#### DarMM

Science Advisor

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The standard Wigner's friend scenario is an example, the canonical example.Hm, that sounds also strange. Either the observer has obtained a fixed result or not. No superobserver can change it, because "fixed" means it's irreversible. I guess, it's hard to discuss without a concrete example.

Imagine a lab that consists of a microscopic spin-1/2 system, a device that can measure it and the rest of the lab environment.

Somebody inside this lab measures a particle in the state:

$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)$$

they observe ##\uparrow## let's say and thus their device shows ##D_{\uparrow}## let's say, just symbolic for however the ##\uparrow## outcome is displayed on the set up.

Wigner outside the lab models the lab as:

$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)\otimes|D_{0}\rangle\otimes |L_{0}\rangle$$

where ##D_{0}## represents the device in the "ready" state before any readings and ##L_{0}## is the initial state of the lab.

After under unitary evolution the entire lab evolves for Wigner as:

$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle + |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right)$$

where ##\{|D_{\downarrow}\rangle ,|D_{\uparrow}\rangle\}## represents indicator states of the device and ##\{|L_{\downarrow}\rangle ,|L_{\uparrow}\rangle\}## are corresponding states of the lab.

Wigner can then measure the entire lab system in the basis:

$$\left\{\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle + |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right),\\

\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle - |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right)\right\}$$

corresponding to some observable ##\mathcal{X}##. Assigning ##\{E_{+},E_{-}\}## to indicate the two outcomes for brevity, he will have

$$P(E_{+}) = 1$$

in seeming contradiction with the device having recorded some result.