# [time 878] Re: Your assumption

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Mon, 4 Oct 1999 12:08:23 +0300 (EET DST)

On Mon, 4 Oct 1999, Hitoshi Kitada wrote:

> Dear Matti,
>
>
> Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:
>
> > [time 876] Message for time
> > Sender: owner-time
> > Precedence: bulk
> >
> >
> >
> > Dear Hitoshi,
> >
> > The previous version was contained still errors. The following formulas
> > provide more correct version. I bet that this is totally trivial for you
> > and also I realized that the introduction of P=(1/2*pi)Int_cdz(1/L_0+iz)
> > is nothing but representing the projector P in elegant manner.
> > In any case, I want to represent the correct formulas.
> >
> > a) Inner product between on mass shell state and scattering state
> > can be defined in the following manner.
> >
> > I write
> >
> > |m(z)> =|m_0> + |m_1(z)>
> >
> > =|m_0> + (1/(L_0+iz))L_0(int)|m_0>.
>
> Is L_0 = L_0(free) or L_0(tot)?
>
> >
> > z-->0 limit must be taken in suitable manner.
> >
> >
> > b) Projector P to the on mass shell states is represented as
> >
> > P= (1/2pi)Int_C dz/(L_0+iz).
> >
> >
> > b) It seems S-matrix can be written
> >
> > S_(m,n)= <m_0|n> = (1/2pi)<m_0|Int_C (dz/(L_0+iz)|n(z)>
> > =<m_0|n_0> + (1/2pi) <m_0|Int_C(1/(L_0+iz)|n_1(z)>.
>
> 1) This (i.e. S_(m,n)= <m_0|n>) is not S-matrix, but wave operator expressed
> on energy shell, as I stated several times.

<m_0|n> is S-matrix and <m_0|n_1> is wave operator as I defined it:

S= 1+T.

What I have done only substitution of P to <m_0||n>=<m_0|P|n>.
That P acts as projector is best seen by simply operating
with it to state. Residy integral picks up only eigenvalue
L_0=0.

>
> 2) How is your equality in the second equation above:
>

P |n> = (1/2pi) Int_C (dz/(L_0+iz)|n(z)>
>
> (= P|m_0> + (1/2pi) Int_C (dz/(L_0+iz)(1/(L_0+iz))L_0(int)|n_0> (by a))
>
>
> proved from a)?
>

I already told the proof. Application to state and calculation
of integral by residy calculus. Only L_0=0 contribution remains as it
should.

> >
> >
> > Here one has
> >
> > |n_1(z)> = sum_(n>0) X^n |n_0>,
> >
> > X(z)= (1/L_0+iz) L_0(int).
> >
> > C is small curve encircling origin and Int is integral over this.
> > The integral gives nonvanishing result if there is pole contribution.
> > This requires that the Laurent expansion of inner product <m_0|n_1(z)>
> > contains
> > constant term. This formulation adds nothing to the previous: it only
> > makes
> > it more elegant and rigorous.
> >
> >
> > c) Consider now unitarity conditions.
> >
> > I must find under what conditions one has <m|n>= <m_0|n_0>:
> >
> > <m,n> = <m_0|n_0> + (1/2*pi)* Int_C (1/L_0+iz) [<m_0|n(z)> +<m(z^*)|n_0>]
> >
> > + (1/2*pi)^2* Int_C Int_C dz* dz (1/L_0-iz^*) (1/L_0+iz)
> > <m_1(z^*)|n_1(z)>.
> >
> >
> > The first two terms give opposite results which cancel each
> > other.
> >
> >
> > The third term vanishes if one has
> >
> >
> > (1/2*pi) Int_C dz L_0(int)(1/L_0+iz) |m(z)>=0.
> >
> > Thus the condition says that the residy of the pole of |m(z)>
> > at z=0 is annihilated by L_0(int). This condition is equivalent with
> > the earlier condition so that nothing new is introduced: Int_C...
> > is only an elegant manner to represent projection operator.
> >
> >
> > With Best,
> > MP
> >
> >
> >
> >
>
> It seems that you start with the space of |n> and the free space of |n_0>.
> If 1/(L_0+iz) is a resolvent, L_0 (=L_0(free) or L_(tot)?) must be defined as
> an operator from \HH into itself. What is your Hilbert space \HH? It must not
> be the space of the free space P\HH corresponding to zero energy nor the space
> of scattering states |n>.
>
> Only after the Hilbert space is identified, we can argue.
>
> Best wishes,
> Hitoshi
>
>

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