Hitoshi Kitada (hitoshi@kitada.com)
Thu, 7 Oct 1999 15:57:23 +0900
Dear Matti,
I understand the third part as follows:
----- Original Message -----
From: Hitoshi Kitada <hitoshi@kitada.com>
To: Matti Pitkanen <matpitka@pcu.helsinki.fi>
Cc: <time@kitada.com>
Sent: Thursday, October 07, 1999 3:35 PM
Subject: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity
finally understood!
> Dear Matti,
>
> Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:
>
> Subject: [time 890] Re: [time 888] Re: [time 886] Unitarity finally
> understood!
>
>
> >
> >
> > On Thu, 7 Oct 1999, Hitoshi Kitada wrote:
> >
> > > Dear Matti,
> > >
> > > Let me make a question at each posting for the time being.
> > >
> > > Matti Pitkanen <matpitka@pcu.helsinki.fi> wrote:
> > >
> > > Subject: [time 888] Re: [time 886] Unitarity finally understood!
> > >
> > >
> > > >
> > > >
> > > > Dear Hitoshi,
> > > >
> > > > The posting did not containg any proof of unitarity. I attach a latex
> file
> > > > with proof.
> > > >
> > > > Best,
> > > > MP
>
> skip
> > > >
> > > > \begin{eqnarray}
> > > > \frac{1}{1+X^{\dagger}} P\frac{1}{1+X}&=&G \per ,\nonumber\\
> > > > \nonumber\\
> > > > G(m,n) &=&\delta (m,n) \langle m\vert m\rangle\per .
> > > > \end{eqnarray}
> > >
> > > >From where the projection P comes?
> > >
> > > My understanding:
> > >
> > > |m> = |m_0> + |m_1> = |m_0> - X|m>,
> > >
> > > X=(L_0+iz)^{-1}V, V=L_0(int),
> > >
> > > thus
> > >
> > > |m> = (1+X)^{-1}|m_0>.
> > >
> > > So unitarity
> > >
> > > <m|n> = <m_0|(1+X^\dagger)^{-1}(1+X)^{-1}|n_0> = <m_0|n_0>
> > >
> > This should be written
> >
> > <m|Pn> = <m_0|n_0>
> >
> > since it is projections P|m> which form outgoing states.
>
> Then what you want to prove is
>
> <m|P|n> = <m_0|n_0>.
>
> If so I understand the formula (1) gives what you want.
>
>
> Next let me see:
>
> > a) Consider first the last term appearing at the
> > left hand side:
> >
> > \begin{eqnarray}
> > \langle m_1\vert P \vert n_1\rangle &=&
> > \oint_C d\bar{z} \langle m_0\vert
> > \sum_{k>0}\left[L_0(int)\frac{1}{L_0(free)-i\bar{z}}\right]^k
> > \frac{1}{L_0(free)-i\bar{z}} P \vert n_1\rangle \per .
> > \nonumber\\
> > \end{eqnarray}
> >
> >
> > \noindent The first thing to observe is that
> > $\langle m_1\vert$ has operator
> > $L_0(int)\frac{1}{L_0(free) -i\bar{z}}P$
> > outmost to the right. Since projection operator
> > effectively forces
> > $L_0$ to zero, one can commute $L_0(int)$ past
> > the operators $1/(L_0(free)-i\bar{z})$
> > so that it acts directly to $P\vert n_1\rangle$. But
> > by the proposed condition $L_0(int)P\vert n_1\rangle=0$
> > vanishes!
> >
>
> You here assume the condition
>
> L_0(int)P\vert n_1\rangle=0,
>
> i.e.
>
> VP|n_1>=0.
>
>
> By
>
> |m>=|m_0>+|m_1>=(1+X)^{-1}|m_0>,
>
> one has
>
> |m_1>=\sum_{k>0}(-X)^k |m_0>.
>
> Thus
>
> <m_1|P|n_1> = \sum_{k>0}<m_0|(-X^*)^k P|n_1>
>
> = - \sum_{k>0}<m_0|(-X)^{k-1} X^*P |n_1>
>
> = - \sum_{k>0}<m_0| (-X^*)^{k-1} V(L_0+iz^*)^{-1}P |n_1>
>
> = - \sum_{k>0}<m_0| (-X^*)^{k-1} (iz^*)^{-1} VP|n_1>
>
> = 0 (by the assumption VP|n_1>=0).
>
> OK, I understand this.
