Chapter 13

Addition of Angular Momenta

13.1  INTRODUCTION

In Chapters 9 and 11, we studied about the eigenvalues and eigenstates of orbital angular momentum operator of an electron in the symmetric (Coulomb) field. In Chapter 12, we studied similar quantities of the spin angular momentum operator and components. In fact, in addition to the knowledge of separate orbital and spin angular momenta, we need to know about the total (orbital + spin) angular momentum of an electron or even about total angular momentum of two or more electrons in the atom.

In this chapter, we proceed to study the eigenvalues and eigenstates of the total angular momentum of many particle system. Also we shall study the relation between the angular momentum of the total system and that of its constituent particles.

13.2  ADDITION OF TWO ANGULAR MOMENTA

Let us consider a particle moving in the field of a central potential V(| r |)—the potential being specified with respect to the origin of coordinate system, 0. The central force field may be assumed to be created by a massive particle put at the origin 0. The (orbital) angular momentum of the particle L = r × P, with respect to the fixed origin 0, does not change with the time, that is, L is conserved. The angular momentum operator of the particle commutes with the Hamiltonian of the particle, .

Let us now consider two non-interacting particles 1 and 2 subject to the same central force field. The Hamiltonian of the system may be written as

 

The three components of angular momentum of operator of particle 1 commute with Ĥ₁,

And as all observables relating to particle 1 commute with all those relating to particle 2, commutes with Ĥ₂ as well. So

Equations (13.3) dictate that

Similarly,

If we consider two particles interacting with each other through a potential V(|r₁ – r₂|), which depends only on the distance between them, the Hamiltonian of the system is

 

 

It may be seen that though , do not commute with Ĥ, the total angular momentum operator given as

has all its components commuting with Ĥ. That means, the three components of total orbital angular momentum operator ,

have the commutation relations

Moreover, the commutation relations amongst the components are

In the above example of two particles, we have assumed that the particles do not have spin and, therefore, we considered only the commutation relations of the total orbital angular momentum operator with the Hamiltonian. There is an equally important another case, where a particle has both the orbital angular momentum as well as the spin angular momentum; for example, an electron in hydrogen atom. Let us assume that the electron is subject only to the central potential V(|r|). The Hamiltonian of electron is

The orbital angular momentum operator is expressed in variables (θ, ϕ) and its components commute with Ĥ₀ (as discussed in Chapter 10). Also, as the spin operator is expressed in spin-variable, and its components commute with Ĥ₀ and . So, for the Hamiltonian Ĥ₀ [Eq. (13.10)], , and their components are constants of motion.

We now include the spin–orbit interaction, so the Hamiltonian becomes

 

 

where Ĥ₀ is given by (13.10) and

 

Let us see if the components and commute with the Hamiltonian Ĥ. We have

Similarly,

So, obviously, and do not commute with Ĥ. However, if we define

We easily see from Eqs (13.12) that

is the z-component of the total angular momentum

Similarly, other components of may be shown to be commuting with Ĥ.

Let us start calling the two partial angular momenta of the above two examples as and and ( and represent and of first example, while represent and of second example). So in each case, the total angular momentum operator

has commutation relations

Also, it can be seen in Exercise (13.2) that

where

Now the operators and commute with each other, so we may have a basis of eigenstates common to all these operators.

Let ψ₁( j₁, m₁) be the simultaneous eigenstate of and , so

and

Similarly, let ψ₂(j₂, m₂) be the simultaneous eigenstate of and . Then, the product of the above mentioned two eigenstates, ψ₁(j₁, m₁) and ψ₂(j₂, m₂) may be written as

These form a basis of eigenstates common to all operators . We may also note here that for given values of quantum numbers of j₁ and j₂, there are (2j₁ + 1) and (2j₂ + 1) possible values of m₁ and m₂, respectively; giving in all (2j₁ + 1) (2j₂ + 1) number of possible eigenstates (13.22).

It may, however, be noted in the above case that the operators and do not commute with the total Hamiltonian Ĥ and so are not constants of motion. On the other hand, the total angular momentum operator is a constant of motion [see Eqs (13.14) and (13.17)]. In fact, both the operators and are constants of motion and have simultaneous eigenstates. Therefore, it may be interesting to construct using the basis of Eq. (13.22), a new basis formed by simultaneous eigenstates of and . Now it can be easily seen that the operators and commute with each other [e.g. Exercise (13.2)]. So the new basis states ϕ( j₁, j₂, j, m) are simultaneous eigenstates of operators and with corresponding eigenvalues j₁ (j₁ + 1) ħ², j₂ (j₂ + 1) ħ², j (j + 1) ħ², and mħ, respectively.

