Subsections

### D.36 Photon wave function derivations

The rules of engagement are as follows:

• The Cartesian axes are numbered using an index , with 1, 2, and 3 for , , and respectively.
• 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 . In that case, a more general expression for the energy is:
The minus sign for the terms appears because this is really a dot product of relativistic four-vectors. The zeroth components in such a dot product acquire a minus sign, chapter 1.2.4 and 1.3.2. In derivation {D.32} it was shown that each of the four integrals is constant. That is because each component satisfies the Klein-Gordon equation. So their sum is constant too.

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

Since and :

From now on, it will be understood that there is a summation over and and that everything has a . Therefore these will no longer be shown.

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 is according to the Lorenz condition equal to , so

Using the Klein-Gordon equation, , and then another integration by parts on the first two terms and renotating by gives the terms in (1).

Now consider the integral of in (2). You get, multiplying out,

Now the first and last terms in the right hand side summed over produce all terms in (1) in which and are different. That leaves the middle terms. An integration by parts yields

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 terms in (1) in which and are equal. That leaves the first part. The term in parentheses is the divergence , so the first part is

Perform an integration by parts

Recognizing once more the divergence, this gives the final term in (1)

#### 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 , while stays . 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 harmonic may 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 are zero at the large radius at which the domain is assumed to terminate. That makes the scalar modes a complete set; any scalar function can be written as a combination of them. (That is because they are the eigenfunctions of the Laplacian inside the sphere, and the zero boundary condition on the sphere surface makes the Laplacian Hermitian. This will not be explicitly proved since it is very standard.)

The Bessel function of the scalar modes satisfy the ordinary differential equation, {A.6}

The following integral is needed (note that is real):
This is valid for large , which applies since is large and the values of interest are finite. The above result comes from the integral of the square two-di­men­sion­al Bessel functions , and a recurrence relation, [40, 27.18,88], using , [1, p 437, 10.1.1], and the asymptotic behavior of the Bessel function you get from {A.6} (A.19). To get the leading asymptotic term, each time you have to differentiate the trigonometric function. And where the trigonometric function in is zero at because of the boundary condition, the one in has magnitude 1.

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

In other words, the integral is only 1 if and and otherwise it is zero. Further

##### D.36.2.2 Basic observations and eigenvalue problem

For any function

This follows from writing out the right hand side

the latter since and are different indices.

The electric modes

are solenoidal because gives zero. The magnetic modes

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 in the electric modes since differentiations commute.

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

because for a solenoidal function, (D.1).

##### D.36.2.3 Spherical form and net angular momentum

In spherical coordinates, the magnetic mode is

and then the electric mode is
from [40, 20.74,76,82] and for the component of the eigenvalue problem of chapter 4.2.3.

Now note that the dependence of is through a simple factor , chapter 4.2.3. Therefore it is seen that if the coordinate system is rotated over an angle around the -​axis, it produces a factor in the vectors. First of all that means that the azimuthal quantum number of net angular momentum is , {A.19}. But it also means that, {A.19},

because either way the vector gets multiplied by for the modes. And if it is true for all the modes, then it is true for any function . Since the -​axis is not special for general , the same must hold for the and angular momentum operators. From that it follows that the modes are also eigenfunctions of net square angular momentum, with azimuthal quantum number .

At the cut-off , 0, which gives:

Also needed is, differentiating (10):
which used (3) to get rid of the second order derivative of .

##### 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

where is the surface of the sphere . The second last line can be verified by differentiating out and the last line is the divergence theorem.

The first and second line in (13) show that the Laplacian is Hermitian if all unequal modes are orthogonal (or have equal values, but orthogonality should be shown anyway.). For unequal values orthogonality may be shown by showing that the final surface integral is zero.

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 , respectively direction and are then zero because is zero on the spherical surface . The final two terms summed over can be renotated by shifting the summation index one unit down, respectively up in the cyclic sequence to give

the latter because of the form of , the fact that for a solenoidal vector, and the energy eigenvalue problem established for . The final term is zero because is.

Next consider the orthogonality of the magnetic modes for different quantum numbers. For or , the orthogonality follows from (9) and (6). For , the orthogonality follows from the final line in (13) since the magnetic modes are zero at , (11).

Finally the electric modes. For or , the orthogonality follows from (10), (5), and (6). For , the orthogonality follows from the final line in (13). To see that, recognize that is the radial derivative of ; therefore using (11) and (12), the integrand vanishes.

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 0, you would generally speaking expect two different types of modes described by scalar functions. The electric and magnetic modes seem to fit that bill. But that does not mean that there could not be say a few more special modes. What is needed is to show completeness. That means to show that any smooth vector field satisfying 0 can be written as a sum of the electric and magnetic modes, and nothing else.

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 satisfying a zero boundary condition at . Then completeness would follow from the fact that they are eigenfunctions of the Hermitian Laplacian. Or from more rigorous arguments that you can find in mathematical books on partial differential equations. But how to do something similar here is not obvious, at least not to this author.

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

where and are scalar functions. Completeness then follows since the modes provide a complete description of any arbitrary function and .

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 does not have a radial component, so it will be of this form. And the inverse Fourier transform of is of the form , compare any book on Fourier transforms and (7).

The proof that must be of the stated form is by construction. Note that automatically, the radial component of the two terms is zero. Writing out the gradients in spherical coordinates, [40, 20.74,82], multiplying out the cross products and equating components gives at any arbitrary radius

Now decompose this in Fourier modes in the direction:

For 0, and follow by integration. Note that the integrands are periodic of period and antisymmetric about the -​axis. That makes and periodic of period too,

For 0 make a coordinate transform from to

Note that . If anybody is actually reading this, send me an email. The system becomes after cleaning up

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 must satisfy

and must the derivative of . The solution satisfying the required regularity at is, [40, 19.8],

That finishes the construction, but you may wonder about potential nonexponential terms in the first integral at and the second integral at . Those would produce weak logarithmic singularities in the physical and . You could simply guess that the two right hand sides will combine so that these terms drop out. After all, there is nothing special about the chosen direction of the -​axis. If you choose a different axis, it will show no singularities at the old -​axis, and the solution is unique.

For more confidence, you can check the cancellation explicitly for the leading order, 1 terms. But there is a better way. If the right hand sides are zero within a nonzero angular distance from the -​axis, there are no singularities. And it is easy to split off a part of that is zero within of the axis and then changes smoothly to the correct in an angular range from to from the axis. The remainder of can than be handled by using say the -​axis as the axis of the spherical coordinate system.

##### D.36.2.6 Density of states

The spherical Bessel function is for large arguments proportional to or . Either way, the zeros are spaced apart. So there is one state 1 in an interval . The ratio gives the stated density of states.

##### 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 by . The parity of something is 1 or even if it does not change, and 1 or odd if it changes sign. Under a parity transformation, the operators and flip over the parity of what they act on. On the other hand, has the same parity as ; the spatial components flip over, but so do the unit vectors that multiply them. And the parity of is even if is even and odd if is odd. The stated parities follow.

##### D.36.2.8 Orbital angular momentum of the states

In principle a state of definite net angular momentum and definite spin 1 may involve orbital angular momentum , and , chapter 7.4.2. But states of definite parity restrict that to either only odd values or only even values, {A.20}. To get the stated parities, for magnetic states and or for electric ones.

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