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

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

.

We have similar expressions for the other Cartesian components.

For orbital angular momentum we have * L*=

.

.

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* (

.

If we define the operator *J ^{2}=J_{x}^{2}+J_{y}^{2}+J_{z}^{2}*,
then we can show, using , that .
This is often written as .

We can find simultaneous eigenstates of *J ^{2}* and

We now want to find the **simultaneous eigenstates **of *J ^{2}* and

.

*J ^{2}* and

- (a) The eigenvalues of
*J*are ³0.^{2}Let

*|y>*be an eigenstate of*J*.^{2}*<y|y>=||*|*y**>||*³0, .^{2}But we also have

.

Therefore . We write and denote the eigenstates of

*J*and^{2}*J*by_{z}*|k,j,m>*. We have . - (b) The eigenvalues of
*J*are_{z}*m*h with -*j£m£j*.Let us introduce the operators . These operators commute with

*J*, but not with each other and with^{2}*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*. If m¹-j then_{-}|k,j,-j>=0*J*µ_{-}|kjm>*|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

*J*with eigenvalue .^{2}is an eigenvector of

*J*_{z}^{ }with eigenvalue . Therefore . - (d) If
*m=j*then*J*. If m¹j then_{+}|k,j-j>=0*J*µ_{+}|kjm>*|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

*J*with eigenvalue .^{2}is an eigenvector of

*J*_{z}^{ }with eigenvalue . Therefore . - (e) The eigenvalues of
*J*are , where^{2}*j*is a non negative integer divided by 2.Let

*|k,j,m>*be a non zero eigenvector of*J*and^{2}*J*. ._{z}if . If but then

must be zero, since an eigenstate with

*m’=m-p-*1*<-j*is not allowed. Butis 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. Butis 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.

Let * J* be an arbitrary angular momentum operator, obeying the commutation
relations . If denotes the eigenvalues of

- (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.

Consider a pair of eigenvalues of *J ^{2}* and

{*|k,j,m>; k=1,2,...,g(j,m)*}, with *<k,j,m|k’,j,m>=d _{k,k’}* and in E(

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+*1*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*.

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

in E.

- Solution:
The subspace E(

*l*,*m*) is globally invariant under the action of*H*. If*|y*is an eigenvector of_{lm}>*L*and^{2}*L*in E(_{z}*l,m*) then*H|y*is still an eigenvector of_{lm}>*L*and^{2}*L*in E(_{z}*l*,*m)*..

We can diagonalize

*H*inside the subspace E(*l*,*l*). Denote by*E*the_{kl}*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*associated with the eigenvalues_{kl}*|k,l,m>*are independent of*m*.