1. Quantum Two 1

2. 2

3. Angular Momentum and Rotations 3

4. Reducible and Irreducible Tensor Operators 4

5. We have seen that it is possible to classify observables of a system in terms of their transformation properties. Thus, scalar observables are, by definition, invariant under rotations. The components of a vector observable, on the other hand, transform into well defined linear combinations of one another under an arbitrary rotation. As it turns out, these two examples constitute special cases of a more general classification scheme involving the concept of tensor operators. Definition: a collection of n operators {Q ₁, Q ₂, ⋯, Q n } comprise an n -component rotational tensor operator Q if they transform under rotations into linear combinations of themselves. In other word, for each rotation R, there exists a set of coefficients D ji (R) such that 5

6. We have seen that it is possible to classify observables of a system in terms of their transformation properties. Thus, scalar observables are, by definition, invariant under rotations. The components of a vector observable, on the other hand, transform into well defined linear combinations of one another under an arbitrary rotation. As it turns out, these two examples constitute special cases of a more general classification scheme involving the concept of tensor operators. Definition: a collection of n operators {Q ₁, Q ₂, ⋯, Q n } comprise an n -component rotational tensor operator Q if they transform under rotations into linear combinations of themselves. In other word, for each rotation R, there exists a set of coefficients D ji (R) such that 6

7. We have seen that it is possible to classify observables of a system in terms of their transformation properties. Thus, scalar observables are, by definition, invariant under rotations. The components of a vector observable, on the other hand, transform into well defined linear combinations of one another under an arbitrary rotation. As it turns out, these two examples constitute special cases of a more general classification scheme involving the concept of tensor operators. Definition: a collection of n operators {Q ₁, Q ₂, ⋯, Q n } comprise an n -component rotational tensor operator Q if they transform under rotations into linear combinations of themselves. In other word, for each rotation R, there exists a set of coefficients D ji (R) such that 7

8. We have seen that it is possible to classify observables of a system in terms of their transformation properties. Thus, scalar observables are, by definition, invariant under rotations. The components of a vector observable, on the other hand, transform into well defined linear combinations of one another under an arbitrary rotation. As it turns out, these two examples constitute special cases of a more general classification scheme involving the concept of tensor operators. Definition: a collection of n operators {Q ₁, Q ₂, ⋯, Q n } comprise an n -component rotational tensor operator Q if they transform under rotations into linear combinations of themselves. In other word, for each rotation R, there exists a set of coefficients D ji (R) such that 8

9. We have seen that it is possible to classify observables of a system in terms of their transformation properties. Thus, scalar observables are, by definition, invariant under rotations. The components of a vector observable, on the other hand, transform into well defined linear combinations of one another under an arbitrary rotation. As it turns out, these two examples constitute special cases of a more general classification scheme involving the concept of tensor operators. Definition: a collection of n operators {Q ₁, Q ₂, ⋯, Q n } comprise an n -component rotational tensor operator Q if they transform under rotations into linear combinations of themselves. In other word, for each rotation R, there exists a set of coefficients D ji (R) such that 9

10. We have seen that it is possible to classify observables of a system in terms of their transformation properties. Thus, scalar observables are, by definition, invariant under rotations. The components of a vector observable, on the other hand, transform into well defined linear combinations of one another under an arbitrary rotation. As it turns out, these two examples constitute special cases of a more general classification scheme involving the concept of tensor operators. Definition: a collection of n operators {Q ₁, Q ₂, ⋯, Q n } comprise an n -component rotational tensor operator Q if they transform under rotations into linear combinations of themselves. In other word, for each rotation R, there exists a set of coefficients D ji (R) such that 10

