So in the thought experiment Wigners friend the paradox is ultimately due to a difference of descriptions of density matrices.

If the physical variable that is measured of the spin system is denoted by $z$, where $z$ takes the possible outcome values $0$ or 1, the above Wigner's friend situation is modelled in the RQM context as follows: $F$ models the situation as the before-after-transition:

$$ \alpha | 0 \rangle_S + \beta | 1 \rangle_S \to |1 \rangle _S $$

of the state of $S$ relative to him (here it was assumed that $F$ received the outcome $z = 1$ in his measurement of $S$). In RQM language, the fact $z = 1$ for the spin of $S$ actualized itself relative to $EF$ during the interaction of the two systems.

A different way to model the same situation is again an outside (Wigner's) perspective. From that viewpoint, a measurement by one system ($F$) of another ($S$) results in a correlation of the two systems. The state displaying such a correlation is equally valid for modelling the measurement process. However, the system with respect to which this correlated state is valid changes. Assuming that Wigner ($W$) has the information that the physical variable $z$ of $S$ is being measured by $F$, but not knowing what $F$ received as result, $W$ must model the situation as:

$$ (\alpha |0 \rangle_S + \beta |1 \rangle_S) | \perp \rangle_F \to \alpha ( | 0 \rangle_S \otimes | 0 \rangle_F ) + \beta (|1\rangle_S \otimes |1 \rangle_F) $$

where ${F}$ is considered the state of $F$ before the measurement, and ${\displaystyle |1\rangle _{F}}{\displaystyle |1\rangle _{F}}$ and ${\displaystyle |0\rangle _{F}}{\displaystyle |0\rangle _{F}}$ are the states corresponding to $F$'s state when he has measured $1$ or $0$ respectively. This model is depicting the situation as relative to $W$, so the assigned states are relative states with respect to the Wigner system. In contrast, there is no value for the $z$ outcome that actualizes with respect to $W$, as he is not involved in the measurement.

The assumptions boil down to those assumptions are

(Q): Quantum theory is correct. (C): Agent's predictions are information-theoretically consistent. (S): A measurement yields only one single outcome.

Now, while decoherence does not solve the measurement problem it everyone seems to agree that it is a step in the right direction. Does de-coherence add value to the discussion when people question axiom $(S)$?

To make more explicit the time evolution can be reformulated in terms of density matrices to remove axiom C. Let's say my wavefunction is in superposition of

$$|\psi \rangle_S = \frac{1}{\sqrt{2}}(\alpha | 0 \rangle_S + \beta | 1 \rangle_S)$$

The corresponding density matrix is:

$$ \rho_S = \begin{bmatrix} |\alpha|^2 & \alpha^* \beta \\ \beta^* \alpha & |\beta|^2 \end{bmatrix} $$

Now, the time average of the density matrix is decoherence of the density matrix which happens due to decoherence is given by:

$$ \langle \rho_S \rangle_t = \begin{bmatrix} |\alpha|^2 & 0 \\ 0 & |\beta|^2 \end{bmatrix} $$

And proceeds to use the density matrix $\rho'$

$$\rho' = \langle \rho_S \rangle_t \tag{D}$$

Via this maneuver I reformulate the problem and still get the contradiction $\rho' \neq \rho$ (replacing axiom (S) with (D)). To me its very obvious one of them is using approximations statement $(D)$. More explicitly:

One of them is using:

$$i \hbar \frac{\partial \rho }{\partial t} = [H,\rho]$$

and the other:

$$i \hbar \frac{\partial \langle \rho \rangle_t}{\partial t} = [H,\langle \rho \rangle_t]$$

The real problem I encounter when I try to explain myself is $(D)$ is a valid replacement of $(S)$. What happens if I put both Wigner and his friend in box (the entire setup).

Edit: In response to Mateus Araújo's answer:

"if the description of the measurement becomes the same if the Friend's decoherence reaches Wigner, i.e., if the Friend is not inside a decoherence-proof box."

Yes, this does happen. Once, decoherence reaches the Wigner. Then Wigner is also using the equation:

$$i \hbar \frac{\partial \langle \rho \rangle_t}{\partial t} = [H,\langle \rho \rangle_t]$$

As a result neither of them see any discrepancy now for both of them their descriptions of the density matrices match.

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    $\begingroup$ It would be nice if you gave more context to the problem. $\endgroup$
    – Rammus
    Commented Dec 25, 2022 at 10:03
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    $\begingroup$ Close-voters: you are misusing the "opinion-based" reasoning. $\endgroup$ Commented Dec 27, 2022 at 4:16
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    $\begingroup$ @MarkS the question asks whether or not decoherence improves the situation. This should be quantifiable mathematically, and hence "opinion-based question" is an inappropriate close reason. Two out of two of the current close votes used that reason. If you think the question is out of scope, you can vote that it's "off-topic". $\endgroup$ Commented Dec 27, 2022 at 13:40
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    $\begingroup$ @MateusAraújo the moment I commit to what a solution would even be it will be closed for being opinion based question $\endgroup$ Commented Dec 28, 2022 at 10:57
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    $\begingroup$ @MoreAnonymous you don't need to "commit to a solution". The problem is explaining better what the question is in the first place. What you mean with "does decoherence improve the situation", what situation is there to improve, etc $\endgroup$
    – glS
    Commented Dec 28, 2022 at 11:09

1 Answer 1


Ok, now I understand your question. You were saying that the problem is that the description of the Friend's measurement was different from the Friend's and Wigner's points of view, and were asking if the description of the measurement becomes the same if the Friend's decoherence reaches Wigner, i.e., if the Friend is not inside a decoherence-proof box.

The answer is yes, it does. But that's not an interesting answer, as you're simply denying the premise of the argument, that the Friend is inside a decoherence-proof box.

A more interesting answer is to assert that it is in principle impossible to do the experiment, that the Friend's decoherence will always engulf Wigner. That would require changing the laws of quantum mechanics, though, as it predicts no such inescapable decoherence. When people go this route they usually postulate some objective collapse model (which, by the way, has never been made compatible with relativity).

  • $\begingroup$ Doesn't decoherence come from thermodynamics? Thanks for answering by the way :) $\endgroup$ Commented Dec 31, 2022 at 12:36
  • $\begingroup$ Also how does one perform a measurement in a de-coherence free box?? $\endgroup$ Commented Dec 31, 2022 at 14:11
  • $\begingroup$ The idea of a decoherence-proof box (sorry, not decoherence-free) is that you make a measurement anyway you want, the box prevents decoherence from spreading outside (to Wigner). $\endgroup$ Commented Dec 31, 2022 at 21:31
  • $\begingroup$ I find your "interesting answer" ... Interesting because it that would mean there is no means by which one can build a quantum computer? But experimentalists seem to be achieving that even if for short time scales $\endgroup$ Commented Dec 31, 2022 at 21:38
  • $\begingroup$ It depends on the details of how this non-quantum decoherence would work, but the most straightforward way to do it would indeed kill quantum computing. Which is one of the many reasons most physicists don't take that seriously. $\endgroup$ Commented Jan 1, 2023 at 1:41

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