**Matti Pitkanen** (*matpitka@pcu.helsinki.fi*)

*Thu, 7 Oct 1999 11:48:19 +0300 (EET DST)*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Matti Pitkanen: "[time 896] V|m_1>=0 instead of VP|m_1>=0"**Previous message:**Matti Pitkanen: "[time 894] Re: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**In reply to:**Hitoshi Kitada: "[time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"

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:
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*>
*

*> 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:
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*>
*

*> Psi=Psi_0 - R_0(z)VPsi, H_0=L_0=L_0(free), H=H_0+V, V=L_0(int).
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*>
*

*> This correspond to your notation
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*>
*

*> Psi = |m>=(1+X)^{-1}Psi_0 =\sum_{k=0}^\infty (-X)^k Psi_0,
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*>
*

*> Psi_0 = |m_0>, X=R_0(z)V,
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*>
*

*> |m_1> = Psi - Psi_0 =-R_0(z)VPsi
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*>
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*> = \sum_{k>0}(-X)^k Psi_0.
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*>
*

*> And your assumption VP|m_1>=0 is
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*>
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*> VPR_0(z)VPsi=0. (A)
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*>
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*> Noting that
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*>
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*> PH_0 = 0
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*>
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*> yields
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*>
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*> PR_0(z) = (-z)^{-1},
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*>
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*> we have (A) is equivalent to
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*>
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*> (-z)^{-1}V^2Psi=0.
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*>
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*> If z not = 0, then
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*>
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*> V^2 Psi=0,
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*>
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*> i.e.
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*>
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*> L_0(int)^2 |m>=0.
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*>
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*> In particular
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*>
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*> <m|L_0(int)^*|L_0(int)|m>=0,
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*>
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*> which means
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*>
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*> L_0(int)|m>=0.
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*>
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*> This implies
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*>
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*> H|m>=L_0(tot)|m>=(L_0(free)+L_0(int))|m>
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*>
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*> = L_0(free)|m> = H_0|m>.
*

*>
*

*> This gives
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*>
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*> (H-z)|m>=(H_0-z)|m>,
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*>
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*> which further yields
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*>
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*> |m>=R(z)(H_0-z)|m>= |m> -R_0(z)V|m>.
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*>
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*> Thus
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*>
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*> R_0(z)V|m>=0.
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*>
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*> This means
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*>
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*> |m_1> = -R_0(z)VPsi=0
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*>
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*> and
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*>
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*> |m>=|m_0>
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*>
*

*> Thus again there is no scattering.
*

*>
*

*>
*

*>
*

*> Best wishes,
*

*> Hitoshi
*

*>
*

*>
*

**Next message:**Matti Pitkanen: "[time 896] V|m_1>=0 instead of VP|m_1>=0"**Previous message:**Matti Pitkanen: "[time 894] Re: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**In reply to:**Hitoshi Kitada: "[time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"

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