Quantum Two

Angular Momentum and Rotations

Angular Momentum and Rotations Rotational Invariance

As we have seen, the behavior exhibited by a quantum system under rotations can usually be connected to a property related to the total angular momentum of the system. In this segment, we consider in some detail the implications and consequences of the notion of rotational invariance, as it applies to quantum states, quantum observables, and which will lead ultimately to the important and useful exercise of identifying irreducible rotationally-invariant subspaces of the 3DRG. We begin by considering what it means for a quantum state |ψ〉 to be rotationally invariant.

As we have seen, the behavior exhibited by a quantum system under rotations can usually be connected to a property related to the total angular momentum of the system. In this segment, we consider in some detail the implications and consequences of the notion of rotational invariance, as it applies to quantum states, quantum observables, and which will lead ultimately to the important and useful exercise of identifying irreducible rotationally-invariant subspaces of the 3DRG. We begin by considering what it means for a quantum state |ψ〉 to be rotationally invariant.

As we have seen, the behavior exhibited by a quantum system under rotations can usually be connected to a property related to the total angular momentum of the system. In this segment, we consider in some detail the implications and consequences of the notion of rotational invariance, as it applies to quantum states, quantum observables, and which will lead ultimately to the important and useful exercise of identifying irreducible rotationally-invariant subspaces of the 3DRG. We begin by considering what it means for a quantum state |ψ〉 to be rotationally invariant.

As we have seen, the behavior exhibited by a quantum system under rotations can usually be connected to a property related to the total angular momentum of the system. In this segment, we consider in some detail the implications and consequences of the notion of rotational invariance, as it applies to quantum states, quantum observables, and which will lead ultimately to the important and useful exercise of identifying irreducible rotationally-invariant subspaces of the 3DRG. We begin by considering what it means for a quantum state |ψ〉 to be rotationally invariant.

As we have seen, the behavior exhibited by a quantum system under rotations can usually be connected to a property related to the total angular momentum of the system. In this segment, we consider in some detail the implications and consequences of the notion of rotational invariance, as it applies to quantum states, quantum observables, and which will lead ultimately to the important and useful exercise of identifying irreducible rotationally-invariant subspaces of the 3DRG. We begin by considering what it means for a quantum state |ψ〉 to be rotationally invariant.

By definition, a physical state |ψ〉 is rotationally invariant if the rotated state Is equal to the un-rotated state for all rotations . Since finite rotations can always be built up out of infinitesimal ones, to be rotationally invariant, it is sufficient for |ψ〉 to be invariant under all infinitesimal rotations, for which Thus, the state |ψ〉 is rotationally invariant if and only if for all

By definition, a physical state |ψ〉 is rotationally invariant if the rotated state Is equal to the un-rotated state for all rotations . Since finite rotations can always be built up out of infinitesimal ones, to be rotationally invariant, it is sufficient for |ψ〉 to be invariant under all infinitesimal rotations, for which Thus, the state |ψ〉 is rotationally invariant if and only if for all

By definition, a physical state |ψ〉 is rotationally invariant if the rotated state Is equal to the un-rotated state for all rotations . Since finite rotations can always be built up out of infinitesimal ones, to be rotationally invariant, it is sufficient for |ψ〉 to be invariant under all infinitesimal rotations, for which Thus, the state |ψ〉 is rotationally invariant if and only if for all

By definition, a physical state |ψ〉 is rotationally invariant if the rotated state Is equal to the un-rotated state for all rotations . Since finite rotations can always be built up out of infinitesimal ones, to be rotationally invariant, it is sufficient for |ψ〉 to be invariant under all infinitesimal rotations, for which Thus, the state |ψ〉 is rotationally invariant if and only if for all

By definition, a physical state |ψ〉 is rotationally invariant if the rotated state Is equal to the un-rotated state for all rotations . Since finite rotations can always be built up out of infinitesimal ones, to be rotationally invariant, it is sufficient for |ψ〉 to be invariant under all infinitesimal rotations, for which Thus, the state |ψ〉 is rotationally invariant if and only if for all

By definition, a physical state |ψ〉 is rotationally invariant if the rotated state Is equal to the un-rotated state for all rotations . Since finite rotations can always be built up out of infinitesimal ones, to be rotationally invariant, it is sufficient for |ψ〉 to be invariant under all infinitesimal rotations, for which Thus, the state |ψ〉 is rotationally invariant if and only if for all

