D.40 Quantization of radiation derivations

This gives various derivations for the addendum of the same name.

It is to be shown first that

so the left-hand integral becomes

Now the curl, , is Hermitian, {D.10}, so the second curl can be pushed in front of the first curl. Then curl curl acts as because is solenoidal and the standard vector identity (D.1). And the eigenvalue problem turns into .

Note incidentally that the additional surface integral in {D.10} is zero even for the photon modes of definite angular momentum, {A.21.7}, because for them either is zero on the surface or is. Also note that the integrals become equal instead of opposite if you push complex conjugates on the first factors in the integrands.

Now the Hamiltonian can be worked out. Using Using (A.152)
and (A.162), it is

When that is multiplied out and integrated, the and terms drop out because of (1). The remaining multiplied-out terms in the Hamiltonian produce the stated Hamiltonian after noting the wave function normalization (A.158).

The final issue is to identify the relationships between the coefficients , and as given in the text. The most important question is here under what circumstances and can get very close to the larger value .

The coefficient was defined as

To estimate this, consider the infinite-dimensional vectors and with coefficients

Note that above is the inner product of these two vectors. And an inner product is less in magnitude than the product of the lengths of the vectors involved.

By changing the notations for the summation indices, (letting and ), the sums become the expectation values of , respectively . So

The final equality is by the definition of . The second inequality already implies that is always smaller than . However, if the expectation value of is large, it does not make much of a difference.

In that case, the bigger problem is the inner product between the
vectors and . Normally it is smaller than
the product of the lengths of the vectors. For it to become equal,
the two vectors have to be proportional. The coefficients of
must be some multiple, call it , of
those of :

For larger values of the square roots are about the same. Then the above relationship requires an exponential decay of the coefficients. For small values of , obviously the above relation cannot be satisfied. The needed values of for negative do not exist. To reduce the effect of this

start-upproblem, significant coefficients will have to exist for a considerable range of values.

In addition to the above conditions, the coefficient has to
be close to . Here the coefficient was defined as

Using the same manipulations as for , but with

gives

To bound this further, define

By expanding the square root in a Taylor series,

where is the expectation value of the linear term in the Taylor series; the inequalities express that a square root function has a negative second order derivative. Multiplying these two expressions shows that

Since it has already been shown that the expectation value of must be large, this inequality will be almost an equality, anyway.

In any case,

This is less than

The big question is now how much it is smaller. To answer that, use the shorthand

where is the expectation value of the square root and is the deviation from the average. Then, noting that the expectation value of is zero,

The second-last term is the bound for as obtained above. So, the only way that can be close to is if the final term is relatively small. That means that the deviation from the expectation square root must be relatively small. So the coefficients can only be significant in some limited range around an average value of . In addition, for the vectors and in the earlier estimate for to be almost proportional,

where is some constant. That again means an exponential dependence, like for the condition on . And will have to be approximately . And will have to be about 1, because otherwise start and end effects will dominate the exponential part. That gives the situation as described in the text.