The Parameter-Shift Rule

Training a variational quantum circuit means adjusting its gate angles to minimize a cost function. To do that efficiently we need gradients — but how do you differentiate with respect to an angle that lives inside a quantum gate? The answer is the parameter-shift rule, an exact, hardware- compatible recipe that requires only two circuit evaluations per parameter.

Expectation values as the loss

A variational circuit $U(\boldsymbol{\theta})$ acting on $|0\rangle^{\otimes n}$ produces the state

|\psi(\boldsymbol{\theta})\rangle = U(\boldsymbol{\theta})|0\rangle^{\otimes n}.

The training signal is an expectation value of some observable $O$ (often a Pauli operator):

E(\boldsymbol{\theta}) = \langle \psi(\boldsymbol{\theta}) | O | \psi(\boldsymbol{\theta}) \rangle.

For the simplest case — a single $R_y(\theta)$ gate followed by a $Z$ measurement — the expectation is

E(\theta) = \langle 0 | R_y(\theta)^\dagger \, Z \, R_y(\theta) | 0 \rangle = \cos\theta.

You can verify this directly: $R_y(\theta)|0\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle$ , so $P(0) = \cos^2(\theta/2)$ and $P(1) = \sin^2(\theta/2)$ , giving $E = P(0) - P(1) = \cos^2(\theta/2) - \sin^2(\theta/2) = \cos\theta$ .

The shift formula

Any generator that appears in a rotation gate of the form $e^{-i\theta G}$ with $G^2 = I/4$ (for example $G = Y/2$ in $R_y$ ) satisfies the following identity (Mitarai et al., 2018; Schuld et al., 2019):

\frac{\partial E}{\partial \theta} = \frac{1}{2}\!\left[E\!\left(\theta + \tfrac{\pi}{2}\right) - E\!\left(\theta - \tfrac{\pi}{2}\right)\right].

This is not a finite-difference approximation — it is exact. The shift $\pm\pi/2$ is not a small $\varepsilon$ chosen for numerical stability; it comes from the eigenvalues of the generator.

Why it works

Write $E(\theta) = \langle \psi | e^{i\theta G} O e^{-i\theta G} | \psi \rangle$ . Differentiating gives $i\langle [G, O] \rangle$ . For $G = Y/2$ the commutator identity and $G^2 = I/4$ lead to

\frac{d}{d\theta} E(\theta) = \frac{E(\theta+\pi/2) - E(\theta-\pi/2)}{2},

which can be evaluated by running the same circuit at two shifted angles and subtracting.

The shift $\pi/2$ can be scaled to any $s$ for generators with different eigenvalue gaps $\pm\lambda$ . The general form (Schuld et al., 2019) is

$\frac{\partial E}{\partial\theta} = \frac{E(\theta+s) - E(\theta-s)}{2\sin(2\lambda s)}.$

For $R_y$ the generator is $G = Y/2$ with eigenvalues $\lambda = \pm 1/2$ . Choosing $s = \pi/2$ gives $2\sin(2\cdot\tfrac{1}{2}\cdot\tfrac{\pi}{2}) = 2\sin(\tfrac{\pi}{2}) = 2$ , recovering the familiar $1/2$ prefactor. For gates whose generator has eigenvalues $\pm 1$ (e.g.\ $R_z$ written as $e^{-i\theta Z}$ ), the natural choice is $s = \pi/4$ , which again gives denominator $2$ .

Applying it to $R_y(\theta)$ at $\theta = \pi/4$

With $E(\theta) = \cos\theta$ the analytic derivative is $dE/d\theta = -\sin\theta$ . At $\theta = \pi/4$ :

\frac{dE}{d\theta}\bigg|_{\theta=\pi/4} = -\sin\!\left(\tfrac{\pi}{4}\right) = -\tfrac{1}{\sqrt{2}} \approx -0.7071.

The parameter-shift formula gives the same result using only the shifted evaluations:

\frac{1}{2}\left[\cos\!\left(\frac{\pi}{4}+\frac{\pi}{2}\right) - \cos\!\left(\frac{\pi}{4}-\frac{\pi}{2}\right)\right] = \frac{1}{2}\left[\cos\frac{3\pi}{4} - \cos\!\left(-\frac{\pi}{4}\right)\right] = \frac{1}{2}\!\left[-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right] = -\frac{1}{\sqrt{2}}.

The two methods agree exactly, confirming the rule.

Try it

Use the parameter-shift rule to compute the gradient of $E(\theta) = \cos\theta$ at $\theta = \pi/4$ . Evaluate $E$ at the two shifted angles and apply the formula. The grader checks that your returned value equals $-1/\sqrt{2}$ to machine precision.

Run your code to see the quantum state.

Expectation values as the loss

The shift formula

Why it works

Applying it to Ry(θ)R_y(\theta)Ry​(θ) at θ=π/4\theta = \pi/4θ=π/4

Try it

Applying it to $R_y(\theta)$ at $\theta = \pi/4$