1. Quantum Two

3. Angular Momentum and Rotations

4. Angular Momentum and Rotations The Angular Momentum Addition Theorem

5. In the last segment we saw that the problem of reducing a direct product space S = S₁⊗S₂
associated with two subsystems with angular momenta and into irreducible invariant subspaces can be broken down into a series of subtasks, by reducing each invariant, but generally reducible, direct product subspace
S(j₁) ⊗ S(j₂)
formed from irreducible invariant subspaces from S₁ and S₂ with fixed j₁ and j₂ into direct sums of their irreducible components.
In this segment we state and then prove the Angular Momentum Addition Theorem which shows exactly how this can be done.

6. In the last segment we saw that the problem of reducing a direct product space S = S₁⊗S₂
associated with two subsystems with angular momenta and into irreducible invariant subspaces can be broken down into a series of subtasks, by reducing each invariant, but generally reducible, direct product subspace
S(j₁) ⊗ S(j₂)
formed from irreducible invariant subspaces from S₁ and S₂ with fixed j₁ and j₂ into direct sums of their irreducible components.
In this segment we state and then prove the Angular Momentum Addition Theorem which shows exactly how this can be done.

7. In the last segment we saw that the problem of reducing a direct product space S = S₁⊗S₂
associated with two subsystems with angular momenta and into irreducible invariant subspaces can be broken down into a series of subtasks, by reducing each invariant, but generally reducible, direct product subspace
S(j₁) ⊗ S(j₂)
formed from irreducible invariant subspaces from S₁ and S₂ with fixed j₁ and j₂ into direct sums of their irreducible components.
In this segment we state and then prove the Angular Momentum Addition Theorem which shows exactly how this can be done.

8. The Angular Momentum Addition Theorem:
The (2j₁+1)(2j₂+1) dimensional space S(j₁, j₂) = S(j₁) ⊗ S(j₂) contains exactly one irreducible rotationally invariant subspace S(j) for each value of j in the sequence
j = j₁ + j₂, j₁ + j₂ - 1, ⋯ , |j₁ - j₂|.
In other words, the subspace S(j₁, j₂) = S(j₁) ⊗ S(j₂) formed from the direct product of two irreducible invariant subspaces can be reduced into the following direct sum
S(j₁) ⊗ S(j₂) = S(j₁+ j₂) ⊕ S(j₁+ j₂-1) ⊕ ⋯ ⊕ S(|j₁ - j₂|)
of irreducible invariant subspaces, where each space S(j) is spanned by 2j+1 basis vectors {|j, m〉 | m = -j, … , j}.

9. The Angular Momentum Addition Theorem:
The (2j₁+1)(2j₂+1) dimensional space S(j₁, j₂) = S(j₁) ⊗ S(j₂) contains exactly one irreducible rotationally invariant subspace S(j) for each value of j in the sequence
j = j₁ + j₂, j₁ + j₂ - 1, ⋯ , |j₁ - j₂|.
In other words, the subspace S(j₁, j₂) = S(j₁) ⊗ S(j₂) formed from the direct product of two irreducible invariant subspaces can be reduced into the following direct sum
S(j₁) ⊗ S(j₂) = S(j₁+ j₂) ⊕ S(j₁+ j₂-1) ⊕ ⋯ ⊕ S(|j₁ - j₂|)
of irreducible invariant subspaces, where each space S(j) is spanned by 2j+1 basis vectors {|j, m〉 | m = -j, … , j}.

10. The Angular Momentum Addition Theorem:
The (2j₁+1)(2j₂+1) dimensional space S(j₁, j₂) = S(j₁) ⊗ S(j₂) contains exactly one irreducible rotationally invariant subspace S(j) for each value of j in the sequence
j = j₁ + j₂, j₁ + j₂ - 1, ⋯ , |j₁ - j₂|.
In other words, the subspace S(j₁, j₂) = S(j₁) ⊗ S(j₂) formed from the direct product of two irreducible invariant subspaces can be reduced into the following direct sum
S(j₁) ⊗ S(j₂) = S(j₁+ j₂) ⊕ S(j₁+ j₂-1) ⊕ ⋯ ⊕ S(|j₁ - j₂|)
of irreducible invariant subspaces, where each space S(j) is spanned by 2j+1 basis vectors {|j, m〉 | m = -j, … , j}.

11. Thus, for example, if we combine an irreducible subspace with j₁ = 3 and j₂ = 1, we get three irreducible invariant subspaces, i.e.,
S(3) ⊗ S(1) = S(4) ⊕ S(3) ⊕ S(2)
starting with j = j₁ + j₂ = 4, and descending down to j = |j₁ - j₂| = 2.
For a spin-½ particle moving on the unit sphere, if we combine the orbital angular momentum j₁ = ℓ of the particle with the spin j₂ = s = ½, then we get for each ℓ > 0, two irreducible invariant subspaces, i.e.,
S(ℓ) ⊗ S(½)=S(ℓ + ½) ⊕ S(ℓ - ½)
corresponding to values of total j = ℓ ± ½.
For the state with zero orbital angular momentum, i.e., ℓ = 0, we only get one irreducible subspace corresponding to j = ½.

