Bob_for_short said:
So, starting from H(A[j]), please write down the problem solution \Psi.
I can't wait to compare the physical and "non-physical" expressions for \Psi.
Patience, like politeness, is a virtue.
But I would have written something similar to Eugene's post #140,
so I won't repeat it.
Bob_for_short said:
[...]A formula like the original one for Ψ but in
terms of A and their vacuums. Isn't it the formula given in my post #110?
Similar, but with some caveats as explained below.
I consider the a-operators as physical. They carry energy-momentum,
spin, etc., as dictated by their antenna j(r,t)
You keep assuming that j(x) is an electric current, or similar. But in
this problem j(x) is just a time-independent external background field.
No extra assumptions concerning its nature were made in the original
problem statement.
"Physical energy" is only defined by reference to the "physical" Hamiltonian.
The "physical" vacuum is defined as the eigenstate of the full Hamiltonian
with lowest eigenvalue, i.e.,
<br />
|0\rangle_z ~:=~ exp(za^* - \bar{z}a)|0\rangle<br />
~=~(?)~ exp(-|z|^2/2) exp(za^*)|0\rangle <br />
The za^* is shorthand for an integral like
<br />
\int\!\! dk \; z(k) \, a^*(k) ~.<br />
|0\rangle_z is annihilated by the A(k) operators.
The A^*(k) act on |0\rangle_z} to generate a Hilbert
space \mathcal{H}_z, just as the a^*(k) act on
|0\rangle to generate a Hilbert space \mathcal{H}_0.
The question is then whether \mathcal{H}_z is unitarily equivalent to \mathcal{H}_0. I.e., can any vector in
\mathcal{H}_z be written as a linear combination of vectors in
\mathcal{H}_0?
Let's recall the definition of z(k), i.e.,
<br />
z(k) ~:=~ C \, \frac{j(k)}{\omega(k)^{3/2}}<br />
(where several constants have been written simply as "C").
The point is this: if z(k) is not square-integrable but divergent, we have
<br />
exp(-|z|^2/2) ~\to~ 0<br />
from which it follows that the two vacua are orthogonal to each other.
Further tedious calculations also show that
every vector in
\mathcal{H}_0 is orthogonal to |0\rangle_z.
Depending on the details of j(x), it may or may not be true that
z(k) is square-integrable.
If j(k) = j\,\delta(k-k_0) (the simplest monochromatic source),
the exact solution |\Psi\rangle is an eigenstate of the operator
a(k_0) which is a coherent state by definition.
The z(k) corresponding your j(k) above is not square-integrable. We cannot even say
that |z|^2 is "divergent". Rather it's simply undefined, involving the square
of a delta distribution. Therefore the solution cannot be expressed as a linear
combination of vectors from \mathcal{H}_0. Nevertheless, the solution
can be expressed in another unitarily-inequivalent Hilbert space,
which is the point of all this.