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Next: Approximation methods Up: Angular momentum Previous: Spin greater than one-half

Addition of angular momentum

Consider a hydrogen atom in an $l=1$ state. The electron possesses orbital angular momentum of magnitude $\hbar$, and spin angular momentum of magnitude $\hbar/2$. So, what is the total angular momentum of the system?

In order to answer this question, we are going to have to learn how to add angular momentum operators. Let us consider the most general case. Suppose that we have two sets of angular momentum operators, ${\bf J}_1$ and ${\bf J}_2$. By definition, these operators are Hermitian, and obey the fundamental commutation relations

$\displaystyle {\bf J}_1\times {\bf J}_1$ $\textstyle =$ $\displaystyle {\rm i} \hbar {\bf J}_1,$ (547)
$\displaystyle {\bf J}_2\times {\bf J}_2$ $\textstyle =$ $\displaystyle {\rm i} \hbar {\bf J}_2.$ (548)

We assume that the two groups of operators correspond to different degrees of freedom of the system, so that
\begin{displaymath}[J_{1i}, J_{2j}]= 0,
\end{displaymath} (549)

where $i, j$ stand for either $x$, $y$, or $z$. For instance, ${\bf J}_1$ could be an orbital angular momentum operator, and ${\bf J}_2$ a spin angular momentum operator. Alternatively, ${\bf J}_1$ and ${\bf J}_2$ could be the orbital angular momentum operators of two different particles in a multi-particle system. We know, from the general properties of angular momentum, that the eigenvalues of $J_1^{ 2}$ and $J_2^{ 2}$ can be written $j_1 (j_1+1) \hbar^2$ and $j_2 (j_2+1)  \hbar^2$, respectively, where $j_1$ and $j_2$ are either integers, or half-integers. We also know that the eigenvalues of $J_{1z}$ and $J_{2z}$ take the form $m_1 \hbar$ and $m_2 \hbar$, respectively, where $m_1$ and $m_2$ are numbers lying in the ranges $j_1, j_1-1,\cdots,
-j_1+1, -j_1$ and $j_2, j_2-1,\cdots,
-j_2+1, -j_2$, respectively.

Let us define the total angular momentum operator

\begin{displaymath}
{\bf J} = {\bf J}_1 + {\bf J}_2.
\end{displaymath} (550)

Now ${\bf J}$ is an Hermitian operator, since it is the sum of Hermitian operators. According to Eqs. (307) and (310), ${\bf J}$ satisfies the fundamental commutation relation
\begin{displaymath}
{\bf J} \times {\bf J} = {\rm i} \hbar  {\bf J}.
\end{displaymath} (551)

Thus, ${\bf J}$ possesses all of the expected properties of an angular momentum operator. It follows that the eigenvalue of $J^{ 2}$ can be written $j (j+1) \hbar^2$, where $j$ is an integer, or a half-integer. The eigenvalue of $J_z$ takes the form $m \hbar$, where $m$ lies in the range $j, j-1,\cdots,
-j+1, -j$. At this stage, we do not know the relationship between the quantum numbers of the total angular momentum, $j$ and $m$, and those of the individual angular momenta, $j_1$, $j_2$, $m_1$, and $m_2$.

Now

\begin{displaymath}
J^{ 2} = J_1^{ 2} + J_2^{ 2} + 2 {\bf J}_1 \!\cdot\! {\bf J}_2.
\end{displaymath} (552)

We know that
$\displaystyle [J_1^{ 2}, J_{1i} ]$ $\textstyle =$ $\displaystyle 0,$ (553)
$\displaystyle [J_2^{ 2}, J_{2i} ]$ $\textstyle =$ $\displaystyle 0,$ (554)

and also that all of the $J_{1i}$ operators commute with the $J_{2i}$ operators. It follows from Eq. (552) that
\begin{displaymath}[J^{ 2}, J_1^{ 2}]= [J^{ 2}, J_2^{ 2}] = 0.
\end{displaymath} (555)

This implies that the quantum numbers $j_1$, $j_2$, and $j$ can all be measured simultaneously. In other words, we can know the magnitude of the total angular momentum together with the magnitudes of the component angular momenta. However, it is clear from Eq. (552) that
$\displaystyle [J^2, J_{1z}]$ $\textstyle \neq$ $\displaystyle 0,$ (556)
$\displaystyle [J^2, J_{2z}]$ $\textstyle \neq$ $\displaystyle 0.$ (557)

This suggests that it is not possible to measure the quantum numbers $m_1$ and $m_2$ simultaneously with the quantum number $j$. Thus, we cannot determine the projections of the individual angular momenta along the $z$-axis at the same time as the magnitude of the total angular momentum.