>
>
> Next:
>
> > b) Consider next second and third terms at the left hand side
> > of the unitarity condition. The sum of these terms can
> > be written as
> >
> > \begin{eqnarray}
> > \begin{array}{l}
> > \langle m_0\vert P \vert n_1\rangle+ \langle m_1\vert P \vert n_0\rangle
> > \\
> > \\
> > = \frac{1}{2\pi}\oint_C dz
> > \langle m_0 \vert \frac{1}{L_0(free)+iz} \sum_{k>0} X^k
> > \vert n_0\rangle \\
> > \\
> > + \frac{1}{2\pi}\oint_C d\bar{z} \langle m_0
> > \vert \sum_{k>0}
> > (X^{\dagger})^k L_0(int)\vert \frac{1}{L_0(free)-i\bar{z}}
> > \vert n_0\rangle \per .\\
> > \end{array}
> > \end{eqnarray}
> > \vm
> >
> >
> > c) One can
> > project out on mass shell contribution to see what kind
> > of contributions one obtains: what happens that the conditions
> > $L_0(int)P\vert m_1\rangle=$ guarantees that these contributions
> > vanish! Consider the second term in the sum to see how this happens.
> > The on mass shell contributions from
> > terms $X^k\vert n_0\rangle$ can be grouped by the following
> > criterion. Each on mass shell contribution
> > can be characterized by an integer $r$
> > telling how many genuinely off mass shell powers of $X$ appear
> > before it. The on mass shell contributions which
> > come after r:th X can be written in the form $X^r PX^{k-r}$
> > The sum over all these terms coming from $\sum_{n>0} X^n$
> > is obviously given by
> >
> > $$ X^r P\sum_{k>r} X^{k-r}\vert m_0\rangle = X^r P\vert m_1\rangle
> > =0
> > $$
> >
> > \noindent and vanishes
> > sinces $X^r$ is of form $...L_0(int)$ and hence
> > annihilates $\vert m_1\rangle$.
> > Thus the condition implying unitarity also implies that
> > on mass shell states do not contribute to the perturbative
> > expansion.
>
> Here, is your assertion the following?
>
> P\sum_{k>0}X^k|m_0>=0 for k>0.
>
> If this would hold, it is obvious
>
> <m_0|P|n_1>+<m_1|P|n_0>
>
> = <m_0|P\sum_{k>0}(-X)^k|n_0>
>
> + <m_0|\sum_{k>0}(-X^*)^k P|n_0>
>
> = 0.
>
> But is your thought not that the sum
>
> <m_0|P\sum_{k>0}(-X)^k|n_0>
>
> + <m_0|\sum_{k>0}(-X^*)^k P|n_0>
>
> cancels each other?
>
>
>
>
> My understanding:
>
> <m_0|P|n_1>+<m_1|P|n_0>
>
> = <m_0|P\sum_{k>0}(-X)^k|n_0>
>
> + <m_0|\sum_{k>0}(-X^*)^k P|n_0>
>
> = - <m_0|PX\sum_{k>0}(-X)^{k-1}|n_0> (NB: \sum_{k>0}(-X)^{k-1}=(1+X)^{-1})
>
> - <m_0|\sum_{k>0}(-X^*)^{k-1}(X^*)P|n_0>
>
> = - <m_0|PX(1+X)^{-1}|n_0>
>
> - <m_0|(1+X^*)^{-1}(X^*)P|n_0>
>
> = - <m_0|PX|n> - <m|(X^*)P|n_0>. (not |n_1> nor |m_1> but |n> and |m>)
>
> Here
>
> PX=P(L_0+iz)^{-1}V=(iz)^{-1}PV,
>
> (X^*)P=-(iz^*)^{-1}VP.
>
> Thus by your present assumption that |m>=|Pm>
>
> <m_0|P|n_1>+<m_1|P|n_0>
>
> = - (iz)^{-1}<m_0|PV|n> + (iz^*)<m|VP|n_0>
>
> = -(iz)^{-1}<m_0|PV|Pn> + (iz^*)<Pm|VP|n_0>
>
> = -(iz)^{-1}<m_0|PVP|n> + (iz^*)<m|PVP|n_0>.
>
> Here no term like VP|n_1> seems appear?
>
Here
VP|n>=VP|n_0>+VP|n_1>=VP|n_0>
and
<m|PV = <m_0|PV.
Thus the RHS cancels out.
Is this OK?
Best wishes,
Hitoshi
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