Angular Momentum

The total angular momentum of an isolated physical system is a constant of motion.  Classically and quantum mechanically, conservation of angular momentum is a consequence of the isotropy of space.  The angular momentum of a system which is not isolated may also be conserved in certain cases.  For example, the orbital angular momentum of a point particle moving in a central potential is conserved.

Classically, angular momentum is defined about a point, it is orbital angular momentum.  In quantum mechanics we associate the observable L (Lx,Ly,Lz) with the orbital angular momentum of a system.  The Hamiltonian of a point particle moving in a central potential commutes with Lx, Ly, and LzL is therefore a constant of motion for this physical system.

In quantum mechanics a particle can have intrinsic angular momentum, for which there exists no classical analog.  We associate with this spin angular momentum the observable S (Sx,Sy,Sz).  We associate with the total angular momentum of a system the observable J (Jx,Jy,Jz).  For a system of N particles

.

We have similar expressions for the other Cartesian components.

Commutation Relations

For orbital angular momentum we have L=RP.  We therefore have

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.

In general we have

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For spin particles we have already shown that

.

We now generalize and define as angular momentum in quantum mechanics any observable J (Jx, Jy, Jz) which satisfies the commutation relations

.

If we define the operator J2=Jx2+Jy2+Jz2, then we can show, using , that .  This is often written as  .

We can find simultaneous eigenstates of J2 and Jz, or J2 and Jx, or J2 and Jy, but not of J2, Jz, Jx, and Jy, since Jz, Jx, and Jy do not commute.


We now want to find the simultaneous eigenstates of J2 and Jz.  We may label these eigenstates by their eigenvalue. Let {lh2} be the set of eigenvalues of J2 and {mh} be the set of eigenvalues of Jz.  We label the eigenstates by |k,l,m>.  We have

.

J2 and Jz do not, in general, constitute a complete set of commuting observables, i.e. the eigenvalues of J2 and Jz do not completely specify the state.  For example, for a spinless particle in a central potential H, L2 and Lz form a C.S.C.O..  Only if we specify the eigenvalues of H, L2 and Lz are the eigenstates no longer degenerate.  The index k is introduced to distinguish between degenerate eigenvectors with the same l and m.

Summary

Let J be an arbitrary angular momentum operator, obeying the commutation relations . If denotes the eigenvalues of J2 and mh denote the eigenvalues of Jz then


Basis states

Consider a pair of eigenvalues of J2 and Jz, and mh .  The eigenvectors associated with this pair of eigenvalues form a subspace E(j,m) of the state space E.  We denote the dimension of this subspace by g(j,m). Let us choose an orthonormal basis for E(j,m),

{|k,j,m>; k=1,2,...,g(j,m)}, with <k,j,m|k’,j,m>=dk,k’ and in E(j,m).  If mj then there exists another subspace associated with the eigenvalues and .  Similarly, if m-j then there exists another subspace associated with the eigenvalues and (m-1)h.  We will now construct an orthonormal basis for

E(j, m+1) and E(j, m-1) starting with the basis { |k,j,m>; k=1,2,...,g(j,m) } chosen for E(j,m).

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Therefore .  The vectors

are orthonormal.  Are they a basis for E(j,m+1)?

Assume they are not. Then there exists a vector orthogonal to all |k,j,m+1>.

is not the null vector, since m+1-j.  But since  .
is orthogonal to all |k,j,m> but is not the null vector.
This is impossible, since { |k,j,m> } form a basis for E(j,m).  The assumption must therefore be wrong.

Similarly we show that the vectors

form an orthonormal basis for E(j,m-1).  The dimensions of E(j,m-1), E(j,m). and E(j,m+1) are the same, i.e. the dimension of E(j,m) is g(j,m)=g(j), independent of m.

Construction of an orthonormal basis for the state space E

Choose an arbitrary orthonormal basis for each subspace E(j,j) for each value of j found in the problem, { |k,j,j>; k=1, 2, g(j) }.  Apply J- to construct the bases for the other 2j subspaces E(j,m).  Thus arrive at the standard basis for E.

in E.

Problem:

Assuming [H,Li]=0, i = x, y, z, show that with H|E,l,m>=E|E,l,m>, E(m)=E(m 1), i.e. E is independent of the Lz eigenvalue m.