Hitoshi Kitada (hitoshi@kitada.com)
Sat, 25 Sep 1999 20:03:51 +0900
Dear Matti,
Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:
Subject: [time 803] Re: [time 801] Re: [time 799] Still about construction of
U
>
>
> On Sat, 25 Sep 1999, Hitoshi Kitada wrote:
>
> > Dear Matti,
> >
> > I have several questions on your construction of S-matrix.
> >
> > 1.
> >
> > > 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
> >
> > >\begin{eqnarray}
> > >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 .
> > >\end{eqnarray}
> >
> > is wrong or cannot be derived by the former note?
>
> As such the equation is probably wrong. The point is
> that I was forced to make *ad hoc* replacement
> of Diff^4 invariant translation generator p_+ with id/dt
>
> p_+--> id/dt
>
>
> in order to obtain Schrodinger equation. The introduction
> of time t leads to potential problems with Poincare invariance,
> which however could be avoided. But this is the main
> ad hoc element of construction.
>
>
> Starting directly from Super Virasoro conditions and just
> writing "scattering solution" for them one avoids
> all ad hoc hypothesis and manifest Poincare invariance is achieved.
> One however loses Schrodinger equation but this is not needed
> since informational "time evolution" is
> totally characterized by S-matrix. Thus I am tending to believe that
> Heisenberg was right: S-matrix has nothing to do with time evolution with
> respect to geometric time.
I will show at the bottom that there is a relation between S-matrix
formulation and the time-dependent formulation. Of course, this is possible
when there are scattering states associated with the Hamiltonian H.
On the point that the total energy is zero is equivalent to non-existence of
time, consider the case:
H\Psi = E\Psi with E not = 0,
where \Psi belongs to the total Hilbert space. That is, \Psi is an eigenstate
of H with non-zero energy.
Then time evolution is
exp(-itH)\Psi = exp(-itE)\Psi.
This means there is no QM motion, i.e. there is no time. That is, even if
there is non zero energy state, it happens that the universe has no time.
>
>
>
> >
> > 2.
> >
> > >\begin{eqnarray}
> > >\Psi&=&\Psi_0 + \frac{V}{E-H_0-V+i\epsilon} \Psi \per .
> > >\end{eqnarray}
> >
> > (This is equation (1) of your note.)
> >
> Sorry: this equation is mistyped:
>
> \Psi&=&\Psi_0 + \frac{V}{E-H_0+i\epsilon} \Psi \per .
>
> The presence of V in denominator would make it to diverge.
> The ordering is also important: V is to the right.
Either equation leads to a correct time dependent formulation. Just by
exchanging roles of some factors.
>
>
> >
> > >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.
>
> In TGD framework single particle Virasoro generators L_0(n) define
> propagators
>
> 1/(p^2-L_0(vib)+i*epsilon)
>
> appearing in stringy diagrams. L_0(vib) is integer valued and gives rise
> to the universal non-negative integer valued mass squared spectrum of
> string models (in suitable units).
>
> In present case i*epsilon is completely equivalent with
> the presence of i*epsilon in the propagators of relativistic quantum field
> theory: epsilon term guarantees that momentum spacetime integration
> over virtual momenta is performed correctly in case that one
> is forced to integrate over pole of propagator.
>
> As far as I can understand this has nothing to do with time but I
> could of course be wrong.
That time does not exist follows from the eigenequation
H\Psi = E\Psi.
But it does not follow from the form of propagators. If H has a continuous
spectral subspace (this space is sometimes called scattering space of H),
then H can have time.
In fact in my formulation,
H \Psi = 0
implies non-existence of time of the universe. But if we want to consider a
scattering state \Phi as the total satte that is orthogonal to eigenspace,
hence to \Psi, then one can recover time.
The local time of a local system arises in the same way. A state \psi of a
local system L is considered a kind of a part of the total state \Psi. Then
it can be shown that \psi can be a scattering state of the local Hamiltonian
H_L. This gives the local time t_L of the system L, but at the same time
there is no time of the total universe.
