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It seems that a coin flip game is a decent metaphor for a 2-level system. Until 1 of the 2 players picks heads or tails, even if the coin has already been flipped, the win/loss wave form has not yet collapsed.

Would rock paper scissors be a good metaphor for qutrits? Where the number of players corresponds w/ the number of qutrits (eg. $3^n$ possible states).

Would a standard pack of 52 cards be a good metaphor for a 52-level quantum system (the game being guessing correctly a card selected from the deck at random)?

Particularly interested in game based metaphors because of the easy correlation to combinitorial game theory & computation complexity.

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Your metaphor can be chosen as $N-1$ identical coins, such that the outcome state vector corresponds to the sum of the heads. Thus we wind up in the state $|k\rangle$ when we have $k$ heads and $N-1-k$ tails.

In this approach you can actually associate your metaphor with a classical phase space, as a generalization of the association of a qubit with the Bloch sphere whose generator can be thought in the coin case as the direction of the normal to the coin's face.

In the multiple coin case, we have $N-1$ such directions. But since in our Hilbert space we don't mind the order of the coins only their summed result, in the classical phase space we should not distinguish states in which the direction of the normal to the coins faces switch.

What I just described to you in words is actually the Majorana representation (some call it the Majorana star representation), which is based on the isomorphism of the symmetrized tensor product of $N-1$ spheres to the complex projective space $CP^{N-1}$.

$$\otimes_{\mathrm{sym}} ^{N-1} S^2 \cong CP^{N-1}$$

The geometric quantization of this complex projective space is the $N-$ level system. There is some renewed interest in this representation, partially motivated by holonomic quantum computation, please see for example Liu and Fu.

Now, the outcome of the coin flipping in the case of a single qubit can be described as a measurement of the angular momentum component of the coin in the $z$ direction $j_z$. It is not hard to see that the corresponding operator in the multiple coin flip is the sum of the individual angular momenta:

$$J_z = \sum_1^{N-1} j^{(i)}_z$$

(The above equation is just standard shorthand used by physicists since the operators act on different components in the tensor product Hilbert space).

(In the $N\times N$ matrix representation, this operator can be chosen as: $J_z = \mathrm{diag}[N-1, N-3, .,.,., -N+1]$).

This representation of the $N-$ qubit system has a further analogy. As in the case of a single coin or qubit, the distribution function of the operator $j_z$ is Bernoullian for any state (density matrix) in which the system is prepared; the distribution function of $J_z$ in the multiple coin case is Binomial, for any choice of the density matrix of the system. Thus in both cases this observable returns the classical distribution function.

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There is always a difference between a quantum system and a classical metaphor. If a system is a qubit in a pure state, then there always exists a measurement basis (or alternatively a proper unitary gate for the standard measurement basis) such that the measurement outcome is 100% predictable, and a measurement basis with measurement outcome 50%-50%. You can't demonstrate this feature using a classical metaphor - in classical physics random is random and deterministic is deterministic.

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A coin is not a great analogy for a quantum system. A (slightly) better one is a box that contains 3 coins. There are 3 windows, labelled x, y and z. The box is rigged so that you can only open one window at a time. When you open a window, you can see the heads/tails state of the corresponding coin, but the other two coins get flipped, and you can’t see what happens to them (unless you open another window, but then you know nothing anymore about the window you just had open because that coin has been flipped again).

I’m not sure that this attempt at an analogy scales very well to larger dimensional systems, because you probably can’t, in general, describe the algebra of the system using a set of mutually anti-commuting observables, so your rules for how the different coins flip would have to be more complicated. The qudit analogy has $d^2-1$ coins. For example, 2 qubits have 15 coins, each corresponds to a tensor product of 2 Paulis. You can open sets of windows that correspond to commuting observables. The other coins get flipped upon opening, but there are some consistency conditions that some outcomes of coin flips are entirely determined by the outcomes of other coin flips. It becomes messy very quickly...


Further explanation (expanding some of the Mathematics)

Any density matrix of a qudit is described by a $d\times d$ matrix. You can pick any basis of matrices that you like to decompose that (Hermitian) matrix. You'll need $d^2$ of them, but one of those terms is always $\mathbb{I}/d$, which I don't need to count. For example, a basis for the qubit are the Pauli matrices $X$, $Y$ and $Z$ (and $\mathbb{I}$). If you correspond each of these basis elements to measurement operators, because there are 2 distinct eigenvalues, you get two measurement outcomes (like head/tails on a coin). If two operators commute, you can simultaneously know the two measurement outcomes. If the two observables anti-commute, the corresponds to maximal uncertainty between the two observables. In other words, you measure one observable (say $Z$), and the other observables ($X$ and $Y$, because both anticommute with $Z$) are complete uncertain, i.e. the coins are flipped.

For qubits, all 3 observables pair-wise anticommute: $\{X,Y\}=\{Y,Z\}=\{X,Z\}=0$, so whichever measurement you choose, the other two observables reset.