It is clear, from the preceding discussion, that we can form two alternate groups of mutually commuting operators. The first group is $J_1^{ 2}, J_2^{ 2}, J_{1z}$, and $J_{2z}$. The second group is $J_1^{ 2}, J_2^{ 2}, J^{ 2},$ and $J_z$. These two groups of operators are incompatible with one another. We can define simultaneous eigenkets of each operator group. The simultaneous eigenkets of $J_1^{ 2}, J_2^{ 2}, J_{1z}$, and $J_{2z}$ are denoted $\vert j_1,j_2; m_1,m_2\rangle$, where

$\displaystyle J_1^{ 2} \vert j_1,j_2; m_1,m_2\rangle$ $\textstyle =$ $\displaystyle j_1 (j_1+1) \hbar^2 \vert j_1,j_2; m_1,m_2\rangle,$ (558)
$\displaystyle J_2^{ 2} \vert j_1,j_2; m_1,m_2\rangle$ $\textstyle =$ $\displaystyle j_2 (j_2+1) \hbar^2 \vert j_1,j_2; m_1,m_2\rangle,$ (559)
$\displaystyle J_{1z} \vert j_1,j_2; m_1,m_2\rangle$ $\textstyle =$ $\displaystyle m_1 \hbar \vert j_1,j_2; m_1,m_2\rangle,$ (560)
$\displaystyle J_{2z} \vert j_1,j_2; m_1,m_2\rangle$ $\textstyle =$ $\displaystyle m_2 \hbar \vert j_1,j_2; m_1,m_2\rangle.$ (561)

The simultaneous eigenkets of $J_1^{ 2}, J_2^{ 2}, J^{ 2}$ and $J_z$ are denoted $\vert j_1, j_2; j, m\rangle$, where
$\displaystyle J_1^{ 2} \vert j_1,j_2; j,m\rangle$ $\textstyle =$ $\displaystyle j_1 (j_1+1) \hbar^2 \vert j_1,j_2; j,m\rangle,$ (562)
$\displaystyle J_2^{ 2} \vert j_1,j_2; j,m\rangle$ $\textstyle =$ $\displaystyle j_2 (j_2+1) \hbar^2 \vert j_1,j_2; j,m\rangle,$ (563)
$\displaystyle J^2 \vert j_1,j_2; j,m\rangle$ $\textstyle =$ $\displaystyle j (j+1) \hbar^2 \vert j_1,j_2; j,m\rangle,$ (564)
$\displaystyle J_{z} \vert j_1,j_2; j,m\rangle$ $\textstyle =$ $\displaystyle m \hbar \vert j_1,j_2; j,m\rangle.$ (565)

Each set of eigenkets are complete, mutually orthogonal (for eigenkets corresponding to different sets of eigenvalues), and have unit norms. Since the operators $J_1^{ 2}$ and $J_2^{ 2}$ are common to both operator groups, we can assume that the quantum numbers $j_1$ and $j_2$ are known. In other words, we can always determine the magnitudes of the individual angular momenta. In addition, we can either know the quantum numbers $m_1$ and $m_2$, or the quantum numbers $j$ and $m$, but we cannot know both pairs of quantum numbers at the same time. We can write a conventional completeness relation for both sets of eigenkets:
$\displaystyle \sum_{m_1}\sum_{m_2 }\vert j_1,j_2; m_1, m_2\rangle \langle j_1,j_2; m_1, m_2\vert$ $\textstyle =$ $\displaystyle 1,$ (566)
$\displaystyle [0.5ex]
\sum_{j}\sum_{m} \vert j_1,j_2; j, m\rangle \langle j_1,j_2; j, m\vert$ $\textstyle =$ $\displaystyle 1,$ (567)

where the right-hand sides denote the identity operator in the ket space corresponding to states of given $j_1$ and $j_2$. The summation is over all allowed values of $m_1$, $m_2$, $j$, and $m$.