>
>
> [In string models similar expansion is derived from
> stringy perturbation theory in which time parameter corresponds
> to string coordinate. The problem in string models is
> that this gives only first quantized theory and one should
> second quantize. In TGD this problem is not encountered. Super
> Virasoro generators are of second quantized form from
> beginning. Also topological description of particle reactions
> emerges automatically via the decomposition of L_0(tot) to
> a sum of single particle Virasoro generators L_0(n). This
> decomposition was described in later posting.]
>
>
>
>
>
> >
> > The Schrodinger equation (1) in your note should be written as:
> >
> > \Psi = \Psi_0 - V ((E+i \epsilon)-H)^{-1} \Psi, (1)
>
> Yes with H replaced with free Hamiltonian: I am sorry for my typo.
> I hope that it does not affect you conclusions.
No, it does not affect my conclusion at all.
>
>
> >
> > or
> >
> > \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.
>
> Yes. In my equation 1/(E-H_0 +iepsilon)*V is meant to be the
> correct ordering. This is easy to see: by operating with E-H_0 to both
> sides of the equation one obtains (E-H_0) Psi= (E-H_0)Psi+ VPsi =VPsi
> which is just Schrodinger equation. Provided that order is
> correct.
>
>
> > 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)'
> >
> > or
> >
> > (H-z)^{-1}\Psi = (H_0-z)^{-1}\Psi_0 (1)"
> >
> > or
> >
> > \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
> >
> > \Psi_0
> >
> > = (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)'''
> >
>
>
>
> Certainly one cannot get rid of time in ordinary wave mechanics.
As I stated, it is possible to have time locally without the total time.
> If you have energy, you have also time!
Even when an energy is not zero, if that energy is eigenenergy, then there is
no time.
>
>
> In TGD approach one has
>
> L_0(tot) Psi=0 rather than HPsi = EPsi! No energy, no time!!
>
> By the way, this condition is analogous to your condition
> that entire universe has vanishing energy
>
> HPsi=0
>
> Thus there is something common between our approaches!
Then you agree with that there is no time for the total universe?
>
>
> Also in general Relativity Hamiltonian vanishes as a constraint.
> In TGD However General Coordinate invariant
> as gauge invariance is replaced by Super Virasoro invariance
> which operates at lightcone boundary and H is replaced with L_0.
>
>
> Best,
>
> MP
>
>
Note:
This note is to show a relation between time independent method and time
dependent method. I attached a LaTeX file for your convenience:
Consider the scattering operator:
S f = lim_{t->\infty} exp(itH_0)exp(-2itH) exp(itH_0) f
= (W_+)^* (W_-) f (1)
and compute the inner product:
(S f - f, g)
= ((W_+)^* ((W_-) - (W_+))f, g)
= (((W_-) - (W_+))f, (W_+) g)
= - (lim_{t->\infty} \int_{-t}^t (d/ds) [exp(isH) exp(-isH_0)f], (W_+) g) ds
= - lim_{t->\infty} \int_{-t}^t i (exp(isH) V exp(-isH_0)f, (W_+) g) ds. (2)
We note the intertwining property of wave operators W = W_+ or W_-:
exp(-isH) W = W exp(-isH_0).
By this the RHS of (2) is equal to
= -i lim_{t->\infty} \int_{-t}^t (V exp(-isH_0)f, (W_+) exp(-isH_0)g) ds. (3)
Similarly, we have
(W_+) exp(-isH_0)g
= exp(-isH_0)g + lim_{s->\infty}i\int_0^s exp(iuH)Vexp(-i(s+u)H_0)g du.
