Where does Porter-Thomas Distribution come from?

• These are some notes I wrote while trying to understand why Porter-Thomas distributions appear in Random Quantum Circuits.

• Simulation of random quantum circuit is a task that is central to demonstrating quantum advantage¹.
• We apply a random circuit to an initial state and study the probability distribution generated by the computer.
• If the circuit has no noise, the probability distribution thus produced has a Porter-Thomas distribution.
• I first learned of this fact from the supremacy proposal paper¹
• The paper they cite is a 1956 paper by Porter and Thomas² on flucations of nuclear reation width (whatever that means).
• I did not fully understand where the Porter-Thomas distribution came from.
• Searching for Porter-Thomas on Google brings up the Wikipedia page on Random Matrix Theory.
• I wanted to understand where the Porter-Thomas distribution comes from.

Distribution of Measurement Outcomes

• Assume the random quantum circuit on $n$ qubits implements a random unitary $U$ of dimension $N \times N$ with $N=2^n$.
• After the circuit is applied, the probability of measuring a particular bitstring $z$ is given by $$|\langle z | U | 0^n\rangle|^2 = |U_{z0}|^2$$
• The distribution of these measurement probabilities is, therefore, directly related to the distribution of the norm of matrix elements $|U_{z0}|^2$.

Distribution of Measurement Probabilities

• We now try to understand the distribution of meaurement probabilities (not to be confused with distributionof meaurement outcomes) by studying the matrix elements of the random unitary $U$.
• To make sure $U$ is unitary, we need that $U^\dagger U = I$. $$U^\dagger U = I \implies \sum_{j=1}^N U_{kj}^\dagger U_{jk} = 1 \implies \forall k, \sum_{k=1}^N |U_{ij}|^2 = 1. $$
• The joint probabilility of the elements in a single row of $U$ is constrained by this normalization. $$P(U_{k1},\ldots, U_{k2}, U_{kN}) \propto\delta\left(1-\sum_{j=1}^N |U_{kj}|^2\right).$$
• We can integrate, for example, the last element $U_{iN}$, to get a marginal on the remainder of the matrix elements. $$P(U_{k1},\ldots, U_{k2}, U_{k(N-1)}) = \int d(x) P(U_{k1},\ldots, U_{k2}, U_{k(N-1)}, x).$$
• Realizing that the integral of delta function is the Heaviside Theta function, we get something like the following $$P(U_{k1},\ldots, U_{k2}, U_{k(N-1)}) \propto \theta\left(1-\sum_{j=1}^{N-1} |U_{kj}|^2\right).$$
• Calculating further marginals is a bit tricky.
• We can verify the following property for Heaviside-theta functions. $$ \int_{0}^{\infty} \theta(1-x^2-a^2) dx \propto \sqrt{1-\alpha^2}\theta(1-\alpha^2)$$
• Using this, we can integrate over one more matrix element $U_{k(N-1)}$, we get something of the form $$P(U_{k1},\ldots, U_{k2}, U_{k(N-2)}) \propto \left(1-\sum_{j=1}^{N-2} |U_{kj}|^2\right)\theta\left(1-\sum_{j=1}^{N-2} |U_{kj}|^2\right)$$
• We can iteratively integrate out $N-1$ matrix elements to calculate the marginal of a single matrix element $U_{k1}$. $$P(U_{k1}) \propto \left(1-|U_{k1}|^2\right)^{N-2}$$
• There is a linear correspondence between probability of observing the matrix element $U_{k1}$ and the probabilty of obeserving the norm $|U_{k1}|^2$. We relabel $x=|U_{k1}|^2$ to get $$P(x) \propto \left(1-x\right)^{N-2}$$
• The distribution $(1-x)^{N-2}$ penalizes matrix elements with large norms.
• In the limit $N \gg 1$ and for small $x$, this approximates the exponential distribution $$P(x) \propto \left(1-\frac{Nx}{N}\right)^{N-2} \approx e^{-Nx}$$
• To get the proportionality constant, we make sure the probability adds up to 1 $$\int_{0}^1 P(x) = 1 \implies = C\int_{0}^{1} e^{-Nx} = 1\implies C \approx N$$
• Giving the final distribution $$ P(x) \approx Ne^{-Nx}$$
• And this is the famous Porter-Thomas distribution.

References