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

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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 (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_z with 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_z 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 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:

• 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.