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Author Topic: Should the Observer have her Own Wave Function?  (Read 4377 times)

Offline Mr. Scientist

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Should the Observer have her Own Wave Function?
« on: 20/10/2009 06:43:00 »
Previously, i made a suggestion where choice was in fact not only resultant from a wave function collapse, but could itself be a function or a part of what we call the wave function itself. Thinking about it with some detail, we might even start to think of a particular wave function ψ for the human being or better said, the observer denoted as α. From here, it seemed to be quite simple to formulate actual wave function collapses where we describe the observer and the observed under the same system 1).

(1) αψ'(βψ,k)=|(αβ)ψ|^2

Those familiar with statistical mechanics and quantum field theory will know the in's and out's of the technicalities of the mathematics, which are generally-classed here as not too difficult. In equation 1, we find the description of two states, state α and state β, which both resemble seemingly independant values, that is, until the seperabiity of the two are naturally tended to in quantum mechanics. These are the systems of the observer and the observed.

In quantum mechanics, we often deal with observables; quantum eigenstates of a system arise when a sufficient measurement has taken place. In this arguement, the observer and her state vector (of wave description) are often forgotten, in lets say a random physical state of A, we would find that in this description A|ψ>=λ|ψ> we would find λ being an eigenstate of A, however, where is the observer in all this? The random appearance of observables are often unpredictable, for we do not know how quantum system appear to chose the eigenstates it has, and what would appear non-deterministically from a wave function of possibilities.

I cannot answer this paradox, but i can suffice some kind of description for the observers role, and how her wave function may be entangled as the same description of the external system (the observed). In equation 1, the observer α and the observed β have a closed connection upon the square of the wave function, which means information in α and β described by their wave fields ψ have collapsed.

More important relations arise. Our experience of observation occurs in real linear time. In fact, we may describe this in a form using the usual notation of A^n+1=Δt, and never has a negative value because of real-time effects caused by observation. Having the left hand-side of equation 1 being a function of a change in time gives us the integral of the the time lapse with the boundary given as ω and let D=αβ;

∫|DΨ|^2=((αβ)ψ,k)Δt (2)
ω

The usage of k in equation 1 and equation 2 measure the field-strength in which ψ' (the so-called outgoing field)
and ψ (the incoming field) interact.

Because the observer experiences real time, the unobserved object cannot be experienced in real time until the collapse of the wave forms ψ' and ψ. This means that they exist in probability given:

ξ(α,β)=∫Vψ'Vψ^(e^i ∫dt_n(α,ψ')+(β,ψ)k) (3)

The part denoted as Vψ'Vψ is what's called a vertex. The vertex crosses over the fields to each other, so that field ψ' interacts with field ψ. The V is actually for the representation of a linear change with the two states.

Here we calculate the change in which would occur


∏ dψ(α_n,β_n)=dψ'(α1,β1)dψ(α2,β2) (4)
i=1

In this instance, i have arbitrarily-chosen to use a feyman path integral, but representing it for the abstractual use of explaining the parameter's of the observers wave function αψ'. But then, mathematics is an abstractual tool anyhow, so it doesn't really matter. All which matters is that it represents the linear changes of states α and β from α_1β_1 to α_2β_2. The shift can be construded as a sucession of time (between the interacted states of α and β) under the statistical description of |(αβ)ψ|^2, given that the observation must result in a real time scenario, and the amplitude gives a positive value of 1 thus that |x_n|^2=1.

In equation 3 we also find the use of the symbol k, which is also found scattered in the previous equations. k is the field-strength, as you might remember, and such a coupling might occur as:

k=(t_1-t_2)(t_2+t_1)/∫ α,β dt

But how such a coupling happens is pretty trivial. So, should the observer have her own wave descriptions, and have such descriptions have a mathematical meaning within the equations of phyics?

1) The observer and the observed have been often described by physicists as having ''a profound relationship.'' Whilst i do not agree consciousness in general has a massive effect on the outside world, the idea has however given rise the interconnectivity of the fundamental world (a hype i don't share in the same light) http://www.starstuffs.com/physcon2/index.html.
« Last Edit: 20/10/2009 07:26:38 by Mr. Scientist »


 

Offline Vern

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Should the Observer have her Own Wave Function?
« Reply #1 on: 20/10/2009 12:42:43 »
I have seen arguments on this forum where observables must be considered as a system consisting of the observer and the observable. These both constitute the wave function. In this sense different observers considering the same observable may experience different outcomes.
 

Offline Mr. Scientist

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Should the Observer have her Own Wave Function?
« Reply #2 on: 21/10/2009 00:41:28 »
I have seen arguments on this forum where observables must be considered as a system consisting of the observer and the observable. These both constitute the wave function. In this sense different observers considering the same observable may experience different outcomes.

Exactly. If a state on a quantum system (those available eigenstates) are not biased towards two different wave function possibilities, then either:

a) There is multiple possibilities for an object when its not being observed

b) All other extreme possibilities arise in different universes

c) There is only one exclusively correct definition of a quantum system when it's not being observed i.e. pilot waves - from the Bohmian Interpretation.

d) That quantum states actually take all possible states in superpositioning before a resolution is made in the wave function, and all these different possible states means that two observers can equally expect different outcomes, albiet it would be small nonetheless.