The operator group $J_1^{ 2}$, $J_2^{ 2}$, $J^{ 2}$, and $J_z$ is incompatible with the group $J_1^{ 2}$, $J_2^{ 2}$, $J_{1z}$, and $J_{2z}$. This means that if the system is in a simultaneous eigenstate of the former group then, in general, it is not in an eigenstate of the latter. In other words, if the quantum numbers $j_1$, $j_2$, $j$, and $m$ are known with certainty, then a measurement of the quantum numbers $m_1$ and $m_2$ will give a range of possible values. We can use the completeness relation (566) to write

\begin{displaymath}
\vert j_1,j_2;j,m\rangle = \sum_{m_1}\sum_{m_2} \langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle
\vert j_1,j_2;m_1,m_2\rangle.
\end{displaymath} (568)

Thus, we can write the eigenkets of the first group of operators as a weighted sum of the eigenkets of the second set. The weights, $\langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle$, are called the Clebsch-Gordon coefficients. If the system is in a state where a measurement of $J_1^{ 2}, J_2^{ 2}, J^{ 2}$, and $J_z$ is bound to give the results $j_1 (j_1+1) \hbar^2, j_2 (j_2+1) \hbar^2, j (j+1) \hbar^2$, and $j_z \hbar$, respectively, then a measurement of $J_{1z}$ and $J_{2z}$ will give the results $m_1 \hbar$ and $m_2 \hbar$ with probability $\vert\langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle\vert^2$.

The Clebsch-Gordon coefficients possess a number of very important properties. First, the coefficients are zero unless

\begin{displaymath}
m = m_1 + m_2.
\end{displaymath} (569)

To prove this, we note that
\begin{displaymath}
(J_z - J_{1z} - J_{2z}) \vert j_1,j_2; j, m\rangle =0.
\end{displaymath} (570)

Forming the inner product with $\langle j_1, j_2; m_1, m_2\vert$, we obtain
\begin{displaymath}
(m-m_1-m_2) \langle j_1, j_2; m_1, m_2\vert j_1,j_2; j, m\rangle=0,
\end{displaymath} (571)

which proves the assertion. Thus, the $z$-components of different angular momenta add algebraically. So, an electron in an $l=1$ state, with orbital angular momentum $\hbar$, and spin angular momentum $\hbar/2$, projected along the $z$-axis, constitutes a state whose total angular momentum projected along the $z$-axis is $3\hbar/2$. What is uncertain is the magnitude of the total angular momentum.

Second, the coefficients vanish unless

\begin{displaymath}
\vert j_1-j_2\vert \leq j \leq j_1+j_2.
\end{displaymath} (572)

We can assume, without loss of generality, that $j_1\geq j_2$. We know, from Eq. (569), that for given $j_1$ and $j_2$ the largest possible value of $m$ is $j_1+j_2$ (since $j_1$ is the largest possible value of $m_1$, etc.). This implies that the largest possible value of $j$ is $j_1+j_2$ (since, by definition, the largest value of $m$ is equal to $j$). Now, there are $(2 j_1+1)$ allowable values of $m_1$ and $(2 j_2+1)$ allowable values of $m_2$. Thus, there are $(2 j_1+1) (2 j_2+1)$ independent eigenkets, $\vert j_1,j_2; m_1,m_2\rangle$, needed to span the ket space corresponding to fixed $j_1$ and $j_2$. Since the eigenkets $\vert j_1, j_2; j, m\rangle$ span the same space, they must also form a set of $(2 j_1+1) (2 j_2+1)$ independent kets. In other words, there can only be $(2 j_1+1) (2 j_2+1)$ distinct allowable values of the quantum numbers $j$ and $m$. For each allowed value of $j$, there are $2 j+1$ allowed values of $m$. We have already seen that the maximum allowed value of $j$ is $j_1+j_2$. It is easily seen that if the minimum allowed value of $j$ is $j_1-j_2$ then the total number of allowed values of $j$ and $m$ is $(2 j_1+1) (2 j_2+1)$: i.e.,
\begin{displaymath}
\sum_{j=j_1-j_2}^{j_1+j_2} (2 j+1) \equiv (2 j_1+1) (2 j_2+1).
\end{displaymath} (573)

This proves our assertion.