By definition, a physical state |ψ〉 is rotationally invariant if the rotated state Is equal to the un-rotated state for all rotations . Since finite rotations can always be built up out of infinitesimal ones, to be rotationally invariant, it is sufficient for |ψ〉 to be invariant under all infinitesimal rotations, for which Thus, the state |ψ〉 is rotationally invariant if and only if for all i = 1, 2, 3

But if then for all i = 1, 2, 3 and so Thus, a state |ψ〉 is rotationally invariant if and only if it is a state of zero angular momentum, i.e., if it is an angular momentum eigenstate with j = m = 0. Such a state is often said to be an s-state. What properties do s-states have? Well, since the state is rotationally invariant, so are all of its external physical properties. If you look at it from any angle, it always looks the same.

But if then for all i = 1, 2, 3 and so Thus, a state |ψ〉 is rotationally invariant if and only if it is a state of zero angular momentum, i.e., if it is an angular momentum eigenstate with j = m = 0. Such a state is often said to be an s-state. What properties do s-states have? Well, since the state is rotationally invariant, so are all of its external physical properties. If you look at it from any angle, it always looks the same.

But if then for all i = 1, 2, 3 and so Thus, a state |ψ〉 is rotationally invariant if and only if it is a state of zero angular momentum, i.e., if it is an angular momentum eigenstate with j = m = 0. Such a state is often said to be an s-state. What properties do s-states have? Well, since the state is rotationally invariant, so are all of its external physical properties. If you look at it from any angle, it always looks the same.

But if then for all i = 1, 2, 3 and so Thus, a state |ψ〉 is rotationally invariant if and only if it is a state of zero angular momentum, i.e., if it is an angular momentum eigenstate with j = m = 0. Such a state is often said to be an s-state. What properties do s-states have? Well, since the state is rotationally invariant, so are all of its external physical properties. If you look at it from any angle, it always looks the same.

But if then for all i = 1, 2, 3 and so Thus, a state |ψ〉 is rotationally invariant if and only if it is a state of zero angular momentum, i.e., if it is an angular momentum eigenstate with j = m = 0. Such a state is often said to be an s-state. What properties do s-states have? Well, since the state is rotationally invariant, so are all of its external physical properties. If you look at it from any angle, it always looks the same.

But if then for all i = 1, 2, 3 and so Thus, a state |ψ〉 is rotationally invariant if and only if it is a state of zero angular momentum, i.e., if it is an angular momentum eigenstate with j = m = 0. Such a state is often said to be an s-state. What properties do s-states have? Well, since the state is rotationally invariant, so are all of its external physical properties. If you look at it from any angle, it always looks the same.

But if then for all i = 1, 2, 3 and so Thus, a state |ψ〉 is rotationally invariant if and only if it is a state of zero angular momentum, i.e., if it is an angular momentum eigenstate with j = m = 0. Such a state is often said to be an s-state. What properties do s-states have? Well, since the state is rotationally invariant, so are all of its external physical properties. It always appears the same, looked at from any angle,.

Consider, e.g., the mean value of an arbitrary observable Q taken with respect to a rotationally invariant state. For such a state, under any rotation i.e., Where is the rotated version of the observable Q. If |ψ〉 is rotationally invariant this is true for all observables, even when Q′ ≠ Q. The physical properties of |ψ〉 do not have any orientation. Thus, for example, the charge distribution r = r(r) of any isolated collection of particles in a state of zero total angular momentum must be spherically symmetric, as it is with closed shell atoms in their ground state.

Consider, e.g., the mean value of an arbitrary observable Q taken with respect to a rotationally invariant state. For such a state, under any rotation i.e., Where is the rotated version of the observable Q. If |ψ〉 is rotationally invariant this is true for all observables, even when Q′ ≠ Q. The physical properties of |ψ〉 do not have any orientation. Thus, for example, the charge distribution r = r(r) of any isolated collection of particles in a state of zero total angular momentum must be spherically symmetric, as it is with closed shell atoms in their ground state.

Consider, e.g., the mean value of an arbitrary observable Q taken with respect to a rotationally invariant state. For such a state, under any rotation i.e., Where is the rotated version of the observable Q. If |ψ〉 is rotationally invariant this is true for all observables, even when Q′ ≠ Q. The physical properties of |ψ〉 do not have any orientation. Thus, for example, the charge distribution r = r(r) of any isolated collection of particles in a state of zero total angular momentum must be spherically symmetric, as it is with closed shell atoms in their ground state.

