## Homework Statement

The conjecture implies the eigenstate of the observer and observed given by alpha and beta. The problem i am reaching at the end, is whether the well-known Wigners Friend Paradox is answerable by describing the observer and the observed, and then a second observer in terms of Dual 4-Space modelling.

## The Attempt at a Solution

$\alpha (\beta) = \beta$

where alpha is the observer, beta is the oberved and act as an eigentstate equation above. Beta on the left handside operates as a ''point reference''.

It is found that:

$\alpha (\beta) \in R^3$

$R^3$ is the flat spacetime metric which has no time description, yet.

$\zeta_{\alpha} (\psi (t))=(\alpha ,t)$

$\zeta_{\beta} (\psi^{\dagger} (t))=(\beta ,t)$

These two equations act as scalar reference elements. The first equation for instance is not so much a function of alpha on psi, but depends on psi. The second is its complex conjugate transpose. This can be given a retarded and advanced solution:

$| \psi > \in V$

Where V is the acting linear space and $\overline{V}$ is its dual space:

$< \psi | \in \overline{V}$

Elementary one can write:

$\alpha (\beta (t))=\beta(t+1)$

$\rightarrow \alpha (\beta (\Delta t)=\alpha (\beta (A^{n+1}))$

where $A^{n+1}= \Delta t$

I now derive the relationship:

$\alpha \psi(k \beta \psi^{\dagger}( \Delta t))= \int |\alpha , \beta (\psi)|^2$

For the probability of the measurement in conjecture with the original eigenstate equation we used.

k is proportional to the coupling:

$k= \frac{ (t_1 - t_2)(t_2+t_1) }{ \int \alpha, \beta dt}$

For now we could assume the coupling is where the wave function collapses when professor Wigner observes the particle system. We also assume that the two conjugate psi waves describing both the prof. and the particle are describe under a final state vector $< 0 >$ ~ If Wigners friend enters the room it must indicate that there is a retrocausal event in the wave function which no longer states that the wave state on the professor and the observed particle is all there is. A change occurs and the state vector now covers both the mind of Wigner and his counterpart.

If we allow

${\alpha_1, \alpha_2, ..., \alpha_n}$ to be the basis of $V$ then let ${\beta_1, \beta_2, ... , \beta_n}$ be its dual basis. Then ${\beta_1*, \beta_2*, ..., \beta_n*}$ is the dual basis of ${\beta_1, \beta_2, ... , \beta_n}$.

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In effect, during the combination of the first eigenstate equation, we can state:

$\begin{bmatrix}1 \iff \alpha(\beta)\\0 otherwise \end{bmatrix}$

The interpretation suggests that when the observed and the observer (connect) in measurement is a real value of $\mathbb{1}$ but when the system is not being attentively watched or associated with the observer $\alpha(\beta) \ne \beta$ it is zero. The connection between the observer and the observed is made in real time measurements $t_{\mathbb{R_{4}}}$.

This leads to the identification of:

$\sum^{n}_{i=1} |\alpha_{i} \beta_{j} \psi(t)|^2$

this means that $\alpha_{i} \beta_{j}$ are direct products between the linear vector component $V$ and its dual space vector component $\overline{V}$ which makes a complex tranpose between two half valued retarded and advanced wave functions, making a single state vector of the current condition of $\alpha(\beta)$.

This makes them members

$\alpha_{i} \beta_{j}=<\alpha_i (\psi(t))|\beta_j (\psi^{\dagger}(t))>$

$\begin{bmatrix} \hat{\alpha} \\ \hat{\beta} \end{bmatrix}$

Where the identities $\hat{\alpha}$ and $\hat{\beta}$ are vector quantities with the values:

$\alpha_1,\alpha_2,\alpha_3,\alpha_4$ [*]

$\beta_1, \beta_2, \beta_3$

It is best noted that [*] is in fact the observer with a geometry of a four dimensional mapping $\mathbb{R^4}$. The scaling of the following is the observed system, and has geometric properties, but does not have a time parameter. It would have a time parameter if it satisfies $V \otimes \overline{V}$ where $\alpha(\beta) \in \mathbb{R^4}$ ~ Notice how the identity relationship now has a four dimensional manifold signature $R^4$.

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Since one cannot have a cross product in a four dimensional manifold, the description of the observer experiencing a real local time, we can use an exterior product giving a new identification of:

$\alpha{\mathbb{R^4} \wedge \beta$ would i have calculated make a seven dimensional linear space belonging in the dual space relation $V \otimes \overline{V}$.

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So before we delve into even more geometry as suggested by post 3, we can identify more Cartesian Products to help bring together a better understanding of the matrix solutions of the direct product relationships $\alpha_{i}\beta_{j}$.