Third, the sum of the modulus squared of all of the Clebsch-Gordon coefficients is unity: i.e.,

\begin{displaymath}
\sum_{m_1}\sum_{m_2} \vert\langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle\vert^2 =1.
\end{displaymath} (574)

This assertion is proved as follows:
    $\displaystyle \langle j_1, j_2; j, m\vert j_1, j_2; j, m\rangle =\mbox{\hspace{7cm}}$  
    $\displaystyle \sum_{m_1}\sum_{m_2} \langle j_1, j_2; j, m\vert j_1, j_2; m_1, m_2\rangle
\langle j_1, j_2; m_1, m_2\vert j_1, j_2; j, m\rangle$  
    $\displaystyle   =\sum_{m_1}\sum_{m_2} \vert\langle j_1,j_2;m_1,m_2\vert j_1,j_2;j,m\rangle\vert^2 =1,$ (575)

where use has been made of the completeness relation (566).

Finally, the Clebsch-Gordon coefficients obey two recursion relations. To obtain these relations we start from

$\displaystyle J^{\pm} \vert j_1,j_2;j,m\rangle$ $\textstyle =$ $\displaystyle (J_1^\pm + J_2^\pm )$ (576)
    $\displaystyle \times\sum_{m_1'}\sum_{m_2'} \langle j_1, j_2; m_1', m_2'\vert j_1, j_2; j, m\rangle
\vert j_1, j_2; m_1', m_2'\rangle.$  

Making use of the well-known properties of the shift operators, which are specified by Eqs. (351)-(352), we obtain
$\displaystyle \sqrt{j (j+1)- m (m\pm 1)}  \vert j_1,j_2;j,m\pm 1\rangle =$      
$\displaystyle \sum_{m_1'}\sum_{m_2'} \left( \sqrt{j_1 (j_1+1)- m_1' (m_1'\pm 1)} 
\vert j_1, j_2; m_1'\pm 1, m_2'\rangle\right.$      
$\displaystyle \left.+\sqrt{ j_2 (j_2+1)- m_2' (m_2'\pm 1)} 
\vert j_1, j_2; m_1', m_2'\pm 1\rangle\right)$      
$\displaystyle \times\langle j_1, j_2; m_1', m_2'\vert j_1, j_2; j, m\rangle.$     (577)

Taking the inner product with $\langle j_1, j_2; m_1, m_2\vert$, and making use of the orthonormality property of the basis eigenkets, we obtain the desired recursion relations:
$\displaystyle \sqrt{j (j+1)- m (m\pm 1)}  \langle j_1, j_2; m_1, m_2\vert j_1,j_2;j, m\pm 1\rangle=$      
$\displaystyle \sqrt{j_1 (j_1+1) - m_1 (m_1\mp 1)}  
\langle j_1, j_2; m_1\mp 1, m_2\vert j_1,j_2;j, m\rangle$      
$\displaystyle + \sqrt{j_2 (j_2+1) - m_2 (m_2\mp 1)}  
\langle j_1, j_2; m_1, m_2\mp 1\vert j_1,j_2;j, m\rangle.$     (578)

It is clear, from the absence of complex coupling coefficients in the above relations, that we can always choose the Clebsch-Gordon coefficients to be real numbers. This is a convenient choice, since it ensures that the inverse Clebsch-Gordon coefficients, $\langle j_1, j_2; j, m\vert j_1, j_2; m_1, m_2\rangle$, are identical to the Clebsch-Gordon coefficients. In other words,
\begin{displaymath}
\langle j_1, j_2; j, m\vert j_1, j_2; m_1, m_2\rangle =
\langle j_1, j_2; m_1, m_2\vert j_1, j_2; j, m\rangle.
\end{displaymath} (579)

