OBSERVABILITY OF THE S MATRIX
David N. Williams
November 17, 2011

  In a recent discussion involving cross sections, I asserted
the following well-known fact:

  The S matrix is observable, up to an overall phase factor in
  each superselection sector.

Upon realizing that I remembered neither a reference nor the
proof, I figured I'd better work it out and write it down, as a
crutch for my aging brain.  Here's the result.

  Assume that all absolute squares,

    |<f|S|g>|² =  <f|S|g><g|S*|f>

are observable, where |f> and |g> are normalized, scattering
state vectors in the same superselection sector, and where S is
unitary on each sector.  By polarization, all expressions of the
form

    <f|S|g><g|S*|h> = <g|S*|h><f|S|g>

are then observable, because |h><f| can be written as a sum of
terms of the form |f'><f'|.  For example, John James three-term
polarization gives:

    |h><f| = ½ |h+f><h+f| - ¼(1-i) |h+if><h+if| - ¼(1+i) |h-if><h-if|

It follows that all matrix elements of the operator

    S|g><g|S*

in a sector are observable, for all |g> in the sector.

  Now let S' be another operator, unitary on the same super-
selection sectors, such that S'|g><g|S'* has the same matrix
elements as before; i.e.,

    S'|g><g|S'* = S|g><g|S*

for all |g> in each sector.  Then

    S*S'|g><g| = |g><g|S*S'.

Since all projection operators are sums of sets of |g><g|'s,
S*S' commutes at least with all bounded, selfadjoint operators
on each sector.  Therefore on each sector

    S*S' = exp(iφ) I,

where φ is a numerical phase, constant on the sector. ∎