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

*Mon, 4 Oct 1999 12:08:23 +0300 (EET DST)*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Matti Pitkanen: "[time 879] Re: [time 877] Re: Your assumption"**Previous message:**Hitoshi Kitada: "[time 877] Re: Your assumption"**In reply to:**Matti Pitkanen: "[time 876]"**Next in thread:**Matti Pitkanen: "[time 879] Re: [time 877] Re: Your assumption"

On Mon, 4 Oct 1999, Hitoshi Kitada wrote:

*> Dear Matti,
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*>
*

*> Comments are below.
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*>
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*> Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:
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*>
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*> > [time 876] Message for time
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*> > Sender: owner-time
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*> > Precedence: bulk
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*> >
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*> >
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*> >
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*> > Dear Hitoshi,
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*> >
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*> > The previous version was contained still errors. The following formulas
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*> > provide more correct version. I bet that this is totally trivial for you
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*> > and also I realized that the introduction of P=(1/2*pi)Int_cdz(1/L_0+iz)
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*> > is nothing but representing the projector P in elegant manner.
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*> > In any case, I want to represent the correct formulas.
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*> >
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*> > a) Inner product between on mass shell state and scattering state
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*> > can be defined in the following manner.
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*> >
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*> > I write
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*> >
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*> > |m(z)> =|m_0> + |m_1(z)>
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*> >
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*> > =|m_0> + (1/(L_0+iz))L_0(int)|m_0>.
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*>
*

*> Is L_0 = L_0(free) or L_0(tot)?
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*>
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*> >
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*> > z-->0 limit must be taken in suitable manner.
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*> >
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*> >
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*> > b) Projector P to the on mass shell states is represented as
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*> >
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*> > P= (1/2pi)Int_C dz/(L_0+iz).
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*> >
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*> >
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*> > b) It seems S-matrix can be written
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*> >
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*> > S_(m,n)= <m_0|n> = (1/2pi)<m_0|Int_C (dz/(L_0+iz)|n(z)>
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*> > =<m_0|n_0> + (1/2pi) <m_0|Int_C(1/(L_0+iz)|n_1(z)>.
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*>
*

*> 1) This (i.e. S_(m,n)= <m_0|n>) is not S-matrix, but wave operator expressed
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*> 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:
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*>
*

The equation belows should read

P |n> = (1/2pi) Int_C (dz/(L_0+iz)|n(z)>

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

I already told the proof. Application to state and calculation

of integral by residy calculus. Only L_0=0 contribution remains as it

should.

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

*> Hitoshi
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*>
*

*>
*

**Next message:**Matti Pitkanen: "[time 879] Re: [time 877] Re: Your assumption"**Previous message:**Hitoshi Kitada: "[time 877] Re: Your assumption"**In reply to:**Matti Pitkanen: "[time 876]"**Next in thread:**Matti Pitkanen: "[time 879] Re: [time 877] Re: Your assumption"

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