[time 876]

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Mon, 4 Oct 1999 07:07:24 +0300 (EET DST)

[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>.

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)>.

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)>
constant term. This formulation adds nothing to the previous: it only
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)

The first two terms give opposite results which cancel each

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,

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