[time 802] Re: [time 799] Still about construction of U

Hitoshi Kitada (hitoshi@kitada.com)
Sat, 25 Sep 1999 15:01:30 +0900

I send this again, since the former post involved a period at a top of a line
and was cut intermediately.


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)'''

This has the same form as that of (1)''' at least formally (i.e. if we
neglect some topological properties).

Thus the relation between \Psi_0 and \Psi reduces to the equations

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

Now note that for \ep > 0, (H-z)^{-1} = (H-(E+i\ep))^{-1} is given by a
Laplace transform of the unitary propagator exp(-it(H-(E+i\ep))):

(H-(E+i\ep))^{-1} = i \int_0^\infty exp( -it(H-(E+i\ep)) ) dt, (3)

i.e. with z = E+i\ep

(H - z)^{-1} = i \int_0^\infty exp( -it(H-z) ) dt. (4)

A similar formula holds for the unperturbed (free) Hamiltonian H_0.

Thus (1)" is written as

f(z) = i \int_0^\infty exp( -is(H-z) ) \Psi(z) ds

      = i \int_0^\infty exp( -is(H_0-z) ) \Psi_0 ds

      = i \int_0^\infty exp( -is(H_0-E) ) exp(-\ep s) \Psi_0 ds, (5)

when \ep>0.

Taking the Fourier transform of both sides of (5) wrt E, we obtain for t > 0

\int_{-\infty}^\infty exp(-itE) f(E+i \ep) dE

= \int_{-\infty}^\infty exp(-itE) \times

    i \int_0^\infty exp( -is(H_0-E) ) exp(-\ep s) \Psi_0 ds dE

= 2 i \pi exp(-itH_0) exp(-\ep t) \Psi_0. (6)

The treatment of the LHS requires some other arguments, but now let us assume
it gives a result like (6):

the LHS = 2 i \pi exp(-itH) exp(-\ep t) \Psi (7).

Then your equation (1) for \ep = 0 is equivalent to considering the time
dependent limit:

\Psi = \lim_{t -> \infty} exp(itH) exp(-itH_0) \Psi_0. (8)

This formula is not exact, by the approximation that replaced \Psi(z) by
\Psi. But another argument can show that to consider the limit of (1) as
\epsilon -> 0 is equivalent to considering the limits

W_+ = lim_{t -> + \infty} exp(itH) exp(-itH_0). (9)

W_- = lim_{t -> - \infty} exp(itH) exp(-itH_0).

These are called wave operators. The scattering operator is then given by

S = (W_+)* W_-

= lim_{t -> \infty} exp(itH_0) exp(-2itH) exp(itH_0). (10)

Note: Your definition of scattering operator as the limit of U(t,-t) = exp
(-2itH) does not exist. It just oscillates forever. To eliminate the
oscillation and get a convergence needs to compare its behavior to the free
evolution exp(-2itH_0). This is the definition of wave operators.

S-matrix is a Fourier representation of S:

S^ = FSF^{-1}, F being Fourier transform. (11)

Time independent method or stationary method looks simple but it involves the
limit as \ep -> +0 or -0 of the resolvent (H-E-i\ep)^{-1}, which corresponds
to the limit t -> +\infty or -\infty of the unitary propagator exp(-itH).
There are technical differences between the two, which might make treatments
possible in one method but not in the other. But the idea is the same in
either method.


>Unitarity of the S-matrix
>follows automatically, when scattering solutions are properly normalized
>and provided that free and interacting Virasoro generators
>$L_0$ can be regarded as Hermitian operators. Potential
>difficulties are caused by the fact that normalization constants
>can diverge: this is indeed what they do in quantum field field
>theories typically.

Even if normalization constant does not diverge, it is usually a hard task to
prove the unitarity of S-matrix. Physicists' conclusion is merely a guess,
not a proof. Also the Hermiticity of Hamiltonians is not enough to assure
the arguments are valid. It usually requires us to prove that the
Hamiltonians are self-adjoint in the first place at least.

Note: Hermiticity (or symmetricity) requires H to satisfy

H^* is an extension of H,

while self-adjointness requires

H^* = H.

These two are the same in the finite dimensional Hilbert space, but the
difference is crucial in the infinite dimensional Hilbert space.

There are many other neglects of rigorous arguments in physicists' guess.
This makes one to neglect their papers and/or any arguments they claim to
have proved. Physicists must restart from learning the very beginning of

No Nobel prize could have been given if the referees were familiar with
mathematics. This is not restricted to physics, but recently is applied to
math: Witten's results are based on the assumption of the existence of some
Feyman path integrals, which must not be assumed but has to be proved: His
results are vacant if his assumptions are invalid.

Best wishes,

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