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

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We have similar expressions for the other Cartesian components.

#### Commutation Relations

For orbital angular momentum we have L=R´P.  We therefore have

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In general we have

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For spin ½ particles we have already shown that

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

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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

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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.

• (a) The eigenvalues of J2 are ³0.

Let |y> be an eigenstate of J2

<y|y>=|| |y>||2 ³0,

But we also have

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Therefore . We write and denote the eigenstates of J2 and Jz by |k,j,m>.  We have .

• (b) The eigenvalues of Jz are mh with -j£m£j.

Let us introduce the operators . These operators commute with J2, but not with each other and with Jz.

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For the products we find

and

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The norm of the vectors is ³0. Therefore

, and

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But we also have

and

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Therefore we have

and

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To simultaneously satisfy both conditions wee need  .

• (c) If m=-j then J-|k,j,-j>=0. If m¹-j then J-|kjm>µ|k,j,m-1>.

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The norm of a vector is zero only if the vector is the null vector.

If then is a non zero vector.

is an eigenvector of J2 with eigenvalue .

is an eigenvector of Jz with eigenvalue .  Therefore .

• (d) If m=j then J+|k,j-j>=0. If m¹j then J+|kjm>µ|k,j,m+1>.

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The norm of a vector is zero only if the vector is the null vector.

If then is a non zero vector.

is an eigenvector of J2 with eigenvalue .

is an eigenvector of Jz with eigenvalue . Therefore .

• (e) The eigenvalues of J2 are , where j is a non negative integer divided by 2.

Let |k,j,m> be a non zero eigenvector of J2 and Jz. .

if . If but then

must be zero, since an eigenstate with m’=m-p-1<-j is not allowed.  But

is only zero if m-p=-j.

Similarly, if . If but then

must be zero, since an eigenstate with m’=m+q+1>j is not allowed.  But

is only zero if m+q=j. We therefore have m+q-m+p=2j,  j=(q+p)/2;

j is a non negative integer divided by two.

#### 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

• (a) the only possible values for j are the non negative half integers, .
• (b) for a fixed j, the only possible values for m are the (2j+1) numbers ; m is an integer if j is an integer, and m is a half integer if j is a half integer.

#### 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 m¹j then there exists another subspace associated with the eigenvalues and .  Similarly, if m¹-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+1¹-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.
• Solution:

The subspace E(l,m) is globally invariant under the action of H. If |ylm> is an eigenvector of L2 and Lz in E(l,m) then H|ylm> is still an eigenvector of L2 and Lz in E(l,m).

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We can diagonalize H inside the subspace E(l,l). Denote by Ekl the g(l) eigenvalues found this way, and by |k,l,l> the associated eigenvectors.
{|k,l,l>; k=1,2,...,g(l) } is an orthonormal basis for E(l,l ).  Construct the orthonormal basis for E(l,m) by applying L-.

.

,

since [H,L]=0.

The eigenvalues Ekl associated with the eigenvalues |k,l,m> are independent of m.