**Hitoshi Kitada** (*hitoshi@kitada.com*)

*Thu, 7 Oct 1999 16:53:11 +0900*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Matti Pitkanen: "[time 894] Re: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**Previous message:**Hitoshi Kitada: "[time 892] Re: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**In reply to:**Matti Pitkanen: "[time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**Next in thread:**Matti Pitkanen: "[time 895] Re: still a question"

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

**Next message:**Matti Pitkanen: "[time 894] Re: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**Previous message:**Hitoshi Kitada: "[time 892] Re: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**In reply to:**Matti Pitkanen: "[time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**Next in thread:**Matti Pitkanen: "[time 895] Re: still a question"

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