1
$\begingroup$

Suppose that in his isolated box, Wigner's friend measures a qbit in state $|→⟩=\dfrac{|↑⟩+|↓⟩}{\sqrt{2}}$ along the vertical axe. Then, he sends Wigner (who remains outside the box) a qbit in the state |↑⟩ or |↓⟩, according to the outcome of the measurement.

If Wigner considers the box as a quantum system, then his friend is in a superposition of states, and the outgoing qbit is thus in the state $\dfrac{|↑⟩+|↓⟩}{\sqrt{2}}=|→⟩$. Thus Wigner would expect to observe |→⟩, if he measures the outgoing qbit along the horizontal axe. But from his friend's point of view, the outgoing qbit is either in the state |↑⟩ or |↓⟩, and in either case he would think that $|←⟩$ is a possible result of Wigner's measurement.

According to non collapse interpretations of quantum mechanics, such as Everett's or Bohm's one, who is right: Wigner or his friend, that is, is $|←⟩$ a possible outcome of Wigner's measurement?

$\endgroup$
2
$\begingroup$

No, Wigner wouldn't expect to observe only $\vert \rightarrow\rangle$ because he was a smart cookie. ;-)

Let's closely examine the state of the rest of the universe as seen by Wigner at various stages of the experiment.

Initially...

Trivially, Wigner sees

$$\vert \textrm{friend}\rangle\otimes\vert \rightarrow\rangle$$

After the Friend Has Measured the Qubit Along $z$...

Now, the friend has been entangled with the spin, and the total state-vector as seen by Wigner would be \begin{align} \frac{1}{\sqrt{2}}\vert \textrm{friend sees}\uparrow\rangle\otimes\vert \uparrow\rangle+\frac{1}{\sqrt{2}}\vert \textrm{friend sees}\downarrow\rangle\otimes\vert \downarrow\rangle \end{align}

After the Friend Has Sent the Signal...

\begin{align}&\frac{1}{\sqrt{2}}\vert \textrm{outcoming }\uparrow\rangle\otimes\vert \textrm{friend sends}\uparrow\rangle\otimes\vert \uparrow\rangle\\+&\frac{1}{\sqrt{2}}\vert \textrm{outcoming }\downarrow\rangle\otimes\vert \textrm{friend sends}\downarrow\rangle\otimes\vert \downarrow\rangle \end{align}

We will call this state $\vert\psi\rangle$. Notice that Wigner does not see the state of the outcoming qubit to be $\vert \rightarrow\rangle$. He sees the outcoming qubit to be in an entangled state with the friend and the measured qubit.

When Wigner Measures!

Now, when Wigner measures the $x$ spin of the outcoming qubit, the state of the system would collapse to

Option $1$ \begin{align} \mathbb{P}_{\rightarrow}\vert\psi\rangle = \vert \mathrm{outcoming }\rightarrow\rangle\otimes\Big(&\langle\mathrm{outcoming }\rightarrow\vert \mathrm{outcoming }\uparrow\rangle\vert\mathrm{friend\ sent }\uparrow\rangle\otimes\vert\uparrow\rangle\\+&\langle\mathrm{outcoming }\rightarrow\vert \mathrm{outcoming }\downarrow\rangle\vert\mathrm{friend\ sent }\downarrow\rangle\otimes\vert\downarrow\rangle\Big) \end{align} which is a state with $+1$ eigenvalue of $x$ spin of the outcoming qubit (up to normalization). Wigner would observe this with a probability \begin{align} \vert\mathbb{P}_{\rightarrow}\vert\psi\rangle\vert^2 = \frac{1}{2}(\vert \langle\rightarrow\vert\uparrow\rangle\vert^2+\vert \langle\rightarrow\vert\downarrow\rangle\vert^2) = \frac{1}{2} \end{align} Option $2$

Similarly, the other option is that Wigner would observe the state \begin{align} \mathbb{P}_{\leftarrow}\vert\psi\rangle = \vert \mathrm{outcoming }\leftarrow\rangle\otimes\Big(&\langle\mathrm{outcoming }\leftarrow\vert \mathrm{outcoming }\uparrow\rangle\vert\mathrm{friend\ sent }\uparrow\rangle\otimes\vert\uparrow\rangle\\+&\langle\mathrm{outcoming }\leftarrow\vert \mathrm{outcoming }\downarrow\rangle\vert\mathrm{friend\ sent }\downarrow\rangle\otimes\vert\downarrow\rangle\Big) \end{align} which is a state with $-1$ eigenvalue of $x$ spin of the outcoming qubit (up to normalization). Wigner would observe this with a probability \begin{align} \vert\mathbb{P}_{\leftarrow}\vert\psi\rangle\vert^2 = \frac{1}{2}(\vert \langle\leftarrow\vert\uparrow\rangle\vert^2+\vert \langle\leftarrow\vert\downarrow\rangle\vert^2) = \frac{1}{2} \end{align}


So, indeed, Wigner would measure the $x$ spin of the particle to be left--pointing half of the times and Wigner wouldn't expect otherwise. As you can see, we derived this conclusion by doing everything w.r.t. Wigner from the beginning to the end.

Physically, what is going on is that the initial superposition $\vert \uparrow\rangle + \vert \downarrow\rangle$ of the first qubit does not get translated to the same superposition $\vert \uparrow\rangle + \vert \downarrow\rangle$ for the outcoming qubit. Rather, first, the friend and then the outcoming qubit gets entangled with the first qubit and the superposition is of the form \begin{align}&\frac{1}{\sqrt{2}}\vert \textrm{outcoming }\uparrow\rangle\otimes\vert \textrm{friend sends}\uparrow\rangle\otimes\vert \uparrow\rangle\\+&\frac{1}{\sqrt{2}}\vert \textrm{outcoming }\downarrow\rangle\otimes\vert \textrm{friend sends}\downarrow\rangle\otimes\vert \downarrow\rangle\end{align}

This makes it so that the $\vert\uparrow\rangle$ and $\vert\downarrow\rangle$ of the outcoming qubit are not in a coherent superposition and thus, the expectation that what would reach Wigner would be $\vert \rightarrow\rangle$ is naive/wrong.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for your detailed answer! I understand my mistake $\endgroup$
    – amblaf
    Aug 7 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.