$\prod^{n}_{k+1} \alpha_{k}(\psi(t))$

Since $\psi(t)$ has a complex conjugate given in the previous given mathematical postulates, $\psi^{\dagger}(t)$ ~ $\prod^{n}_{k+1} \beta_{k}(\psi^{\dagger}(t))$

This means that it satisfies

$\prod^{4}_{n=1} \mathbp{R}= \mathbp{R}x \mathbp{R}x \mathbp{R}x \mathbp{R} = \mathbb{R^4}$

$\prod^{n}_{k+1} \alpha, \beta_{k}(\psi^{\dagger}(t))=\sum \alpha_i \beta_j {\psi}'(t)$ [*]

First $\alpha_i \beta_j = \frac{\partial}{\partial t}\psi (t)$ ~ this means that the equation [*] is then substituted to provide
$\prod^{n}_{k+1} (\alpha, \beta_{k}(\psi^{\dagger}(t)))=\sum^{4} \alpha_i \beta_j (\frac{\partial}{\partial t} \psi (t))(\frac{\partial}{\partial t}\psi^{\dagger} (t))$

The interactions are working out the derivative of $\int |\psi|^2$ since it takes the form of $\alpha_i \beta_{J}=<\alpha_i (\psi(t))|(\psi^{\dagger}(t))\beta_{j}>$ and so act retrocausally when the systems state vector $|\Phi>=\alpha |\psi>$ and $<\Phi|=<\psi|\beta$ where $\Phi$ becomes a reference-frame in the mathematical configurations:

$[\zeta_{\Phi}(\alpha, \beta(\psi(t)))]$

The reference element $\Phi$ acts as the ''association'' but not depending on the psi function which is evidently time-dependant. A function of time itself,

$\zeta_{\Phi}(\psi(t))$ focusing on $\psi(t)$ It's interesting to note that its not always this is true when taken through diffeomorphisms invariants, which leads to identifications of $\hat{H}|\psi>=0$.

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$\prod^{n}_{k+1} (\alpha, \beta_{k}(\psi^{\dagger}(t)))=\sum^{4} \alpha_i \beta_j (\frac{\partial}{\partial t} \psi (t))(\frac{\partial}{\partial t}\psi^{\dagger} (t))$

Has a time description which remains static by definition, but vanishes mathematically;
$\prod^{n}_{k+1} (\alpha, \beta_{k}(\psi^{\dagger}(t)))=\sum^4 \alpha_i \beta_j (\frac{\partial}{\partial t} \psi)(\frac{\partial}{\partial t}\psi^{\dagger})$

The idea is that the dynamical arrow $\frac{\partial}{\partial t}$ is conserved due to the fundamental processes, but the time-dependance of the wave function (or mathematically-speaking) the psi-function being a dependance on a function of time disappears in a $\hat{H}|\psi> \ne \Delta Mc^2$.

The loss of this dependancy in the time function of the wave function must mean an indication of an absent observer, since the evolution of time is purely local, and intricate to the appearance of a recording machine - just like ourselves, who experience a linear flow to time even if the linear experience of time does not exist fundamentally.

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Since the observer experiences a local real time which is linear by perception, it has a description which can be explained in terms of tangent vector analysis of a manifold. For the state of $\Delta t$ then it can be expressed as:

$\mathbp{R^3}(\chi_{(1,2)} ,t')f=<\nabla f, \mathbp{R^3} (\chi_{(1,2)} , t')|_{t}$

By mapping out a visual representation on this manifold, the direction of time which is represented by the change in position along the tangent vector line $\chi_{(1)t} \rightarrow \chi_{(2)t'}$ we can by preference treat the change in time as one experienced by the observer, which has been denoted so far as simply $\alpha$.

Two forms don't exist for the identity of $\alpha \wedge \beta$, because the observer in this interpretation has not only the visual experience of three-dimensions, but is interpretated to be intimately related on the observed through the perception of measurements made in real time. So to solve the no-two form one can have:

$\mathbp{R^4} \triangleq d \alpha_1 \wedge d \alpha _2 \wedge d \alpha_3 \wedge d \alpha_4$

Knowing we can make the identification of vector bundles on the manifold, we can speculate to treat $\alpha$ as a p one-form will invariantly make it decomposable $\Phi=\alpha_1 \wedge ... \wedge \alpha_p$. If we treat $\mathbp{R^4} \triangleq d \alpha_1 \wedge d \alpha _2 \wedge d \alpha_3 \wedge d \alpha_4$ as a subspace, a decomposable p-form corresponding to the subspace gives us a tensor contraction.

$i (i( \beta) \Phi) \Phi =0$

These are contractions made remember for the condition of $\alpha(\beta) \in V \otimes \overline{V}$. This means it can be defined to be decomposable in its dual space through:

$\beta \in \wedge^{p+1} \overline{V}$

This leads to the condition of

$F_{[\alpha, \beta (\psi)]} = \psi \frac{1}{2}( F_{[\alpha, \beta]} + F_{[\alpha, \beta]})$

and a second solution on the manifold

$F_{[\alpha, \beta (\psi^{\dagger})]} = \psi^{\dagger} \frac{1}{2}( F_{[\alpha, \beta]} - F_{[\alpha, \beta]})$

This immediately suggests that there is no preferred symmetric or antisymmetric tensor properties satisfying the identity of $\alpha(\beta) \in V \otimes \overline{V}$.

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