orbital angular mmentum

1- Orbital angular momentum

Consider a particle described by the Cartesian coordinates (x; y; z) _ r and their conjugate momenta (px; py; pz) ? p. The classical definition of the orbital angular momentum of such a particle about the origin is L = r × p , giving

Lx = y pz - z py (1)
Ly = z px - x pz (2)
Lz = x py - y px (3)


Let us assume that the operators (Lx; Ly; Lz) ? L which represent the components of orbital angular momentum in quantum mechanics can be defined in an analogous manner to the corresponding components of classical angular momentum. In other words, we are going to assume that the above equations specify the angular momentum operators in terms of the position and linear momentum operators. Note that Lx, Ly, and Lz are Hermitian, so they represent things which can, in principle, be measured. Note, also, that there is no ambiguity regarding the order in which operators appear in products on the right-hand sides Eqs. (1).(3), since all of the products consist of operators which commute.

The fundamental commutation relations satisfied by the position and linear momentum operators are:

[xi, xj] = 0 (4)
[pi, pj] = 0 (5)
[xi, pj] = i (6)
where i and j stand for either x, y, or z. Consider the commutator of the operators Lx and Lz:

[Lx, Ly] = [(y pz - z py); (z px - x pz)] = y [pz, z] px + x py [z, pz]
= i (-y px + x py) = i Lz


The cyclic permutations of the above result yield the fundamental commutation relations satisfied by the components of an angular momentum:

[Lx; Ly] = i Lz (8)
[Ly; Lz] = i Lx (9)
[Lz; Lx] = i Ly (10)


These can be summed up more succinctly by writing:

L × L = i L (11)

The three commutation relations (8).(10) are the foundation for the whole theory of angular momentum in quantum mechanics. Whenever we encounter three operators having these commutation relations, we know that the dynamical variables which they represent have identical properties to those of the components of an angular momentum (which we are about to derive). In fact, we shall assume that any three operators which satisfy the com mutation relations (8). (10) represent the components of an angular momentum. Suppose that there are N particles in the system, with angular momentum vectors Li (where i runs from 1 to N). Each of these vectors satisfies Eq. (11), so that

Li × Li = i Li (12)

However, we expect the angular momentum operators belonging to different particles to commute, since they represent different degrees of freedom of the system. So, we can write

Li ×Lj + Lj ×Li = 0 (13)


Consider the magnitude squared of the angular momentum vector,
L2 ? L 2 x + L 2 y + L 2 z . The commutator of L2 and Lz is written

[L2; Lz] = [L2X,LZ] + [L2y,Lz] + [L 2 z ; Lz] (15)

It is easily demonstrated that

[x2,Lz]= -i (Lx Ly + Ly Lx) (16)
[L2y,Lz] = +i (Lx Ly + Ly Lx) (17)
[L 2z ; Lz] = 0 (18)
so
[L2; Lz] = 0 (19)

Since there is nothing special about the z-axis, we conclude that L2 also commutes with Lx and Ly. It is clear from Eqs. (8).(10) and (19) that the best we can do in quantum mechanics is to specify the magnitude of an angular momentum vector along with one of its components (by convention, the z-component). It is convenient to define the shift operators L+ and L-:

L+ = Lx + i Ly (20)
L- = Lx - i Ly (21)
Note that
[L+; Lz] = - L+ (22)
[L-; Lz] = + L- (23)
[L+; L-] = 2 Lz (24)
Note, also, that both shift operators commute with L2.