Hitoshi Kitada (hitoshi@kitada.com)
Thu, 7 Oct 1999 15:35:55 +0900
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?
Best wishes,
Hitoshi
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