Consider, e.g., the mean value of an arbitrary observable Q taken with respect to a rotationally invariant state. For such a state, under any rotation i.e., Where is the rotated version of the observable Q. If |ψ〉 is rotationally invariant this is true for all observables, even when Q′ ≠ Q. The physical properties of |ψ〉 do not have any orientation. Thus, for example, the charge distribution r = r(r) of any isolated collection of particles in a state of zero total angular momentum must be spherically symmetric, as it is with closed shell atoms in their ground state.

Consider, e.g., the mean value of an arbitrary observable Q taken with respect to a rotationally invariant state. For such a state, under any rotation i.e., where is a rotated version of the observable Q. If |ψ〉 is rotationally invariant this is true for all observables, even when Q′ ≠ Q. The physical properties of |ψ〉 do not have any orientation. Thus, for example, the charge distribution r = r(r) of any isolated collection of particles in a state of zero total angular momentum must be spherically symmetric, as it is with closed shell atoms in their ground state.

Consider, e.g., the mean value of an arbitrary observable Q taken with respect to a rotationally invariant state. For such a state, under any rotation i.e., where is a rotated version of the observable Q. If |ψ〉 is rotationally invariant this is true for all observables, even when Q′ ≠ Q. The physical properties of |ψ〉 do not have any orientation. Thus, for example, the charge distribution r = r(r) of any isolated collection of particles in a state of zero total angular momentum must be spherically symmetric, as it is with closed shell atoms in their ground state.

Consider, e.g., the mean value of an arbitrary observable Q taken with respect to a rotationally invariant state. For such a state, under any rotation i.e., where is a rotated version of the observable Q. If |ψ〉 is rotationally invariant this is true for all observables, even when Q′ ≠ Q. The physical properties of |ψ〉 do not have any orientation. Thus, for example, the charge distribution r = r(r) of any isolated collection of particles in a state of zero total angular momentum must be spherically symmetric, as it is with closed shell atoms in their ground state.

Consider, e.g., the mean value of an arbitrary observable Q taken with respect to a rotationally invariant state. For such a state, under any rotation i.e., where is a rotated version of the observable Q. If |ψ〉 is rotationally invariant this is true for all observables, even when Q′ ≠ Q. The physical properties of |ψ〉 do not have any orientation. Thus, for example, the charge distribution r = r (r) of any isolated collection of particles in a state of zero total angular momentum must be spherically symmetric, as it is with closed shell atoms in their ground state.

In contrast to quantum states, as we have already seen, an observable Q that is rotationally invariant, satisfies the relation for all and is what we have referred to as a scalar with respect to rotations. The necessary and sufficient condition for Q to be a scalar is that it satisfy the scalar commutation relations for all i = 1, 2, 3 But that then implies, also, that Since Q commutes with and , it is then possible to construct a standard representation of common eigenstates states {|τ, q, j, m〉} common to J , , and Q, where τ is an index that distinguishes different possible basis states with the same values of q, j, and m.

In contrast to quantum states, as we have already seen, an observable Q that is rotationally invariant, satisfies the relation for all and is what we have referred to as a scalar with respect to rotations. The necessary and sufficient condition for Q to be a scalar is that it satisfy the scalar commutation relations for all i = 1, 2, 3 But that then implies, also, that Since Q commutes with and , it is then possible to construct a standard representation of common eigenstates states {|τ, q, j, m〉} common to J , , and Q, where τ is an index that distinguishes different possible basis states with the same values of q, j, and m.

In contrast to quantum states, as we have already seen, an observable Q that is rotationally invariant, satisfies the relation for all and is what we have referred to as a scalar with respect to rotations. The necessary and sufficient condition for Q to be a scalar is that it satisfy the scalar commutation relations for all i = 1, 2, 3 But that then implies, also, that Since Q commutes with and , it is then possible to construct a standard representation of common eigenstates states {|τ, q, j, m〉} common to J , , and Q, where τ is an index that distinguishes different possible basis states with the same values of q, j, and m.

