[time 895] Re: still a question


Matti Pitkanen (matpitka@pcu.helsinki.fi)
Thu, 7 Oct 1999 11:48:19 +0300 (EET DST)


On Thu, 7 Oct 1999, Hitoshi Kitada wrote:

> Dear Matti,
>
> Now I think I understand your proof at least formally.
>
> Now I am considering about your assumption:
>
> VP|m_1>=0.

Yes. Although I failed to notice that P is between! I really
thought that it is V|m_1>=0 which guarantees the result!

I cannot follow the details of you argument. But certainly, in real
context it is true. The conditions T+T^dagger=0 and T^daggerT=0 imply
that T vanishes. The point is that diagonal elements of T^daggerT are
positive definite and physically represent total reaction cross
section: the only conlusion is that T vanishes.

Here come however p-adics into the game. The explicit
constuction shows that there is infinite category of
iT matrices satisfyuing hermitian nilpotency conditions.
In fact, tensor product of any hermitian matrix with
hermitian nilponent iT is new hermitian nilpotent iT.

Best,
MP

>
> 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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