- D.36.1 Rewriting the energy integral
- D.36.2 Angular momentum states
- D.36.2.1 About the scalar modes
- D.36.2.2 Basic observations and eigenvalue problem
- D.36.2.3 Spherical form and net angular momentum
- D.36.2.4 Orthogonality and normalization
- D.36.2.5 Completeness
- D.36.2.6 Density of states
- D.36.2.7 Parity
- D.36.2.8 Orbital angular momentum of the states

D.36 Photon wave function derivations

The rules of engagement are as follows:

- The Cartesian axes are numbered using an index
, with1, 2, and 3 for , , andrespectively. - Also,
indicates the coordinate in the direction, , , or. - Derivatives with respect to a coordinate
are indicated by a simple subscript . - If the quantity being differentiated is a vector, a comma is used to separate the vector index from differentiation ones.
- Index
is the number immediately following in the cyclic sequence ...123123...and is the number immediately preceding . - Time derivatives are indicated by a subscript t.
- A bare
integral sign is assumed to be an integration over all space, or over the entire box for particles in a box. The is normally omitted for brevity and to be understood. - A superscript
indicates a complex conjugate.

D.36.1 Rewriting the energy integral

As given in the text, the energy in an electromagnetic field in free
space that satisfies the Coulomb-Lorenz gauge is, writing out the
square magnitudes and individual components,

However, a bit more general expression is desirable. If only the Lorenz condition is satisfied, there may also be an electrostatic potential

The claim to verify now is that the same energy can be obtained from integrating the electric and magnetic fields as

From now on, it will be understood that there is a summation over

Start with the electric field integral. It is, using the above expressions
and multiplying out,

The first term already gives the vector-potential time derivatives in (1). That leaves the final three terms. Perform an integration by parts on the first two. It will always be assumed that the potentials vanish at infinity or that the system is in a periodic box. In that case there are no boundary terms in an integration by parts. So the three terms become

However, the divergence

Using the Klein-Gordon equation,

Now consider the integral of

Now the first and last terms in the right hand side summed over

Renotate the indices cyclically to get

(If you want, you can check that this is the same by writing out all three terms in the sum.) This is equivalent to

as you can see from differentiating and multiplying out. The final term gives after integration by parts the

Perform an integration by parts

Recognizing once more the divergence, this gives the final

D.36.2 Angular momentum states

The rules of engagement listed at the start of this section apply. In addition:

- The quantum numbers
and will be renotated by and , whilestays . That is easier to type. - The quantum numbers are not shown unless needed. For example,
stands for . - A bar on a quantity, like in
, means the complex conjugate. In addition, the (unlisted) quantum numbers of spherical harmonicmay in general be different from those of and are indicated by bars too. - The symbols
and are used as generic scalar functions. They often stand in particular for the scalar modes. - An integral in spherical coordinates takes the form
where .

D.36.2.1 About the scalar modes

The scalar modes are the

It will be assumed that the

The Bessel function

The spherical harmonics are orthonormal on the unit sphere, {D.14.4}

D.36.2.2 Basic observations and eigenvalue problem

For any function

the latter since

The electric modes

are solenoidal because

are solenoidal for the same reason, after noting (7) above.

The Laplacian commutes with the operators in front of the scalar
functions in the electric and magnetic modes. That can be seen for
the magnetic ones from

and the final two terms cancel. And the Laplacian also commutes with the additional

From this it follows that the energy eigenvalue problem is satisfied
because by definition of the scalar modes

D.36.2.3 Spherical form and net angular momentum

In spherical coordinates, the magnetic mode is

Now note that the

because either way the vector gets multiplied by

At the cut-off

D.36.2.4 Orthogonality and normalization

Whether the modes are orthogonal, and whether the Laplacian is
Hermitian, is not obvious because of the weird boundary conditions at

In general the important relations here

The first and second line in (13) show that the Laplacian is Hermitian
if all unequal modes are orthogonal (or have equal

It is convenient to show right away that the electric and magnetic
modes are always mutually orthogonal:

The first two terms in the right hand side can be integrated in the

the latter because of the form of

Next consider the orthogonality of the magnetic modes for different
quantum numbers. For

Finally the electric modes. For

The integral of the absolute square integral of a magnetic mode is,
using (9), (6), and (4),

The integral of the absolute square integral of an electric mode is,
using (10), (5), and (6),

Apply an integration by parts on the second integral,

and then use (3) to get

The normalizations given in the text follow.

D.36.2.5 Completeness

Because of the condition

This author does not know any simple way to do that. It would be
automatic without the solenoidal condition; you would just take each
Cartesian component to be a combination of the scalar modes

What will be done is show that any reasonable solenoidal vector can be
written in the form

where

But to show the above does not seem easy either, so what will be
actually shown is that any vector without radial component can be
written in the form

That is sufficient because the Fourier transform of

The proof that

Now decompose this in Fourier modes

For

For

Note that

It is now easiest to solve the above equations for each of the two right hand sides separately. Here the first right hand side will be done, the second right hand side goes similarly.

From the two equations it is seen that

and

That finishes the construction, but you may wonder about potential
nonexponential terms in the first integral at

For more confidence, you can check the cancellation explicitly for the
leading order,

D.36.2.6 Density of states

The spherical Bessel function is for large arguments proportional to

D.36.2.7 Parity

Parity is what happens to the sign of the wave function under a parity
transformation. A parity transformation inverts the positive
direction of all three Cartesian axes, replacing any position vector

D.36.2.8 Orbital angular momentum of the states

In principle a state of definite net angular momentum

woof.