12. Thus, for example, if we combine an irreducible subspace with j₁ = 3 and j₂ = 1, we get three irreducible invariant subspaces, i.e.,
S(3) ⊗ S(1) = S(4) ⊕ S(3) ⊕ S(2)
starting with j = j₁ + j₂ = 4, and descending down to j = |j₁ - j₂| = 2.
For a spin-½ particle moving on the unit sphere, if we combine the orbital angular momentum j₁ = ℓ of the particle with the spin j₂ = s = ½, then we get for each ℓ > 0, two irreducible invariant subspaces, i.e.,
S(ℓ) ⊗ S(½)=S(ℓ + ½) ⊕ S(ℓ - ½)
corresponding to values of total j = ℓ ± ½.
For the state with zero orbital angular momentum, i.e., ℓ = 0, we only get one irreducible subspace corresponding to j = ½.

13. Thus, for example, if we combine an irreducible subspace with j₁ = 3 and j₂ = 1, we get three irreducible invariant subspaces, i.e.,
S(3) ⊗ S(1) = S(4) ⊕ S(3) ⊕ S(2)
starting with j = j₁ + j₂ = 4, and descending down to j = |j₁ - j₂| = 2.
For a spin-½ particle moving on the unit sphere, if we combine the orbital angular momentum j₁ = ℓ of the particle with the spin j₂ = s = ½, then we get for each ℓ > 0, two irreducible invariant subspaces, i.e.,
S(ℓ) ⊗ S(½)=S(ℓ + ½) ⊕ S(ℓ - ½)
corresponding to values of total j = ℓ ± ½.
For the state with zero orbital angular momentum, i.e., ℓ = 0, we only get one irreducible subspace corresponding to j = ½.

14. To prove this result we first simplify the notation, make a few general observations, and then follow up with what is essentially a proof-by-construction.
Regarding the notation, we will use the fact that j₁ and j₂ are assumed fixed and introduce the simpler notation
|m₁, m₂〉 = |j₁, m₁ ; j₂, m₂〉
for the original direct product states, which are eigenvectors of J₁² and J₂² with eigenvalues j₁(j₁+1) and j₂(j₂+1), respectively, and of J1z and J2z with eigenvalues m₁ and m₂.
Given the statement of the theorem, the sought-after common eigenstates of the total angular momentum (i.e., of J² and Jz) in this subspace we will denote by
|j, m〉 = | j₁, j₂, j, m〉
to be formed as linear combinations of the direct product states |m₁, m₂〉.

15. To prove this result we first simplify the notation, make a few general observations, and then follow up with what is essentially a proof-by-construction.
Regarding the notation, we will use the fact that j₁ and j₂ are assumed fixed and introduce the simpler notation
|m₁, m₂〉 = |j₁, m₁ ; j₂, m₂〉
for the original direct product states, which are eigenvectors of J₁² and J₂² with eigenvalues j₁(j₁+1) and j₂(j₂+1), respectively, and of J1z and J2z with eigenvalues m₁ and m₂.
Given the statement of the theorem, the sought-after common eigenstates of the total angular momentum (i.e., of J² and Jz) in this subspace we will denote by
|j, m〉 = | j₁, j₂, j, m〉
to be formed as linear combinations of the direct product states |m₁, m₂〉.

16. To prove this result we first simplify the notation, make a few general observations, and then follow up with what is essentially a proof-by-construction.
Regarding the notation, we will use the fact that j₁ and j₂ are assumed fixed and introduce the simpler notation
|m₁, m₂〉 = |j₁, m₁ ; j₂, m₂〉
for the original direct product states, which are eigenvectors of J₁² and J₂² with eigenvalues j₁(j₁+1) and j₂(j₂+1), respectively, and of J1z and J2z with eigenvalues m₁ and m₂.
Given the statement of the theorem, the sought-after common eigenstates of the total angular momentum (i.e., of J² and Jz) in this subspace we will denote by
|j, m〉 = | j₁, j₂, j, m〉
to be formed as linear combinations of the direct product states |m₁, m₂〉.

17. As we will see the states |j, m〉 will be eigenstates of J₁², J₂², J², and Jz, but except for special cases they will not be eigenstates of J1z and J2z.
Then, to proceed, we note first that the space S(j₁, j₂) contains states | m₁, m₂〉 with
m₁ = -j₁, … , j₁ and m₂ = - j₂, … , j₂
Thus, the eigenvalues m = m₁ + m₂ of Jz within this subspace take on all values in the sequence
m₁ = - (j₁ + j₂), … , (j₁ + j₂)
this implies that the eigenvalues of J² must be labeled by values of j such that
(j₁ + j₂) ≥ j.

18. As we will see the states |j, m〉 will be eigenstates of J₁², J₂², J², and Jz, but except for special cases they will not be eigenstates of J1z and J2z.
Then, to proceed, we note first that the space S(j₁, j₂) contains states | m₁, m₂〉 with
m₁ = -j₁, … , j₁ and m₂ = - j₂, … , j₂
Thus, the eigenvalues m = m₁ + m₂ of Jz within this subspace take on all values in the sequence
m₁ = - (j₁ + j₂), … , (j₁ + j₂)
this implies that the eigenvalues of J² must be labeled by values of j such that
(j₁ + j₂) ≥ j.

19. As we will see the states |j, m〉 will be eigenstates of J₁², J₂², J², and Jz, but except for special cases they will not be eigenstates of J1z and J2z.
Then, to proceed, we note first that the space S(j₁, j₂) contains states | m₁, m₂〉 with
m₁ = -j₁, … , j₁ and m₂ = - j₂, … , j₂
Thus, the eigenvalues m = m₁ + m₂ of Jz within this subspace take on all values in the sequence
m = - (j₁ + j₂), … , (j₁ + j₂)
this implies that the eigenvalues of J² must be labeled by values of j such that
(j₁ + j₂) ≥ j.