There are four specific arguements which tend to arise to answer how the wave function entangled between two observers can yield a different probability value in each case.
 

Offline Mr. Scientist

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Should the Observer have her Own Wave Function?
« Reply #3 on: 21/10/2009 01:41:41 »
Taking that arguement again vern, if there is such a description of two wave functions like αψ'(βψ,k), then the wave state ψ concerning the observed object β may have a defined value which may influence the wave state of ψ' concerning the oberver α. If this is correct, then there would not only be a coupling function of k determining the strength of interaction, but perhaps a dimensionless coefficient may be required to evaluate the feasibility of βψ and αψ'. This way, the given state of the observer does not only have a direct impact on what is observed, but the observed itself inherently holds information ψ which seems relative to the observer(s) in question. The feasibility of ψ to yield a possibility may depend intrinsically on the [parameters] of ψ' and vice versa.
« Last Edit: 21/10/2009 01:43:30 by Mr. Scientist »
 

Offline Mr. Scientist

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Should the Observer have her Own Wave Function?
« Reply #4 on: 21/10/2009 03:12:20 »
I could use retarded and advanced wave-forms that interact according to the wave functions of ψ' and ψ. They are just conjugate values of the one ''thing'' they can square to create. It's the advanced field of wave mechanics and the Transactional Interpretation which takes such wave solutions seriously. One way to derive two different wave solutions which together can either multiply and create another value, or they can cancel out comes in this form:

αψ'(βψ,k)=|(αβ)ψ|^2

and knowing that the effect of ψ' on ψ  and vice versa would mean that a change in α would almost certainly mean a change in β, so:

∂^2ψ'/∂α = ∂^2ψ/∂β

and can have a solution of

ψ(α,β)=f(α-β)+ g(α+β)

where f and g are arbitrary functions, and it has two solutions of:

f(α-β)= αψ'

g(α+β)= βψ

In the description of αψ'(βψ,k), the two wave solutions are split according to the two equations above. But the observer may not observe whatever she may measure due to any randomness, so the wave solution would not need to take two forms, and αψ' would be an independant vector state of a particular given βψ. Suppose now that we assume that there is a feasibility value of f(α-β) to interact with g(α+β), then the dimensionless coefficient can be recognised as ζ, and this is the coefficient which measures the feasibility of something to occur between the two states of αψ' and βψ - or you can imagine the coefficient as measuring the possible states of the two fields over a given time. The expression αψ'(βψ,k) is now given as Δαψ'ζ(Δβψ,k). In this, where the coefficient measures the change in α relative to the change in β and their respective fields, the mathematical propeties in this model are treating both αψ'and βψ as two waves eigenstates which can converge (or multiply) to create a single binded description.

So finally, the last equation is the coefficient and changes in its fullest form:

Δαψ'ζ(Δβψ,k)=(α2-α1)Ψ'(β2-β1)ΨĽk

where integrating α leads to:

(α2-α1)=ζΔβ
 

Offline Vern

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Should the Observer have her Own Wave Function?
« Reply #5 on: 24/10/2009 03:23:16 »
Either you or I have been eating too many green apples. :) Maybe it is I. I didn't see how your response related to the assertion that wave function collapse might entail systems of observer and observed and that each observer and observed must be taken as a system. Two such systems involving the same observable might obtain different results. :)

I saw this assertion on this forum. I suspect that it is not what is real in nature. But it is a ponder-able possibility if we allow the concept of wave function collapse; which I also suspect is not what is real in nature.
 

Offline Mr. Scientist

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Should the Observer have her Own Wave Function?
« Reply #6 on: 24/10/2009 04:42:54 »
Either you or I have been eating too many green apples. :) Maybe it is I. I didn't see how your response related to the assertion that wave function collapse might entail systems of observer and observed and that each observer and observed must be taken as a system. Two such systems involving the same observable might obtain different results. :)

I saw this assertion on this forum. I suspect that it is not what is real in nature. But it is a ponder-able possibility if we allow the concept of wave function collapse; which I also suspect is not what is real in nature.

Well,two observers either exist within states independant of each other or they don't. This would mean that the collapse of state within the consciousness of one individual would give the possibility of an observable yielding two different states when a second observer is involved.

 

Offline Vern

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Should the Observer have her Own Wave Function?
« Reply #7 on: 25/10/2009 14:35:45 »
What you say seems so to me on the condition we think in terms of the Copenhagen interpretation of Quantum physics. I have drifted away from that interpretation over the last twenty years or so.
 

Offline Mr. Scientist

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Should the Observer have her Own Wave Function?
« Reply #8 on: 25/10/2009 21:16:33 »
Yes - its definitely copenhagansistic.
 

Offline demografx

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Should the Observer have her Own Wave Function?
« Reply #9 on: 07/11/2009 03:12:21 »


No such word.
 

Offline Vern

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Should the Observer have her Own Wave Function?
« Reply #10 on: 07/11/2009 15:23:11 »
But we know what it means; so it is a valid communication morsel; and it was effective. :)
 

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Should the Observer have her Own Wave Function?
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