11. It is straightforward to show that under these circumstances the matrices {[ D(R) ]} with matrix elements D ji (R) form an n -dimensional representation of the 3D rotation group. As an example, we note that a scalar operator Q 0 forms a 1-component tensor, since which is of the form written above, with all of the matrices [ D 0 (R) ] just equal to the 1×1 identity matrix [ D 0 (R) ] = 1. As a more familiar example, we note that the Cartesian components { V x, V y, V z } of a vector operator form the components of a 3-component tensor V. Indeed, under an arbitrary rotation, the operator is transformed into where indicates the direction obtained by performing the rotation R on the vector. 11

12. It is straightforward to show that under these circumstances the matrices {[ D(R) ]} with matrix elements D ji (R) form an n -dimensional representation of the 3D rotation group. As an example, we note that a scalar operator Q 0 forms a 1-component tensor, since which is of the form written above, with all of the matrices [ D 0 (R) ] just equal to the 1×1 identity matrix [ D 0 (R) ] = 1. As a more familiar example, we note that the Cartesian components { V x, V y, V z } of a vector operator form the components of a 3-component tensor V. Indeed, under an arbitrary rotation, the operator is transformed into where indicates the direction obtained by performing the rotation R on the vector. 12

13. It is straightforward to show that under these circumstances the matrices {[ D(R) ]} with matrix elements D ji (R) form an n -dimensional representation of the 3D rotation group. As an example, we note that a scalar operator Q 0 forms a 1-component tensor, since which is of the form written above, with all of the matrices [ D 0 (R) ] just equal to the 1×1 identity matrix [ D 0 (R) ] = 1. As a more familiar example, we note that the Cartesian components { V x, V y, V z } of a vector operator form the components of a 3-component tensor V. Indeed, under an arbitrary rotation, the operator is transformed into where indicates the direction obtained by performing the rotation R on the vector. 13

14. It is straightforward to show that under these circumstances the matrices {[ D(R) ]} with matrix elements D ji (R) form an n -dimensional representation of the 3D rotation group. As an example, we note that a scalar operator Q 0 forms a 1-component tensor, since which is of the form written above, with all of the matrices [ D 0 (R) ] just equal to the 1×1 identity matrix [ D 0 (R) ] = 1. As a more familiar example, we note that the Cartesian components { V x, V y, V z } of a vector operator form the components of a 3-component tensor V. Indeed, under an arbitrary rotation, the operator is transformed into where indicates the direction obtained by performing the rotation R on the vector. 14

15. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 15

16. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 16

17. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 17

18. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 18

19. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 19

20. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 20

21. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 21

22. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 22

23. Thus, If corresponds to the Cartesian unit vector, then and so that i.e., This shows that the Cartesian components of transform under rotations into linear combinations of one another, with coefficients given by the 3×3 rotation matrices A R of SO(3), pretty much as we would expect. 23

24. As another illuminating example, if and are both vector operators and is a scalar operator, then the operators form a seven component tensor, since under an arbitrary rotation R they are taken onto the following linear combinations of themselves, and thus satisfies the definition of a tensor operator. 24

25. As another illuminating example, if and are both vector operators and is a scalar operator, then the operators form a seven component tensor, since under an arbitrary rotation R they are taken onto the following linear combinations of themselves, and thus satisfies the definition of a tensor operator. 25

26. As another illuminating example, if and are both vector operators and is a scalar operator, then the operators form a seven component tensor, since under an arbitrary rotation R they are taken onto the following linear combinations of themselves, and thus satisfies the definition of a tensor operator. 26

27. As another illuminating example, if and are both vector operators and is a scalar operator, then the operators form a seven component tensor, since under an arbitrary rotation R they are taken onto the following linear combinations of themselves, and thus satisfies the definition of a tensor operator. 27

28. As another illuminating example, if and are both vector operators and is a scalar operator, then the operators form a seven component tensor, since under an arbitrary rotation R they are taken onto the following linear combinations of themselves, and thus satisfies the definition of a tensor operator. 28

29. As another illuminating example, if and are both vector operators and is a scalar operator, then the operators form a seven component tensor, since under an arbitrary rotation R they are taken onto the following linear combinations of themselves, and thus satisfies the definition of a tensor operator. 29

