**Required math: algebra, calculus**

**Required physics: quantum mechanics, angular momentum**

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec. 4.3.2 & Problem 4.21.

We have seen how to calculate the eigenvalues of the angular momentum operators and using only the commutation relations for the three components of . This doesn’t tell us the eigenfunctions, however, so we’ll consider that here. The procedure unfortunately involves a lot of calculation, so we just have to get on with it.

It is easiest to work in spherical coordinates, so we first need to express in spherical coordinates. We have

The radius vector is and the cross product of any vector with itself is zero, so the angular momentum operator becomes

using the cross product relations for the spherical unit vectors and .

We need to extract from this formula the values for the three components of , so we need to convert this equation so that the vectors are the cartesian unit vectors. To do this, we can use the formulas

Plugging these into the equation for and collecting terms, we get

The raising and lowering operators can be worked out from these formulas, and we get

We now have all the bits we need to calculate . We use the formula

We work out the first term by applying the two operators to a test function :

Subscripts on the test function represent derivatives with respect to the subscripted variable. Using the trig identity , we get the required result after removing the test function:

We now work out :

using the trig identity to simplify the second derivative term in .

The eigenfunction must satisfy

The eigenfunction equation for is

This equation is equivalent to

since they both have the solution , where is some function of (the precise function doesn’t matter since it cancels out on the left hand side).

Looking back at the solution to the angular part of the SchrÃ¶dinger equation in three dimensions, we see that these differential equations are the same as the ones we encountered there when we used separation of variables. The solutions there were the spherical harmonics, so we see that the eigenfunctions of are in fact spherical harmonics.

One final comment is worth making. When we worked out the eigenvalues of we saw that they had the form where could be any non-negative integer or half-integer. In the solution of the differential equations that give rise to the spherical harmonics, however, only integer values of are allowed. You might think that these extra values of are not physically significant, but when we come to study spin, we see that since the eigenstates of spin have no relation to spatial coordinates so that spherical harmonics are not involved, there is no reason to discard the half-integer values. In fact they turn out to be fundamental, since half-integer spins are common amongst the elementary particles (such as protons, neutrons and electrons).

## Trackbacks

[…] can use the results obtained in expressing the angular momentum operators in spherical coordinates to get a general formula for […]

[…] can use the raising operator to find from this. In spherical coordinates, the raising operator […]

[…] can use the raising operator to find from this. In spherical coordinates, the raising operator […]

[…] is directly proportional to , the eigenfunctions are just the spherical harmonics, and state therefore has degeneracy […]

[…] raising and lowering operators in spherical coordinates […]

[…] the series expansion of the exponential, and the form of in spherical coordinates, , we […]

[…] symmetric potential, their angular factors are the spherical harmonics, which are also the eigenfunctions of the orbital angular momentum operators and meaning that this hamiltonian commutes with these operators, which in turn means […]

[…] that is the product of a radial function and a spherical harmonic. Since spherical harmonics are eigenfunctions of angular momentum, we can use the commutation relations between the angular momentum operators and the position […]