Hitoshi Kitada (hitoshi@kitada.com)
Mon, 4 Oct 1999 14:36:13 +0900
Dear Matti,
Comments are below.
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.
2) How is your equality in the second equation above:
|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)?
>
>
> 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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