30. As another illuminating example, if and are both vector operators and is a scalar operator, then the operators form a seven component tensor, since under an arbitrary rotation R they are taken onto the following linear combinations of themselves, and thus satisfies the definition of a tensor operator. 30

31. As another illuminating example, if and are both vector operators and is a scalar operator, then the operators form a seven component tensor, since under an arbitrary rotation R they are taken onto the following linear combinations and thus satisfy the definition of a tensor operator. 31

32. Clearly in this case, however, the set of seven components can be partitioned into 3 separate sets of operators,, and that independently transform into linear combinations of one another. One says, then, that this seven component tensor can be reduced into two 3- component tensors and one single-component tensor. Note also that in this circumstance the matrices [ D 0 (R) ] = governing the transformation of these operators is block diagonal. 32

33. Clearly in this case, however, the set of seven components can be partitioned into 3 separate sets of operators,, and that independently transform into linear combinations of one another. One says, then, that this seven component tensor can be reduced into two 3- component tensors and one single-component tensor. Note also that in this circumstance the matrices [ D 0 (R) ] = governing the transformation of these operators is block diagonal. 33

34. Clearly in this case, however, the set of seven components can be partitioned into 3 separate sets of operators,, and that independently transform into linear combinations of one another. One says, then, that this seven component tensor can be reduced into two 3- component tensors and one single-component tensor. Note also that in this circumstance the matrices [ D 0 (R) ] = governing the transformation of these operators is block diagonal. 34

35. Thus this representation of the rotation group is really a combination (a direct sum in fact) of three separate representations, one of which is 1-dimensional and two of which are 3-dimensional. This leads to the idea of reducible and irreducible tensors. Definition: A tensor operator T is said to be reducible if its components {T ₁, T ₂, ⋯, T n }, or any set of n linear combinations thereof, can be partitioned into two or more tensors each having a smaller number of components than T. A tensor that cannot be so partitioned is said to be irreducible. For example, if is a vector operator, the set of operators { V x, V y } do not form a two component tensor, because a rotation of V x about the y axis through π/2 takes it onto V z, which cannot be expressed as a linear combination of V x and V y. 35

36. Thus this representation of the rotation group is really a combination (a direct sum in fact) of three separate representations, one of which is 1-dimensional and two of which are 3-dimensional. This leads to the idea of reducible and irreducible tensors. Definition: A tensor operator T is said to be reducible if its components {T ₁, T ₂, ⋯, T n }, or any set of n linear combinations thereof, can be partitioned into two or more tensors each having a smaller number of components than T. A tensor that cannot be so partitioned is said to be irreducible. For example, if is a vector operator, the set of operators { V x, V y } do not form a two component tensor, because a rotation of V x about the y axis through π/2 takes it onto V z, which cannot be expressed as a linear combination of V x and V y. 36

37. Thus this representation of the rotation group is really a combination (a direct sum in fact) of three separate representations, one of which is 1-dimensional and two of which are 3-dimensional. This leads to the idea of reducible and irreducible tensors. Definition: A tensor operator T is said to be reducible if its components {T ₁, T ₂, ⋯, T n }, or any set of n linear combinations thereof, can be partitioned into two or more tensors each having a smaller number of components than T. A tensor that cannot be so partitioned is said to be irreducible. For example, if is a vector operator, the set of operators { V x, V y } do not form a two component tensor, because a rotation of V x about the y axis through π/2 takes it onto V z, which cannot be expressed as a linear combination of V x and V y. 37

