# How is a single qubit fundamentally different from a classical coin spinning in the air?

I had asked this question earlier in the comment section of the post: What is a qubit? but none of the answers there seem to address it at a satisfactory level.

The question basically is:

How is a single qubit in a Bell state $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ any different from a classical coin spinning in the air (on being tossed)?

The one-word answer for difference between a system of 2 qubits and a system of 2 classical coins is "entanglement". For instance, you cannot have a system of two coins in the state $\frac{1}{\sqrt 2}|00\rangle+\frac{1}{\sqrt 2}|11\rangle$. The reason is simple: when two "fair" coins are spinning in air, there is always some finite probability that the first coin lands heads-up while the second coin lands tails-up, and also the vice versa is true. In the combined Bell state $\frac{1}{\sqrt 2}|00\rangle+\frac{1}{\sqrt 2}|11\rangle$ that is not possible. If the first qubit turns out to be $|0\rangle$, the second qubit will necessarily be $|1\rangle$. Similarly, if the first qubit turns out to be $|1\rangle$, the second qubit will necessarily turn out to be $|1\rangle$. At, this point someone might point out that if we use $2$ "biased" coins then it might be possible to recreate the combined Bell state. The answer is still no (it's possible to mathematically prove it...try it yourself!). That's because the Bell state cannot be decomposed into a tensor product of two individual qubit states i.e. the two qubits are entangled.

While the reasoning for the 2-qubit case is understandable from there, I'm not sure what fundamental reason distinguishes a single qubit from a single "fair" coin spinning in the air.

This answer by @Jay Gambetta somewhat gets at it (but is still not satisfactory):

This is a good question and in my view gets at the heart of a qubit. Like the comment by @blue, it's not that it can be an equal superposition as this is the same as a classical probability distribution. It is that it can have negative signs.

Take this example. Imagine you have a bit in the $0$ state and then you apply a coin flipping operation by some stochastic matrix $\begin{bmatrix}0.5 & 0.5 \\0.5 & 0.5 \end{bmatrix}$ this will make a classical mixture. If you apply this twice it will still be a classical mixture.

Now lets got to the quantum case and start with a qubit in the $0$ state and apply a coin flipping operation by some unitary matrix $\begin{bmatrix}\sqrt{0.5} & \sqrt{0.5} \\\sqrt{0.5} & -\sqrt{0.5} \end{bmatrix}$. This makes an equal superposition and you get random outcomes like above. Now applying this twice you get back the state you started with. The negative sign cancels due to interference which cannot be explained by probability theory.

Extending this to n qubits gives you a theory that has an exponential that we can't find efficient ways to simulate.

This is not just my view. I have seen it shown in talks of Scott Aaronson and I think its best to say quantum is like “Probability theory with Minus Signs” (this is a quote I seen Scott make).

I'm not exactly sure how they're getting the unitary matrix $\begin{bmatrix}\sqrt{0.5} & \sqrt{0.5} \\\sqrt{0.5} & -\sqrt{0.5} \end{bmatrix}$ and what the motivation behind that is. Also, they say: "The negative sign cancels due to interference which can not be explained by probability theory." The way they've used the word interference seems very vague to me. It would be useful if someone can elaborate on the logic used in that answer and explain what they actually mean by interference and why exactly it cannot be explained by classical probability. Is it some extension of Bell's inequality for 1-qubit systems (doesn't seem so based on my conversations with the folks in the main chat though)?

How is a single qubit in a Bell state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ any different from a classical coin spinning in the air (on being tossed)?

For both of them, the probability of getting heads is 1/2 and getting tails is also 1/2 (we can assume that heads$\equiv|1\rangle$ and tails$\equiv|0\rangle$ and that we are "measuring" in the heads-tails basis).

For any 1-qubit state $|\psi\rangle$, if all you do is measure it in the computational basis, you will always be able to explain it in terms of a probability distribution p(heads)$=|\langle 0|\psi\rangle|^2$ and p(tails)$=|\langle 1|\psi\rangle|^2$. The key differences are in using different bases and/or performing unitary evolutions.

The classic example is the Mach-Zehnder interferometer. Think of it this way: any 1-bit probabilistic operation is described by a $2\times 2$ stochastic matrix (i.e. all columns sum to 1). Call it $P$. It is easy enough to show that there is no $P$ such that $P^2=X$, where $X$ is the Pauli matrix (in other words, a NOT gate). Thus, there is no probabilistic gate that can be considered the square-root of NOT. On the other hand, we can build such a device. A half-silvered mirror performs the square-root of not action.

