# Confusion about the output distribution of Haar random quantum states

Consider a Haar random quantum state $$|\psi \rangle$$. I was confused between two facts about $$|\psi \rangle$$, which appear related:

1. Consider the output distribution of a particular $$n$$-qubit $$|\psi \rangle$$. For a large enough $$n$$, the probability of what fraction of strings in the output distribution of $$|\psi \rangle$$ lie between $$\frac{1}{2^{n}}$$ and $$\frac{2}{2^{n}}$$ (or between any two numbers)? According to the formulas in the supplement of the Google paper (section IV C, page 9) and this answer, the answer is $$\int_\frac{1}{2^{n}}^\frac{2}{2^{n}} 2^n e^{-2^np} dp.$$ How to prove this? Also, is this statement true for any Haar-random $$|\psi\rangle$$, or only with high probability over a particular choice of a Haar-random $$|\psi\rangle$$? For example, $$|\psi\rangle$$ could be something trivial like $$|0\rangle^{\otimes n}$$ and this statement would not hold.

2. We know that for a particular output string $$z \in \{0, 1\}^{n}$$, if we define $$p_z = |\langle z| \psi \rangle|^{2}$$, then the random variable $$p_z$$ (for a fixed $$z$$, but for $$|\psi\rangle$$ chosen uniformly at random from the Haar measure) follows the Porter-Thomas distribution, for every such fixed $$z$$. The probability density function of the Porter-Thomas distribution is given by $$$$\text{PorterThomas}(p) \sim 2^{n} e^{-2^{n}p}.$$$$ The same density function also appears inside the integration in the first item. Is this just a coincidence, or are these facts related? The two situations do not seem related (we are interested in a particular $$|\psi\rangle$$ for the first one and a particular $$z$$ for the second one) and I do not see an obvious way of going from one to another.

The two facts are connected in that they both arise as a result of rotational invariance of the Haar measure.

We will derive them in the case of large $$n$$ since this is when the Porter-Thomas distribution takes the exponential form given in the question. Also, this case admits an intuitive proof backed by a geometric picture. For small $$n$$, Porter-Thomas distribution is a Beta distribution. In this case, the proof turns into a lengthy calculation.

Consider a Haar-random quantum state $$|\psi\rangle$$ of $$n$$ qubits. Let $$N=2^n$$. Commonly, $$|\psi\rangle$$ is thought of as a complex vector $$(a_0+ib_0, a_1+ib_1, \dots, a_{N-1}+ib_{N-1})^T \in \mathbb{C}^{N}$$ of unit norm, but we will think of it as a real vector

$$v = \sqrt{2N}(a_0, b_0, a_1, b_1, \dots, a_{N-1}, b_{N-1})^T\in\mathbb{R}^{2N}.$$

Since $$|\psi\rangle$$ is drawn from the Haar measure, $$v$$ is uniformly distributed over a sphere of radius $$\sqrt{2N}$$ in $$\mathbb{R}^{2N}$$. We would like to characterize the distribution of the real coordinates $$a_j$$ and $$b_j$$.

An easy observation is that all coordinates $$a_j$$ and $$b_j$$ have the same distribution. This follows from the fact that Haar measure is unitarily invariant and permutations are unitary matrices. We can say more using the following

Theorem (Diaconis-Freedman). The first $$k=o(d)$$ coordinates of a point uniformly distributed over the surface of the $$d$$-sphere of radius $$\sqrt{d}$$ are independent standard normal variables in the limit of $$d\to\infty$$.[1]

In our case, permutation invariance means that any $$k=o(N)$$ coordinates of $$v$$ are independent standard normal variables, $$a_j\sqrt{2N} \sim \mathcal{N}(0, 1)$$ and $$b_j\sqrt{2N} \sim \mathcal{N}(0, 1)$$. Consequently,

$$2Np_j = 2N|\langle j|\psi\rangle|^2 = \left(a_j\sqrt{2N}\right)^2 + \left(b_j\sqrt{2N}\right)^2$$

is the sum of squares of two standard normal variables. In other words, $$2Np_j$$ is a chi-square random variable with two degrees of freedom which is the same distribution as the exponential distribution with rate parameter $$\lambda = \frac{1}{2}$$. Thus, the probability density function of $$2Np_j$$ is $$\frac{1}{2}\exp(-\frac{1}{2}p)$$, so probability density function of $$p_j$$ is $$N\exp(-Np)$$, establishing fact 1 in the question.

Now suppose we independently draw two quantum states $$|\psi_1\rangle$$ and $$|\psi_2\rangle$$ from the Haar measure. Fix an output bitstring $$j \in \{0, 1\}^n$$ and consider the output probabilities $$|\langle j|\psi_1\rangle|^2$$ and $$|\langle j|\psi_2\rangle|^2$$. Since $$|\psi_1\rangle$$ and $$|\psi_2\rangle$$ have been drawn independently from the Haar measure, the two probabilities are independent from each other. Moreover, by the arguments above both have the same distribution with density function $$N\exp(-Np)$$, establishing fact 2 in the question.

The key point to explain the symmetry between facts 1 and 2 is the independence of different coordinates of a uniformly distributed point on a sphere. As long as we only have access to a small number $$k=o(2^n)$$ of output probabilities of a Haar-random quantum state they are all independent and identically distributed random variables with Porter-Thomas distribution. In other words, for any quantum states $$|\psi_j\rangle$$ chosen independently from the Haar measure and any $$k=o(2^n)$$ bitstrings $$z_i\in\{0, 1\}^n$$, also chosen independently, each of the probabilities $$|\langle z_i|\psi_j\rangle|^2$$ constitutes an independent sample from the same distribution with density function $$N\exp(-Np)$$. This highlights the very high degree of symmetry of the Haar measure.

The observation that some states with non-zero probability density, such as $$|0\rangle^{\otimes n}$$, do not exhibit Porter-Thomas output distribution is correct. One might be tempted to dismiss this case as a zero-measure set. However, there is a small, but positive measure set of states in the vicinity of $$|0\rangle^{\otimes n}$$ that also do not exhibit Porter-Thomas output distribution.

The key point is that typical Haar-random states do. This can probably be made more rigorous by deriving an appropriate concentration inequality, e.g. bounding the probability that the total variation distance between the output distribution of a Haar-random state and the Porter-Thomas distribution exceeds a threshold. However, the following informal argument illustrates the point. One way to think of the process of drawing a Haar-random state is as a long sequence of draws of the real and imaginary parts of each amplitude from (approximate) standard normal distribution: $$a_0, b_0, a_1, b_1, \dots$$. A quantum state in the vicinity of $$|0\rangle^{\otimes n}$$ can be thought of as a sequence in which $$a_0$$ is approximately $$1$$ and all other numbers are approximately $$0$$. Probability of obtaining such a result from a sequence of draws from (approximate) standard normal distribution is extremely small, because the dimension of the Hilbert space and thus the length of the sequence is very large.

[1] Persi Diaconis and David Freedman. A dozen de Finetti-style results in search of a theory. Ann. Inst. H. Poincar´e Probab. Statist., 23(2, suppl.), p.397–423, 1987.

• Thanks for an excellent answer! Feb 2 '21 at 19:19
• You're welcome! I'm glad it's helpful! Thank you for asking deep, interesting questions! :-) Feb 2 '21 at 19:20