38. Thus this representation of the rotation group is really a combination (a direct sum in fact) of three separate representations, one of which is 1-dimensional and two of which are 3-dimensional. This leads to the idea of reducible and irreducible tensors. Definition: A tensor operator T is said to be reducible if its components {T ₁, T ₂, ⋯, T n }, or any set of n linear combinations thereof, can be partitioned into two or more tensors each having a smaller number of components than T. A tensor that cannot be so partitioned is said to be irreducible. For example, if is a vector operator, the set of operators { V x, V y } do not form a two component tensor, because a rotation of V x about the y axis through π/2 takes it onto V z, which cannot be expressed as a linear combination of V x and V y. 38

39. Thus this representation of the rotation group is really a combination (a direct sum in fact) of three separate representations, one of which is 1-dimensional and two of which are 3-dimensional. This leads to the idea of reducible and irreducible tensors. Definition: A tensor operator T is said to be reducible if its components {T ₁, T ₂, ⋯, T n }, or any set of n linear combinations thereof, can be partitioned into two or more tensors each having a smaller number of components than T. A tensor that cannot be so partitioned is said to be irreducible. For example, if is a vector operator, the set of operators { V x, V y } do not form a two component tensor, because a rotation of V x about the y axis through π/2 takes it onto V z, which cannot be expressed as a linear combination of V x and V y. 39

40. Thus, a vector operator cannot be reduced into smaller subsets of operators that independently transform into linear combinations of themselves. Since all vectors transform the same way, it follows that all vector operators are irreducible 3-component tensors. Note that the language used to describe the components of tensor operators is very similar to that describing the behavior of the basis vectors associated with rotationally invariant subspaces of a quantum mechanical Hilbert space, some of which are reducible into smaller sets of irreducible components. Indeed, in a certain sense the basic problem of reducing tensors into their irreducible components has already been solved in the context of adding angular momenta. So we already have a model for sets of objects that transform irreducibly into linear combinations of themselves under rotations. 40

41. Thus, a vector operator cannot be reduced into smaller subsets of operators that independently transform into linear combinations of themselves. Since all vectors transform the same way, it follows that all vector operators are irreducible 3-component tensors. Note that the language used to describe the components of tensor operators is very similar to that describing the behavior of the basis vectors associated with rotationally invariant subspaces of a quantum mechanical Hilbert space, some of which are reducible into smaller sets of irreducible components. Indeed, in a certain sense the basic problem of reducing tensors into their irreducible components has already been solved in the context of adding angular momenta. So we already have a model for sets of objects that transform irreducibly into linear combinations of themselves under rotations. 41

42. Thus, a vector operator cannot be reduced into smaller subsets of operators that independently transform into linear combinations of themselves. Since all vectors transform the same way, it follows that all vector operators are irreducible 3-component tensors. Note that the language used to describe the components of tensor operators is very similar to that describing the behavior of the set of basis vectors associated with rotationally invariant subspaces of a quantum mechanical Hilbert space, some of which are reducible into smaller irreducible sets. Indeed, in a certain sense the basic problem of reducing tensors into their irreducible components has already been solved in the context of adding angular momenta. So we already have a model for sets of objects that transform irreducibly into linear combinations of themselves under rotations. 42

43. Thus, a vector operator cannot be reduced into smaller subsets of operators that independently transform into linear combinations of themselves. Since all vectors transform the same way, it follows that all vector operators are irreducible 3-component tensors. Note that the language used to describe the components of tensor operators is very similar to that describing the behavior of the set of basis vectors associated with rotationally invariant subspaces of a quantum mechanical Hilbert space, some of which are reducible into smaller irreducible sets. So we already have a model for sets of objects that transform irreducibly into linear combinations of themselves under rotations. 43

44. Thus, a vector operator cannot be reduced into smaller subsets of operators that independently transform into linear combinations of themselves. Since all vectors transform the same way, it follows that all vector operators are irreducible 3-component tensors. Note that the language used to describe the components of tensor operators is very similar to that describing the behavior of the set of basis vectors associated with rotationally invariant subspaces of a quantum mechanical Hilbert space, some of which are reducible into smaller irreducible sets. So we already have a model for sets of objects that transform irreducibly into linear combinations of themselves under rotations. Thus, after some reflection, one is led quite naturally to introduce a very useful class of irreducible tensors, referred to as spherical tensors. 44

