**Matti Pitkanen** (*matpitka@pcu.helsinki.fi*)

*Thu, 7 Oct 1999 11:41:48 +0300 (EET DST)*

**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ]**Next message:**Matti Pitkanen: "[time 895] Re: still a question"**Previous message:**Hitoshi Kitada: "[time 893] still a question"**In reply to:**Matti Pitkanen: "[time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**Next in thread:**Hitoshi Kitada: "[time 892] Re: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"

On Thu, 7 Oct 1999, Hitoshi Kitada wrote:

*> 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.
*

*>
*

OK. This IS the condition. I have talked about V|n_1>=0 so that I was

still slightly wrong in the formulation of my condition! The action of

vertex operator on projection P|m> is same as on |m_0>. In this sense

virtual cloud does not contribute to the vertex.

*>
*

*> 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.
*

XP\sum_{k>0}X^k|m_0>=0 because it is just

VP|m_1>

which must vanish! The argument is exactly the same as in showing that

T^daggerT type term vanishes. X is of form 1/(L_0+ iz)V so that

XP|m_1> is proportional to

VP[m_1> and vanishes.

Note that the idea is to separate all intermediate

states to on mass shell and off mass shell contributions.

For instance X^2 is written as XPX +X(1-P)X. Same applies

to however powers X^k in the series for 1/(1+X)

I get form typically terms like

X(1-p)X(1-P)...X(1-P)X P \sum_(m>0) X^m = X(1-P)...X P|m_1>

1-P appears r times before P. There is sum over all values of r

boiling down to O|m_1>. Now I notice that XP|m_1> is proportional

to VP|m_1> and vanishes by my assumption.

*>
*

*> 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?
*

*>
*

I try desperately to figure out what you have typed with my

old eyes..

My though is that T+T^dagger term, that is

the term above, vanishes in unitarity

condition. The first part of argument was for T^daggerT term.

*>
*

*>
*

*>
*

*> 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>
*

This is not assumption. I distinguish between |m> and P|m>.

Latter contains no off mass shell components.

But: in the slightly incorrect formulation V|m_1>=0

I failed to make this distinction: VP|m_1>=0.

*>
*

*> <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?
*

*>
*

*>
*

As I explained, the terms emerge when I separate the dangerous

on mass shell contributions from the powers series \sum X^r.

It is these terms which can spoil the very naive concluiosn that

the two terms above automatically cancel each other.

Best,

MP

**Next message:**Matti Pitkanen: "[time 895] Re: still a question"**Previous message:**Hitoshi Kitada: "[time 893] still a question"**In reply to:**Matti Pitkanen: "[time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"**Next in thread:**Hitoshi Kitada: "[time 892] Re: [time 891] Re: [time 890] Re: [time 888] Re: [time 886] Unitarity finally understood!"

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