In the previous article, we showed that the total -component spin angular momentum operator is , which is a special case of the general form:

(where ) is the total angular momentum component operator. and are component operators of and respectively. and are operators of two sources of angular momentum, where they may be: 1) the orbital angular momentum operator of particle 1 and orbital angular momentum operator of particle 2 respectively; 2) the spin angular momentum operator of particle 1 and spin angular momentum operator of particle 2 respectively; 3) the orbital angular momentum operator and spin angular momentum operator respectively of a particle.

Also mentioned in the previous article is that and . It is therefore easy to accept the validity of points 1) and 2). For point 3, the proposal that may seem untenable. However, spin angular momentum, like orbital angular momentum, is a form of angular momentum. In fact, the total angular momentum of a system is defined as the vector sum . If point 3) is valid, must satisfy the same commutation relations as described by eq99, eq100 and eq101.

###### Question

Show that satisfies the same commutation relations as described by eq99, eq100 and eq101.

###### Answer

Expanding the RHS of the above equation and noting that and commute because they act on different vector spaces, we have

With reference to eq99, eq100, eq101, eq165, eq166 and eq167, , where and is the ** Levi-Civita symbol**. So,

Similarly, we have and .

Since the total angular momentum component operators satisfy the form of commutation relations as described by eq99, eq100 and eq101, the raising and lowering operators also apply to the total angular momentum operator . We would therefore expect

You’ll realise from the workings of the above Q&A that we can simplify the notation of eq205 as

To show that commutes with , where can be either or , we have and . Therefore, the eigenstate of is simultaneously the eigenstates of and . This implies that the eigenvalues of are the sum of the eigenvalues of and , i.e. , or

In other words, the allowed values of the total magnetic quantum number are the sum of the allowed values of the two contributing magnetic quantum numbers. As for the allowed values of the total angular momentum quantum number , let’s further define the eigenvalues of and as and respectively. This allows us to work with the quantum numbers and .

Now, the maximum value of in eq207 is . Since the maximum value of a magnetic quantum number is the angular momentum quantum number (i.e. and , where ), the highest value of is

Furthermore, **for a particular value of **** **in the coupled representation, there are values of and therefore states. So has states. These states are . The states for the next lower value of (denoted by ) are . The same logic applies for states all the way to the lowest value of .

###### Question

Show that the total number of states in the uncoupled representation is .

###### Answer

In eq193, the total number of states in the uncoupled representation is the number of ways to form Kronecker products of basis vectors from each vector space. Since there are basis vectors in the 1^{st} vector space and basis vectors in the 2^{nd} vector space,

To determine the lower values of , we consider the lower values of , the first being . There are two possible ways to obtain this value, with and , or and . Since each state is characterised by a unique value of for a particular value of , one of the two possibilities is accounted for by the state . The remaining possibility must be due to . Since , we must have . Furthermore, because , we have . The state is therefore .

For , there are three possible ways to obtain it. Again, one of the possible ways is accounted for by and the second way by . The remaining possibility must be due to the state .

Therefore, the allowed values of are

or

To determine , we note that the total number of states for the system can be written as because there are states associated with each value of . Since , we can further split the sum as:

###### Question

Show that and hence .

###### Answer

For the 2^{nd} term on RHS of 2^{nd} equality of eq211, , which if written in the reverse order becomes . Adding the two sums, we have

For the 3^{rd} term on RHS of 2^{nd} equality of in eq211

Substitute eq214 and eq215 back in eq211, we have . Let , we have

Using eq216, where for , and for , eq210 becomes

The total number of states (energy levels) of a system must be independent of the chosen representation. Substituting eq209 in LHS of eq217 and eq208 in RHS of eq217 and simplifying,

Eq218 is equivalent to because , , , and may be a larger value than . Therefore, for a given value of and a given value of , the allowed values of the total angular momentum quantum number are:

which is called the * Clebsch-Gordan series*.

###### Question

Write all the eigenstates (in the form of ) and basis states (in the form of ) of a system with two sources of angular momentum, and .

###### Answer

There are a total of 15 eigenstates and also 15 basis states. The allowed values of are 3, 2 and 1. The eigenstates are , , , , , , , , , , , , , and . The basis states are , , , , , , , , , , , , , and . Each spin eigenstate of the system is a linear combination of the 15 basis states.

What we have described so far pertains to a system with two sources of angular momentum. If the system has more than two sources of angular momentum, the Clebsch-Gordan series is applied repeatedly, i.e. a first series is written with and , and then the Clebsch-Gordan procedure is again applied to each value of this series with to form a second resultant series, and the procedure is repeated until a final resultant series is developed with . For example, a system with three sources of angular momentum, , and , has the following allowed values of :

1st series using and ,

2nd and final series using and ,

For this system, there are 8 basis states, whose explicit forms can be expressed as follows:

###### Question

What are the allowed angular momenta of a system with three sources of angular momentum, , and , and how many basis states are there in total?

###### Answer

. The total number of basis states is 27.