45. Definition: a collection of 2j  1 operators form the components of an irreducible spherical tensor T j of rank j if they transform under rotations into linear combinations of themselves in the same way as the basis vectors | τ, j, m 〉 of an irreducible invariant subspace S(j, τ). Specifically, this means that under a rotation R, each of the operators is taken onto which is to be compared to where the are matrix elements of the rotation matrix associated with an irreducible invariant subspace S(j, τ) with this value of j. 45

46. Definition: a collection of 2j  1 operators form the components of an irreducible spherical tensor T j of rank j if they transform under rotations into linear combinations of themselves in the same way as the basis vectors | τ, j, m 〉 of an irreducible invariant subspace S(j, τ). Specifically, this means that under a rotation R, each of the operators is taken onto which is to be compared to where the are matrix elements of the rotation matrix associated with an irreducible invariant subspace S(j, τ) with this value of j. 46

47. Definition: a collection of 2j  1 operators form the components of an irreducible spherical tensor T j of rank j if they transform under rotations into linear combinations of themselves in the same way as the basis vectors | τ, j, m 〉 of an irreducible invariant subspace S(j, τ). Specifically, this means that under a rotation R, each of the operators is taken onto which is to be compared to where the are matrix elements of the rotation matrix associated with an irreducible invariant subspace S(j, τ) with this value of j. 47

48. Since the basis vectors | τ, j, m 〉 transform irreducibly, it is not hard to see that the components of spherical tensors of this sort do so as well. It is also not hard to see, I hope, that according to this definition, a scalar observable is an irreducible spherical tensor of rank zero, i.e., its transformation law R[Q₀⁰]=Q₀⁰ is the same as that of the single basis vector | 0, 0 〉 associated with an irreducible invariant subspace S(0) of zero angular momentum, for which R[|0,0 〉 ]=|0,0 〉. Similarly, a vector operator, which defines an irreducible three component tensor, can be viewed as a spherical tensor V ₁ of rank one, which we can express either in terms either of its Cartesian components, or its spherical components. 48

49. Since the basis vectors | τ, j, m 〉 transform irreducibly, it is not hard to see that the components of spherical tensors of this sort do so as well. It is also not hard to see, I hope, that according to this definition, a scalar observable is an irreducible spherical tensor of rank zero, i.e., its transformation law R[Q₀⁰]=Q₀⁰ is the same as that of the single basis vector | 0, 0 〉 associated with an irreducible invariant subspace S(0) of zero angular momentum, for which R[|0,0 〉 ]=|0,0 〉. Similarly, a vector operator, which defines an irreducible three component tensor, can be viewed as a spherical tensor V ₁ of rank one, which we can express either in terms either of its Cartesian components, or its spherical components. 49

50. Since the basis vectors | τ, j, m 〉 transform irreducibly, it is not hard to see that the components of spherical tensors of this sort do so as well. It is also not hard to see, I hope, that according to this definition, a scalar observable is an irreducible spherical tensor of rank zero, i.e., its transformation law R[Q₀⁰]=Q₀⁰ is the same as that of the single basis vector | 0, 0 〉 associated with an irreducible invariant subspace S(0) of zero angular momentum, for which R[|0,0 〉 ]=|0,0 〉. Similarly, a vector operator, which defines an irreducible three component tensor, can be viewed as a spherical tensor V ₁ of rank one, which we can express either in terms either of its Cartesian components, or its spherical components. 50