The inverse Clebsch-Gordon coefficients are the weights in the expansion of the $\vert j_1,j_2; m_1,m_2\rangle$ in terms of the $\vert j_1, j_2; j, m\rangle$:
\begin{displaymath}
\vert j_1,j_2; m_1,m_2\rangle = \sum_{j}\sum_m \langle j_1, ...
...j, m\vert j_1, j_2; m_1, m_2\rangle \vert j_1,j_2; j,m\rangle.
\end{displaymath} (580)

It turns out that the recursion relations (578), together with the normalization condition (574), are sufficient to completely determine the Clebsch-Gordon coefficients to within an arbitrary sign (multiplied into all of the coefficients). This sign is fixed by convention. The easiest way of demonstrating this assertion is by considering some specific examples.

Let us add the angular momentum of two spin one-half systems: e.g., two electrons at rest. So, $j_1=j_2=1/2$. We know, from general principles, that $\vert m_1\vert \leq 1/2$ and $\vert m_2\vert\leq 1/2$. We also know, from Eq. (572), that $0\leq j\leq 1$, where the allowed values of $j$ differ by integer amounts. It follows that either $j=0$ or $j=1$. Thus, two spin one-half systems can be combined to form either a spin zero system or a spin one system. It is helpful to arrange all of the possibly non-zero Clebsch-Gordon coefficients in a table:

\begin{displaymath}
\begin{tabular}{\vert\vert c\vert c\vert\vert c\vert c\vert ...
...ptstyle j_2=1/2$ & $m$ & 1 & 0 & -1 & 0 \ \hline
\end{tabular}\end{displaymath}

The box in this table corresponding to $m_1=1/2, m_2=1/2, j=1, m=1$ gives the Clebsch-Gordon coefficient $\langle 1/2, 1/2; 1/2, 1/2
\vert 1/2, 1/2; 1, 1\rangle$, or the inverse Clebsch-Gordon coefficient $\langle 1/2, 1/2; 1, 1\vert
1/2, 1/2; 1/2, 1/2
\rangle$. All the boxes contain question marks because we do not know any Clebsch-Gordon coefficients at the moment.

A Clebsch-Gordon coefficient is automatically zero unless $m_1+m_2=m$. In other words, the $z$-components of angular momentum have to add algebraically. Many of the boxes in the above table correspond to $m_1+m_2\neq m$. We immediately conclude that these boxes must contain zeroes: i.e.,

\begin{displaymath}
\begin{tabular}{\vert\vert c\vert c\vert\vert c\vert c\vert ...
...ptstyle j_2=1/2$ & $m$ & 1 & 0 & -1 & 0 \ \hline
\end{tabular}\end{displaymath}

The normalization condition (574) implies that the sum of the squares of all the rows and columns of the above table must be unity. There are two rows and two columns which only contain a single non-zero entry. We conclude that these entries must be $\pm 1$, but we have no way of determining the signs at present. Thus,

\begin{displaymath}
\begin{tabular}{\vert\vert c\vert c\vert\vert c\vert c\vert ...
...ptstyle j_2=1/2$ & $m$ & 1 & 0 & -1 & 0 \ \hline
\end{tabular}\end{displaymath}

Let us evaluate the recursion relation (578) for $j_1=j_2=1/2$, with $j=1$, $m=0$, $m_1=m_2=\pm 1/2$, taking the upper/lower sign. We find that

\begin{displaymath}
\langle 1/2, -1/2\vert 1, 0\rangle
+ \langle -1/2, 1/2\vert ...
...angle =\sqrt{2} \langle 1/2, 1/2\vert 1,1\rangle=\pm\sqrt{2},
\end{displaymath} (581)

and
\begin{displaymath}
\langle 1/2, -1/2\vert 1, 0\rangle
+ \langle -1/2, 1/2\vert...
... =\sqrt{2}  \langle -1/2, -1/2\vert 1,-1\rangle
=\pm\sqrt{2}.
\end{displaymath} (582)

