[time 323] Re: [time 320] Re: [time 319] Re: [time 318] correction to [time 317]

Matti Pitkanen (matpitka@pcu.helsinki.fi)
Tue, 18 May 1999 08:51:20 +0300 (EET DST)

On Tue, 18 May 1999, Hitoshi Kitada wrote:

> Dear Matti,
>
> Although I am still at chapt.3, I will try to give a brief explanation ...
>
> ----- Original Message -----
> From: Matti Pitkanen <matpitka@pcu.helsinki.fi>
> Sent: Monday, May 17, 1999 1:44 PM
> Subject: [time 319] Re: [time 318] correction to [time 317]
>
>
> >
> >
> > Dear Hitoshi,
> >
> > could you explain how scalar wave equation results in Frieden's theory.
>
> He assumes a priori that space-time coordinates is
> (ict,x,y,z)=(x_0,x_1,x_2,x_3). So Lorentz transformation follows from
> (exactly, is sufficient to assure) the invariance of Fisher information
> under change of reference frames. In this context his wave equation is
> Klein-Gordon one. Shroedinger equation follows from this as an approximation
> as c tends to infinity.
>
> Frieden considers the wave function \psi(t,x,y,z) as a complex valued
> function. Fisher infromation I for this is
>
> I = const \int_{R^4} dx \nabla \psi* \cdot \nabla \psi,
>
> where
>
> dx = |dx_0| dx_1 dx_2 dx_3,
>
> \nabla=(\partial/\partial x_0, ... ,\partial/\partial x_3).

>
> The bound information J is
>
> J = const \int dr dt \psi* \psi
>
> with r = (x,y,z).
>
> Then the information loss K which should be extrematized (usually minimized)
> is
>
> K = I - J
>
> = \int dr dt [ - \nabla\psi*\cdot\nabla\psi +
> (1/c^2) (\partial \psi*/\partial t) (\partial \psi/\partial t)
> - (m^2 c^2 / \hbar^2) \psi* \psi].
>
> Thus the Euler-Lagrange equation for this variational problem is the
> Klein-Gordon equation.
>
> Frieden's points are in that the Fisher information I becomes the free
> energy part by some simplification of the expression of the information and
> in the argument that the bound information J should be introduced. The
> latter point is in chapt. 4 and I have to postpone it. The Fisher
> information I for one dimensional case is defined as
>
> I = \int dx (\partial (\log p)/ \partial \tehat)^2 p
>
> for a probability density p=p(x|\theta)=p(x,\theta)/p(\theta) (conditional
> probability, here p is assumed real-valued).
>
> In the case of translation invariant p such that p(x|\theta)=p(y-\theta), I
> is equal to
>
> I = \int dx p'(x)^2/p(x)

relateed with time derivative term comes. Is Fisher information
still in question when one uses imaginery coordinate x0 =it?

>
> If p is expressed by real amplitude function q(x) as
>
> p(x)= q(x)^2,

Thank You! It was just this step which I did not really
understand.

>
> then
>
> I = 4 \int dx q'(x)^2.
>
> This is extended to complex valued amplitude function with introducing
> multiple probability density functions, which gives the I for \psi above.
>
>

Best,

MP

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