# How is Q superposition in Q information different from a classical dice?

Let's consider a dice with 1 or -1 on its faces, opposites faces adding up to 0.

The point of this post is to relate the quantum coherence of a single qubit to something "weird", i.e., incapable of being reproduced by classical probability.

All the faces of the dice are covered, and we are repeatedly going to uncover and read the top face, and then cover it again. A $\sigma_z$ measurement will do simply this, while a $\sigma_x$ measurement turns the dice 90 degrees to the right, uncovers the top face, reads and covers it, and finally rotates the dice back, but assume that a device (maybe yourself) MUST roll the dice by keeping its vertical axis fix right after covering the face, similar to spinning a top. Note that this protocol tries to reensemble the incompatibility of the two measurements in QM.

The state preparation is what we do to the dice before giving it to someone else.

By simply rolling the dice, performing a $\sigma_z$ or $\sigma_x$ measurement will give a -1 or 1 with half probability, as for the quantum state {{1/2,0},{0,1/2}}.

By rolling the dice and then making sure that the left face has a "1", you will get the same output as if I gave you the quantum state {{1/2,1/2},{1/2,1/2}} for both $\sigma_z$ and $\sigma_x$ measurements!

After this analysis, I see that it is possible to reproduce the quantum superposition in Q information by classical means, which then makes it not "weird" at all. The only "special" feature I see is that nature itself automatically does the covering of the faces and the spinning at the measurements for you, but this is not my point, so...

...without going into non-locality, Bell inequalities and so on (which I understand are truly quantum and needed for Q cryptography), is there anything "weird" in the coherence alone of Q states in Q information that I can use to say "this is not classical" after looking at the measurements (which is all I can look at)? Am I missing something? If not, then I don't understand how a single qubit can be somehow better than a single bit in the same sense as a quantum superposition state can be more powerful than a single (maybe tricked and in a state not prepared by you) dice.

• What about applying an arbitrary unitary rotation and then measuring in some basis? – Joel Klassen Oct 21 '15 at 5:29
• @Joel QM relies on the fact that the commutator of $\sigma_x$ and $\sigma_z$ is not 0 (same for the $y$ direction). I think if I can simulate this, then I could do it also for any arbitrary rotation and measurement of a correspondingly "tricked" and state-prepared dice. – Rol Oct 21 '15 at 7:45

So what you are describing is a hidden variable theory of quantum mechanics, and I believe the answer is ultimately no, there is nothing fundamentally non-classical in a single qubit. You can construct hidden variable theories of quantum mechanics as long as they are non-local, and since there is no notion of locality in a single qubit, that last caveat should not be a concern.

Bell wrote a paper in 1966 on this topic:

The question at issue is whether the quantum mechanical states can be regarded as ensembles of states further specified by additional variables, such that given values of these variables together with the state vector determine precisely the results of individual measurements.

He concluded that indeed such models can be constructed, but that they have a non-local nature:

The curious feature is that the trajectory equations for the hidden variables have in general a grossly nonlocal character.

With regard to a qubit being more powerful than a bit, we must be careful about the comparisons we are making. A hidden variable model of a qubit will have more states than a classical bit, so in that sense already a qubit is greater than a bit, although that is not particularly compelling. Additionally, I do not know of any useful quantum computations that can be performed with a single qubit.

The more interesting question is why are collections of qubits, which presumably as a group should have some non-local but classical hidden variable description, able to perform computations which a classical computer can not? I do not have a comprehensive answer to that question, but there is a paper written by Scott Aaronson which claims that for any hidden variable theory, if you had access to the hidden variables you would have even more computational power than a quantum computer. So it seems in some sense that computational power does not come from non-classicality exclusively. We must remember that computers are physical systems, and when you lose non-locality you change the physical rules under which they operate, and thus what they are capable of doing.

Let's consider a dice with 1 or -1 on its faces, opposites faces adding up to 0.

The point of this post is to relate the quantum coherence of a single qubit to something "weird", i.e., incapable of being reproduced by classical probability.

It is my understanding from what little I have messed with the problem that, in order to truly get to a system which you cannot explain classically, you need at least two qubits (Bell's inequality violations) to see a statistical difference, and that you need three (the GHZ experiment) to get an absolute difference beyond the statistical.

I like to phrase this GHZ experiment as a simple game for a 3-player team.

