Quantum Two 1. 2 Angular Momentum and Rotations 3.

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Presentation transcript:

Quantum Two 1

2

Angular Momentum and Rotations 3

The tensor product of tensors 4

Tensor Products In the last segment, we 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. However in many cases, many-component tensor operators naturally arise from the product of two or more scalars, vectors, or tensors of higher rank. 5

Tensor Products Indeed, it turns out that the product of two tensors is, itself, generally a tensor. For example, if T j ₁ and Q j ₂ represent spherical tensors of rank j ₁ and j ₂, respectively, then the set of (2 j ₁ +1) (2 j ₂ +1) products, form a tensor. In general, however, such a tensor is reducible into tensors of smaller rank. Indeed, since the components T j ₁ m ₁ and Q j ₂ m ₂ of each tensor transform as the basis vectors |j ₁, m ₁〉 and | j ₂, m ₂〉 of irreducible subspaces S(j ₁ ) and S(j ₂ ), the set of products {T j ₁ m ₁ Q j ₂ m ₂ } transform just like the direct product basis states |j ₁, j ₂, m ₁, m ₂〉 of a direct product space S(j ₁ ) ⊗ S(j ₂ ) But we know how to form linear combinations of the direct product states that transform “like” the basis states |j, m 〉 of irreducible invariant subspaces S(j). 6

Tensor Products Indeed, it turns out that the product of two tensors is, itself, generally a tensor. For example, if T j ₁ and Q j ₂ represent spherical tensors of rank j ₁ and j ₂, respectively, then the set of (2 j ₁ +1) (2 j ₂ +1) products, form a tensor. In general, however, such a tensor is reducible into tensors of smaller rank. Indeed, since the components T j ₁ m ₁ and Q j ₂ m ₂ of each tensor transform as the basis vectors |j ₁, m ₁〉 and | j ₂, m ₂〉 of irreducible subspaces S(j ₁ ) and S(j ₂ ), the set of products {T j ₁ m ₁ Q j ₂ m ₂ } transform just like the direct product basis states |j ₁, j ₂, m ₁, m ₂〉 of a direct product space S(j ₁ ) ⊗ S(j ₂ ) But we know how to form linear combinations of the direct product states that transform “like” the basis states |j, m 〉 of irreducible invariant subspaces S(j). 7

Tensor Products Indeed, it turns out that the product of two tensors is, itself, generally a tensor. For example, if T j ₁ and Q j ₂ represent spherical tensors of rank j ₁ and j ₂, respectively, then the set of (2 j ₁ +1) (2 j ₂ +1) products, form a tensor. In general, however, such a tensor is reducible into tensors of smaller rank. Indeed, since the components T j ₁ m ₁ and Q j ₂ m ₂ of each tensor transform as the basis vectors |j ₁, m ₁〉 and | j ₂, m ₂〉 of irreducible subspaces S(j ₁ ) and S(j ₂ ), the set of products {T j ₁ m ₁ Q j ₂ m ₂ } transform just like the direct product basis states |j ₁, j ₂, m ₁, m ₂〉 of a direct product space S(j ₁ ) ⊗ S(j ₂ ) But we know how to form linear combinations of the direct product states that transform “like” the basis states |j, m 〉 of irreducible invariant subspaces S(j). 8

Tensor Products Indeed, it turns out that the product of two tensors is, itself, generally a tensor. For example, if T j ₁ and Q j ₂ represent spherical tensors of rank j ₁ and j ₂, respectively, then the set of (2 j ₁ +1) (2 j ₂ +1) products, form a tensor. In general, however, such a tensor is reducible into tensors of smaller rank. Indeed, since the components T j ₁ m ₁ and Q j ₂ m ₂ of each tensor transform as the basis vectors |j ₁, m ₁〉 and | j ₂, m ₂〉 of irreducible subspaces S(j ₁ ) and S(j ₂ ), the set of products {T j ₁ m ₁ Q j ₂ m ₂ } transform just like the direct product basis states |j ₁, j ₂, m ₁, m ₂〉 of a direct product space S(j ₁ ) ⊗ S(j ₂ ) But we know how to form linear combinations of the direct product states that transform “like” the basis states |j, m 〉 of irreducible invariant subspaces S(j). 9

