Matti Pitkanen (matpitka@pcu.helsinki.fi)
Thu, 7 Oct 1999 11:41:48 +0300 (EET DST)
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
This archive was generated by hypermail 2.0b3 on Sun Oct 17 1999 - 22:40:46 JST