# My favorite game: Betrayal.

"Betrayal" is played by a cooperating team of three people, during which they will undergo $N$ separate trials for some large $N$ (around 50 or 100 or so): if they succeed at all trials then all members each get some very large sum of money. (Note that $N$ is also chosen to be small enough to allow for some reasonable quantum incoherence to creep in -- we just want to limit the chance that luck alone accounts for the results, since each trial can conceivably be passed by luck alone). Even though the team is expected to cooperate however they can, the rules of the game attempt to set up one as a traitor against the other two; the team wins if they can recover from this betrayal gracefully.

Each trial works as follows: The three people can take whatever measurement apparatus they want, and are put into sealed boxes and placed at relativistic distances for the duration of the trial, to preclude any sort of communication. The boxes they are in will each have a screen and two buttons labeled $0$ and $1$. The three people must each press exactly one button exactly one time during the trial or else they lose the trial (and hence the game); if they don't answer in time, they lose. The screen contains instructions concerning the sum of the three numbers that are pressed, and a countdown timer telling them how much time they have left before they lose by default.

With probability $1/4$, the trial will be a control trial: on all screens we show the instruction "make the sum even", and the team passes the trial if their sum is even.

With probability $3/4$, we select one of the 3 team members at random to be the "traitor" and on their screen we display the instruction "make the sum even." The two other team members receive the instruction "make the sum odd," and the team passes the trial if their sum is odd.

# Classical analysis

Any classical strategy is a joint probability distribution on 6 random variables for each of the three members $A, B, C$ and each of the occurrences that they are asked to accomplish, $0$ for "make the sum even" and $1$ for "make the sum odd." Hence for a sure solution we need to simultaneously guarantee four modular equivalences,\begin{align} A_0 + B_0 + C_0 \equiv 0 &\;(\operatorname{mod}2)\\ A_0 + B_1 + C_1 \equiv 1 &\;(\operatorname{mod}2)\\ A_1 + B_0 + C_1 \equiv 1 &\;(\operatorname{mod}2)\\ A_1 + B_1 + C_0 \equiv 1 &\;(\operatorname{mod}2).\end{align}However this is impossible as summing them all up guarantees $$2(A_0 + B_0 + C_0 + A_1 + B_1 + C_1) \equiv 3 \equiv 1 \;(\operatorname{mod}2),$$which is impossible (the left hand side is transparently even, the right hand side is transparently odd). I believe that this also limits the maximum probability of success to $3/4,$ essentially they need to always choose one to violate; and this can be attained for instance by $A_0 = B_0 = C_0 = 1$ and $A_1 = B_1 = C_1 = 0:$ but I've not actually done a computation to prove it. If that's correct then you only need to pass about 50 trials in order for us to be sure (to something like $29\sigma$) that you are doing something non-classical.

# Quantum analysis

Let $|+\rangle = \sqrt{1/2}\big(\;|0\rangle + |1\rangle\;\big)$ and $|-\rangle = \sqrt{1/2}\big(\;|0\rangle - |1\rangle\;\big);$ notice that $$\sqrt{8} |+++\rangle = |000\rangle + |001\rangle + |010\rangle + |011\rangle + \dots + |111\rangle,$$ while $|---\rangle$ is similar but every state that has an odd number of $1$ bits has a preceding $-$ sign.

Therefore, before each trial, our team entangles three qubits into the $\sqrt{1/2}\big(\;|+++\rangle + |---\rangle\;\big)$ state, and they each take one qubit into the room with them.

Any team member who sees the prompt to "make the sum even" measures their qubit in the computational basis.

Any team member who sees the prompt to "make the sum odd" first performs the $X$-basis controlled-phase rotation by 90 degrees, \begin{align}|+\rangle \mapsto&\;\;|+\rangle\\ |-\rangle \mapsto&\; i |-\rangle,\end{align}and then measures in the computational basis.

In the control rounds, we therefore get a measurement of the state $|+++\rangle + |---\rangle$ which is distributed only over the states with an even number of $1$ bits. However in all of the traitor rounds, we get a measurement of the state $|+++\rangle - |---\rangle,$ which is distributed only over the states with an odd number of $1$ bits. Therefore the team is locally able to accomplish a classically-impossible feat: winning the trial with (in principle) 100% success rate.