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 (L_x,L_y,L_z) with the orbital angular momentum of a system. The Hamiltonian of a point particle moving in a central potential commutes with L_x, L_y, and L_z. L 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 (S_x,S_y,S_z). We associate with the total angular momentum of a system the observable J (J_x,J_y,J_z). For a system of N particles

We have similar expressions for the other Cartesian components.

Commutation Relations

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

In general we have

For spin ½ particles we have already shown that

We now generalize and define as angular momentum in quantum mechanics any observable J (J_x, J_y, J_z) which satisfies the commutation relations

If we define the operator J²=J_x²+J_y²+J_z², then we can show, using , that . This is often written as .

We can find simultaneous eigenstates of J² and J_z, or J² and J_x, or J² and J_y, but not of J², J_z, J_x, and J_y, since J_z, J_x, and J_y do not commute.

We now want to find the simultaneous eigenstates of J² and J_z. 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 J_z. We label the eigenstates by |k,l,m>. We have

J² and J_z do not, in general, constitute a complete set of commuting observables, i.e. the eigenvalues of J² and J_z do not completely specify the state. For example, for a spinless particle in a central potential H, L² and L_z form a C.S.C.O.. Only if we specify the eigenvalues of H, L² and L_z 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 . Therefore . We write and denote the eigenstates of J² and J_z by \|k,j,m>. We have .
	(b) The eigenvalues of J_z are mh with -j£m£j. Let us introduce the operators . These operators commute with J², but not with each other and with J_z. . For the products we find and . The norm of the vectors is ³0. Therefore , and . But we also have and . Therefore we have and . 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>. . 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 J_zwith eigenvalue . Therefore .
	(d) If m=j then J₊\|k,j-j>=0. If m¹j then J₊\|kjm>µ\|k,j,m+1>. . 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 J_zwith 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 J_z. . 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 J_z 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 J_z, 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>=d_k,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).

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,L_i]=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 L_z eigenvalue m.

Solution:

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

We can diagonalize H inside the subspace E(l,l). Denote by E_kl 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 E_kl associated with the eigenvalues |k,l,m> are independent of m.

	(a) The eigenvalues of J² are ³0. Let \|y> be an eigenstate of J². <y\|y>=\|\| \|y>\|\|² ³0, . But we also have . Therefore . We write and denote the eigenstates of J² and J_z by \|k,j,m>. We have .
	(b) The eigenvalues of J_z are mh with -j£m£j. Let us introduce the operators . These operators commute with J², but not with each other and with J_z. . For the products we find and . The norm of the vectors is ³0. Therefore , and . But we also have and . Therefore we have and . 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>. . 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 J_zwith eigenvalue . Therefore .
	(d) If m=j then J₊\|k,j-j>=0. If m¹j then J₊\|kjm>µ\|k,j,m+1>. . 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 J_zwith 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 J_z. . 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.