Inserting this into (3) we have
(S f - f, g)
= -i lim_{t->\infty}\int_{-t}^t (V exp(-isH_0)f, exp(-isH_0)g) ds
+ i lim_{t->\infty} i \int_{-t}^t \int_0^s
(V exp(-isH_0)f, exp(iuH)Vexp(-i(s+u)H_0)g) du ds. (4)
This time limit is equal to the Abelian limit
(S f - f, g)
= lim_{\ep,\ep'->+0}
[ i \int_0^\infty exp(-\ep u) i \int_{-\infty}^\infty
exp(-\ep'|s|)(exp(i(s+u)H_0)Vexp(-iuH)Vexp(-isH_0)f, g) ds du
-i \int_{-\infty}^\infty exp(-\ep'|s|)(exp(isH_0)V exp(-isH_0)f,g) ds ]. (5)
Letting E_0(\lambda) be the spectral measure for the self-adjoint operator
H_0, we rewrite the inner products (with \lam = \lambda):
(exp(i(s+u)H_0)V exp(-iuH)V exp(-isH_0)f, g)
= \int_{-\infty}^\infty
exp(i(s+u)\lam) (dE_0(\lam)V exp(-iuH)V exp(-isH_0)f, g),
and
(exp(isH_0)V exp(-isH_0)f,g)
= \int_{-\infty}^\infty exp(is\lam) (dE_0(\lam)V exp(-isH_0)f,g).
Then (5) becomes
(S f - f, g)
= lim_{\ep,\ep'->+0}
[ i\int_0^\infty i \int_{-\infty}^\infty
\int_{-\infty}^\infty (dE_0(\lam)V exp(-iu(H-\lam-i\ep))
V exp(-is(H_0-\lam)-|s|\ep')f, g) ds du
-i \int_{-\infty}^\infty \int_{-\infty}^\infty
(dE_0(\lam)V exp(-is(H_0-\lam)-|s|\ep')f, g) ds ]. (6)
Here
\int_0^\infty exp(-iu(H-\lam-i\ep)) du = (1/i) R(\lam+i\ep),
where R(z) = (H-z)^{-1} is the resovent of H, and
\int_{-\infty}^\infty exp(-is(H_0-\lam)-|s|\ep') ds
= (1/i)[R_0(\lam+i\ep') - R_0(\lam-i\ep')].
Thus (6) becomes
(S f - f, g)
= lim_{\ep,\ep'->0} \int_{-\infty}^\infty
(dE_0(\lam)V R(\lam+i\ep)V [R_0(\lam+i\ep')-R_0(\lam-i\ep')]f, g)
- lim_{\ep'->0} \int_{-\infty}^\infty
(dE_0(\lam)V [R_0(\lam+i\ep')-R_0(\lam-i\ep')]f,g). (7)
If H_0 is simple enough such that H_0 has only absolutely continuous
spectrum, then one has
E'_0(\lam)=dE_0/d\lam(\lam) = (2i\pi)^{-1}[R_0(\lam+i0)-R_0(\lam-i0)],
where
R_0(\lam+i0) = lim_{\ep'->0} R_0(\lam+i\ep'), etc.
as an operator between Hilbert spaces with suitable topologies. Thus
(Sf - f, g)
= 2i\pi \int_{-\infty}^\infty (E'_0(\lam)V R(\lam+i0) VE'_0(\lam)f,g)d\lam
-2i\pi \int_{-\infty}^\infty (E'_0(\lam)VE'_0(\lam)f,g)d\lam. (8)
This shows that if f is restricted to the spectrum \lam wrt H_0, then the
image is also restricted to \lam wrt H_0. In particular, S is decomposable
with respect to the spectrum of H_0, or in other words, S is decomposable wrt
the direct integral expression wrt H_0 of the Hilbert space on which H_0 is
defined.Thus S is expressed as wrt to this direct integral decomposition
S = \int S(\lam) d\lam, (9)
where each S(\lam) is an operator in the fiber H(\lam) of the direct integral
expression of the Hilbert space H:
H = \int H(\lam) d\lam.