51. Since the basis vectors | τ, j, m 〉 transform irreducibly, it is not hard to see that the components of spherical tensors of this sort do so as well. It is also not hard to see, I hope, that according to this definition, a scalar observable is an irreducible spherical tensor of rank zero, i.e., its transformation law R[Q₀⁰]=Q₀⁰ is the same as that of the single basis vector | 0, 0 〉 associated with an irreducible invariant subspace S(0) of zero angular momentum, for which R[|0,0 〉 ]=|0,0 〉. Similarly, a vector operator, which defines an irreducible three component tensor, can be viewed as a spherical tensor V ₁ of rank one, which we can express either in terms either of its Cartesian components, or its spherical components. 51

52. Since the basis vectors | τ, j, m 〉 transform irreducibly, it is not hard to see that the components of spherical tensors of this sort do so as well. It is also not hard to see, I hope, that according to this definition, a scalar observable is an irreducible spherical tensor of rank zero, i.e., its transformation law R[Q₀⁰]=Q₀⁰ is the same as that of the single basis vector | 0, 0 〉 associated with an irreducible invariant subspace S(0) of zero angular momentum, for which R[|0,0 〉 ]=|0,0 〉. Similarly, a vector operator, which defines an irreducible three component tensor, can be viewed as a spherical tensor V ₁ of rank one, which we can express either in terms either of its Cartesian components, or its spherical components. 52

53. Definition: the spherical components of a vector operator are given as the following linear combinations of its Cartesian components: In this representation we see that the raising and lowering operators can be expressed in terms of the spherical tensor components of the angular momentum operator through the relation 53

54. Definition: the spherical components of a vector operator are given as the following linear combinations of its Cartesian components: In this representation we see that the raising and lowering operators can be expressed in terms of the spherical tensor components of the angular momentum operator through the relation 54

55. Definition: the spherical components of a vector operator are given as the following linear combinations of its Cartesian components: In this representation we see that the raising and lowering operators can be expressed in terms of the spherical tensor components of the angular momentum operator through the relation 55

56. Definition: the spherical components of a vector operator are given as the following linear combinations of its Cartesian components: In this representation we see that the raising and lowering operators can be expressed in terms of the spherical tensor components of the angular momentum operator through the relation 56

57. To see that the spherical components of a vector define an irreducible tensor of unit rank we must show that they transform appropriately. To this end it suffices to demonstrate the transformation properties for any vector operator, since the transformation law will clearly be the same for all vector operators (as it is for the Cartesian components). Consider, then, in the space of a single particle, the vector operator which has the effect in the position representation of multiplying the wavefunction at by the radial unit vector, i.e., so that 57

58. To see that the spherical components of a vector define an irreducible tensor of unit rank we must show that they transform appropriately. To this end it suffices to demonstrate the transformation properties for any vector operator, since the transformation law will clearly be the same for all vector operators (as it is for the Cartesian components). Consider, then, in the space of a single particle, the vector operator which has the effect in the position representation of multiplying the wavefunction at by the radial unit vector, i.e., so that 58

59. To see that the spherical components of a vector define an irreducible tensor of unit rank we must show that they transform appropriately. To this end it suffices to demonstrate the transformation properties for any vector operator, since the transformation law will clearly be the same for all vector operators (as it is for the Cartesian components). Consider, then, in the space of a single particle, the vector operator which has the effect in the position representation of multiplying the wavefunction at by the radial unit vector, i.e., so that 59

60. To see that the spherical components of a vector define an irreducible tensor of unit rank we must show that they transform appropriately. To this end it suffices to demonstrate the transformation properties for any vector operator, since the transformation law will clearly be the same for all vector operators (as it is for the Cartesian components). Consider, then, in the space of a single particle, the vector operator which has the effect in the position representation of multiplying the wavefunction at by the radial unit vector, i.e., so that 60

61. To see that the spherical components of a vector define an irreducible tensor of unit rank we must show that they transform appropriately. To this end it suffices to demonstrate the transformation properties for any vector operator, since the transformation law will clearly be the same for all vector operators (as it is for the Cartesian components). Consider, then, in the space of a single particle, the vector operator which has the effect in the position representation of multiplying the wavefunction at by the radial unit vector, i.e., so that 61

