What experiments would tell us about Wigner's friend?

Question

I am trying to understand the Wigner's friend thought experiment and I can see what the problem is, but I'm not really understanding the implications (I've read this and related questions, but something is still unclear to me). It looks like this is an experiment that could be made in practice, so I'm wandering what an experiment would tell us.

Suppose for example we have two electrons in a singlet state $|\psi\rangle = \frac{1}{\sqrt2}(|+\rangle_A |-\rangle_B - |-\rangle_A |+\rangle_B)$ and Wigner's friend performes a measure on the state of spin of particle $A$.

Now the wave function collapses for the friend, but what happens to the wave function in Wigner's opinion? Suppose both Wigner and his friend now measure spin along a different axis: such a measure would return different values based on the state of the system. Wigner and his friend have never talked to each other, which one is correct?

1. Wigner's wave function is $|\psi\rangle = \frac{1}{\sqrt2}(|+\rangle_A |-\rangle_B - |-\rangle_A |+\rangle_B)$
2. Wigner's wave function is a mixed state $\frac{1}{2}(|+\rangle_A\langle+|_A\otimes|-\rangle_B\langle-|_B + |-\rangle_A\langle-|_A \otimes |+\rangle_B\langle+|_B)$
3. Wigner's wave function is the same as his friend's

Without wasting too much time on calculations, let's suppose 1., 2. and 3. give different values for a certain operator (for instance the spin along $x$ axis), then we should be able to experimentally verify which one is the actual Wigner's wave function.

Is this experiment doable? Has it been done? Which is the correct wave function?

Answer 1

From Wigner's point of view, the relevant object under consideration is the combined system of both lab and particle. Here I'll get rid of the singlet because it's unneeded for the full "paradox" and just go with a single quantum bit as it keeps the math simpler. The "particle" subsystem has basis states $\mathcal{B}_P := \{ |0\rangle, |1\rangle \}$, while the "friend" subsystem has states - for simplicity -

$$\mathcal{B}_F := \{\ |\text{"I haven't seen anything yet"}\rangle, |\text{"I saw a '0'"}\rangle, |\text{"I saw a '1'"}\rangle\ \}$$

The composite system tensors these two, i.e. it has basis $\mathcal{B}_F \otimes \mathcal{B}_P$.

Hence, at the beginning of the experiment, say time $t_0$, Wigner regards the joint system to be in the state

$$|\psi\rangle(t_0) := \underbrace{|\text{"I haven't seen anything yet"}\rangle}_{\text{"friend" part}} \otimes \underbrace{\left(\frac{1}{\sqrt{2}} [|0\rangle + |1\rangle]\right)}_{\text{"particle" part}}$$

Namely, the "particle" subsystem is in the pure state

$$|\psi_P\rangle(t_0) := \frac{1}{\sqrt{2}} [|0\rangle + |1\rangle]$$

. The inner measurement now occurs at time $t_M$. At time $t > t_M$, Wigner regards the whole system now as in

$$|\psi\rangle(t) = \frac{1}{\sqrt{2}} \left(|\text{"I saw a '0'"}\rangle \otimes |0\rangle + |\text{"I saw a '1'"}\rangle \otimes |1\rangle\right)$$

Now, note that this is a tensor that is not factorizable. Thus we cannot disentangle the particle subsystem; the state is quantum-entangled. That means that Wigner must consider the particle subsystem to be a mixed state, because that's what happens when you try to do the factorization anyways. Namely,

$$\rho_P(t) = \frac{1}{2} \left(|0\rangle \langle 0| + |1\rangle \langle 1|\right)$$

is the mixed state of the particle. This mixture is maximal, and thus this means Wigner can say for sure an outcome was obtained at this point. Hence (2), in your list, is the correct answer. Finally, the "paradox" now is this:

1. Wigner regards the whole joint system is still in a pure state,
2. Wigner's friend presumably has seen only one outcome, given the mixed assignment on the particle,
3. But that suggests that Wigner should really take the state of the whole system as being one of $\left\{\ |\text{"I saw a '0'"}\rangle \otimes |0\rangle,\ |\text{"I saw a '1'"}\rangle \otimes |1\rangle\ \right\}$, and while a 50/50 random mixture of those and $|\psi\rangle(t)$ give the same statistics when the "usual" measurement is made, i.e. go in and ask the friend "What did you see?" ...
4. ... if Wigner could instead measure the joint system in the strange basis $$\mathcal{B}_\mathrm{wtf} := \begin{align}\left\{\frac{1}{\sqrt{2}} \left(|\text{"I saw a '0'"}\rangle \otimes |0\rangle + |\text{"I saw a '1'"}\rangle \otimes |1\rangle\right), \\ \frac{1}{\sqrt{2}} \left(|\text{"I saw a '0'"}\rangle \otimes |0\rangle - |\text{"I saw a '1'"}\rangle \otimes |1\rangle\right)\right\} \end{align} $$ he could tell whether the state is a mixture of those two or "really" $|\psi\rangle(t)$, as he will get different statistics on this weird basis.
5. Experiments show simpler versions of this system (i.e. super simple "friends"), where we can make that measurement, do indeed have state $|\psi\rangle(t)$,
6. The Universe can be regarded as an ever expanding series of Wigner boxes, since information propagates outward no faster than light speed, so at any time we should be in a $|\psi\rangle(t)$ for some suitably-remote "Wigner",
7. and yet here we are, always experiencing singular outcomes!