(H here is not the operator H above.)
The family {S(\lam)} of S(\lam) in (9) is called S-matrix.
Summary: We started with the time dependent definition of the scattering
operator S, and ariived at its stationary definiton.
\beq
S f&=& (W_+)^* (W_-) f= \lim_{t\to\infty} \exp(itH_0)\exp(-2itH) \exp(itH_0)
f\nonumber\\
&=& I+2i\pi \int_{-\infty}^\infty E'_0(\lam)V R(\lam+i0) VE'_0(\lam)d\lam
-2i\pi \int_{-\infty}^\infty E'_0(\lam)VE'_0(\lam)d\lam.
\ene
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\begin{document}
{\bf A Relation between time dependent and stationary representations}
\normalsize
\BP
\F
Consider the scattering operator:
\beq
S f &=& \lim_{t\to\infty} \exp(itH_0)\exp(-2itH) \exp(itH_0) f\nonumber\\
&=& (W_+)^* (W_-) f
\ene
and compute the inner product:
\beq
(S f - f, g)&
=& (W_+^* (W_- - W_+)f, g)\nonumber\\
&=& ((W_- - W_+)f, W_+ g)\nonumber\\
&=& - \left(\lim_{t\to\infty} \int_{-t}^t (d/ds) [\exp(isH) \exp(-isH_0)f],
W_+ g\right) ds \nonumber\\
&=& - \lim_{t\to\infty} \int_{-t}^t i (\exp(isH) V \exp(-isH_0)f, W_+ g) ds.
\ene
We note the intertwining property of wave operators $W = W_+$ or $W_-$:
\beq
\exp(-isH) W = W \exp(-isH_0).\nonumber
\ene
By this the RHS of (2) is equal to
\beq
= -i \lim_{t\to\infty} \int_{-t}^t (V \exp(-isH_0)f, W_+ \exp(-isH_0)g) ds.
\ene
Similarly, we have
\beq
(W_+) \exp(-isH_0)g
= \exp(-isH_0)g + \lim_{s\to\infty}i\int_0^s \exp(iuH)V\exp(-i(s+u)H_0)g
du.\nonumber
\ene
Inserting this into (3) we have
\beq
&&(S f - f, g)\nonumber\\
&=& -i \lim_{t\to\infty}\int_{-t}^t (V \exp(-isH_0)f, \exp(-isH_0)g)
ds\nonumber\\
&&+ i \lim_{t\to\infty} i \int_{-t}^t \int_0^s
(V \exp(-isH_0)f, \exp(iuH)V\exp(-i(s+u)H_0)g) du ds.
\ene
This time limit is equal to the Abelian limit
\beq
&&(S f - f, g)\nonumber\\
&=& \lim_{\ep,\ep'\to+0}
[ i \int_0^\infty \exp(-\ep u) i \nonumber\\
&&\times\int_{-\infty}^\infty
\exp(-\ep'|s|)(\exp(i(s+u)H_0)V\exp(-iuH)V\exp(-isH_0)f, g) ds du\nonumber\\
&&-i \int_{-\infty}^\infty \exp(-\ep'|s|)(\exp(isH_0)V \exp(-isH_0)f,g) ds ].
\ene
Letting $E_0(\lambda)$ be the spectral measure for the self-adjoint operator
$H_0$, we rewrite the inner products:
\beq
&&(\exp(i(s+u)H_0)V \exp(-iuH)V \exp(-isH_0)f, g)\nonumber\\
&=& \int_{-\infty}^\infty
\exp(i(s+u)\lam) (dE_0(\lam)V \exp(-iuH)V\exp(-isH_0)f, g),\nonumber
\ene
and
\beq
(\exp(isH_0)V \exp(-isH_0)f,g)
= \int_{-\infty}^\infty \exp(is\lam) (dE_0(\lam)V \exp(-isH_0)f,g).\nonumber
\ene
Then (5) becomes
\beq
&&(S f - f, g)\nonumber\\
&&= \lim_{\ep,\ep'\to+0} \nonumber\\
&&[ i\int_0^\infty i \int_{-\infty}^\infty
\int_{-\infty}^\infty (dE_0(\lam)V \exp(-iu(H-\lam-i\ep))
V \exp(-is(H_0-\lam)-|s|\ep')f, g) ds du\nonumber\\
&&-i \int_{-\infty}^\infty \int_{-\infty}^\infty
(dE_0(\lam)V \exp(-is(H_0-\lam)-|s|\ep')f, g) ds ].
