Hitoshi Kitada (hitoshi@kitada.com)
Fri, 8 Oct 1999 15:09:34 +0900
Dear Matti,
Rethinking about your proof, I found an alternative simple proof of unitarity
of S-matrix. Of course this is a proof on a formal level.
The proof:
Set
R_0(z)=(H_0-z)^{-1}, z: non-real, H_0=L_0(free), V=L_0(int),
P = projection onto the eigenspace of H_0 with eigenvalue 0.
Let m_0 in P\HH (=eigenspace of H_0 with eigenvalue 0). Then the scattering
state m(z) satisfies (I omit the bracket notation)
m(z)=m_0+m_1(z)=m_0-R_0(z)Vm(z).
Thus
Pm(z)=Pm_0+Pm_1(z)=m_0+z^{-1}PVm(z). (1)
Therefore
PVm(z)=zPm_1(z). (2)
Your assumption is
VPm_1(z)=0. (A)
Thus (2) and (A) yield
VPVm(z)=zVPm_1(z)=0, which means
<m(z)|VPV|m(z)>=0,
hence
PVm(z)=0.
This and (2) imply
Pm_1(z)=z^{-1}PVm(z)=0.
Thus by (1) we have
Pm(z)=m_0.
Therefore
<m(z)|P|n(z)>=<m_0|n_0>: unitarity. (U)
Comment 1. As Im z -> 0,
m(z) = m_0 - R_0(z)Vm(z) = (1+R_0(z)V)^{-1}m_0
would be outside the Hilbert space \HH. Then (U) might lose its meaning as Im
z -> 0. This would require the introduction of some larger space \HH_-.
Comment 2. The present formulation of yours uses P explicitly. Namely m=m(z)
may be outside of P\HH. Thus it is free of my criticism in [time 882].
Best wishes,
Hitoshi
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