In contrast to quantum states, as we have already seen, an observable Q that is rotationally invariant, satisfies the relation for all and is what we have referred to as a scalar with respect to rotations. The necessary and sufficient condition for Q to be a scalar is that it satisfy the scalar commutation relations for all i = 1, 2, 3 But that then implies, also, that Since Q commutes with and , it is then possible to construct a standard representation of common eigenstates states {|τ, q, j, m〉} common to J , , and Q, where here τ is an index that distinguishes different possible basis states with the same values of q, j, and m.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that |τ, q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue.

It is not hard to see, then, that the eigenstates of Q associated with a given eigenvalue q necessarily come in 2j + 1 -fold degenerate multiplets. To prove this rotational or essential degeneracy, suppose that | q, j, m〉 is a basis state of such a standard representation such that Now, since [J_{i},Q]=0, it follows that [J_{±},Q]=0, , and so showing that each of the 2j + 1 states obtained by repeated application of is also an eigenstate of Q with the same eigenvalue q.

As an example, for the hydrogen atom, the Hamiltonian is a scalar with respect to rotations, since the quantum kinetic energy is proportional to the scalar operator , and the spherically symmetric potential energy is a scalar-valued function of the scalar operator Thus, for this system there exists a standard representation {|n,ℓ,m〉} of eigenstates of H, ℓ², and ℓz that exhibit just such a rotational degeneracy. For a given value of the principal quantum number n, and a given value of the total angular momentum quantum number ℓ, there are 2ℓ+1 degenerate states {|n,ℓ,m〉 | m = -ℓ, ⋯ , ℓ} with the same energy εn. From the general arguments presented above, it is clear that this will be true of the bounds states of any spherically symmetric potential V(r).

As an example, for the hydrogen atom, the Hamiltonian is a scalar with respect to rotations, since the quantum kinetic energy is proportional to the scalar operator , and the spherically symmetric potential energy is a scalar-valued function of the scalar operator Thus, for this system there exists a standard representation {|n,ℓ,m〉} of eigenstates of H, ℓ², and ℓz that exhibit just such a rotational degeneracy. For a given value of the principal quantum number n, and a given value of the total angular momentum quantum number ℓ, there are 2ℓ+1 degenerate states {|n,ℓ,m〉 | m = -ℓ, ⋯ , ℓ} with the same energy εn. From the general arguments presented above, it is clear that this will be true of the bounds states of any spherically symmetric potential V(r).

As an example, for the hydrogen atom, the Hamiltonian is a scalar with respect to rotations, since the quantum kinetic energy is proportional to the scalar operator , and the spherically symmetric potential energy is a scalar-valued function of the scalar operator Thus, for this system there exists a standard representation {|n,ℓ,m〉} of eigenstates of H, ℓ², and ℓz that exhibit just such a rotational degeneracy. For a given value of the principal quantum number n, and a given value of the total angular momentum quantum number ℓ, there are 2ℓ+1 degenerate states {|n,ℓ,m〉 | m = -ℓ, ⋯ , ℓ} with the same energy εn. From the general arguments presented above, it is clear that this will be true of the bounds states of any spherically symmetric potential V(r).

As an example, for the hydrogen atom, the Hamiltonian is a scalar with respect to rotations, since the quantum kinetic energy is proportional to the scalar operator , and the spherically symmetric potential energy is a scalar-valued function of the scalar operator Thus, for this system there exists a standard representation {|n,ℓ,m〉} of eigenstates of H, ℓ², and ℓz that exhibit just such a rotational degeneracy. For a given value of the principal quantum number n, and a given value of the total angular momentum quantum number ℓ, there are 2ℓ+1 degenerate states {|n,ℓ,m〉 | m = -ℓ, ⋯ , ℓ} with the same energy εn. From the general arguments presented above, it is clear that this will be true of the bounds states of any spherically symmetric potential V(r).

As an example, for the hydrogen atom, the Hamiltonian is a scalar with respect to rotations, since the quantum kinetic energy is proportional to the scalar operator , and the spherically symmetric potential energy is a scalar-valued function of the scalar operator Thus, for this system there exists a standard representation {|n,ℓ,m〉} of eigenstates of H, ℓ², and ℓz that exhibit just such a rotational degeneracy. For a given value of the principal quantum number n, and a given value of the total angular momentum quantum number ℓ, there are 2ℓ+1 degenerate states {|n,ℓ,m〉 | m = -ℓ, ⋯ , ℓ} with the same energy εn. From the general arguments presented above, it is clear that this will be true of the bounds states of any spherically symmetric potential V(r).

