Page 757 Lecture 45: The Eigenvalue Problem of L z and L 2 in Three Dimensions, ct d: Operator Method Date Revised: 2009/02/17 Date Given: 2009/02/11
The Eigenvector-Eigenvalue Problem of L z and L 2 Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 758 Operator Method We ve found the eigenfunctions and eigenvalues in the standard pedestrian way. Let s now use some clever operator methods that recall how we used raising and lowering operators to determine the eigenvalues of the SHO without having to explicitly find the eigenfunctions. We shall see that this method leads to a simpler way to find the eigenfunctions too, just as we were able to obtain all the eigenfunctions of the SHO by applying the raising operator in the position basis to the simple Gaussian ground-state wavefunction. Let s assume we know nothing about the eigenvalue spectrum of L 2 and L z except that the operators commute so they have simultaneous eigenvectors. Denote an eigenstate of L 2 and L z with eigenvalues α and β by α, β. That is L 2 α, β = α α, β L z α, β = β α, β
Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 759 We define angular momentum raising and lowering operators: L ± = L x ± i L y They are named this way because they satisfy so that [L z, L ± ] = ± L ± L z (L ± α, β ) = (± L ± + L ± L z ) α, β = (± + β) (L ± α, β ) That is, when α, β has L z eigenvalue β, the state obtained by applying a raising or lowering operator in the state, L ± α, β, is an eigenvector of L z with eigenvalue β ±. The raising and lowering operators commute with L 2, [L 2, L ± ] = 0 so we are assured that α, β and L ± α, β have the same eigenvalue α of L 2.
Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 760 So, our space will break down into subspaces that are eigenspaces of L 2, which will be further decomposed into subspaces that are eigenspaces of L z. L ± moves between these subspaces of a particular L 2 eigenspace. Explicitly, we have L ± α, β = C ± (α, β) α, β ± We run into the same problem we had with the SHO raising and lowering operators, which is that we so far have no condition that puts a lower or upper limit on the L z eigenvalue β. Heuristically, it would be unphysical to have β 2 > α. This can be seen rigorously as follows: α, β `L 2 L 2 z α, β = α, β `L2 x + L 2 y α, β The latter expression is nonnegative because the eigenvalues of L 2 x and L 2 y are all nonnegative because the eigenvalues of L x and L y are real because they are Hermitian. So we see α β 2 0, or α β 2 as desired.
Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 761 So, we require there to be states α, β max and α, β min that satisfy L + α, β max = 0 L α, β min = 0 where by 0 we mean the null vector, usually referred to as 0, which may be confusing in this situation. We need to rewrite these expressions in terms of L 2 and L z to further reduce them; let s apply L and L + to do this: L L + α, β max = 0 L +L α, β min = 0 `L2 L 2 z L z α, βmax = 0 `L2 L 2 z + L z α, βmin = 0 `α β 2 max β max α, βmax = 0 β max (β max + ) = α `α β 2 min + β min α, βmin = 0 β min (β min ) = α which implies β min = β max
Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 762 In order for the raising chain begun at β min and the lowering chain begun at β max to terminate, it is necessary that there be a k + and k such that Therefore (L +) k++1 α, β min α, β max (L ) k +1 α, β max α, β min β min + k + = β max β max k = β min So we have k + = k k β max β min = k Since β min = β max, we then have β max = k 2 α = β max (β max + ) = 2 k 2 «k 2 + 1 k = 0, 1, 2,... For k even, we recover the allowed eigenvalues we obtained via the differential equation method. The k odd eigenvalues are a different beast, though, and are associated with spin, a degree of freedom that behaves like angular momentum in many ways but is not associated with orbital motion of a particle.
Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 763 The last point is a very important one: the k odd values arose only from the assumption of the angular momentum operator commutation relations. They did not come from the differential equations, which is what ties all of this to the behavior of spatial wavefunctions; the differential equations method does not permit k odd. This is the source of our statement that the k odd values are not associated with orbital angular momentum. In detail, the restriction to k even comes from the requirement that the wavefunction be single-valued in φ, which is required by Hermiticity of L z. Such a requirement would not hold for a particle spin s z-component operator because there will be no spatial wavefunction to consider. Thus, the above proof tells us which values of k are allowed, and then other restrictions can further reduce the set. Unlike Shankar, who gives a bit more detailed of a hint at what is meant by spin, we will delay discussion until we have time to do it thoroughly. For now it is not important to have a physical picture of the states that result in half-integral values of L z.
Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 764 Given that the spectrum of eigenvalues we have derived is more general than just orbital angular momentum L, we will follow standard notation and use J instead of L. We will denote the eigenvalues as follows: We will denote by j the value of k/2. j may take on any nonnegative integral or half-integral value. The J 2 eigenvalue is α = 2 j(j + 1). However, we will replace α in α, β by j for brevity. The J z eigenvalue β can take on values from j to j in steps of size. We define m = β/. We will replace β in α, β by m for consistency with the notation we developed via the differential equation method. Therefore, simultaneous eigenstates of J 2 and J z will be denoted by j, m and will have J 2 eigenvalue α = 2 j (j + 1) and J z eigenvalue β = m.
Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 765 Summary Let us take a step back and see what we have done and where we should go. What we have done: We are considering problems in two or three spatial dimensions in cylindrical and spherical coordinates with an eye toward working with Hamiltonians that are invariant under rotations and hence depend on the cylindrical coordinate ρ or the radial coordinate r. Since a continuous symmetry transformation of a Hamiltonian derives from a generator operator that commutes with the Hamiltonian, we knew it would be useful to find the generator and its eigenvalues and eigenvectors to help us reduce or solve the eigenvector-eigenvalue problem of the full Hamiltonian. This led us to write explicit forms for L and L 2 and to obtain their eigenvectors and eigenfunctions, both in the position basis and in the more natural basis of their eigenstates. We have thus been able to organize the Hilbert space into subspaces of specific values of the angular momentum magnitude.
Section 14.5 Rotations and Orbital Angular Momentum: The Eigenvector-Eigenvalue Problem of L z and L 2 Page 766 We have two important tasks left: To understand the full structure of the Hilbert space in terms of the eigenstates of J 2 and J z ; i.e., let s write down explicit forms for all the operators we have considered: J x, J y, J +, J and rotation operators. To understand the connection between the { j, m } basis and the position basis eigenstates essentially, to show that we can obtain the position basis eigenstates from the structure of the Hilbert space in terms of the { j, m } basis. We consider these tasks next.