Tensor Products Indeed, it turns out that the product of two tensors is, itself, generally a tensor. For example, if T j ₁ and Q j ₂ represent spherical tensors of rank j ₁ and j ₂, respectively, then the set of (2 j ₁ +1) (2 j ₂ +1) products, form a tensor. In general, however, such a tensor is reducible into tensors of smaller rank. Indeed, since the components T j ₁ m ₁ and Q j ₂ m ₂ of each tensor transform as the basis vectors |j ₁, m ₁〉 and | j ₂, m ₂〉 of irreducible subspaces S(j ₁ ) and S(j ₂ ), the set of products {T j ₁ m ₁ Q j ₂ m ₂ } transform just like the direct product basis states |j ₁, j ₂, m ₁, m ₂〉 of a direct product space S(j ₁ ) ⊗ S(j ₂ ). But we know how to form linear combinations of the direct product states that transform “like” the basis states |j, m 〉 of irreducible invariant subspaces S(j). 10

Tensor Products Indeed, it turns out that the product of two tensors is, itself, generally a tensor. For example, if T j ₁ and Q j ₂ represent spherical tensors of rank j ₁ and j ₂, respectively, then the set of (2 j ₁ +1) (2 j ₂ +1) products, form a tensor. In general, however, such a tensor is reducible into tensors of smaller rank. Indeed, since the components T j ₁ m ₁ and Q j ₂ m ₂ of each tensor transform as the basis vectors |j ₁, m ₁〉 and | j ₂, m ₂〉 of irreducible subspaces S(j ₁ ) and S(j ₂ ), the set of products {T j ₁ m ₁ Q j ₂ m ₂ } transform just like the direct product basis states |j ₁, j ₂, m ₁, m ₂〉 of a direct product space S(j ₁ ) ⊗ S(j ₂ ). But we already we know how to form linear combinations of the direct product states |j ₁, j ₂, m ₁, m ₂〉 that transform “like” the basis states |j, m 〉 of irreducible invariant subspaces S(j). 11

Indeed, given the equivalence of the transformation laws, we deduce from the angular momentum addition theorem the following Tensor Product Reduction Theorem: From the (2 j ₁ +1) (2 j ₂ +1) components {T j ₁ m ₁ Q j ₂ m ₂ } defining the tensor product T j ₁ ⊗ Q j ₂ one can form, for each j in the sequence, j  j ₁  j ₂, …, |j ₁  j ₂ | exactly one irreducible spherical tensor W j with components which is to be compared to: 12

Indeed, given the equivalence of the transformation laws, we deduce from the angular momentum addition theorem the following Tensor Product Reduction Theorem: From the (2 j ₁ +1) (2 j ₂ +1) components {T j ₁ m ₁ Q j ₂ m ₂ } defining the tensor W = T j ₁ ⊗ Q j ₂ one can form, for each j in the sequence, j  j ₁  j ₂, …, |j ₁  j ₂ | exactly one irreducible spherical tensor W j with components which is to be compared to: 13

Indeed, given the equivalence of the transformation laws, we deduce from the angular momentum addition theorem the following Tensor Product Reduction Theorem: From the (2 j ₁ +1) (2 j ₂ +1) components {T j ₁ m ₁ Q j ₂ m ₂ } defining the tensor W = T j ₁ ⊗ Q j ₂ one can form, for each j in the sequence, j  j ₁  j ₂, …, |j ₁  j ₂ | exactly one irreducible spherical tensor W j with components which is to be compared to: 14

Indeed, given the equivalence of the transformation laws, we deduce from the angular momentum addition theorem the following Tensor Product Reduction Theorem: From the (2 j ₁ +1) (2 j ₂ +1) components {T j ₁ m ₁ Q j ₂ m ₂ } defining the tensor W = T j ₁ ⊗ Q j ₂ one can form, for each j in the sequence, j  j ₁  j ₂, …, |j ₁  j ₂ | exactly one irreducible spherical tensor W j with components which is to be compared to: 15

Indeed, given the equivalence of the transformation laws, we deduce from the angular momentum addition theorem the following Tensor Product Reduction Theorem: From the (2 j ₁ +1) (2 j ₂ +1) components {T j ₁ m ₁ Q j ₂ m ₂ } defining the tensor W = T j ₁ ⊗ Q j ₂ one can form, for each j in the sequence, j  j ₁  j ₂, …, |j ₁  j ₂ | exactly one irreducible spherical tensor W j with components which is to be compared to: 16

