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, and .
By definition, these operators are Hermitian, and obey the fundamental commutation
relations

(547) | |||

(548) |

We assume that the two groups of operators correspond to different degrees of freedom of the system, so that

(549) |

Let us define the total angular momentum operator

(550) |

(551) |

Now

(553) | |||

(554) |

and also that all of the operators commute with the operators. It follows from Eq. (552) that

(555) |

(556) | |||

(557) |

This suggests that it is not possible to measure the quantum numbers and simultaneously with the quantum number . Thus, we cannot determine the projections of the individual angular momenta along the -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
, and
. The second group is
and . These two
groups of operators are incompatible with one another. We can define simultaneous
eigenkets of each operator group. The simultaneous eigenkets of
, and
are denoted
, where

(558) | |||

(559) | |||

(560) | |||

(561) |

The simultaneous eigenkets of and are denoted , where

(562) | |||

(563) | |||

(564) | |||

(565) |

Each set of eigenkets are complete, mutually orthogonal (for eigenkets corresponding to different sets of eigenvalues), and have unit norms. Since the operators and are common to both operator groups, we can assume that the quantum numbers and 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 and , or the quantum numbers and , 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:

where the right-hand sides denote the identity operator in the ket space corresponding to states of given and . The summation is over all allowed values of , , , and .

The operator group , , , and
is incompatible with the group , , , and .
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 , , , and are known with
certainty, then a measurement of the quantum numbers and will
give a range of possible values. We can use the completeness relation
(566) to write

(568) |

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

(570) |

(571) |

Second, the coefficients vanish unless

(573) |

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

(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

(576) | |||

Making use of the well-known properties of the shift operators, which are specified by Eqs. (351)-(352), we obtain

(577) |

Taking the inner product with , and making use of the orthonormality property of the basis eigenkets, we obtain the desired recursion relations:

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, , are identical to the Clebsch-Gordon coefficients. In other words,

(579) |

(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, . We know, from general principles,
that
and . We also know, from Eq. (572),
that , where the allowed values of differ by integer amounts.
It follows that either or . 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:

The box in this table corresponding to gives the Clebsch-Gordon coefficient , or the inverse Clebsch-Gordon coefficient . 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
. In other words, the -components of angular momentum have to
add algebraically. Many of the boxes in the above table correspond to
. We immediately conclude that these boxes must contain zeroes:
*i.e.*,

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 , but we have no way of determining the
signs at present. Thus,

Let us evaluate the recursion relation (578) for , with
, ,
, taking the upper/lower sign. We
find that

(581) |

(582) |

(583) |

(584) |

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

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 .
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

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

(585) |

(586) |

Note that our table is really a combination of two sub-tables, one involving
states, and one involving states. The Clebsch-Gordon coefficients
corresponding to two different choices of are completely independent:
*i.e.*, there is no recursion relation linking Clebsch-Gordon coefficients
corresponding to different values of . Thus, for every choice of , ,
and we can construct a table of Clebsch-Gordon coefficients corresponding
to the different allowed values of , , and (subject to the
constraint that ). 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
and ) in many standard reference books.