However, for two qubits, the relationships are not nearly so simple. You have all possible terms $$ \mathbb{I}\otimes X \qquad \mathbb{I}\otimes Y \qquad \mathbb{I}\otimes Z \\ X\otimes \mathbb{I} \qquad Y\otimes \mathbb{I} \qquad Z\otimes \mathbb{I} \\ X\otimes X \qquad X\otimes Y \qquad X\otimes Z \\ Y\otimes X \qquad Y\otimes Y \qquad Y\otimes Z \\ Z\otimes X \qquad Z\otimes Y \qquad Z\otimes Z $$ You can see that not all of these pair-wise anticommute, because $\mathbb{I}\otimes Z$ and $Z\otimes \mathbb{I}$ commute, for example. So, we could simultaneously measure the set of observables $\mathbb{I}\otimes Z$, $Z\otimes \mathbb{I}$ and $Z\otimes Z$, but all other observables are completely uncertain. Overall, you'd have 15 coins and there are sets of 3 windows that you can open simultaneously, and all other coins are flipped at that instant.

If you want to describe the same thing for qutrits, it gets more messy because you can use a basis where everything gives 2 answers, but there's not a perfect division into whether operators commute or anticommute, so you get partial connections between coins which are messier to give a classical equivalent of.

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  • $\begingroup$ "mutually anti-commuting observables" returns 6 results via google. could u (briefly) explain? if there is more than will fit in a comment, perhaps a new question is in order? $\endgroup$ – meowzz Aug 4 '18 at 12:22
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A coin is an extremely bad and highly misleading analogy for a qubit. You shouldn't use it by any means.

Yet, if you want an equally bad and misleading analogy for a qu-$d$-it, you should use something which is random and has $d$ possible outcome. For $d=3$, this might be rock-paper-scissors. On the other hand, a deck of cards with $52$ cards can have $52!$ possibilities, so it is an equally bad analogy for a qu-$(52!)$-it.


It has been suggested to add an explanation why I think this is a horrible analogy. The key point is that it pretends that quantum mechanics is merely classical randomness -- something which can be easily understood with our classical intuition, and thereby implying that all that talking about quantum mechanics being special is just talking. Indeed, the coin analogy already fails for a single qubit, when one performs measurements in more than one basis. This is for instance explained in DaftWullie's answer. Now one could argue that his answer also provides a way to model this with coins, but (1) these are coins which are rigged in a very weird way, and (2) it is still incomplete -- I just need to toss in measurement in yet another basis to make the whole description break down, and to yield an even more complex pattern of rigged coins (even worse, those coins would not get tossed after looking at another coins as in DW's answer, but they would get tossed in a biased way depending on which measurement I did).

Of course, since quantum theory is a theory about measurements -- which have classical inputs, i.e. measurement settings, and classical outputs, i.e. measurement outcomes -- we can always model this with classical objects which follow some odd distribution. However, the point is exactly that this distribution cannot be modeled by coins any more in any even remotely reasonable way (even more so once we consider two spatially separated qubits).

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  • $\begingroup$ The question is what a good metaphor would be. If there is no classical metaphors, how about one that is quantum. $\endgroup$ – meowzz Aug 7 '18 at 15:43
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    $\begingroup$ @meowzz I'd say explaining people QM takes a bit more patience on both sides than just talking about a classical coin. If either them or you are not willing to invest that time, then so it be. I think a good way to explain it would be to start from one measurement, which is random and can be seen as a coin (which is tossed once, and then we can look at it as often as we want), and then toss in more measurements (=coins) and then explain how these affect each other, see DaftWullie's answer. Another great way of explaining it is the way in which Preskill in his Lecture explains ... $\endgroup$ – Norbert Schuch Aug 7 '18 at 15:47
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    $\begingroup$ ... Bell inequalities (in the original setting), by having Alice & Bob have three boxes with a coin each. It will make people understand that quantum mechanics is not classical, and that is exactly the point. $\endgroup$ – Norbert Schuch Aug 7 '18 at 15:47
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    $\begingroup$ @meowzz But rather than asking for that here, why don't you ask a question about "what are good analogies", maybe specifying the context (one qubit, entanglement, ...), or maybe just in a general context? This might (well, hopefully at least) generate more interesting answers than this question, which is based on a misleading premise to start with. $\endgroup$ – Norbert Schuch Aug 7 '18 at 15:49
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    $\begingroup$ @meowzz I think it makes more sense to ask a new question. Otherwise, you will invalidate half of the answers (as you will change the meaning of the question), which is not really fair towards the people who answered. $\endgroup$ – Norbert Schuch Aug 7 '18 at 16:01
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Rock paper scissors seems a good one. Now for the cards pack, I would not see the collapse metaphor here. While a player has not shown his cards, the cards are already set even if you do not see them. But I guess it is a point of view.

Maybe just pick a MMORPG with 52 monsters appearing randomly.

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  • $\begingroup$ I suppose the game w/ the deck would be "pick a card" $\endgroup$ – meowzz Aug 1 '18 at 23:22
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    $\begingroup$ A quantum MMORPG, so cool ;). $\endgroup$ – FSic Aug 2 '18 at 9:12
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A rolling n-sided die would be a good analogy that follows the coin example very closely. Until the die settles you can think of it as having not "collapsed", and die can come in whatever side number desired (at least in your mind, if you're trying to make a physical example that might be difficult.)

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    $\begingroup$ This would be the best analogy to the classical coin flipping, as far as I can see. $\endgroup$ – user1271772 Aug 2 '18 at 9:24

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