20. As we will see the states |j, m〉 will be eigenstates of J₁², J₂², J², and Jz, but except for special cases they will not be eigenstates of J1z and J2z.
Then, to proceed, we note first that the space S(j₁, j₂) contains states | m₁, m₂〉 with
m₁ = -j₁, … , j₁ and m₂ = - j₂, … , j₂
Thus, the eigenvalues m = m₁ + m₂ of Jz within this subspace take on all values in the sequence
m = - (j₁ + j₂), … , (j₁ + j₂)
This implies that the eigenvalues of J² must be labeled by values of j such that
(j₁ + j₂) ≥ j.

21. The second thing to note is that the value of m = m₁ + m₂ will be integral if m₁ and m₂ are both integral or if both are half-integral.
The value of m will be half-integral if m₁ is integral and m₂ is half-integral, or the other way around.
Since the integral or half integral character of m, m₁, and m₂ is determined by the character of j, j₁ and j₂, we deduce that

22. The second thing to note is that the value of m = m₁ + m₂ will be integral if m₁ and m₂ are both integral or if both are half-integral.
The value of m will be half-integral if m₁ is integral and m₂ is half-integral, or the other way around.
Since the integral or half integral character of m, m₁, and m₂ is determined by the character of j, j₁ and j₂, we deduce that

23. The second thing to note is that the value of m = m₁ + m₂ will be integral if m₁ and m₂ are both integral or if both are half-integral.
The value of m will be half-integral if m₁ is integral and m₂ is half-integral, or the other way around.
Since the integral or half integral character of m, m₁, and m₂ is determined by the character of j, j₁ and j₂, we deduce that

24. The second thing to note is that the value of m = m₁ + m₂ will be integral if m₁ and m₂ are both integral or if both are half-integral.
The value of m will be half-integral if m₁ is integral and m₂ is half-integral, or the other way around.
Since the integral or half integral character of m, m₁, and m₂ is determined by the character of j, j₁ and j₂, we deduce that

25. With these preliminary observations out of the way, we now proceed to observe the following:
Fact 1. In the subspace S(j₁, j₂) there is only one direct product state |m₁, m₂〉 in which the value of m = m₁ + m₂ takes on its largest value, m = j₁ + j₂, that is, the state |m₁, m₂〉 in which
m₁ = j₁ and m₂ = j₂
We now note that, because
there is only one of these in the entire space, and because
states of total angular momentum j always come in 2j + 1 fold multiplets,
this single eigenstate of Jz with the largest value of m must be a maximally aligned state |j, j〉 i.e., a state of total angular momentum (j, m), such that m = j = j₁ + j₂.

26. With these preliminary observations out of the way, we now proceed to observe the following:
Fact 1. In the subspace S(j₁, j₂) there is only one direct product state |m₁, m₂〉 in which the value of m = m₁ + m₂ takes on its largest value, m = j₁ + j₂, that is, the state |m₁, m₂〉 in which
m₁ = j₁ and m₂ = j₂
We now note that, because
there is only one of these in the entire space, and because
states of total angular momentum j always come in 2j + 1 fold multiplets,
this single eigenstate of Jz with the largest value of m must be a maximally aligned state |j, j〉 i.e., a state of total angular momentum (j, m), such that m = j = j₁ + j₂.

27. With these preliminary observations out of the way, we now proceed to observe the following:
Fact 1. In the subspace S(j₁, j₂) there is only one direct product state |m₁, m₂〉 in which the value of m = m₁ + m₂ takes on its largest value, m = j₁ + j₂, that is, the state |m₁, m₂〉 in which
m₁ = j₁ and m₂ = j₂
We now note that, because
there is only one of these in the entire space, and because
states of total angular momentum j always come in 2j + 1 fold multiplets,
this single eigenstate of Jz with the largest value of m must be a maximally aligned state |j, j〉 i.e., a state of total angular momentum (j, m), such that m = j = j₁ + j₂.

28. With these preliminary observations out of the way, we now proceed to observe the following:
Fact 1. In the subspace S(j₁, j₂) there is only one direct product state |m₁, m₂〉 in which the value of m = m₁ + m₂ takes on its largest value, m = j₁ + j₂, that is, the state |m₁, m₂〉 in which
m₁ = j₁ and m₂ = j₂
We now note that, because
there is only one of these in the entire space, and because
states of total angular momentum j always come in 2j + 1 fold multiplets,
this single eigenstate of Jz with the largest value of m must be a maximally aligned state |j, j〉 i.e., a state of total angular momentum (j, m), such that m = j = j₁ + j₂.

29. With these preliminary observations out of the way, we now proceed to observe the following:
Fact 1. In the subspace S(j₁, j₂) there is only one direct product state |m₁, m₂〉 in which the value of m = m₁ + m₂ takes on its largest value, m = j₁ + j₂, that is, the state |m₁, m₂〉 in which
m₁ = j₁ and m₂ = j₂
We now note that, because
there is only one of these in the entire space, and because
states of total angular momentum j always come in 2j + 1 fold multiplets,
this single eigenstate of Jz with the largest value of m must be a maximally aligned state |j, j〉 i.e., a state of total angular momentum (j, m), such that m = j = j₁ + j₂.

