4.2: Total orbital angular momentum

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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.

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

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

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

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

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