A half-silvered mirror has two inputs (labelled 0 and 1) and two outputs (also labelled 0 and 1). Each input is a photon coming in a different direction, and it is either reflected or transmitted. If you just look at one half-silvered mirror, then whatever input you give, the output is 50:50 reflected or transmitted. It seems just like the coin you're talking about. However, if you put two of them together, if you input 0, or always get the output 1, and vice versa. The only way to explain this is with probability amplitudes, and a transition matrix that looks like $$U=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right).$$ In quantum mechanics, the square-root of not gate exists.

• It's hard to pin down a why. I'm just using it to show very explicitly that there is a difference. And, more to the point, to show that classical is insufficient to describe what actually happens in the experiment. So, you need a broader formalism. The idea of probability amplitudes gives you that broader formalism. It's like if you restrict to real numbers only, the square root is not well defined, because you need complex numbers to be able to explain it. That's basically what we're doing here. – DaftWullie Jun 25 '18 at 8:00
• I get the transition matrix by saying that for one use, you get 50:50 outputs, so all 4 matrix elements must have a mod-square equal to 1/2. What $2\times 2$ complex matrices $U$ are there, satisfying that constraint, such that $U\cdot U=X$? Any answer will do. – DaftWullie Jun 25 '18 at 8:02
• What do you mean by "fundamental"? Mathematically, it's because we have to describe quantum mechanics using a richer mathematical structure than classical (as proven by this square root of not, not that this gate is particularly special: you can replace the X by any stochastic matrix with a negative eigenvalue). In terms of physics, well physics is just the working theory that describes experimental outcomes (such as square root of not). In terms of some underlying explanation of why the world is the way that it is, who knows? – DaftWullie Jun 25 '18 at 8:27
• You might also be interested in the Kochen-Specker Theorem. It only applies to qutrits and higher, but may help to cover what you want. – DaftWullie Jun 25 '18 at 8:31
• The main problem in this answer is that, although the transition matrix in the case you mentioned i.e. $$U=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right)$$ doesn't occur in the classical case, it doesn't mean the effects of the transition cannot be replicated for a classical coin. After all, the transition matrix cannot be measured directly. All we can measure is the outcome probabilities! – Sanchayan Dutta Jun 26 '18 at 17:42

The analogy between qubits and coin flips is popular but can be misleading. (See, for example, this video: https://www.youtube.com/watch?v=lypnkNm0B4A) A coin spinning in the air and landing on the ground is not truly random, though we may describe it as such. The key point is how you measure it.

At any point in time the coin has a definite orientation, though it may be unknown to us. Likewise, qubits have a definite state at any time, which we can describe by a point on the surface of a sphere (the so-called Bloch sphere). Mathematically, a coin's orientation and a qubit's state are equivalent. While in the air, the coin may undergo deterministic and reversible motion (e.g., spinning and falling). Likewise, prior to measurement a qubit may undergo deterministic and reversible transformations (e.g., unitary gate operations on a quantum computer).

Measurement represents an irreversible process. For a coin, it is a series of inelastic collisions with the ground, bouncing and spinning until it comes to rest. If we are completely ignorant of the initial conditions of the coin, the two final orientations (heads or tails) will appear equally likely, but this is not always the case. If I drop it oriented "heads up" from a short height, it will land flat with "heads up" with near certainty. But suppose I was standing next to a large magnetic wall and did this. The coin would hit edge-on and would likely land with either heads or tails showing, with equal probability. One could imagine doing this experiment with various initial orientations of the coin and orientations of the magnetic wall (upright, flat, slanted, etc.). You can imagine that the probability of getting heads or tails will be different, depending on the relative orientations of the coin and wall. (In theory it's all completely deterministic, but in practice we never know the initial conditions that precisely.)

