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

*Fri, 8 Oct 1999 15:09:34 +0900*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Matti Pitkanen: "[time 908] Re: [time 907] A proof of unitarity"**Previous message:**Matti Pitkanen: "[time 906] Re: [time 902] IBM TeX plugin"**Next in thread:**Matti Pitkanen: "[time 908] Re: [time 907] A proof of unitarity"

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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