Now, we may choose either the orthonormal set of eigenstates

or the orthonormal set of eigenstates

as the basis states.

We have found above that the angular momentum eigenstates for a system of two particles may be characterized by either of the two sets of quantum numbers; the eigenstate characterized by set of quantum numbers {j₁, j₂, m₁, m₂} and the eigenstate characterized by the set {j₁, j₂, j, m}. In eigenstate the z-component and the magnitude of angular moment of individual particles are specified. This will be called as uncoupled representation, where J_1z and J_2z (alongwith and ) are specified. In eigenstate the z-component and the magnitude of angular momentum of the composite system of two particles are specified. This representation is called as coupled representation, where J² and J_z (along with and ) are specified.

The set of eigenstates ϕ may be obtained from the eigenstates ψ through the transformation

or, in ket notation

where the coefficients

are known as Clebsch–Gordon coefficients. The eigenstates ψ and ϕ and the Clebsch–Gordon coefficients C in Eqs (13.24) and (13.25) are written for fixed (values of) quantum numbers j₁ and j₂. Generally, we do not write j₁ and j₂ in designating ψ, ϕ, and C. So, Eqs (13.24) and (13.25) are conventionally written as (with the understanding that j₁ and j₂ are the maximum values of m₁ and m₂)

or

and

We shall use abbreviation for Clebsch–Gordon coefficients as C–G coefficients.

13.3  RECURSION RELATIONS FOR THE C–G COEFFICIENTS

As discussed in Section 13.2, the information about the total angular momentum of two-particle system may be obtained in two different basis states, either in the form of the orthonormal set of eigenstates ψ(j₁, j₂, m₁, m₂) , called as uncoupled representation or in the form of the orthonormal set of states ϕ(j₁, j₂, j, m) , called as coupled representation. These two sets of states are connected with each other through the C–G coefficients [Eqs (13.26)]. Therefore, it is very important to know the C–G coefficients to go from ψ representation of eigenstates to ϕ representation, and vice versa.

To find C–G coefficients, firstly we proceed to find the recursion relations between C–G coefficients. For this, we pick up Eq. (13.24b), which can be re-written as

Let us consider the total angular momentum operator component . It is clear that the operator has eigenstate ϕ(j₁, j₂, j, m) and eigenvalue equation

The eigenvalue equations of operators and are

and

which we may re-write as

and

But we know that , therefore, we have

and

Adding Eqs (13.31a and 13.31b), we get

From Eqs (13.28) and (13.32), we may write

and

Taking Hermitian conjugate of Eq. (13.33b), we have

It may be noted here that is an Hermitian operator. Comparing Eqs (13.33a) and (13.33c), it can be easily concluded that quantum numbers m are subject to the selection rule

 

 

and

Now, let us consider the operator . Its operation on eigenkets of Eq. (13.24b) gives

Using Eq. (11.50a), the above equation gives

or

or

As mentioned earlier, all the above expansions have been written for fixed (given values of) j₁ and j₂. So, we generally omit the symbols j₁, j₂ and re-write the above equation as

Similarly, when operating with on eigenkets of Eq. (13.24b), we get

Here

Combining (13.35a) and (13.35b), we get recursion relations for C–G coefficients as

As emphasized in the beginning of this section, if we know all the C–G coefficients for a given set of values of quantum numbers j₁ and j₂, we can easily find out the orthonormal set of states ϕ(j₁, j₂, j, m) (the coupled representation) from the orthonormal set of states ψ(j₁, j₂, m₁, m₂) (the uncoupled representation) and vice versa. The recursion relations for C–G coefficients [Eq. (13.36)] shall prove to be very useful in finding C–G coefficients.

13.4  THE POSSIBLE VALUES OF j

Let us start with (fixed) known values of j₁ and j₂. We know that the corresponding magnetic quantum numbers m₁ and m₂ have values from j₁ to –j₁ and from j₂ to –j₂, respectively. Also, we know that the possible values of the total magnetic quantum number m are given as m = m₁ + m₂.

