Matti Pitkanen (matpitka@pcu.helsinki.fi)
Thu, 7 Oct 1999 19:26:29 +0300 (EET DST)
On Fri, 8 Oct 1999, Hitoshi Kitada wrote:
> Dear Matti,
>
> Thanks for posting your paper. I read it but before going to physical
> justification part I again stumbled on mathematical part: the proof that (19)
> vanishes. As I reread your [time 894], I found it is interesting idea but does
> not seem to work. I calculated like a blind mathematician:
>
> m_1=Ym_0, Y=\sum_{k>0}(-X)^k =X(1+Y),
>
You srat from state |m_1> and manipulate it.
> (BTW note (1+r)^{-1}=\sum_{k=or>0} (-r)^k not r^k !)
>
> = Xm_0 + Xm_1 (1)
>
> = Xm_0 + XPm_1 + X(1-P)m_1
OK
>
> XPm_1 = 0 yields
>
> m_1 = Xm_0 + X(1-P)m_1
>
You are manipulating the state m_1 in the following. OK.
> = Xm_0 + Zm_1, Z=X(1-P),
>
> = Xm_0 + ZXm_0 + ZXm_1 (by (1))
>
> = (1+Z)Xm_0 + ZXPm_1 + ZX(1-P)m_1
>
> =(1+Z)Xm_0 + ZX(1-P)m_1 (by XPm_1=0)
>
> = ....
>
> = (1+Z+Z^2+Z^3+...)Xm_0 +lim_{k->infty}Z^kXm_1
>
> =(1-Z)^{-1}Xm_0 + lim_{k->infty}Z^kXm_1.
>
> This does not seem to vanish in general.
>
This is state |m_1> as far as I can understand and should not vanish.
> If this does vanish, your T is 0, not only that sum T+T^dagger=0: One would
> need to use cancellation.
I must admit that I cannot follow what you mean. You demonstrate that
|m_1> can be written as |m_1> = (1-Z)^(-1)X|m_0>
OK?
Certainly, if T is zero if |m_1> vanishes. But why |m_1> should vanish?
If your argument shows that |m_1> vanishes then I am in trouble
but your argument seems to show that |m_1> does NOT vanish??
It seems that I do not understand your point!
Best,
MP
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