So clearly something is wrong here - and the debate is about just what that must be.

Answer 2

It is option 2. Option 3 doesn't make much sense. According to Wigner, the electrons and his friend are all currently in a big, entangled quantum superposition. In some parts of the superposition, the friend believes "$A$ has collapsed to state $|{+}_A\rangle$" and in others "$A$ has collapsed to state $|{-}_A\rangle$", so there's really no "friend's opinion on the wavefunction" for Wigner to agree with.

Option 1 and option 2 can be distinguished pretty easily. For simplicity, let's first notice that electron $B$ is unnecessary. An electron is prepared into state $|\psi\rangle=\sqrt{\frac{1}{2}}(|{+}\rangle+|{-}\rangle).$ Wigner's friend measures $|\psi\rangle$ in the basis $|{+}\rangle,|{-}\rangle.$ Then,

• Option 1: in Wigner's opinion, the state of the electron continues to be described by $|\psi\rangle=\sqrt{\frac{1}{2}}(|{+}\rangle+|{-}\rangle).$
• Option 2: in Wigner's opinion, the state of the electron can no longer be described by a pure state. It is now described by a density operator, $\rho=\frac{1}{2}(|{+}\rangle\langle{+}|+|{-}\rangle\langle{-}|).$

The experiment Wigner performs to distinguish these is to apply a Hadamard gate to the electron and then measure its spin in the $|{+}\rangle,|{-}\rangle$ basis (or equivalently, he just measures it in the $\sqrt{\frac{1}{2}}(|{+}\rangle+|{-}\rangle),\sqrt{\frac{1}{2}}(|{+}\rangle-|{-}\rangle)$ basis—if the original basis was spin eigenstates along the z axis, I believe these are the states along the x axis.) We'll define this gate by $H|{+}\rangle=\sqrt{\frac{1}{2}}(|{+}\rangle+|{-}\rangle),H|{-}\rangle=\sqrt{\frac{1}{2}}(|{+}\rangle-|{-}\rangle).$ Calculate this out and you find that $H|\psi\rangle=|{+}\rangle,$ while $H\rho H^\dagger=\rho$: in option 1 Wigner gets the same result always, while in option 2 Wigner gets both results with 50/50 odds.

If you allow reducing Wigner's "friend" to a single qubit, you can perform this experiment "yourself" on e.g. IBM's cloud quantum processors by simply running the following circuit.

The "electron" in this system is the qubit on top, the "friend" is the qubit in the middle, "Wigner" is "you"/the classical bit at the bottom, and $|{+}\rangle,|{-}\rangle$ are now relabeled $|0\rangle,|1\rangle.$ The first H gate simply prepares our superposition from the initial $|0\rangle$ state. The following CNOT gate represents the "friend" measuring the "electron"; the important parts of the gate's operation are that $CNOT|0_\mathrm{electron}0_\mathrm{friend}\rangle=|0_\mathrm{electron}0_\mathrm{friend}\rangle,CNOT|1_\mathrm{electron}0_\mathrm{friend}\rangle=|1_\mathrm{electron}1_\mathrm{friend}\rangle$: assuming the "friend" starts in the right state, this gate copies the state of the "electron" into the "friend", simulating a "measurement". Then the following Hadamard and measurement into "Wigner's" state represents the protocol I described above. You should find that "Wigner" gets either result with about 50/50 chances, indicating option 2. Only if you remove the CNOT will "Wigner" then see the qubit always in one state, as option 1 would predict. (As a practical note, if you do decide to perform this second experiment yourself, you should probably introduce a "barrier" in the circuit in place of the CNOT, or else the transpiler might just remove both Hadamards as an optimization.)

Doing this experiment as you describe it where Wigner's friend is an actual person is probably impossible for the next few decades/centuries/millennia/forever?. Macroscopic objects (and people especially) are constantly "talking" (i.e. interacting) with each other via thermal motions etc (there's just usually no way to collect/decipher this information in a usable manner). There's simply no isolation good enough for one person to be able to see another as being in a quantum superposition. Reducing the friend to some quantum particle trapped in a qubit makes it much easier to protect it from decoherence (and even that is pushing our limits!).

Do note that your experiment does not quite get to the heart of the paradox. The paradox comes from considering the electron and the friend together as quantum system that remains in a pure state (according to Wigner), not just the one electron, which (according to Wigner and as evidenced our experiment) is just in a boring mixed state. In principle (again, due to decoherence problems this is not practical for the forseeable future), Wigner can perform certain actions on the electron+friend system that "prove" that the friend was in a superposition: particularly, if he projects the system along the state $|0_\mathrm{electron}0_\mathrm{friend}\rangle+|1_\mathrm{electron}1_\mathrm{friend}\rangle$ he should always measure success. This is philosophically problematic, because the friend presumably does not subjectively experience the superposition in any way. (Also, if you try to implement that measurement in the above quantum circuit, Wigner ends up uncomputing the friend's measurement! Now that's a riddle.)