Global versus local costs on a barren plateau

The product ansatz $V(\theta)=R_X(\theta_1)\otimes\cdots\otimes R_X(\theta_n)$ with $R_X(\theta)=e^{-i\theta X/2}$ acts on $\lvert0\cdots0\rangle$, and the angles $\theta_1,\dots,\theta_n$ are independent and uniform on $[0,2\pi)$. The two costs are

$C_G(\theta)=1-\prod_{j=1}^n\cos^2\frac{\theta_j}2,\qquad C_L(\theta)=1-\frac1n\sum_{j=1}^n\cos^2\frac{\theta_j}2 .$

For $n=6$ give

(a) $\mathrm{Var}[\partial_{\theta_1}C_G]$, the variance of the derivative of the global cost with respect to $\theta_1$;

(b) $\mathrm{Var}[\partial_{\theta_1}C_L]$, the same for the local cost.

Enter two real numbers in this order. A decimal answer is accepted within a relative error of $1\,\%$ (plus an absolute $10^{-7}$); an exact expression such as (1/8)*(3/8)^5 or 1/288 must be correct to within $10^{-12}$.

Complete the TypeScript function

gradientVariance(evaluate: (params: number[]) => number, n: number, samples: number, seed: number): number

What evaluate does. evaluate(p) runs the product circuit $R_X(p_1)\otimes\cdots\otimes R_X(p_n)$ on $\lvert0\cdots0\rangle$ and returns the exact cost $C(p)$, which is either $C_G$ or $C_L$ of the previous step (the tests use both; your code does not need to know which). p must be an array of exactly $n$ finite numbers; p[j-1] is the angle $\theta_j$ of qubit $j-1$. Internally the cost is the expectation value of a Pauli operator whose all-identity term carries the constant: $C_G=(1-2^{-n})\,I-2^{-n}\sum_{S\neq\emptyset}Z_S$ and $C_L=\frac12I-\frac1{2n}\sum_jZ_j$. Every call is counted, as each would be one circuit execution on hardware.

What to compute. Create rng = mulberry32(seed) (import it with import { mulberry32 } from 'quantumcore';). For $s=1,\dots,S$ with $S=$ samples:

1. draw a fresh parameter vector $\theta_j=2\pi\cdot$rng() for $j=1,\dots,n$, in this order;
2. compute $g_s=\partial_{\theta_1}C(\theta)$ with the parameter-shift rule $g_s=\frac12\big[C(\theta+\frac\pi2e_1)-C(\theta-\frac\pi2e_1)\big]$, where only the first entry is shifted: exactly two calls of evaluate per sample, $2S$ in total.

Return the unbiased sample variance $\frac1{S-1}\sum_{s=1}^S(g_s-\bar g)^2$ with $\bar g=\frac1S\sum_sg_s$.

Tests. Each test calls gradientVariance(evaluate, n, 4000, seed). The values check compares your result with the exact variance from the lesson, $\frac18\big(\frac38\big)^{n-1}$ (global) or $\frac1{8n^2}$ (local), with a relative tolerance of $25\,\%$. The calls check runs the same problems again (tests calls-…) and also requires exactly $2S=8000$ calls.

• global-n2 (visible): $C_G$ on $n=2$ qubits, seed $1$, exact value $\frac18\cdot\frac38=0.046875$.
• local-n2, global-n4, local-n4 (hidden): $C_L$ on $2$ qubits, $C_G$ on $4$ qubits and $C_L$ on $4$ qubits. Only their arguments are sent to your browser.

Why $25\,\%$. A Monte-Carlo variance is itself random. Because $\mathbb E[g]=0$, the relative standard error of the sample variance is $\sqrt{(\kappa-1)/S}$ with $\kappa=\mathbb E[g^4]/\mathbb E[g^2]^2$. For the global cost $\kappa=\frac32\big(\frac{35}{18}\big)^{n-1}$, which is $11.0$ at $n=4$: the heavy tail of $\prod_j\cos^2\frac{\theta_j}2$ (rare samples where every factor is close to $1$) makes the estimate noisy, about $5\,\%$ at $S=4000$. The tolerance is five such standard errors; a missing factor $\frac12$ in the shift rule is off by $300\,\%$.