Addition of Angular Momenta Chapter 10 Addition of Angular Momenta

Total angular momentum Let us recall the definition of the classical angular momentum of a system of particles: What is the quantum equivalent of the total angular momentum of the system in which there are parts with their own angular momenta? The answer: What are the eigenvalues and the eigenstates associated with this observable?

Addition of two spin-½’s 10.B.1 Addition of two spin-½’s Let us start with the simplest nontrivial case of two angular momenta being added, each of spin ½: The state space is 4D: The orthonormal basis: These are the eigenstates of this set of operators: It comprises a CSCO

Addition of two spin-½’s 10.B.1 Addition of two spin-½’s The eigneproblem: S is also an angular momentum operator: Similarly: So:

Addition of two spin-½’s 10.B.1 Addition of two spin-½’s Another important operator is:

Addition of two spin-½’s 10.B.1 Addition of two spin-½’s Another important operator is:

Addition of two spin-½’s 10.B.1 Addition of two spin-½’s Important commutation relations: What about this commutator?

Addition of two spin-½’s 10.B.1 Addition of two spin-½’s Let us calculate: Since

Basis change We just showed that these observables commute: Since This set is not the same as the CSCO Is this a CSCO?: In other words, is there a new basis that will satisfy these equations?

10.B.2 Basis change Since One can write: Thus: I.e.,

10.B.2 Basis change So:

10.B.3 Basis change Let us recall: Then:

10.B.3 Basis change Let us recall: Then:

10.B.3 Basis change Let us recall: Then:

10.B.3 Basis change Let us recall: Then:

10.B.3 Basis change Synopsizing: Thus:

+ - Basis change Let us diagonalize this matrix: Secular equation: Roots: + -

10.B.3 Basis change Normalizing:

Basis change Thus, we found all four eigenstates: The corresponding values of M: Thus the corresponding values of S:

Basis change So, we successfully obtained the new orthonormal basis: The first three states with S = 1 are called a triplet, and the fourth state with S = 0 is called a singlet

Basis change So, we successfully obtained the new orthonormal basis: As a result of the basis change the subspace allocation became different:

10.C.4 Basis change The same results can be obtained using the ladder operators Recall:

10.C.4 Basis change The same results can be obtained using the ladder operators Recall:

10.C.4 Basis change On the other hand Similarly:

Case of two arbitrary angular momenta Let us consider addition of two angular momenta:

Case of two arbitrary angular momenta The state space is: Each subspace in the tensor sum has a dimension of (2j1 + 1)·(2j2 + 1)

Case of two arbitrary angular momenta J is also an angular momentum operator: Similarly: So: Moreover:

Case of two arbitrary angular momenta Important commutation relations:

Basis change The eigneproblem: We just showed that these observables commute: Since This set is not the same as: Is there a new state set satisfying these equations?

10.C.2 Basis change If this is the case, the following tensor sum can be considered: The following questions have to be answered: 1) What are the values of J for a given pair of j1 and j2? 2) How can the eigenvectors of J2 and Jz be expanded via the basis of eigenvectors of Ji2 and Jiz?

Eigenvalues of J2 and Jz Let us assume that Thus Since One can write What is the degeneracy of these values?

10.C.3 Eigenvalues of J2 and Jz Consider a case

10.C.3 Eigenvalues of J2 and Jz Consider a case

10.C.3 Eigenvalues of J2 and Jz Consider a case

10.C.3 Eigenvalues of J2 and Jz Consider a case

10.C.3 Eigenvalues of J2 and Jz Consider a case

10.C.3 Eigenvalues of J2 and Jz The following dependence is obvious:

10.C.3 Eigenvalues of J2 and Jz The following dependence is obvious:

10.C.3 Eigenvalues of J2 and Jz The following dependence is obvious:

10.C.3 Eigenvalues of J2 and Jz Thus, for given j1 and j2, the eigenvalues of J2 are: One can associate a single invariant subspace E (J) with each of these values For each value of J and each value of M, compatible with it, there is only one state vector in E (j1, j2) Therefore, J2 and Jz comprise a CSCO in E (j1, j2)

Eigenstates of J2 and Jz As a result: 10.C.4 Eigenstates of J2 and Jz As a result: Let us start with the first subspace E (j1 + j2) Setting: Application of the lowering operator should yield:

10.C.4 Eigenstates of J2 and Jz

10.C.4 Eigenstates of J2 and Jz

10.C.4 Eigenstates of J2 and Jz This procedure can be repeated by further applying the lowering ladder operator until reaching The same approach can be used for Here, the new maximum value of M is: The corresponding eigenvector should be proportional to

Eigenstates of J2 and Jz Assuming With It should be orthogonal to: 10.C.4 Eigenstates of J2 and Jz Assuming With It should be orthogonal to: Thus: Solution:

Eigenstates of J2 and Jz Thereby: 10.C.4 Eigenstates of J2 and Jz Thereby: The rest of the states can be obtained by a repeated application of the lowering ladder operator until reaching Then, the tensor sum is considered: Etc.

Clebsch-Gordan coefficients Recall that inside a given space E (j1, j2) this set is a basis: It is complete, thus the closure relation applies: Using it we can write down the following expansion:

Clebsch-Gordan coefficients These coefficients are called Clebsch-Gordan coefficients Rudolf Friedrich Alfred Clebsch (1833 – 1872) Paul Albert Gordan (1837 – 1912)

Clebsch-Gordan coefficients Conventionally, the Clebsch-Gordan coefficients are chosen to be real On the other hand, since this set is a complete basis: The closure relation applies: Thus:

Clebsch-Gordan coefficients The Clebsch-Gordan coefficients can be calculated by iteration using the ladder operators Usually they are compiled in a tabular form:

Clebsch-Gordan coefficients