We start with Eq. (13.36) (with upper sign) and set m₁ = j₁ and m = j. We get

Use of selection rule (13.34a) in Eq. (13.37) gives m₂ = j − j₁ − 1. Then Eq. (13.37) becomes

Obviously, recursion relation (13.38) dictates that if C−G coefficient is known, we can easily determine another C−G coefficient . Now taking lower sign in Eq. (13.36) and setting m₁ = j₁, m = j − 1, we get

Selection rule (13.34a) when used in Eq. (13.39), gives m₂ = (j − j₁) and then Eq. (13.39) gives

So from the known C–G coefficients and , we can determine the C–G coefficient using Eq. (13.40). Using recursion relations (13.36) for fixed values of j₁, j₂, and j, we can continue in this manner to find all C−G coefficients in terms of just one C−G coefficient, namely.

Now the general expression of C−G coefficient is

along with the selection rule m = m₁ + m₂. Comparing Eq. (13.27) with Eq. (13.41), we have

 

 

 

 

As we know that

 

 

So, we have the range of j given by

 

or

 

In fact, all C−G coefficients could very well be expressed in terms of C−G coefficients (instead of in terms of ). In that case, if we follow the steps corresponding to [Eqs (13.27) and (13.42−(13.44a)], we could have concluded that the range of j be given as

 

 

These two inequalities [Eqs (13.44a) and (13.44b)] taken together, put the range of j as

 

 

Thus the allowed values of j are

 

 

For example,

1. for j₁ = 3, j₂ = 2, allowed values of j are 5, 4, 3, 2, 1
2. for j₁ = 2, , allowed values of j are .

13.5  ADDITION OF TWO SPIN ANGULAR MOMENTA

We know, in the orthonormal set of eigenstates ψ(m₁, m₂) (the uncoupled representation), m₁ and m₂ have, respectively, (2j₁ + 1) and (2j₂ + 1) different values. So there are (2j₁ + 1) (2j₂ + 1) total number of states in the set. In the orthonormal set of states ϕ(j, m) (the coupled representation), the quantum number j takes values from |j₁ – j₂| to (j₁ + j₂) differing by unity. And for each j, the quantum number m has (2j + 1) different values. So, the total number of states in the set is

Thus, the number of ψ(m₁, m₂) states is same as that of ϕ(j₁, m) states; and the C−G coefficients transforming states ψ into ϕ (or vice versa) form a N × N square matrix where N = (2j₁ + 1) (2j₂ + 1).

In the previous section, we found the recursion relations for C−G coefficients. It is straightforward, though lengthy procedure to find values of all C−G coefficients for given j₁ and j₂. In this section, we shall find C−G coefficients for the case not by the rigorous method of using recursion relations (13.36) but using simple ladder operations on eigenstates.

For the case of , the possible values of m₁ and m₂ are , So, we have four different eigenstates ψ(m₁, m₂) (in uncoupled representation).

The allowed values of j are 1 and 0;

1. for j = 1, m = 1, 0 – 1
2. for j = 0, m = 0

So, there are four different eigenstates ϕ(j, m) (in coupled representation).

 

 

We had seen the selection rule dictates that the C−G coefficients are non-zero only if m = m₁ + m₂. This simply means that an eigenstate ψ(m₁, m₂) is related to the eigenstate ϕ(j₁, m), only if m = m₁ + m₂. This fact suggests the following equivalence:

This fact also suggests that each of the remaining two eigenstates ϕ(1, 0) and ϕ(0, 0) are connected with the two eigenstates and . Let us write

Now let the lowering (ladder) operator operate on Eq. (13.47a)

Using Eq. (11.50b), we get

where C_– are given by Eq. (11.51b). Putting values of C_–, C_1–, and C_2–, we get

or

As state ϕ(0, 0) should be orthonormal to ϕ(1, 0), it can be easily seen that ϕ(0, 0) is given by

Equations (13.47) and (13.49) may be put in the matrix form as

The C−G coefficients C(j, m, m₁, m₂) for , case are given by the matrix elements of the 4 × 4 matrix of Eq. (13.50).

Let us now re-look at the (coupled representation) eigenstates ϕ(j, m) of Eqs (13.47) and (13.49). For j = 0, there is only one eigenstate ϕ(0, 0), which is called as singlet state. For j = 1, there are three eigenstates ϕ(1, 1), ϕ(1, 0), and ϕ(1, –1), which are called triplet states. If we look at the expressions of ϕ(0, 0) and ϕ(1, 0), we notice that ϕ(1, 0) is symmetric with respect to the interchange of m₁ and m₂, whereas the state ϕ(0, 0) is antisymmetric in m₁ and m₂. As we know, the value of m represents the z-component of spin angular momentum, so ϕ(0, 0) describes the state with total spin angular momentum having value zero and its z-component with value zero (as j = 0, m = 0). On the other hand, ϕ(1, 0) describes the state with total spin angular momentum having value or and its z-component with value zero (as j = 1, m = 0). In both the states [ϕ(0, 0) and ϕ(1, 0)], the z-component is zero, whereas the magnitude of angular momenta are different. The first case corresponds to the so-called antiparallel spin arrangement and the latter to parallel spin arrangement. In fact, for parallel spin arrangement, where j = 1, the total spin may have three orientations corresponding to m = 1, 0, −1.