30. With these preliminary observations out of the way, we now proceed to observe the following:
Fact 1. In the subspace S(j₁, j₂) there is only one direct product state |m₁, m₂〉 in which the value of m = m₁ + m₂ takes on its largest value, m = j₁ + j₂, that is, the state |m₁, m₂〉 in which
m₁ = j₁ and m₂ = j₂
We now note that, because
there is only one of these in the entire space, and because
states of total angular momentum j always come in 2j + 1 fold multiplets,
this single eigenstate of Jz with the largest value of m must be a maximally aligned state |j, j〉 i.e., a state of total angular momentum (j, m), such that m = j = j₁ + j₂.

31. Thus, this state must be the "top element" element of a 2j + 1 sequence of such states with j = j₁ + j₂.
So S(j₁, j₂) contains at least one irreducible invariant subspace S(j) with j = j₁ + j₂.
In general there would have to be one such vector starting each sequence, for each irreducible invariant subspaces with this value of j.
Since there is only one such state in S(j₁, j₂) of this sort, it must contain only one irreducible subspace S(j) with j = j₁ + j₂ .
Thus we make the identification
|j₁ + j₂, j₁ + j₂ 〉 = | j₁, j₂〉,
j , m m₁, m₂
where the left side of this expression indicates the |j, m〉 state, and the right side indicates the direct product state |m₁, m₂〉.

32. Thus, this state must be the "top element" element of a 2j + 1 sequence of such states with j = j₁ + j₂.
So S(j₁, j₂) contains at least one irreducible invariant subspace S(j) with j = j₁ + j₂.
In general there would have to be one such vector starting each sequence, for each irreducible invariant subspaces with this value of j.
Since there is only one such state in S(j₁, j₂) of this sort, it must contain only one irreducible subspace S(j) with j = j₁ + j₂ .
Thus we make the identification
|j₁ + j₂, j₁ + j₂ 〉 = | j₁, j₂〉,
j , m m₁, m₂
where the left side of this expression indicates the |j, m〉 state, and the right side indicates the direct product state |m₁, m₂〉.

33. Thus, this state must be the "top element" element of a 2j + 1 sequence of such states with j = j₁ + j₂.
So S(j₁, j₂) contains at least one irreducible invariant subspace S(j) with j = j₁ + j₂.
In general there would have to be one such vector starting each sequence, for each irreducible invariant subspace with this value of j.
Since there is only one such state in S(j₁, j₂) of this sort, it must contain only one irreducible subspace S(j) with j = j₁ + j₂ .
Thus we make the identification
|j₁ + j₂, j₁ + j₂ 〉 = | j₁, j₂〉,
j , m m₁, m₂
where the left side of this expression indicates the |j, m〉 state, and the right side indicates the direct product state |m₁, m₂〉.

34. Thus, this state must be the "top element" element of a 2j + 1 sequence of such states with j = j₁ + j₂.
So S(j₁, j₂) contains at least one irreducible invariant subspace S(j) with j = j₁ + j₂.
In general there would have to be one such vector starting each sequence, for each irreducible invariant subspace with this value of j.
Since there is only one such state in S(j₁, j₂) of this sort, it must contain only one irreducible subspace S(j) with j = j₁ + j₂ .
Thus we make the identification
|j₁ + j₂, j₁ + j₂ 〉 = | j₁, j₂〉,
j , m m₁, m₂
where the left side of this expression indicates the |j, m〉 state, and the right side indicates the direct product state |m₁, m₂〉.

35. Thus, this state must be the "top element" element of a 2j + 1 sequence of such states with j = j₁ + j₂.
So S(j₁, j₂) contains at least one irreducible invariant subspace S(j) with j = j₁ + j₂.
In general there would have to be one such vector starting each sequence, for each irreducible invariant subspace with this value of j.
Since there is only one such state in S(j₁, j₂) of this sort, it must contain only one irreducible subspace S(j) with j = j₁ + j₂ .
Thus we make the identification
|j₁ + j₂, j₁ + j₂ 〉 = | j₁, j₂〉,
j , m m₁, m₂
where the left side of this expression indicates the |j, m〉 state, and the right side indicates the direct product state |m₁, m₂〉.

36. The remaining basis states |j, m〉 in this irreducible space S(j) with j = j₁ + j₂ can now be produced by repeated application of the lowering operator
For example, we note that, in the |j, m〉 representation with j and m both equal to j₁ + j₂ ,
But , so this same expression can be written in the |m₁, m₂〉 representation, after some manipulation, as . . .

37. The remaining basis states |j, m〉 in this irreducible space S(j) with j = j₁ + j₂ can now be produced by repeated application of the lowering operator
For example, we note that, in the |j, m〉 representation with j and m both equal to j₁ + j₂ ,
But , so this same expression can be written in the |m₁, m₂〉 representation, after some manipulation, as . . .

38. The remaining basis states |j, m〉 in this irreducible space S(j) with j = j₁ + j₂ can now be produced by repeated application of the lowering operator
For example, we note that, in the |j, m〉 representation with j and m both equal to j₁ + j₂ ,
But , so this same expression can be written in the |m₁, m₂〉 representation, after some manipulation, as . . .

39. The remaining basis states |j, m〉 in this irreducible space S(j) with j = j₁ + j₂ can now be produced by repeated application of the lowering operator
For example, we note that, in the |j, m〉 representation with j and m both equal to j₁ + j₂ ,
But , so this same expression can be written in the |m₁, m₂〉 representation, after some manipulation, as . . .

40. The remaining basis states |j, m〉 in this irreducible space S(j) with j = j₁ + j₂ can now be produced by repeated application of the lowering operator
For example, we note that, in the |j, m〉 representation with j and m both equal to j₁ + j₂ ,
But , so this same expression can be written in the |m₁, m₂〉 representation, after some manipulation, as . . .

