[time 801] Re: [time 799] Still about construction of U

Hitoshi Kitada (hitoshi@kitada.com)
Sat, 25 Sep 1999 14:53:01 +0900

Dear Matti,

I have several questions on your construction of S-matrix.


> Contrary to earlier expectations, it seems that one cannot assign
>explicit Schr\"odinger equation with S-matrix although the
>general structure of the solutions of the Virasoro conditions
>is same as that associated with time dependent perturbation theory
>and S-matrix is completely analogous to that obtained as
>time evolution operator $U(-t,t)$, $t\rightarrow \infty$ in
>the perturbation theory for Schr\"odinger equation.

Does this mean that your former equation

>i\frac{d}{dt}\Psi&=& H\Psi\per , \nonumber\\
>H &\equiv& k\left[-P_T^2 - L_0(vib)-L_0(int)\right]\Psi\per .

is wrong or cannot be derived by the former note?


>\Psi&=&\Psi_0 + \frac{V}{E-H_0-V+i\epsilon} \Psi \per .

(This is equation (1) of your note.)

>Since ordinary Schr\"odinger equation is consistent with the scattering
>matrix formalism avoiding elegantly the difficulties with the
>definition of the limit $U(-t,t)$, $t\rightarrow \infty$, it
>is natural to take this form of Schr\"odinger equation as starting
>point when trying to find Schr\"odinger equation for the 'time' evolution
>operator $U$. One can even forget the assumption
>about time evolution and require only
>that the basic algebraic information guaranteing
>unitarity is preserved. This information boils down to the Hermiticity
> of free and interacting Hamiltonians and
>to the assumption that the spectra
>non-bound states for free and interacting Hamiltonians
>are identical.

It is known that to consider the limit as \epsilon -> 0 in the Schrodinger
equation (1) of your note is equivalent to considering the time limit as t ->
\infty of exp(-itH). So you cannot avoid the difficulty: Below I will try to
show this.

The Schrodinger equation (1) in your note should be written as:

\Psi = \Psi_0 - V ((E+i \epsilon)-H)^{-1} \Psi, (1)


\Psi = \Psi_0 - ((E+i \epsilon)-H)^{-1} V \Psi. (2)

The reason we consider two equations is that the fractional expression in
your note has two interpretations, since the relevant factors are operators,
hence they are noncommutative in general. Also the sign on the RHS (right
hand side) should be minus if the definition of H is H = H_0 + V.

We write z = E+i\ep, where \ep = \epsilon > 0 and E is any real number.

In case (1), the equation is equivalent to

(I - V (H-z)^{-1} ) \Psi = \Psi_0.

This is rewritten as

(H - z - V) (H - z)^{-1} \Psi = \Psi_0,

which is equivalent to

(H_0-z) (H-z)^{-1}\Psi = \Psi_0 (1)'


(H-z)^{-1}\Psi = (H_0-z)^{-1}\Psi_0 (1)"


\Psi = (H-z)(H_0-z)^{-1}\Psi_0

       = \Psi_0 + V(H_0-z)^{-1}\Psi_0 (1)'''

by H = H_0 + V.

Thus if \Psi_0 is a given fixed state function, \Psi depends on z = E+i\ep,
and it should be written as a function of z:

\Psi = \Psi(z).

Note that these hold only when \ep > 0, NOT for an infinitesimal number \ep
because the inverse (H-(E+i\ep))^{-1} does not exist for \ep = 0 in general.

In case (2), the equation is equivalent to


= (I - (H-z)^{-1} V)\Psi (2)'

= (H-z)^{-1} (H-z-V)\Psi

= (H-z)^{-1} (H_0 - z)\Psi.

This is rewritten:

\Psi(z) = (H_0-z)^{-1} (H-z) \Psi_0. (2)"

If we transform \Psi_0 and \Psi to

\Phi_0 = (H-z) \Psi_0,

\Phi = (H-z) \Psi,

the equation (2)" becomes

\Phi(z) = (H-z) (H_0-z)^{-1} \Phi_0. (2)'''

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