As we shall discuss in detail in a later Chapter, for the anti-parallel spin arrangement (i.e., the singlet case), the space part of the (two-particle) total wave function is symmetric in position variables of the two electrons, and the spin wave function is anti-symmetric in the spin variables of the two electrons. This is what we see in the spin wave function ϕ(0, 0) [Eq. (13.49b)] which is anti-symmetric in m₁ and m₂. For the parallel spin arrangement (i.e., the triplet case), the space part of the total wave function is anti-symmetric in position variables and the spin wave function is symmetric in spin variables. This is seen in spin wave function ϕ(1, 0) [Eq. (13.49a)] which is symmetric in m₁ and m₂. To conclude:

1. Singlet case: j = 0, m = 0 (antisymmetric spin eigenstate)
   
2. Triplet case: j = 1, m = 1, 0, –1 (symmetric spin eigenstates)
   
   
   

As mentioned above, the two-electron spin wave function ϕ(j, m) is generally denoted by . Similarly, spin wave function ψ(m₁, m₂) is denoted by . Therefore, the singlet and triplet spin wave functions given above in Eqs (13.51) may be restated as

or

or

and

Sometimes, following notations are used. The singlet state is written as

The triplet states are written as

13.6  ADDITION OF j = 1 AND j = ANGULAR MOMENTA

In this section, we shall find C−G coefficients for the case j₁ = 1 and using simple ladder operations on eigenstates. For j = 1 and , the possible values of m₁ and m₂ are m₁ = 1, 0, −1 and . So in the uncoupled representation, we have six different eigenstates ψ(m₁, m₂):

The allowed values of j extend from j₁ + j₂ to |j₁ – j₂|. So, and are two allowed values.

1. for
2. for

Therefore, in the coupled representation, there are, again, six different eigenstates ϕ(j, m):

Now we know that an eigenstate ψ(m₁, m₂) is related to the eigenstate ϕ(j, m) only if m = m₁ + m₂ (the selection rule). This suggests the following equivalence:

Now operating the lowering (ladder) operator in Eq. (13.55a), we have

Using Eq. (11.50b), we get

where constants C_– are given by Eq. (11.51b). Putting values of C_–, C_1–, and C_2–, we get

or

Operating again on Eq. (13.56a) gives

or

It may be noted here that the last term in above equation gives zero as the lowering operator operating on the state cannot reduce the value of m₂ below its value in . Putting the values of C_–, C_1–, and C_2–, the above equation gives

or

We notice here that the coupled representation state is connected to the two uncoupled representation states, and [Eq. (13.56a)]. [As noticed earlier, a state ϕ(j, m) is connected to the state ψ(m₁, m₂) only in m = m₁ + m₂]. With the same criterion, the state should also be connected to the two states and . But the state has to be orthogonal to the state [given by Eq. (13.56a)]. So the expression of is

Similarly, is expressed in terms of states and moreover, is orthogonal to , so

Expressions (13.55), (13.56), and (13.57) may be put in the matrix form as

The Clebsch–Gordon coefficients for are shown in Table 13.1

Table 13.1 C–G coefficients for

EXERCISES

Exercise 13.1

Quarks are Fermions, and carry spin . Three quarks bind together to make a baryon (such as proto and neutron). A quark and an anti-quark bind together to make a meson such as the pion or kaon. Assume the quarks to be in the ground state so that their orbital angular momentum is zero,

1. What are the possible spins of baryons?
2. What are the possible spins of mesons?

Exercise 13.2

Show that

where

Exercise 13.3

Find out the commutator where . Generalize your result to show that

Exercise 13.4

Show that the total angular momentum of two fermions is always an integer. (In fact, this exercise establishes that two coupled fermions form a composite boson.)

Exercise 13.5

One particle of spin 1 and another of spin 2 are at rest and are in such a configuration that the total spin is 3 and its z-component is ħ.

1. If we measure the z-component of the angular momentum of spin 1 particle, what values might we get and what is the probability of each?
2. If, instead, the measurement is made on spin 2 particle, what will be the answer to the above mentioned question?

