#### DarMM

Science Advisor

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- Wigner's friend seems to lead to certainty in two complimentary contexts

**Summary:**Wigner's friend seems to lead to certainty in two complimentary contexts

This is probably pretty dumb, but I was just thinking about Wigner's friend and wondering about the two contexts involved.

The basic set up I'm wondering about is as follows:

The friend does a spin measurement in the ##\left\{|\uparrow_z\rangle, |\downarrow_z\rangle\right\}## basis, i.e. of ##S_z## at time ##t_1##. And let's say the particle is undisturbed after that.

For experiments outside the lab Wigner considers the lab to be in the basis:

$$\frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle + |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right)$$

He then considers a measurement of the observable ##\mathcal{X}## which has eigenvectors:

$$\left\{\frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle + |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right), \frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle - |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right)\right\}$$

with eigenvalues ##\{1,-1\}## respectively.

At time ##t_2## the friend flips a coin and either he does a measurement of ##S_z## or Wigner does a measurement of ##\mathcal{X}##

However if the friend does a measurement of ##S_z## he knows for a fact he will get whatever result he originally got. However he also knows Wigner will obtain the ##1## outcome with certainty.

However ##\left[S_{z},\mathcal{X}\right] \neq 0##. Thus the friend seems to be predicting with certainty observables belonging to two separate contexts. Which is not supposed to be possible in the quantum formalism.

What am I missing?