Measurements of qubits are quite similar. I can prepare a qubit in the state $\frac{1}{\sqrt{2}}[|0\rangle + |1\rangle]$, measure it in the 0/1 basis $\{|0\rangle, |1\rangle\}$, and get either $|0\rangle$ or $|1\rangle$ with equal probability. If, however, I measure in the +/- basis $\{|+\rangle, |-\rangle\}$ (analogous to using a magnetic wall), I get $|+\rangle$ with near certainty. (I say "near certainty" because, well, nothing in the real world is perfect.) Here, $|\pm\rangle = \frac{1}{\sqrt{2}}[|0\rangle\pm|1\rangle$ are the +/- basis states. For polarized photons, for example, this could be done used polarization filters rotated $45^\circ$.

The difference between preparing the state $\frac{1}{\sqrt{2}}[|0\rangle + |1\rangle]$ and the state $\frac{1}{\sqrt{2}}[|0\rangle - |1\rangle]$ is the difference between preparing a vertically oriented coin with either heads or tails facing away from the wall. (A good picture would really help here.) We can tell which of the two states is prepared based on the outcome of a suitably chosen measurement, which in this case would be a +/- basis (or magnetic wall) measurement.

Jay Gambetta mentions a unitary matrix that is used to represent a Hadamard gate. It corresponds to rotating a coin by $90^\circ$, so a coin that's initially heads up becomes vertically oriented with, say, heads facing away from the wall. If the wall is magnetic and you release the coin, it will stick to it with heads up. If, instead, you started with a coin that's tails up and applied the same rotation, it would be vertical with tails facing away from the wall. If you release it (and the wall is still magnetic), you get tails. On the other hand, if the wall is not magnetic and you drop it, it lands heads or tails with equal probability. Using a "floor" measurement doesn't distinguish between the two vertical orientations, but using a "wall" measurement does. It's not so much whether things are predictable or not, it's the type of measurement you do that distinguishes one quantum state (or coin orientation) from another.

This is the whole of it. The only remaining mystery is that the outcome of the coin measurement is considered to be, in theory, completely deterministic, while that of the qubit is considered to be, except in special cases, "intrinsically random." But that's another discussion...

• Could you also add an explanation for Jay Gambetta's approach using transition matrix (which he/she apparently justifies using quantum interference)? – Sanchayan Dutta Jun 24 '18 at 15:32
• And to summarize, your main point is that while for a coin if we know the initial conditions sufficiently precisely, then the outcome of a measurement is completely predictable. But for a qubit, simply knowing the initial state sufficiently precisely isn't sufficient to predict the outcome of a measurement (which is essentially what the Copenhagen interpretation says). Yes? – Sanchayan Dutta Jun 24 '18 at 15:41
• @Blue, please see my recent edits for an answer to your question. – Brian R. La Cour Jun 24 '18 at 17:04

You have already mentioned the practical differences, such as qubit entanglement, and the negative signs (or more general "phases").

The fundamental reason for this is that allowed quantum states are solutions of the Schrödinger equation, which is a linear differential equation. The sum of solutions to a linear differential equation is always also a solution to that differential equation [1],[2],[3]. Since "solution" to differential equation is synonymous with "allowed quantum state" or "allowed wavefunction", any sum of allowed states is also allowed (ie. superpositions like Bell states are allowed).

That is the fundamental reason why quantum mechanical bits (qubits) can exist in superpositions. In fact, not just any sum, but any linear combination of states is an allowed state because the differential equation is linear. This means we can even add constants (phases of -1 or +1 or $e^{i\theta}$) and still have allowed states.

Bits that follow the rules of quantum physics, for example, the Schrödinger equation, can physically exist in superpositions and with phases, due to linearity (review vector spaces if this is not clear). Classical physics does not give any mechanism for a system to be in more than one state at the same time.

• @Blue: I saw you had a conversation with someone where that person mentioned the need for "boundary conditions". It is not really true, qubits can exist in superposition and with phases because of linearity of the equation describing them. I have given 3 links which prove this fact in many ways. – user1271772 Jun 24 '18 at 18:01
• Boundary conditions are what ensure that the states are discrete. The Schrodinger equation by itself cannot posit that. Moreover, the superposition that you speak of is certainly not something intrinsic to quantum mechanics. For instance, a classical system can be in a harmonic motion which is a superposition of two individual harmonic motions (basis motions), satisfying a certain ODE. – Sanchayan Dutta Jun 24 '18 at 20:23
• "Classical physics does not give any mechanism for a system to be in more than one state at the same time." <--- being in more than one state at the same instant $\neq$ being in a superposition of two basis states. – Sanchayan Dutta Jun 24 '18 at 20:25