Exercise 13.6

The electron in a hydrogen atom is in the combined orbital and spin state

1. If we measure the orbital angular momentum squared L², what values might we get and what is the probability of each?
2. Same question for the z-component of orbital angular momentum L_z.
3. Same question for the spin angular momentum squared S².
4. Same question for the z-component of spin angular momentum S_z.

Exercise 13.7

In the above question, let J = L + S be the total angular momentum. What different values shall we get, if we measure,

1. J²
2. J_z

Exercise 13.8

In the Exercise (13.6)

1. If we measure the position of electron, what is the probability density for finding it at (r, θ, ϕ)?
2. If we measure both the z-component of the spin of the electron and its distance from the origin (these are compatible observables, i.e., corresponding operators are commutating), what is the radial probability density for finding the electron at radius r with spin up?

Exercise 13.9

Consider two particles of spin having spin angular momentum operators and . Find the expectation value of the product S₁ . S₂ in the following states:

1. in the singlet state given by Eq. (13.52a)
2. in the triplet state given by Eq. (13.52c)

Exercise 13.10

For the above case, find the expectation value of S₁ · S₂ in the

1. triplet state given by Eq. (13.52b)
2. triplet state given by Eq. (13.52d)

SOLUTIONS

Solution 13.1

We may visualize the binding of three quarks taking place in two steps, In first step, two quarks bind in singlet and triplet states. In the next step, the two-quark system binds with the third quark. The singlet pair with spin quantum number S = 0 will be bound with the third quark giving net spin quantum number . The triplet pair with spin quantum number S = 1, will combine with the third quark giving net spin quantum number or . So the three quark bound state is either with (a quartet state) or with (a doublet state).

Solution 13.2

We have

and

Now commutes with and commutes with . It is simple to evaluate other commutators and conclude that

Solution 13.3

Similarly, we may find the commutators and . Combining the three commutators, we get

Solution 13.4

Let us consider two fermions: one with orbital angular momentum and spin angular momentum and the other with corresponding operators and .

So the total angular momentum operator is

where

and

The orbital angular momentum quantum number l of two-particle system takes integer values from l₁ + l₂ to |l₁ – l₂|. Now with s₁ = (n₁/2) and s₂ = (n₂/2) (where n₁ and n₂ are odd integers), we find the total spin quantum number s having integer values from [(n₁/2) + (n₂/2)] to | (n₁/2) – (n₂/2)|.

The total angular momentum quantum number j, is then given by

 

Solution 13.5

Here total spin quantum number s = s₁ + s₂ = 3. In the coupled representation, the state is specified by the total spin quantum number s and total magnetic quantum number m_s. The uncoupled representation is specified by the individual states and . In this question, the state in the coupled representation is specified by s = 3 and m_s = 1, that is, the state is .

The coupled state is formed of the uncoupled states with the condition m_s = 1, m_s1 = 1, 0, – 1 and m_s2 = 2 2, 1, 0, –1, –2. The condition m_s1 + m_s2 = 1, is followed by the following combinations:

ms1	ms2	ms = ms1 + ms2
1	0	1
0	1	1
–1	2	1

1. So, if z-component of spin 1 is measured, we will get ħ, 0 –ħ with equal probability.
2. If z-component of spin 2 is measured, we will get 2ħ, ħ, 0 will equal probability.

Solution 13.6

1. L² = 2ħ      with probability 1.
2. L_z = 0      with probability

   L_z = ħ      with probability

3. S² = ħ²      with probability 1

   S_z = – ħ/2      with probability

4. S_z = ħ/2      with probability

Solution 13.7

In the first part of the wave function, we have

so

For the second part of the wave function, we have

so

1.  
   
2.  
   

Solution 13.8

Probability density P(r, θ, ϕ) is given as

Since the spin states χ₊ and χ_– are orthonormalized, the cross term does not contribute. Putting the values of radial and spherical parts, we have

Solution 13.9

We may write

Now

1.  
   
2.  
   

Solution 13.10

1.  
   

   so

   
2. Similarly, we may get
   

 REFERENCES

1. Tinkham, M. 1964. Group Theory and Quantum Mechanics. New York: McGraw-Hill.
2. Rose, M.E. 1957. Elementary Theory of Angular Momentum. NewYork: John Wiley.
3. Schiff, L.I. 1968. Quantum Mechanics. 3rd edn., McGraw-Hill.
4. Gasiorowicz, S. 1996. Quantum Physics. 2nd edn., New York: John Wiley.
5. Bethe, H.A. and Morrison, P. 1956. Elementary Nuclear Physics. New York: John Wiley.