Indeed, given the equivalence of the transformation laws, we deduce from the angular momentum addition theorem the following Tensor Product Reduction Theorem: From the (2 j ₁ +1) (2 j ₂ +1) components {T j ₁ m ₁ Q j ₂ m ₂ } defining the tensor W = T j ₁ ⊗ Q j ₂ one can form, for each j in the sequence, j  j ₁  j ₂, …, |j ₁  j ₂ | exactly one irreducible spherical tensor W j with components which is to be compared to: 17

Indeed, given the equivalence of the transformation laws, we deduce from the angular momentum addition theorem the following Tensor Product Reduction Theorem: From the (2 j ₁ +1) (2 j ₂ +1) components {T j ₁ m ₁ Q j ₂ m ₂ } defining the tensor W = T j ₁ ⊗ Q j ₂ one can form, for each j in the sequence, j  j ₁  j ₂, …, |j ₁  j ₂ | exactly one irreducible spherical tensor W j with components which is to be compared to: 18

Indeed, given the equivalence of the transformation laws, we deduce from the angular momentum addition theorem the following Tensor Product Reduction Theorem: From the (2 j ₁ +1) (2 j ₂ +1) components {T j ₁ m ₁ Q j ₂ m ₂ } defining the tensor W = T j ₁ ⊗ Q j ₂ one can form, for each j in the sequence, j  j ₁  j ₂, …, |j ₁  j ₂ | exactly one irreducible spherical tensor W j with components which is to be compared to: 19

Thus, for example, if and are vector operators, with spherical components and, then from the nine products we can form linear combinations that form irreducible tensors W j of rank j = 2, 1, and 0. The tensor of rank 0 is a scalar that is clearly related to the dot product The tensor W₁ of rank 1 is proportional to the spherical components of the vector and the tensor of rank 2 is a 5 -component tensor, similar to the quadrupole tensor encountered in electrostatics. 20

Thus, for example, if and are vector operators, with spherical components and, then from the nine products we can form linear combinations that form irreducible tensors W j of rank j = 2, 1, and 0. The tensor of rank 0 is a scalar that is clearly related to the dot product The tensor W₁ of rank 1 is proportional to the spherical components of the vector and the tensor of rank 2 is a 5 -component tensor, similar to the quadrupole tensor encountered in electrostatics. 21

Thus, for example, if and are vector operators, with spherical components and, then from the nine products we can form linear combinations that form irreducible tensors W j of rank j = 2, 1, and 0. The tensor of rank 0 is a scalar that is clearly related to the dot product The tensor W₁ of rank 1 is proportional to the spherical components of the vector and the tensor of rank 2 is a 5 -component tensor, similar to the quadrupole tensor encountered in electrostatics. 22

Thus, for example, if and are vector operators, with spherical components and, then from the nine products we can form linear combinations that form irreducible tensors W j of rank j = 2, 1, and 0. The tensor of rank 0 is a scalar that is clearly related to the dot product The tensor W₁ of rank 1 is proportional to the spherical components of the vector and the tensor of rank 2 is a 5 -component tensor, similar to the quadrupole tensor encountered in electrostatics. 23

Note, as with angular momentum addition, the total number of new components is equal to the total number of the original components, i.e., with j ₁ = j ₂ = 1 has components while the have components. In the general case, to prove that the 2j +1 components of the W j constructed in the manner described in the statement of the theorem comprise a spherical tensor of rank j, we must show that they satisfy the appropriate transformation law. 24

Note, as with angular momentum addition, the total number of new components is equal to the total number of the original components, i.e., with j ₁ = j ₂ = 1 has components while the have components. In the general case, to prove that the 2j +1 components of the W j constructed in the manner described in the statement of the theorem comprise a spherical tensor of rank j, we must show that they satisfy the appropriate transformation law. 25

Note, as with angular momentum addition, the total number of new components is equal to the total number of the original components, i.e., with j ₁ = j ₂ = 1 has components while the have components. In the general case, to prove that the 2j +1 components of the W j constructed in the manner described in the statement of the theorem comprise a spherical tensor of rank j, we must show that they satisfy the appropriate transformation law. 26

Note, as with angular momentum addition, the total number of new components is equal to the total number of the original components, i.e., with j ₁ = j ₂ = 1 has components while the have components. In the general case, to prove that the 2j +1 components of the tensors W j constructed in the manner described in the statement of the theorem comprise a spherical tensor of rank j, we must show that they satisfy the appropriate transformation law. 27