41. Equating these last two results gives after a little work
This procedure can obviously be repeated for the remaining basis vectors with this value of j.

42. Equating these last two results gives after a little work
This procedure can obviously be repeated for the remaining basis vectors with this value of j.

43. Equating these last two results gives after a little work
This procedure can obviously be repeated for the remaining basis vectors with this value of j.

44. Equating these last two results gives after a little work
This procedure can obviously be repeated for the remaining basis vectors with this value of j.

45. Equating these last two results gives after a little work
This procedure can obviously be repeated for the remaining basis vectors with this value of j.

46. We now proceed to essentially repeat the argument, by noting
Fact 2: there are exactly two direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its next largest value, m = j₁ + j₂ - 1, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 1〉
and m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ 〉
From these two orthogonal states we can produce any eigenvectors of Jz in S(j₁, j₂) having eigenvalue m = j₁ + j₂ - 1.
Indeed, we have already used these two states to produce the state |j, m〉 with j = j₁ + j₂ and m = j₁ + j₂ - 1.

47. We now proceed to essentially repeat the argument, by noting
Fact 2: there are exactly two direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its next largest value, m = j₁ + j₂ - 1, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 1〉
and m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ 〉
From these two orthogonal states we can produce any eigenvectors of Jz in S(j₁, j₂) having eigenvalue m = j₁ + j₂ - 1.
Indeed, we have already used these two states to produce the state |j, m〉 with j = j₁ + j₂ and m = j₁ + j₂ - 1.

48. We now proceed to essentially repeat the argument, by noting
Fact 2: there are exactly two direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its next largest value, m = j₁ + j₂ - 1, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 1〉
and m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ 〉
From these two orthogonal states we can produce any eigenvectors of Jz in S(j₁, j₂) having eigenvalue m = j₁ + j₂ - 1.
Indeed, we have already used these two states to produce the state |j, m〉 with j = j₁ + j₂ and m = j₁ + j₂ - 1.

49. We now proceed to essentially repeat the argument, by noting
Fact 2: there are exactly two direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its next largest value, m = j₁ + j₂ - 1, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 1〉
and m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ 〉
From these two orthogonal states we can produce any eigenvectors of Jz in S(j₁, j₂) having eigenvalue m = j₁ + j₂ - 1.
Indeed, we have already used these two states to produce the state |j, m〉 with j = j₁ + j₂ and m = j₁ + j₂ - 1.

50. We now proceed to essentially repeat the argument, by noting
Fact 2: there are exactly two direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its next largest value, m = j₁ + j₂ - 1, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 1〉
and m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ 〉
From these two orthogonal states we can produce any eigenvectors of Jz in S(j₁, j₂) having eigenvalue m = j₁ + j₂ - 1.
Indeed, we have already used these two states to produce the state |j, m〉 with j = j₁ + j₂ and m = j₁ + j₂ - 1.

51. We now proceed to essentially repeat the argument, by noting
Fact 2: there are exactly two direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its next largest value, m = j₁ + j₂ - 1, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 1〉
and m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ 〉
From these two orthogonal states we can produce any eigenvectors of Jz in S(j₁, j₂) having eigenvalue m = j₁ + j₂ - 1.
Indeed, we have already used these two states to produce the state |j, m〉 with j = j₁ + j₂ and m = j₁ + j₂ - 1.

52. But we can also produce from these states a vector orthogonal to the one we already constructed, i.e., the vector
But this state is not in our previously constructed irreducible invariant subspace.
So it must be in some other one.
And it is the only state unaccounted for with this large a value of m.
So, this latter state must be the maximally aligned eigenstate | j, j 〉 of J² and Jz with j = m = j₁ + j₂ - 1.
In other words, it is that top vector of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 1.

53. But we can also produce from these states a vector orthogonal to the one we already constructed, i.e., the vector
But this state is not in our previously constructed irreducible invariant subspace.
So it must be in some other one.
And it is the only state unaccounted for with this large a value of m.
So, this latter state must be the maximally aligned eigenstate | j, j 〉 of J² and Jz with j = m = j₁ + j₂ - 1.
In other words, it is that top vector of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 1.

54. But we can also produce from these states a vector orthogonal to the one we already constructed, i.e., the vector
But this state is not in our previously constructed irreducible invariant subspace.
So it must be in some other one.
And it is the only state unaccounted for with this large a value of m.
So, this latter state must be the maximally aligned eigenstate | j, j 〉 of J² and Jz with j = m = j₁ + j₂ - 1.
In other words, it is that top vector of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 1.

55. But we can also produce from these states a vector orthogonal to the one we already constructed, i.e., the vector
But this state is not in our previously constructed irreducible invariant subspace.
So it must be in some other one.
And it is the only state unaccounted for with this large a value of m.
So, this latter state must be the maximally aligned eigenstate | j, j 〉 of J² and Jz with j = m = j₁ + j₂ - 1.
In other words, it is that top vector of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 1.

56. But we can also produce from these states a vector orthogonal to the one we already constructed, i.e., the vector
But this state is not in our previously constructed irreducible invariant subspace.
So it must be in some other one.
And it is the only state unaccounted for with this large a value of m.
So, this latter state must be the maximally aligned eigenstate | j, j 〉 of J² and Jz with j = m = j₁ + j₂ - 1.
In other words, it is that top vector of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 1.