62. In the position representation the Cartesian components of this operator can be written in spherical coordinates (r, θ, ϕ ) in the usual way, i.e., In this same representation the spherical components of this vector operator take the form 62

63. In the position representation the Cartesian components of this operator can be written in spherical coordinates (r, θ, ϕ ) in the usual way, i.e., In this same representation the spherical components of this vector operator take the form 63

64. In the position representation the Cartesian components of this operator can be written in spherical coordinates (r, θ, ϕ ) in the usual way, i.e., In this same representation the spherical components of this vector operator take the form 64

65. In the position representation the Cartesian components of this operator can be written in spherical coordinates (r, θ, ϕ ) in the usual way, i.e., In this same representation the spherical components of this vector operator take the form 65

66. In the position representation the Cartesian components of this operator can be written in spherical coordinates (r, θ, ϕ ) in the usual way, i.e., In this same representation the spherical components of this vector operator take the form 66

67. In the position representation the Cartesian components of this operator can be written in spherical coordinates (r, θ, ϕ ) in the usual way, i.e., In this same representation the spherical components of this vector operator take the form 67

68. Aside from an overall constant, these are equivalent to the spherical harmonics of order one, i.e., which, being eigenfunctions of orbital angular momentum, do indeed transform as the basis vectors of an irreducible invariant subspace with j = 1 68

69. Aside from an overall constant, these are equivalent to the spherical harmonics of order one, i.e., which, being eigenfunctions of orbital angular momentum, do indeed transform as the basis vectors of an irreducible invariant subspace with j = 1 69

70. Aside from an overall constant, these are equivalent to the spherical harmonics of order one, i.e., which, being eigenfunctions of orbital angular momentum, do indeed transform as the basis vectors of an irreducible invariant subspace with j = 1 70

71. Aside from an overall constant, these are equivalent to the spherical harmonics of order one, i.e., which, being eigenfunctions of orbital angular momentum, do indeed transform as the basis vectors of an irreducible invariant subspace with j = 1, i.e., 71

72. Aside from an overall constant, these are equivalent to the spherical harmonics of order one, i.e., which, being eigenfunctions of orbital angular momentum, do indeed transform as the basis vectors of an irreducible invariant subspace with j = 1, i.e., 72

73. from which it follows that the spherical components of transform in the same way, i.e., Thus, this vector operator (and hence all vector operators) define a spherical tensor of rank one. As a natural extension of this, it is not hard to see that the spherical harmonics of order ℓ define an irreducible tensor operator Y ℓ with 2ℓ +1 components {Y_{ℓ}^{m}|m=-ℓ, ⋯,ℓ} which in the position representation The components of this tensor operator arise quite naturally in the multipole expansion of the electric potential energy of a charged particle moving in the field of an arbitrary charge distribution. 73

74. from which it follows that the spherical components of transform in the same way, i.e., Thus, this vector operator (and hence all vector operators) define a rank one spherical tensor. As a natural extension of this, it is not hard to see that the spherical harmonics of order ℓ define an irreducible tensor operator Y ℓ with 2ℓ +1 components {Y_{ℓ}^{m}|m=-ℓ, ⋯,ℓ} which in the position representation The components of this tensor operator arise quite naturally in the multipole expansion of the electric potential energy of a charged particle moving in the field of an arbitrary charge distribution. 74

75. from which it follows that the spherical components of transform in the same way, i.e., Thus, this vector operator (and hence all vector operators) define a rank one spherical tensor. As a natural extension of this, it is not hard to see that the spherical harmonics of order ℓ define an irreducible tensor operator Y ℓ with 2ℓ +1 components {Y_{ℓ}^{m}|m=-ℓ, ⋯,ℓ} which in the position representation The components of this tensor operator arise quite naturally in the multipole expansion of the electric potential energy of a charged particle moving in the field of an arbitrary charge distribution. 75

