Orbital angular momentum


The operator J, whose Cartesian components satisfy the commutation relations is defined as an angular momentum operator.  For such an operator we have [Ji,J2]=0, i.e. the operator J2=Jx2+Jy2+Jz2 commutes with each Cartesian component of J.  We can therefore find an orthonormal basis of eigenfunctions common to J2 and Jz.  We denote this basis by {|k,j,m>}. We have .

The index j can take on only integral and half integral positive values.  Which integral and half integral values of j are allowed depends on the exact nature of the physical problem.  For a given j the index m can take on one of 2j+1 possible values, m=-j,-j+1,... ,j-1,j.

We define the operators J+=Jx+iJy  and J-=Jx-iJy.  We then have and .

The operators J operating on the basis states {|k,j,m>} yield


Orbital angular momentum

The operator L=RP satisfies the commutation relations and is called the orbital angular momentum operator.  We denote the common eigenstates of L2 and Lz by {|k,l,m>}.


A measurement of L2 and Lz for a free particle yields the values l=1 and m=1.  Later a measurement of Ly is made.

(a) What are the possible values for Ly?
(b) Calculate the probabilities for each of the possible values given in part (a).

Let |y> be an arbitrary state vector, and let the result of operating with Lx on |y> be |f>, i.e. Lx |y>=|f>.  In coordinate representation we write




In coordinate representation the operator Lx is therefore written as




It is often more convenient to work in spherical coordinates, r, q, f;

  are the relationships between Cartesian coordinates and spherical coordinates.  Each spherical coordinate is a function of x, y, and z and each Cartesian coordinate is a function of r, q, f.  We can derive the relationships between the partial derivatives

Image3414a.gif (1372 bytes).


Inverting the matrix we find



Image3418a.gif (1677 bytes)



This yields L2=Lx2+Ly2+Lz2,


L+=Lx+iLy, L-=Lx-iLy,


In coordinate representation the eigenfunction associated with the eigenvalues and of L2 and Lz respectively are therefore solutions to the partial differential equations


with l integral or half integral and m=-l,-l+1,...l-1,l.

The variable r does not appear in any differential operator.  We may therefore separate variables and look for solutions of the form

.  We will show that the are unique.

To have a normalized wave function we need


It is convenient to normalize in the following way

, .

The normalized common eigenfunctions of L2 and Lz are called the spherical harmonics.

Properties of the spherical harmonics

The form a complete set of functions of f and the form a complete set of functions of cosq, therefore the form a complete set of functions of angle on the unit sphere.  Orthonormality is expressed through , and completeness is expressed through .

We can expand an arbitrary function f(q,f) in the series

 with .

Most mathematical tables list properties of the spherical harmonics.  You need to be aware of these properties, so that you can look them up as needed.

Recursion Relations


Complex conjugation

.  This can be derived from the expression of the Ylm in terms of the .


real positive number.  This is guarantied by the choice of cl.

The addition theorem

, with .


The parity operator operating on a function of the coordinates replaces in this function the coordinates of any point in space by those of the point which is obtained by reflection through the origin of the coordinate system.

 in Cartesian coordinates,

 in spherical coordinates.



The parity of the spherical harmonics is well defined and depends only on l.


Consider a particle whose state is described by .  We may expand y in terms of the eigenfunctions of L2 and Lz.

Operating on these eigenfunctions with L yields .

We know that the yklm(r) are of the form .

Therefore operating with L also yields  .

Comparison shows that Rklm1(r)=Rklm(r), i.e. that Rklm(r) is independent of m.

We therefore write .

If [H,L]=0, then we can find common eigenfunctions of H, L2, and Lz.  The eigenvalues Ekl of H and the radial function Rkl(r) are then independent of m.