57. But we can also produce from these states a vector orthogonal to the one we already constructed, i.e., the vector
But this state is not in our previously constructed irreducible invariant subspace.
So it must be in some other one.
And it is the only state unaccounted for with this large a value of m.
So, this latter state must be the maximally aligned eigenstate | j, j 〉 of J² and Jz with j = m = j₁ + j₂ - 1.
In other words, it is that top vector of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 1.

58. But we can also produce from these states a vector orthogonal to the one we already constructed, i.e., the vector
But this state is not in our previously constructed irreducible invariant subspace.
So it must be in some other one.
And it is the only state unaccounted for with this large a value of m.
So, this latter state must be the maximally aligned eigenstate | j, j 〉 of J² and Jz with j = m = j₁ + j₂ - 1.
In other words, it is that top vector of a sequence of basis vectors for an irreducible invariant subspace S(j) with j = j₁ + j₂ - 1.

59. Since there are no more vectors with this value of m, there can be only one irreducible subspace for this value of j.
The remaining vectors in this sequence can be constructed by repeated application of

60. Since there are no more vectors with this value of m, there can be only one irreducible subspace for this value of j.
The remaining vectors in this sequence can be constructed by repeated application of

61. To proceed we note
Fact 3: that (unless we have run out of vectors) there are exactly three direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its third largest value, m = j₁ + j₂ - 2, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 2〉
m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ - 1〉
m ₁ = j₁ m₂ = j₂ |j₁ - 2, j₂ 〉
From these states, we have already constructed those states with this value of m in the spaces S(j) with j = j₁ + j₂, and j = j₁ + j₂ - 1.
We can therefore construct one vector orthogonal to those two.
This remaining vector must by the top element (i.e., maximally aligned vector | j, j 〉 of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 2, the remaining elements of which we can obtain using

62. To proceed we note
Fact 3: that (unless we have run out of vectors) there are exactly three direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its third largest value, m = j₁ + j₂ - 2, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 2〉
m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ - 1〉
m ₁ = j₁ m₂ = j₂ |j₁ - 2, j₂ 〉
From these states, we have already constructed those states with this value of m in the spaces S(j) with j = j₁ + j₂, and j = j₁ + j₂ - 1.
We can therefore construct one vector orthogonal to those two.
This remaining vector must by the top element (i.e., maximally aligned vector | j, j 〉 of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 2, the remaining elements of which we can obtain using

63. To proceed we note
Fact 3: that (unless we have run out of vectors) there are exactly three direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its third largest value, m = j₁ + j₂ - 2, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 2〉
m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ - 1〉
m ₁ = j₁ m₂ = j₂ |j₁ - 2, j₂ 〉
From these states, we have already constructed those states with this value of m in the spaces S(j) with j = j₁ + j₂, and j = j₁ + j₂ - 1.
We can therefore construct one vector orthogonal to those two.
This remaining vector must by the top element (i.e., maximally aligned vector | j, j 〉 of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 2, the remaining elements of which we can obtain using

64. To proceed we note
Fact 3: that (unless we have run out of vectors) there are exactly three direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its third largest value, m = j₁ + j₂ - 2, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 2〉
m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ - 1〉
m ₁ = j₁ m₂ = j₂ |j₁ - 2, j₂ 〉
From these states, we have already constructed those states with this value of m in the spaces S(j) with j = j₁ + j₂, and j = j₁ + j₂ - 1.
We can therefore construct one vector orthogonal to those two.
This remaining vector must by the top element (i.e., maximally aligned vector | j, j 〉 of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 2, the remaining elements of which we can obtain using

65. To proceed we note
Fact 3: that (unless we have run out of vectors) there are exactly three direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its third largest value, m = j₁ + j₂ - 2, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 2〉
m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ - 1〉
m ₁ = j₁ m₂ = j₂ |j₁ - 2, j₂ 〉
From these states, we have already constructed those states with this value of m in the spaces S(j) with j = j₁ + j₂, and j = j₁ + j₂ - 1.
We can therefore construct one vector orthogonal to those two.
This remaining vector must by the top element (i.e., maximally aligned vector | j, j 〉 of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 2, the remaining elements of which we can obtain using

66. To proceed we note
Fact 3: that (unless we have run out of vectors) there are exactly three direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its third largest value, m = j₁ + j₂ - 2, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 2〉
m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ - 1〉
m ₁ = j₁ m₂ = j₂ |j₁ - 2, j₂ 〉
From these states, we have already constructed those states with this value of m in the spaces S(j) with j = j₁ + j₂, and j = j₁ + j₂ - 1.
We can therefore construct one vector orthogonal to those two.
This remaining vector must by the top element (i.e., maximally aligned vector | j, j 〉 of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 2, the remaining elements of which we can obtain using

67. To proceed we note
Fact 3: that (unless we have run out of vectors) there are exactly three direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its third largest value, m = j₁ + j₂ - 2, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 2〉
m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ - 1〉
m ₁ = j₁ m₂ = j₂ |j₁ - 2, j₂ 〉
From these states, we have already constructed those states with this value of m in the spaces S(j) with j = j₁ + j₂, and j = j₁ + j₂ - 1.
We can therefore construct one vector orthogonal to those two.
This remaining vector must by the top element (i.e., maximally aligned vector | j, j 〉 of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 2, the remaining elements of which we can obtain using