\ene
Here
\beq
\int_0^\infty \exp(-iu(H-\lam-i\ep)) du = (1/i) R(\lam+i\ep),\nonumber
\ene
where $R(z) = (H-z)^{-1}$ is the resolvent of $H$, and
\beq
\int_{-\infty}^\infty \exp(-is(H_0-\lam)-|s|\ep') ds
= (1/i)[R_0(\lam+i\ep') - R_0(\lam-i\ep')].\nonumber
\ene
Thus (6) becomes
\beq
(S f - f, g)
&&= \lim_{\ep,\ep'\to0} \int_{-\infty}^\infty
(dE_0(\lam)V R(\lam+i\ep)V [R_0(\lam+i\ep')-R_0(\lam-i\ep')]f, g)\nonumber\\
&&- \lim_{\ep'\to0} \int_{-\infty}^\infty
(dE_0(\lam)V [R_0(\lam+i\ep')-R_0(\lam-i\ep')]f,g).
\ene
If $H_0$ is simple enough such that $H_0$ has only absolutely continuous
spectrum, then one has
\beq
E'_0(\lam)=dE_0/d\lam(\lam) =
(2i\pi)^{-1}[R_0(\lam+i0)-R_0(\lam-i0)],\nonumber
\ene
where
\beq
R_0(\lam+i0) = \lim_{\ep'\to0} R_0(\lam+i\ep'), \q \mbox{etc.}\nonumber
\ene
as an operator between Hilbert spaces with suitable topologies. Thus
\beq
(Sf - f, g)
&=& 2i\pi \int_{-\infty}^\infty (E'_0(\lam)V R(\lam+i0)
VE'_0(\lam)f,g)d\lam\nonumber\\
&&-2i\pi \int_{-\infty}^\infty (E'_0(\lam)VE'_0(\lam)f,g)d\lam.
\ene
This shows that if $f$ is restricted to the spectrum $\lam$ wrt $H_0$, then
the image is also restricted to $\lam$ wrt $H_0$. In particular, $S$ is
decomposable with respect to the spectrum of $H_0$, or in other words, $S$ is
decomposable wrt the direct integral expression wrt $H_0$ of the Hilbert
space $\HH$ on which $H_0$ is defined. Thus $S$ is expressed wrt to this
direct integral decomposition as follows:
\beq
S = \int^\oplus \SS(\lam) d\lam,
\ene
where each $\SS(\lam)$ is an operator in the fiber $\HH(\lam)$ of the direct
integral expression of the Hilbert space $\HH$:
\beq
\HH = \int^\oplus \HH(\lam) d\lam.\nonumber
\ene
The family $\{\SS(\lam)\}$ of $\SS(\lam)$ in (9) is called S-matrix.
\BP
\F
{\bf Summary}: We started with the time dependent definition of the
scattering operator $S$, and arrived at its stationary definition:
\beq
S f&=& (W_+)^* (W_-) f= \lim_{t\to\infty} \exp(itH_0)\exp(-2itH) \exp(itH_0)
f\nonumber\\
&=& I+2i\pi \int_{-\infty}^\infty E'_0(\lam)V R(\lam+i0) VE'_0(\lam)d\lam
-2i\pi \int_{-\infty}^\infty E'_0(\lam)VE'_0(\lam)d\lam.
\ene
\end{document}
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