76. from which it follows that the spherical components of transform in the same way, i.e., Thus, this vector operator (and hence all vector operators) define a rank one spherical tensor. As a natural extension of this, it is not hard to see that the spherical harmonics of order ℓ define an irreducible tensor operator Y ℓ with 2ℓ +1 components {Y_{ℓ}^{m}|m=-ℓ, ⋯,ℓ} which in the position representation The components of this tensor operator arise quite naturally in the multipole expansion of the electric potential energy of a charged particle moving in the field of an arbitrary charge distribution. 76

77. from which it follows that the spherical components of transform in the same way, i.e., Thus, this vector operator (and hence all vector operators) define a rank one spherical tensor. As a natural extension of this, it is not hard to see that the spherical harmonics of order ℓ define an irreducible tensor operator Y ℓ with 2ℓ +1 components {Y_{ℓ}^{m}|m=-ℓ, ⋯,ℓ} which in the position representation We note in passing that the components of this tensor operator arise quite naturally in the multipole expansion of the electric potential energy of a charged particle moving in the field of an arbitrary charge distribution. 77

78. So in this segment, we have extended our classification scheme for operators based on the way that they transform under rotations, to include tensor operators, that can have more than 3 components, and that transform under an arbitrary rotation into linear combinations of themselves. We saw, however, that many tensor operators are reducible into two or more irreducible tensor operators. We were led, therefore, to introduce the idea of spherical tensors, whose components transform irreducibly into linear combinations of themselves in the same manner as the basis vectors of an irreducible invariant angular momentum subspace, and gave examples of such operators, based on the spherical harmonics. Finally, we found that the product of two tensors is itself a generally reducible tensor, and proved a theorem that tells us explicitly how to reduce a product of spherical tensors into their irreducible components. 78

79. So in this segment, we have extended our classification scheme for operators based on the way that they transform under rotations, to include tensor operators, that can have more than 3 components, and that transform under an arbitrary rotation into linear combinations of themselves. We saw, however, that many tensor operators are reducible into two or more irreducible tensor operators. We were led, therefore, to introduce the idea of spherical tensors, whose components transform irreducibly into linear combinations of themselves in the same manner as the basis vectors of an irreducible invariant angular momentum subspace, and gave examples of such operators, based on the spherical harmonics. Finally, we found that the product of two tensors is itself a generally reducible tensor, and proved a theorem that tells us explicitly how to reduce a product of spherical tensors into their irreducible components. 79

80. So in this segment, we have extended our classification scheme for operators based on the way that they transform under rotations, to include tensor operators, that can have more than 3 components, and that transform under an arbitrary rotation into linear combinations of themselves. We saw, however, that many tensor operators are reducible into two or more irreducible tensor operators. We were led, therefore, to introduce the idea of spherical tensors, whose components transform irreducibly into linear combinations of themselves in the same manner as the basis vectors of an irreducible invariant angular momentum subspace, and gave examples of such operators, based on the spherical harmonics. Finally, we found that the product of two tensors is itself a generally reducible tensor, and proved a theorem that tells us explicitly how to reduce a product of spherical tensors into their irreducible components. 80

81. So in this segment, we have extended our classification scheme for operators based on the way that they transform under rotations, to include tensor operators, that can have more than 3 components, and that transform under an arbitrary rotation into linear combinations of themselves. We saw, however, that many tensor operators are reducible into two or more irreducible tensor operators. We were led, therefore, to introduce the idea of spherical tensors, whose components transform irreducibly into linear combinations of themselves in the same manner as the basis vectors of an irreducible invariant angular momentum subspace, and gave examples of such operators, based on the spherical harmonics. In the next segment, we show that the the product of two tensors is itself a generally reducible tensor, and prove a theorem that tells us explicitly how to reduce a product of spherical tensors into their irreducible components. 81

82. 82