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 equation of the particle commutes with the Hamiltonian of the particle, equation.

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

 

Ĥ0 = Ĥ1 + Ĥ2        (13.1)
equation

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

equation

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

equation

Equations (13.3) dictate that

equation

Similarly,

equation

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

 

Ĥ = Ĥ0 + V(|r1r2|)        (13.5)

 

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

equation

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

equation
equation
equation

have the commutation relations

equation

Moreover, the commutation relations amongst the components are

equation
equation
equation

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

equation

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

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

 

Ĥ = Ĥ0 + Ĥ′        (13.11a)

 

where Ĥ0 is given by (13.10) and

 

equation
equation

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

equation

Similarly,

equation

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

equation

We easily see from Eqs (13.12) that

equation

equation is the z-component of the total angular momentum

equation

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

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

equation

has commutation relations

equation
equation
equation
equation

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

equation

where

equation

Now the operators equation and equation 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 equation and equation, so

equation

and

equation

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

equation

These form a basis of eigenstates common to all operators equation. 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 equation and equation do not commute with the total Hamiltonian Ĥ and so are not constants of motion. On the other hand, the total angular momentum operator equation is a constant of motion [see Eqs (13.14) and (13.17)]. In fact, both the operators equation and equation 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 equation and equation. Now it can be easily seen that the operators equation and equation commute with each other [e.g. Exercise (13.2)]. So the new basis states ϕ( j1, j2, j, m) are simultaneous eigenstates of operators equation and equation with corresponding eigenvalues j1 (j1 + 1) ħ2, j2 (j2 + 1) ħ2, j (j + 1) ħ2, and , respectively.

Now, we may choose either the orthonormal set of eigenstates

equation

or the orthonormal set of eigenstates

equation

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 equation characterized by set of quantum numbers {j1, j2, m1, m2} and the eigenstate equation characterized by the set {j1, j2, j, m}. In eigenstate equation the z-component and the magnitude of angular moment of individual particles are specified. This will be called as uncoupled representation, where J1z and J2z (alongwith equation and equation) are specified. In eigenstate equation 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 Jz (along with equation and equation) are specified.

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

equation

or, in ket notation

equation

where the coefficients

equation

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)

equation

or

equation

and

equation

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) equation, called as uncoupled representation or in the form of the orthonormal set of states ϕ(j1, j2, j, m) equation, 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

equation

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

equation

The eigenvalue equations of operators equation and equation are

equation

and

equation

which we may re-write as

equation

and

equation

But we know that equation, therefore, we have

equation

and

equation

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

equation

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

equation

and

equation

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

equation

It may be noted here that equation 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

 

m = m1 + m2        (13.34a)

 

and

equation

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

equation

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

equation

or

equation

or

equation

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

equation

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

equation

Here

equation

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

equation

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

equation

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

equation

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

equation

Selection rule (13.34a) when used in Eq. (13.39), gives m2 = (jj1) and then Eq. (13.39) gives

equation

So from the known C–G coefficients equation and equation, we can determine the C–G coefficient equation 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.

equation

Now the general expression of C−G coefficient is

equation

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

 

m1 = j1

 

m2 = j − j1        (13.42)

 

m = j

 

As we know that

 

−j2m2j2        (13.43)

 

So, we have the range of j given by

 

−j2jj1j2

or

j1j2jj1 + j2        (13.44a)

 

In fact, all C−G coefficients could very well be expressed in terms of C−G coefficients equation (instead of in terms of equation). 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

 

j2j1jj2 + j1        (13.44b)

 

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

 

|j1j2| ≤ jj1 + j2        (13.44c)

 

Thus the allowed values of j are

 

j1 + j2, j1 + j2 1, ...., | j1− j2| + 1, |j1 − j2|         (13.45)

 

For example,

  1. for j1 = 3, j2 = 2, allowed values of j are 5, 4, 3, 2, 1
  2. for j1 = 2, equation, allowed values of j are equation.
13.5  ADDITION OF TWO SPIN equation 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 |j1j2| 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

equation

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 equation not by the rigorous method of using recursion relations (13.36) but using simple ladder operations on eigenstates.

