Gradient Measurement

To train a VQE you need the gradient of the energy with respect to every circuit parameter. On a classical simulator you could differentiate the state vector directly, but on real hardware you only ever get expectation values from measurements. The question this lesson answers is: how do you obtain an exact analytic gradient using only circuit evaluations?

Why finite differences fail

The naive approach is the finite-difference estimate

\frac{\partial E}{\partial\theta_k} \approx \frac{E(\theta_k + \varepsilon) - E(\theta_k - \varepsilon)}{2\varepsilon}.

This has two problems on a quantum device. A small $\varepsilon$ makes the numerator a difference of two nearly equal noisy numbers — the shot noise blows up as $1/\varepsilon$ . A large $\varepsilon$ introduces $O(\varepsilon^2)$ truncation bias. There is no good middle ground.

The parameter-shift rule

For gates of the form $U(\theta) = e^{-i\theta P/2}$ with a Pauli generator $P$ (so $P^2 = I$ , which covers $R_X, R_Y, R_Z$ ), the gradient is given exactly by the same circuit evaluated at two macroscopically shifted points:

\frac{\partial E}{\partial\theta_k} = \frac{1}{2}\Bigl[\,E\bigl(\theta_k + \tfrac{\pi}{2}\bigr) - E\bigl(\theta_k - \tfrac{\pi}{2}\bigr)\Bigr].

This is not an approximation. Because the eigenvalues of $P/2$ are $\pm\tfrac{1}{2}$ , the energy as a function of $\theta_k$ is a pure sinusoid $a\cos\theta_k + b\sin\theta_k + c$ , and a sinusoid's derivative is recovered exactly from samples a quarter-period apart. The shift is a fixed $\pi/2$ , so each evaluation is a normal-magnitude expectation value — no division by a tiny $\varepsilon$ , no truncation bias.

A concrete check

Take the simplest case: a single qubit in $|0\rangle$ , apply $R_Y(\theta)$ , and measure $Z$ . Since $R_Y(\theta)|0\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle$ ,

E(\theta) = \langle Z\rangle = \cos^2\!\tfrac{\theta}{2} - \sin^2\!\tfrac{\theta}{2} = \cos\theta.

The true derivative is $-\sin\theta$ . Apply the shift rule:

\frac{1}{2}\bigl[\cos(\theta + \tfrac{\pi}{2}) - \cos(\theta - \tfrac{\pi}{2})\bigr] = \frac{1}{2}\bigl[-\sin\theta - \sin\theta\bigr] = -\sin\theta. \checkmark

The two-evaluation recipe reproduces the analytic gradient exactly.

Try it

Implement $E(\theta) = \cos\theta$ for the $R_Y$ -then- $Z$ circuit, then return the parameter-shift estimate of the gradient at $\theta = 0.8$ radians. The grader checks your returned number against the exact value $-\sin(0.8)$ .

Run your code to see the quantum state.