68. To proceed we note
Fact 3: that (unless we have run out of vectors) there are exactly three direct product states |m₁, m₂〉 in S(j₁, j₂) in which the value of m = m₁ + m₂ takes on its third largest value, m = j₁ + j₂ - 2, namely, the states with
m ₁ = j₁ m₂ = j₂ |j₁, j₂ - 2〉
m ₁ = j₁ m₂ = j₂ |j₁ - 1, j₂ - 1〉
m ₁ = j₁ m₂ = j₂ |j₁ - 2, j₂ 〉
From these states, we have already constructed those states with this value of m in the spaces S(j) with j = j₁ + j₂, and j = j₁ + j₂ - 1.
We can therefore construct one vector orthogonal to those two.
This remaining vector must by the top element (i.e., maximally aligned vector | j, j 〉 of a sequence of basis vectors for an irreducible subspace S(j) with j = j₁ + j₂ - 2, the remaining elements of which we can obtain using

69. The next steps, we hope, are clear: repeat this procedure until we run out of vectors.
The basic idea is that as we move down through each value of j we always encounter just enough linearly-independent direct product states
m ₁ = j₁ m₂ = j₂ - n |j₁, j₂ - n〉
⋮ ⋮ ⋮
m ₁ = j₁ - n m₂ = j₂ |j₁ - n, j₂ 〉
with a given value of m = j₁ + j₂ - n to form the basis vectors associated with those irreducible spaces already generated with higher values of j, as well as one additional vector which can be constructed orthogonal to those from these other irreducible spaces already constructed.
This remaining state forms the beginning vector for a new sequence of 2j + 1 basis vectors associated with the next value of j = j₁ + j₂ - n.

70. The next steps, we hope, are clear: repeat this procedure until we run out of vectors.
The basic idea is that as we move down through each value of j we always encounter just enough linearly-independent direct product states
m ₁ = j₁ m₂ = j₂ - n |j₁, j₂ - n〉
⋮ ⋮ ⋮
m ₁ = j₁ - n m₂ = j₂ |j₁ - n, j₂ 〉
with a given value of m = j₁ + j₂ - n to form the basis vectors associated with those irreducible spaces already generated with higher values of j, as well as one additional vector which can be constructed orthogonal to those from these other irreducible spaces already constructed.
This remaining state forms the beginning vector for a new sequence of 2j + 1 basis vectors associated with the next value of j = j₁ + j₂ - n.

71. The next steps, we hope, are clear: repeat this procedure until we run out of vectors.
The basic idea is that as we move down through each value of j we always encounter just enough linearly-independent direct product states
m ₁ = j₁ m₂ = j₂ - n |j₁, j₂ - n〉
⋮ ⋮ ⋮
m ₁ = j₁ - n m₂ = j₂ |j₁ - n, j₂ 〉
with a given value of m = j₁ + j₂ - n to form the basis vectors associated with those irreducible spaces already generated with higher values of j, as well as one additional vector which can be constructed orthogonal to those from these other irreducible spaces already constructed.
This remaining state forms the beginning vector for a new sequence of 2j + 1 basis vectors associated with the next value of j = j₁ + j₂ - n.

72. The next steps, we hope, are clear: repeat this procedure until we run out of vectors.
The basic idea is that as we move down through each value of j we always encounter just enough linearly-independent direct product states
m ₁ = j₁ m₂ = j₂ - n |j₁, j₂ - n〉
⋮ ⋮ ⋮
m ₁ = j₁ - n m₂ = j₂ |j₁ - n, j₂ 〉
with a given value of m = j₁ + j₂ - n to form the basis vectors associated with those irreducible spaces already generated with higher values of j, as well as one additional vector which can be constructed orthogonal to those from these other irreducible spaces already constructed.
This remaining state forms the beginning vector for a new sequence of 2j + 1 basis vectors associated with the next value of j = j₁ + j₂ - n.

73. The next steps, we hope, are clear: repeat this procedure until we run out of vectors.
The basic idea is that as we move down through each value of j we always encounter just enough linearly-independent direct product states
m ₁ = j₁ m₂ = j₂ - n |j₁, j₂ - n〉
⋮ ⋮ ⋮
m ₁ = j₁ - n m₂ = j₂ |j₁ - n, j₂ 〉
with a given value of m = j₁ + j₂ - n to form the basis vectors associated with those irreducible spaces already generated with higher values of j, as well as one additional vector which can be constructed orthogonal to those from these other irreducible spaces already constructed.
This remaining state forms the beginning vector for a new sequence of 2j + 1 basis vectors associated with the next value of j = j₁ + j₂ - n.

74. The next steps, we hope, are clear: repeat this procedure until we run out of vectors.
The basic idea is that as we move down through each value of j we always encounter just enough linearly-independent direct product states
m ₁ = j₁ m₂ = j₂ - n |j₁, j₂ - n〉
⋮ ⋮ ⋮
m ₁ = j₁ - n m₂ = j₂ |j₁ - n, j₂ 〉
with a given value of m = j₁ + j₂ - n to form the basis vectors associated with those irreducible spaces already generated with higher values of j, as well as one additional vector which can be constructed orthogonal to those from these other irreducible spaces already constructed.
This remaining state forms the beginning vector for a new sequence of 2j + 1 basis vectors associated with the next value of j = j₁ + j₂ - n.

75. The next steps, we hope, are clear: repeat this procedure until we run out of vectors.
The basic idea is that as we move down through each value of j we always encounter just enough linearly-independent direct product states
m ₁ = j₁ m₂ = j₂ - n |j₁, j₂ - n〉
⋮ ⋮ ⋮
m ₁ = j₁ - n m₂ = j₂ |j₁ - n, j₂ 〉
with a given value of m = j₁ + j₂ - n to form the basis vectors associated with those irreducible spaces already generated with higher values of j, as well as one additional vector which can be constructed orthogonal to those from these other irreducible spaces already constructed.
This remaining state forms the beginning vector for a new sequence of 2j + 1 basis vectors associated with the next value of j = j₁ + j₂ - n.