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

equation

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).

 

ϕ(1, 1), ϕ(1, 0), ϕ(1, –1), and ϕ(0, 0)        (13.46b)

 

We had seen the selection rule dictates that the C−G coefficients equation 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:

equation
equation

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

equation
equation

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

equation

Using Eq. (11.50b), we get

equation

where C are given by Eq. (11.51b). Putting values of C, C1, and C2, we get

equation

or

equation

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

equation

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

equation

The C−G coefficients C(j, m, m1, m2) equation for equation, equation 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 equation or equationand 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)
    equation
  2. Triplet case: j = 1, m = 1, 0, –1 (symmetric spin eigenstates)
    equation
    equation
    equation

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

equation

or

equation

or

equation

and

equation
equation
equation

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

equation

The triplet states are written as

equation
equation
equation
13.6  ADDITION OF j = 1 AND j = equation ANGULAR MOMENTA

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

equation

The allowed values of j extend from j1 + j2 to |j1j2|. So, equation and equation are two allowed values.

  1. for equation
  2. for equation

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

equation

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:

equation
equation

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

equation

Using Eq. (11.50b), we get

equation

where constants C are given by Eq. (11.51b). Putting values of C, C1, and C2, we get

equation

or

equation

Operating equation again on Eq. (13.56a) gives

equation

or

equation

It may be noted here that the last term in above equation equation gives zero as the lowering operator equation operating on the state equation cannot reduce the value of m2 below its valueequation in equation. Putting the values of C, C1, and C2, the above equation gives

equation

or

equation

We notice here that the coupled representation state equation is connected to the two uncoupled representation states, equation and equation [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 equation should also be connected to the two states equation and equation. But the state equation has to be orthogonal to the state equation [given by Eq. (13.56a)]. So the expression of equation is

equation

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

equation

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

equation

The Clebsch–Gordon coefficientsequation for equation are shown in Table 13.1

Table 13.1 C–G coefficients Table 13.1 for equation

equation
EXERCISES

Exercise 13.1

Quarks are Fermions, and carry spin equation. 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

equation

where

equation

Exercise 13.3

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

equation

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

equation
  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 Lz.
  3. Same question for the spin angular momentum squared S2.
  4. Same question for the z-component of spin angular momentum Sz.

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. Jz

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 equation having spin angular momentum operators equation and equation. 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 equation. The triplet pair with spin quantum number S = 1, will combine with the third quark giving net spin quantum number equation or equation. So the three quark bound state is either with equation (a quartet state) or with equation (a doublet state).

Solution 13.2

We have

equation

and

equation

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

equation

Solution 13.3

equation

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

equation

Solution 13.4

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

So the total angular momentum operator is

equation

where

equation

and

equation

The orbital angular momentum quantum number l of two-particle system takes integer values from l1 + l2 to |l1l2|. 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

 

j = (l + s) to |ls|,   which are integers.

Solution 13.5

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

The coupled state is formed of the uncoupled states with the condition ms = 1, ms1 = 1, 0, – 1 and ms2 = 2 2, 1, 0, –1, –2. The condition ms1 + ms2 = 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. Lz = 0      with probability equation

    Lz = ħ      with probability equation

  3. S2 = equationħ2      with probability 1

    Sz = – ħ/2      with probability equation

  4. Sz = ħ/2      with probability equation

Solution 13.7

In the first part of the wave function, we have

equation

so

equation

For the second part of the wave function, we have

equation

so

equation
  1.  
    equation
  2.  
    equation

Solution 13.8

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

equation

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

equation

Solution 13.9

We may write

equation

Now

equation
  1.  
    equation
  2.  
    equation

Solution 13.10

  1.  
    equation

    so

    equation
  2. Similarly, we may get
    equation
 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.
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