Thus, we consider We now observe that the product of rotation operator matrix elements in this expression can be re-written, using the Clebsch-Gordon series 28

Thus, we consider We now observe that the product of rotation operator matrix elements in this expression can be re-written, using the Clebsch-Gordon series 29

Thus, we consider We now observe that the product of rotation operator matrix elements in this expression can be re-written, using the Clebsch-Gordon series 30

Thus, we consider We now observe that the product of rotation operator matrix elements in this expression can be re-written, using the Clebsch-Gordon series 31

Thus, we consider We now observe that the product of rotation operator matrix elements in this expression can be re-written, using the Clebsch-Gordon series 32

Thus, we consider We now observe that the product of rotation operator matrix elements in this expression can be re-written, using the Clebsch-Gordon series to obtain... 33

Thus, we consider We now observe that the product of rotation operator matrix elements in this expression can be re-written, using the Clebsch-Gordon series to obtain... 34

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 35

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 36

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 37

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 38

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 39

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 40

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 41

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 42

which shows that the do indeed transform as the components of a tensor of rank j, thus proving the theorem. 43

In this segment, we found that the product of two tensors is itself a tensor. In many instances, however, the tensor so produced is reducible into two or more tensors with fewer components. We then stated and proved a tensor product reduction theorem, based entirely on analogy to the angular momentum addition theorem, that tells us explicitly how to reduce a product of spherical tensors into their irreducible tensor components. We note in passing, that this theorem allows one to implicitly represent products of two or more spherical harmonics as a sum over single spherical harmonics, which is the context in which the Clebsch-Gordan coefficients were originally derived. 44

In the next segment we rap up our discussion of angular momentum and rotations by proving something called the Wigner-Eckart theorem, which makes a very powerful statement about the matrix elements of the components of tensor operators taken with respect to the basis states of a standard representation of angular momentum eigenstates. In a certain sense the Wigner-Eckart theorem says that, aside from some overall multiplicative constants, the matrices representing corresponding components of tensor operators of the same rank, pretty much look exactly alike. This gives rise to selection rules for these operators, which are extremely useful for identifying “forbidden” and “allowed” transitions, i.e., which tell us in advance which matrix elements connecting two states are necessarily zero, and which ones are not. We will see how this comes about in the next segment. 45

In the next segment we rap up our discussion of angular momentum and rotations by proving something called the Wigner-Eckart theorem, which makes a very powerful statement about the matrix elements of the components of tensor operators taken with respect to the basis states of a standard representation of angular momentum eigenstates. In a certain sense the Wigner-Eckart theorem says that, aside from some overall multiplicative constants, the matrices representing corresponding components of tensor operators of the same rank, pretty much look exactly alike. This gives rise to selection rules for these operators, which are extremely useful for identifying “forbidden” and “allowed” transitions, i.e., which tell us in advance which matrix elements connecting two states are necessarily zero, and which ones are not. We will see how this comes about in the next segment. 46

In the next segment we rap up our discussion of angular momentum and rotations by proving something called the Wigner-Eckart theorem, which makes a very powerful statement about the matrix elements of the components of tensor operators taken with respect to the basis states of a standard representation of angular momentum eigenstates. In a certain sense the Wigner-Eckart theorem says that, aside from some overall multiplicative constants, the matrices representing corresponding components of tensor operators of the same rank, pretty much look exactly alike. This gives rise to selection rules for these operators, which are extremely useful for identifying “forbidden” and “allowed” transitions, i.e., which tell us in advance which matrix elements connecting two states are necessarily zero, and which ones are not. We will see how this comes about in the next segment. 47

In the next segment we rap up our discussion of angular momentum and rotations by proving something called the Wigner-Eckart theorem, which makes a very powerful statement about the matrix elements of the components of tensor operators taken with respect to the basis states of a standard representation of angular momentum eigenstates. In a certain sense the Wigner-Eckart theorem says that, aside from some overall multiplicative constants, the matrices representing corresponding components of tensor operators of the same rank, pretty much look exactly alike. This gives rise to selection rules for these operators, which are extremely useful for identifying “forbidden” and “allowed” transitions, i.e., which tell us in advance which matrix elements connecting two states are necessarily zero, and which ones are not. We will see how this comes about in the next segment. 48

49