As an example, for the hydrogen atom, the Hamiltonian is a scalar with respect to rotations, since the quantum kinetic energy is proportional to the scalar operator , and the spherically symmetric potential energy is a scalar-valued function of the scalar operator Thus, for this system there exists a standard representation {|n,ℓ,m〉} of eigenstates of H, ℓ², and ℓz that exhibit just such a rotational degeneracy. For a given value of the principal quantum number n, and a given value of the total angular momentum quantum number ℓ, there are 2ℓ+1 degenerate states {|n,ℓ,m〉 | m = -ℓ, ⋯ , ℓ} with the same energy εn. From the general arguments presented above, it is clear that this will be true of the bounds states of any spherically symmetric potential V(r).

Quantum states |ψ〉 of zero angular momentum and observables Q that are scalars under rotation exhibit what is sometimes referred to as local rotational invariance. The rotationally invariant |ψ〉 state is taken under rotations exactly back onto itself. The same thing happens with any scalar observable Q. In addition to this local invariance that can occur under rotations, it is also possible for sets of quantum objects to exhibit global invariance with respect to rotations, which means that under an arbitrary rotation, the set of rotated objects is the same as the un-rotated set, but not every element of the set is taken right back onto itself. In the next segment we explore this global rotational invariance as it to rotationally invariant subspaces of the 3DRG.

Quantum states |ψ〉 of zero angular momentum, and observables Q that are scalars under rotation, exhibit what is sometimes referred to as local rotational invariance. The rotationally invariant state |ψ〉 is taken under rotations exactly back onto itself. The same thing happens with any scalar observable Q. In addition to this local invariance that can occur under rotations, it is also possible for sets of quantum objects to exhibit global invariance with respect to rotations, which means that under an arbitrary rotation, the set of rotated objects is the same as the un-rotated set, but not every element of the set is taken right back onto itself. In the next segment we explore this global rotational invariance as it to rotationally invariant subspaces of the 3DRG.

Quantum states |ψ〉 of zero angular momentum and observables Q that are scalars under rotation exhibit what is sometimes referred to as local rotational invariance. The rotationally invariant state |ψ〉 is taken under rotations exactly back onto itself. The same thing happens with any scalar observable Q. In addition to this local invariance that can occur under rotations, it is also possible for sets of quantum objects to exhibit global invariance with respect to rotations, which means that under an arbitrary rotation, the set of rotated objects is the same as the un-rotated set, but not every element of the set is taken right back onto itself. In the next segment we explore this global rotational invariance as it to rotationally invariant subspaces of the 3DRG.

Quantum states |ψ〉 of zero angular momentum and observables Q that are scalars under rotation exhibit what is sometimes referred to as local rotational invariance. The rotationally invariant state |ψ〉 is taken under rotations exactly back onto itself. The same thing happens with any scalar observable Q. In addition to this local invariance that can occur under rotations, it is also possible for sets of quantum objects to exhibit global invariance with respect to rotations, which means that under an arbitrary rotation, the set of rotated objects is the same as the un-rotated set, but not every element of the set is taken right back onto itself. In the next segment we explore this global rotational invariance as it to rotationally invariant subspaces of the 3DRG.

Quantum states |ψ〉 of zero angular momentum and observables Q that are scalars under rotation exhibit what is sometimes referred to as local rotational invariance. The rotationally invariant state |ψ〉 is taken under rotations exactly back onto itself. The same thing happens with any scalar observable Q. In addition to this local invariance that can occur under rotations, it is also possible for sets of quantum objects to exhibit global invariance with respect to rotations, which means that under an arbitrary rotation, the set of rotated objects is the same as the un-rotated set, but not every element of the set is taken right back onto itself. In the next segment we explore this global rotational invariance as it to rotationally invariant subspaces of the 3DRG.

Quantum states |ψ〉 of zero angular momentum and observables Q that are scalars under rotation exhibit what is sometimes referred to as local rotational invariance. The rotationally invariant state |ψ〉 is taken under rotations exactly back onto itself. The same thing happens with any scalar observable Q. In addition to this local invariance that can occur under rotations, it is also possible for sets of quantum objects to exhibit global invariance with respect to rotations, which means that under an arbitrary rotation, the set of rotated objects is the same as the un-rotated set, but not every element of the set is taken right back onto itself. In the next segment we explore this global rotational invariance as it applies to rotationally invariant subspaces of the 3DRG.