[ExI] PSI, PSI*, and Psi
The Avantguardian
avantguardian2020 at yahoo.com
Sat Dec 11 00:22:55 UTC 2010
----- Original Message ----
> From: Damien Broderick <thespike at satx.rr.com>
> To: ExI chat list <extropy-chat at lists.extropy.org>
> Sent: Thu, December 9, 2010 10:27:35 AM
> Subject: Re: [ExI] PSI, PSI*, and Psi
> It sounds as if the TI is new to you, Stuart. It seems to me that some TI
>version is the only way to make sense of the empirical world as I understand it
>(including presentiment, etc). But Cramer's approach has been critiqued in,
>e.g., Michael Price's Many World's FAQ:
TI is new to me, I ran across a reference to it while I was googling on
retrocasuality.
> < 5) Transactional model [C]. Explicitly non-local. An imaginative theory,
>based on the Feynman-Wheeler absorber-emitter model of EM, in which advanced and
>
>retarded probability amplitudes combine into an atemporal "transaction" to form
>the Born probability density. It requires that the input and output states, as
>defined by an observer, act as emitters and absorbers respectively, but not any
>internal states (inside the "black box"), and, consequently, suffers from the
>familiar measurement problem of the Copenhagen interpretation.
Huh? They are called internal states precisely because they can't be observed or
measured. This is a problem (or a feature) of the math of QM and not of this
particular interpretation. Furthermore, as far as I understand Cramer's
explanation, any quantum energy transfer has both an emitter and an absorber,
regardless of whether there is an observer to define their I/O states or not.
Therefore, the wavefunction does collapse but does so independently of an
observer and without the necessity of invoking decoherence. Photons of
light exist independently of eyes because if light didn't exist first, then eyes
would never have evolved to see it.
> If the internal states did act as emitters/absorbers then the wavefunction
>would collapse, for example, around one of the double slits (an internal state)
>in the double slit experiment, destroying the observed interference fringes. In
>transaction terminology a transaction would form between the first single slit
>and one of the double slits and another transaction would form between the same
>double slit and the point on the screen where the photon lands. This never
>observed.>
Why would the slits be part of the transaction? The light source is the emitter
and the screen is the absorber. If you replace the screen with a which-way
detector, then that becomes the absorber. Since whatever you place after the
slits sends a signal backwards in time to the emitter, the photon knows, before
it ever starts it's journey, whether it is supposed to go through one slit or
both. You can't fool light.
Another really cool feature about TI that Cramer *doesn't* mention and maybe
hasn't noticed, is that it directly predicts the vacuum energy of QFT. According
to QFT, every point of empty space is vibrating at every possible frequency. The
energy of this vibration is that of the ground state quantum harmonic oscillator
and is given by the formula: energy = (1/2) * hbar * omega.
Since hbar is the reduced plancks constant which is planck's constant divided by
2*pi, and omega is the angular frequency which is the Hertz frequency multiplied
by 2*pi, you can simply cancel out the 2*pi and you get (1/2) * plancks constant
* frequency. This means that every point in space is vibrating with exactly one
half the energy of photons of every possible frequency. And this is precisely
what you would expect if space was filled with unconfirmed offer waves.
Stuart LaForge
"There is nothing wrong with America that faith, love of freedom, intelligence,
and energy of her citizens cannot cure."- Dwight D. Eisenhower
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