76. This argument proceeds until we run out of direct product states.
It is straightforward to show that this occurs after we have constructed the space
S(j) with j = |j₁ - j₂| .
That this must be the case also follows by simply counting the total number of new basis vectors |j, m〉 produced through the procedure outlined.
Indeed, for each value of j = j₁ + j₂, ⋯ , |j₁ - j₂| the procedure generates 2j + 1 basis vectors |j, m〉.
The total number of such basis vectors is then represented by the readily computable sum
which is indeed the dimension of this direct product subspace.

77. This argument proceeds until we run out of direct product states.
It is straightforward to show that this occurs after we have constructed the space
S(j) with j = |j₁ - j₂| .
That this must be the case also follows by simply counting the total number of new basis vectors |j, m〉 produced through the procedure outlined.
Indeed, for each value of j = j₁ + j₂, ⋯ , |j₁ - j₂| the procedure generates 2j + 1 basis vectors |j, m〉.
The total number of such basis vectors is then represented by the readily computable sum
which is indeed the dimension of this direct product subspace.

78. This argument proceeds until we run out of direct product states.
It is straightforward to show that this occurs after we have constructed the space
S(j) with j = |j₁ - j₂| .
That this must be the case also follows by simply counting the total number of new basis vectors |j, m〉 produced through the procedure outlined.
Indeed, for each value of j = j₁ + j₂, ⋯ , |j₁ - j₂| the procedure generates 2j + 1 basis vectors |j, m〉.
The total number of such basis vectors is then represented by the readily computable sum
which is indeed the dimension of this direct product subspace.

79. This argument proceeds until we run out of direct product states.
It is straightforward to show that this occurs after we have constructed the space
S(j) with j = |j₁ - j₂| .
That this must be the case also follows by simply counting the total number of new basis vectors |j, m〉 produced through the procedure outlined.
Indeed, for each value of j = j₁ + j₂, ⋯ , |j₁ - j₂| the procedure generates 2j + 1 basis vectors |j, m〉.
The total number of such basis vectors is then represented by the readily computable sum
which is indeed the dimension of this direct product subspace.

80. This argument proceeds until we run out of direct product states.
It is straightforward to show that this occurs after we have constructed the space
S(j) with j = |j₁ - j₂| .
That this must be the case also follows by simply counting the total number of new basis vectors |j, m〉 produced through the procedure outlined.
Indeed, for each value of j = j₁ + j₂, ⋯ , |j₁ - j₂| the procedure generates 2j + 1 basis vectors |j, m〉.
The total number of such basis vectors is then represented by the readily computable sum
which is indeed the dimension of this direct product subspace.

81. This argument proceeds until we run out of direct product states.
It is straightforward to show that this occurs after we have constructed the space
S(j) with j = |j₁ - j₂| .
That this must be the case also follows by simply counting the total number of new basis vectors |j, m〉 produced through the procedure outlined.
Indeed, for each value of j = j₁ + j₂, ⋯ , |j₁ - j₂| the procedure generates 2j + 1 basis vectors |j, m〉.
The total number of such basis vectors is then represented by the readily computable sum
which is indeed the dimension of this direct product subspace.

82. So in this segment we stated and proved the angular momentum addition theorem, which tells how many irreducible invariant subspaces one gets in the direct product space formed from two such irreducible invariant subspaces themselves, and the proof of which shows exactly how to construct a standard basis of angular momentum eigenstates for this direct product space.
We thus, in principle have two sets of orthonormal basis vectors for the same direct product space.
Presumably, there is a unitary operator which maps one set onto the other.
In the next segment, we investigate properties of the matrix elements of this unitary operator, which are referred to as Clebsch-Gordon coefficients.

83. So in this segment we stated and proved the angular momentum addition theorem, which tells how many irreducible invariant subspaces one gets in the direct product space formed from two such irreducible invariant subspaces themselves, and the proof of which shows exactly how to construct a standard basis of angular momentum eigenstates for this direct product space.
We thus, in principle have two sets of orthonormal basis vectors for the same direct product space.
Presumably, there is a unitary operator which maps one set onto the other.
In the next segment, we investigate properties of the matrix elements of this unitary operator, which are referred to as Clebsch-Gordon coefficients.

84. So in this segment we stated and proved the angular momentum addition theorem, which tells how many irreducible invariant subspaces one gets in the direct product space formed from two such irreducible invariant subspaces themselves, and the proof of which shows exactly how to construct a standard basis of angular momentum eigenstates for this direct product space.
We thus, in principle have two sets of orthonormal basis vectors for the same direct product space.
Presumably, there is a unitary operator which maps one set onto the other.
In the next segment, we investigate properties of the matrix elements of this unitary operator, which are referred to as Clebsch-Gordon coefficients.

85. So in this segment we stated and proved the angular momentum addition theorem, which tells how many irreducible invariant subspaces one gets in the direct product space formed from two such irreducible invariant subspaces themselves, and the proof of which shows exactly how to construct a standard basis of angular momentum eigenstates for this direct product space.
We thus, in principle have two sets of orthonormal basis vectors for the same direct product space.
Presumably, there is a unitary operator which maps one set onto the other.
In the next segment, we investigate properties of the matrix elements of this unitary operator, which are referred to as Clebsch-Gordon coefficients.