Here, the $j_1$ and $j_2$ labels have been suppressed for ease of notation. We also know that
\begin{displaymath}
\langle 1/2, -1/2\vert 1, 0\rangle^2 +
\langle -1/2, 1/2\vert 1, 0\rangle^2 = 1,
\end{displaymath} (583)

from the normalization condition. The only real solutions to the above set of equations are
$\displaystyle \sqrt{2}  \langle 1/2, -1/2\vert 1, 0\rangle$ $\textstyle =$ $\displaystyle \sqrt{2}  \langle -1/2, 1/2\vert 1, 0\rangle$  
  $\textstyle =$ $\displaystyle \langle 1/2,1/2\vert 1,1\rangle = \langle 1/2,1/2\vert 1,-1\rangle = \pm 1.$ (584)

The choice of sign is arbitrary--the conventional choice is a positive sign. Thus, our table now reads

\begin{displaymath}
\begin{tabular}{\vert\vert c\vert c\vert\vert c\vert c\vert ...
...ptstyle j_2=1/2$ & $m$ & 1 & 0 & -1 & 0 \ \hline
\end{tabular}\end{displaymath}

We could fill in the remaining unknown entries of our table by using the recursion relation again. However, an easier method is to observe that the rows and columns of the table must all be mutually orthogonal. That is, the dot product of a row with any other row must be zero. Likewise, for the dot product of a column with any other column. This follows because the entries in the table give the expansion coefficients of one of our alternative sets of eigenkets in terms of the other set, and each set of eigenkets contains mutually orthogonal vectors with unit norms. The normalization condition tells us that the dot product of a row or column with itself must be unity. The only way that the dot product of the fourth column with the second column can be zero is if the unknown entries are equal and opposite. The requirement that the dot product of the fourth column with itself is unity tells us that the magnitudes of the unknown entries have to be $1/\sqrt{2}$. The unknown entries are undetermined to an arbitrary sign multiplied into them both. Thus, the final form of our table (with the conventional choice of arbitrary signs) is

\begin{displaymath}
\begin{tabular}{\vert\vert c\vert c\vert\vert c\vert c\vert ...
...ptstyle j_2=1/2$ & $m$ & 1 & 0 & -1 & 0 \ \hline
\end{tabular}\end{displaymath}

The table can be read in one of two ways. The columns give the expansions of the eigenstates of overall angular momentum in terms of the eigenstates of the individual angular momenta of the two component systems. Thus, the second column tells us that

\begin{displaymath}
\vert 1,0\rangle = \frac{1}{\sqrt{2}} \left( \vert 1/2,-1/2\rangle + \vert-1/2,1/2\rangle \right).
\end{displaymath} (585)

The ket on the left-hand side is a $\vert j,m\rangle$ ket, whereas those on the right-hand side are $\vert m_1, m_2\rangle$ kets. The rows give the expansions of the eigenstates of individual angular momentum in terms of those of overall angular momentum. Thus, the second row tells us that
\begin{displaymath}
\vert 1/2,-1/2\rangle = \frac{1}{\sqrt{2}} \left( \vert 1,0\rangle + \vert,0\rangle \right).
\end{displaymath} (586)

Here, the ket on the left-hand side is a $\vert m_1, m_2\rangle$ ket, whereas those on the right-hand side are $\vert j,m\rangle$ kets.

Note that our table is really a combination of two sub-tables, one involving $j=0$ states, and one involving $j=1$ states. The Clebsch-Gordon coefficients corresponding to two different choices of $j$ are completely independent: i.e., there is no recursion relation linking Clebsch-Gordon coefficients corresponding to different values of $j$. Thus, for every choice of $j_1$, $j_2$, and $j$ we can construct a table of Clebsch-Gordon coefficients corresponding to the different allowed values of $m_1$, $m_2$, and $m$ (subject to the constraint that $m_1+m_2=m$). A complete knowledge of angular momentum addition is equivalent to a knowing all possible tables of Clebsch-Gordon coefficients. These tables are listed (for moderate values of $j_1, j_2$ and $j$) in many standard reference books.


next up previous
Next: Approximation methods Up: Angular momentum Previous: Spin greater than one-half
Richard Fitzpatrick 2006-02-16