In these lecture notes by Scott Aaronson, the author states the following (towards the end of the document, just before the Linearity section):

There's actually another phenomenon with the same "Goldilocks" flavor, which was observed by Bill Wootters -- and this leads to my third reason why amplitudes should be complex numbers. Let's say we choose a quantum state $$\sum_{i=1}^N \alpha_i |i\rangle$$ uniformly at random (if you're a mathematician, under the Haar measure). And then we measure it, obtaining outcome $\lvert i\rangle$ with probability $|\alpha_i|^2$. The question is, will the resulting probability vector also be distributed uniformly at random in the probability simplex? It turns out that if the amplitudes are complex numbers, then the answer is yes. But if the amplitudes are real numbers or quaternions, then the answer is no! (I used to think this fact was just a curiosity, but now I'm actually using it in a paper I'm working on...)

Numerically, I can see that this seems to be the case. Indeed, using $N=3$ as an example, if I generate random elements in the sphere, and look at the resulting distribution of the squares of the coordinates, I get the picture below on the left, while doing the same with complex coefficients (that is, with random states), I get the one on the right:

(* sampling wiht real coefficients *)
#/Norm@# & /@ RandomReal[{-1, 1}, {10000, 3}] // Graphics3D[{Point[#^2]}] &
(* sampling with complex coefficients *)
RandomUnitary[m_] := Orthogonalize @ Function@ Map[
    Function[#[[1]] + I * #[[2]]], #, {2}
] @ RandomReal[NormalDistribution[0, 1], {m, m, 2}];
Table[First @ RandomUnitary @ 3, 10000] // Abs[#]^2 & // Graphics3D[{Point[#]}] &

Notice how different for example the edge are. Sampling with real coefficients clearly gives a much denser distribution towards the edges of the simplex. Plotting the associated histograms shows clearly that the real coefficients do not give a uniform distribution, while the complex ones do.

Why is this the case?

  • $\begingroup$ Have you experimented with $N=4$ and $\alpha_i\in\mathbb{R}$ - that is, two qubits with only real amplitudes chosen uniformly at random (and satisfying the normalization requirement)? I believe your question and Aaronson's comments imply that we do not necessarily measure each of $\vert 00\rangle$, $\vert 01 \rangle$, $\vert 10\rangle$, and $\vert 11\rangle$ with probability $\frac{1}{4}$? $\endgroup$ Nov 9, 2019 at 16:16
  • $\begingroup$ @MarkS I do get average probabilities consistent with $1/4$ if I do it numerically, but I might not be understanding the problem correctly. Isn't this equivalent to taking random points in the hypersphere $S^3$ and averaging the square of the components? Yes I understood the statement as saying that we would not get probabilities equal to $1/4$ $\endgroup$
    – glS
    Nov 9, 2019 at 16:38
  • $\begingroup$ Judging from your answers and comments on this site you are much better mathematician than me, but is your actual state that you get from sampling on the hypersphere $S^3$ such that each of $|\alpha_{00}|^2$, $|\alpha_{01}|^2$, $|\alpha_{10}|^2$, and $|\alpha_{11}|^2$ are about $1/4$, or are the states more likely to be heavy on one $\alpha$ and lighter on another $\alpha$? To me it feels like those old puzzles about the probability of forming a triangle by breaking a stick in three. $\endgroup$ Nov 9, 2019 at 16:45
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    $\begingroup$ @MarkS don't overestimate me =). I think I see what you are saying though. If I compute the averages of the square of the components I get $1/4$, but the statement is about the distributions themselves, not just the averages $\endgroup$
    – glS
    Nov 9, 2019 at 17:46
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    $\begingroup$ See also this thread wherein the relationship to Archimedes' Hat Box theorem is also made. $\endgroup$ Nov 10, 2019 at 13:47

1 Answer 1


We can begin this question by comparing two small cases, a qubit with a "rebit", the latter being a qubit confined to real amplitudes. The set of pure state of a rebit makes a great circle in the Bloch sphere, say a vertical circle. The corresponding measure on the probability interval is given by projecting uniform measure on the circle onto the $z$ axis, i.e., a diameter of the circle. If you draw a picture of this, it leaps out at you that of course it's not uniform, because the circle is parallel to its diameter at the middle, but angled near the poles. If we parametrize the projection interval as $[-1,1]$, which is different from the probability parametrization $[0,1]$ but convenient here, then the projected density is the same as the arc length integrand for a circle, namely it is $\frac{dz}{\pi\sqrt{1-z^2}}$. Just as the geometry suggests, the density goes to infinity at the endpoints.

If you repeat this same thinking for the Bloch sphere of a qubit, then you see a coincidence that has been celebrated for a very long time. Borrowing from spherical coordinates, let $z = \cos \theta$. Then a latitude band at a polar angle of $\theta$ of height $dz$ is tilted to give you its area a factor of $1/(\sin \theta)$, but its radius is $\sin \theta$ rather than 1. The factors cancel! The area of the band is always $2\pi dz$, independent of $\theta$, and the projection onto the $z$ axis is indeed uniform measure on the probability interval. I said that this has been celebrated for a long time, but I didn't say how long. In geometric form, it was first published by Archimedes.

Note that the argument only works for the Bloch sphere surface, i.e., for pure states. For mixed states, the middle of a ball is always fatter than its ends, so uniform measure does not project to uniform measure.

