\bf L}\) is a constant of the motion. This is illustrated schematically below:

    \({\bf L}\) corresponds to the angular momentum of the particle in such a potential field. In practice, this is not a bad assumption since the mass of the proton is approximately 2000 time that of the electron.

    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

    \begin{displaymath}
H = -{\hbar^2 \over 2m}\left(\nabla_1^2 + \nabla_2^2\right) + V(\vert{\bf r}_1-{\bf r}_2\vert)
\end{displaymath}

    where

    \begin{displaymath}
\nabla_1 = {\partial \over \partial {\bf r}_1}\;\;\;\;\;\;\;\;\;\;
\nabla_2 = {\partial \over \partial {\bf r}_2}
\end{displaymath}

    and

    \begin{displaymath}
V(\vert{\bf r}_1-{\bf r}_2\vert) = -{e^2 \over \vert{\bf r}_1-{\bf r}_2\vert}
\end{displaymath}

    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

    \begin{displaymath}
{\bf L}_1 = {\bf r}_1\times {\bf p}_1\;\;\;\;\;\;\;\;\;\;{\bf L}_2 = {\bf r}_2\times {\bf p}_2
\end{displaymath}

    are no longer conserved, i.e.,

    \begin{displaymath}
\left[{\bf L}_1,H\right]\neq 0\;\;\;\;\;\;\;\;\;\;\left[{\bf L}_2,H\right]\neq 0
\end{displaymath}

    To see that this is true, consider the \(z\) components of the angular momentum operators:

    \begin{displaymath}
L_{1z} = {\hbar \over i}\left(x_1{\partial \over \partial y_...
...over \partial y_2} -
y_2 {\partial \over \partial x_2}\right)
\end{displaymath}

    It is straightforward to compute the commutators (left as an exercise for the reader) and it is found that

    \(\displaystyle \left[L_{1z},H\right]\) \(\textstyle =\) $\displaystyle {\hbar \over i}\left(x_1{\partial V \over \partial y_1}-y_1{\partial V \over \partial x_1}
\right)$
    \(\textstyle =\) $\displaystyle {\hbar \over i}
\left[x_1 V'(\vert{\bf r}_1-{\bf r}_2\vert){y_1-y...
...r}_1-{\bf r}_2\vert){x_1-x_2 \over \vert{\bf r}_1-{\bf r}_2\vert}\right] \neq 0$

    Similarly,

    \(\displaystyle \left[L_{2z},H\right]\) \(\textstyle =\) $\displaystyle {\hbar \over i}\left(x_2{\partial V \over \partial y_2}-y_2{\partial V \over \partial x_2}
\right)$
    \(\textstyle =\) $\displaystyle {\hbar \over i}
\left[x_2 V'(\vert{\bf r}_1-{\bf r}_2\vert)\left(...
...bf r}_2\vert)\left(-{x_1-x_2 \over \vert{\bf r}_1-{\bf r}_2\vert}\right)\right]$
    \(\textstyle =\) $\displaystyle -{\hbar \over i}
\left[x_2 V'(\vert{\bf r}_1-{\bf r}_2\vert){y_1-...
...r}_1-{\bf r}_2\vert){x_1-x_2 \over \vert{\bf r}_1-{\bf r}_2\vert}\right] \neq 0$

    However, if we add these together, it can be see that

    \(\displaystyle \left[L_{1z},H\right] + \left[L_{2z},H\right]\) \(\textstyle =\) \(\displaystyle \left[(L_{1z}+L_{2z}),H\right]\)
    \(\textstyle =\) $\displaystyle {\hbar \over i}
\left[V'(\vert{\bf r}_1-{\bf r}_2\vert){(x_1-x_2)...
...bf r}_2\vert){(y_1-y_2)(x_1-x_2) \over \vert{\bf r}_1-{\bf r}_2\vert}\right] =0$

    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

    \begin{displaymath}
{\bf L}= {\bf L}_1+{\bf L}_2
\end{displaymath}

    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

    \begin{displaymath}
H = -\hbar^2 \sum_{i=1}^N {1 \over 2m_i}\nabla_i^2 +
\sum_{i=1}^N\sum_{j=i+1}^N V(\vert{\bf r}_i-{\bf r}_j\vert)
\end{displaymath}

    then the total orbital angular momentum

    \begin{displaymath}
{\bf L}= \sum_{i=1}^N {\bf L}_i
\end{displaymath}

    will be a constant of the motion.


    This page titled 4.2: Total orbital angular momentum is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark E. Tuckerman.

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