Hitoshi Kitada (hitoshi@kitada.com)
Thu, 7 Oct 1999 16:53:11 +0900
Dear Matti,
Now I think I understand your proof at least formally.
Now I am considering about your assumption:
VP|m_1>=0.
In my notation I will write with z =E+ie, E: real, e>0:
Psi=Psi_0 - R_0(z)VPsi, H_0=L_0=L_0(free), H=H_0+V, V=L_0(int).
This correspond to your notation
Psi = |m>=(1+X)^{-1}Psi_0 =\sum_{k=0}^\infty (-X)^k Psi_0,
Psi_0 = |m_0>, X=R_0(z)V,
|m_1> = Psi - Psi_0 =-R_0(z)VPsi
= \sum_{k>0}(-X)^k Psi_0.
And your assumption VP|m_1>=0 is
VPR_0(z)VPsi=0. (A)
Noting that
PH_0 = 0
yields
PR_0(z) = (-z)^{-1},
we have (A) is equivalent to
(-z)^{-1}V^2Psi=0.
If z not = 0, then
V^2 Psi=0,
i.e.
L_0(int)^2 |m>=0.
In particular
<m|L_0(int)^*|L_0(int)|m>=0,
which means
L_0(int)|m>=0.
This implies
H|m>=L_0(tot)|m>=(L_0(free)+L_0(int))|m>
= L_0(free)|m> = H_0|m>.
This gives
(H-z)|m>=(H_0-z)|m>,
which further yields
|m>=R(z)(H_0-z)|m>= |m> -R_0(z)V|m>.
Thus
R_0(z)V|m>=0.
This means
|m_1> = -R_0(z)VPsi=0
and
|m>=|m_0>
Thus again there is no scattering.
Best wishes,
Hitoshi
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