Another case that does not work is a qubit with quaternionic amplitudes, which I guess you could call a "ququbit". The analogue of a Bloch sphere in this case is a 4-sphere in $\mathbb{R}^5$. The beautiful coincidence of Archimedes is now wrecked because the polar regions are too small rather than too large as in the case of a circle. The tilt still gives you a factor of $1/(\sin \theta)$, but a radius of $\sin \theta$ now gives you a factor of $(\sin \theta)^3$ in the 3-dimensional volume, to give you a total answer proportional to $(\sin \theta)^2$ or $(1-z^2)dz$.

You might still wonder how this is going to work for higher-dimensional qudits. There are various ways to argue this. Some of them (like the moment map interpretation in symplectic geometry) are so elegant that they give you a uniform measure coincidence that is vastly more general than the original question; but these arguments tend to pull in more advanced mathematics (like symplectic geometry). One good argument relies on a pair of algorithms, which are valuable tricks in computer software, for generating either a uniformly random point in the unit sphere in $\mathbb{R}^n$ or the regular simplex subtended by the standard basis in $\mathbb{R}^n$.

In the former case, you should choose each coordinate of a vector using Gaussian measure proportional to $\exp(-x^2)$, and then rescale the total vector to have Euclidean length 1. This works because the joint density is proportional to $$ \exp(-x_1^2 - x_2^2 - \cdots - x_n^2),$$ which is constant on concentric sphere surfaces. Similarly, to get a uniformly random point on the simplex of non-negative vectors with coordinate sum 1, you should choose each coordinate with exponential measure $\exp(-x)$, and now rescale to make the coordinate sum 1. This time the joint density is $$ \exp(-x_1 - x_2 - \cdots - x_n),$$ which is constant on parallel simplices in the positive orthant $\mathbb{R}^n_{\ge 0}$.

Now let $n = 2d$ and interpret $\mathbb{R}^n$ as $\mathbb{C}^d$, and look at what happens when you choose each complex coordinate $z = x+iy$ using Gaussian measure $\exp(-x^2-y^2)$ in the complex plane. If $z$ is an amplitude, the corresponding probability is $p = |z|^2 = x^2 + y^2 = r^2$. To convert the measure from $z \in \mathbb{C}$ to $p \in \mathbb{R}_{\ge 0}$, we project onto a radial ray and then square the radius. To do this competently, we need to remember infinitesimal length and area factors from calculus. We get $$ \exp(-x^2-y^2)\, dx\, dy = \exp(-r^2)\, r\, dr\, d\theta. $$ That extra factor of $r$ (remember that the length of a circle is proportional to its radius) is crucial given that $dp = 2r\,dr$. It means that if we ditch $\theta$ and substitute $p$ for $r$, we get a density of $\exp(-p)\, dp$ for the distribution of $p$.

In conclusion, if we pass from a vector of complex amplitudes in $\mathbb{C}^d$ to its corresponding real vector of probabilities in $\mathbb{R}_{\ge 0}^d$, then Gaussian measure $$ \exp(-|z_1|^2-|z_2|^2- \cdots - |z_d|^2)$$ projects to exponential measure $$ \exp(-p_1-p_2- \cdots - p_d).$$ Thus the algorithm to generate a uniformly random point in the unit sphere in $\mathbb{C}^d$ matches the algorithm to generate a uniformly random point in the standard simplex in $\mathbb{R}_{\ge 0}^d$.

  • $\begingroup$ What a lovely mental image! This answer puts mine to shame. I think it also explains the scatterplot images added by @gIS - for a (single) rebit case the projection of the points on a meridian onto the $z$ axis would be less dense in the center of the simplex, and more dense elsewhere. I believe you, but why is it that "the corresponding measure on the probability interval is given by projecting uniform measure on the circle onto the $z$ axis, i.e., a diameter of the circle?" $\endgroup$ Nov 10, 2019 at 12:34
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    $\begingroup$ It's the result of angle doubling from vector states to operator states (as discussed in another QC SE post). If $|\psi\rangle = [\cos \theta,\sin \theta]$, then $|\psi\rangle\langle\psi| = ((\cos 2\theta)Z + (\sin 2\theta)X + I)/2$. Since the probabilities are the diagonals entries of $|\psi\rangle\langle\psi|$, the measure is a linear projection of this second circle. Moreover, if $\theta$ is uniformly random, then so is $2\theta$. The subtlety is that the vector state circle wraps twice around the operator state circle. $\endgroup$ Nov 10, 2019 at 16:46
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    $\begingroup$ Angle-doubling is an ever-present subtlety. In the case of a qubit, the Bloch sphere is an $S^2$, while the unit sphere in $\mathbb{C}^2$ is an $S^3$, and you pass from one to the other with the angle-doubling Hopf fibration. By symmetry, uniform measure on $S^3$ projects to uniform measure on $S^2$. $\endgroup$ Nov 10, 2019 at 16:56
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    $\begingroup$ @glS That is what I am talking about, and it is exactly the correct picture for the case of a qubit. The map to the probabilities is quadratic as a map from vector states, but it is linear as a map from operator states. The space of operator states is where the Bloch sphere lives. The map from the Bloch sphere to the classical state on an interval is projection onto the $z$ axis. Work in the usual basis $\rho = (xX + yY + zZ + I)/2$ and you will see it. $\endgroup$ Nov 12, 2019 at 18:50
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    $\begingroup$ Two small remarks on the formula for $\rho$: (1) I am using the Hermitian version of the operator $Y = \sigma_y$, i.e., a Pauli spin matrix. (2) As already mentioned in the rebit case, for pure states $\rho = |\psi\rangle\langle\psi|$. These are the states that lie on the Bloch sphere surface. $\endgroup$ Nov 12, 2019 at 18:54

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