Lesson 43 · 100 min · advanced

Variational Quantum Circuits: Architecture

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U(x)

Prerequisites: Angle & Phase Encoding, Postulate 3: Projective Measurement & the Born Rule

Objectives and key results

$U(x)$
$f_\theta(x)=\mathrm{Tr}\big(\rho(x)\,W^\dagger(\theta)HW(\theta)\big)$
$f_\theta(x)\in[\lambda_{\min}(H),\lambda_{\max}(H)]$
$f=\cos(x+\theta)$
$(1-f^2)/N$
$M(2P+1)$
$\theta\in\big(-\tfrac\pi2-1.8,\,-\tfrac\pi2-0.9\big)\pmod{2\pi}$

definition · Variational quantum model: $f_\theta(x)=\langle0\vert U^\dagger(x)\,W^\dagger(\theta)\,H\,W(\theta)\,U(x)\vert0\rangle,\qquad\hat y(x)=\mathrm{sign}\,f_\theta(x)$
identity · A VQC is linear in the encoded density operator: $f_\theta(x)=\mathrm{Tr}\big(\rho(x)\,O_\theta\big)=\langle\rho(x),O_\theta\rangle_{HS},\qquad\rho(x)=U(x)\lvert0\rangle\langle0\rvert U^\dagger(x),\quad O_\theta=W^\dagger(\theta)HW(\theta)$
theorem · Output range: $\lambda_{\min}(H)\le f_\theta(x)\le\lambda_{\max}(H)\qquad\text{for all }x,\theta$
formula · One-qubit threshold model: $U(x)=R_Y(x),\ W(\theta)=R_Y(\theta),\ H=Z:\qquad f_\theta(x)=\cos(x+\theta);\qquad U(x)=R_X(x):\ f_\theta(x)=\cos x\cos\theta$
formula · Shot noise of a Pauli expectation value: $\mathrm{Var}[\hat f]=\frac{1-f^2}{N},\qquad N\ge\frac{2\ln(2/\delta)}{\varepsilon^2}\ \Rightarrow\ \Pr\big(\lvert\hat f-f\rvert\ge\varepsilon\big)\le\delta$
algorithm · Cost of one training epoch: $M\,(2P+1)\ \text{circuits}\times N_{\text{shots}}=O\big(M\,P\,N_{\text{shots}}\big)\ \text{shots per epoch}$

Feature map, ansatz and observable#

x\mapsto U(x)\lvert0^n\rangle

Definition

Feature map, ansatz and observable

n

Definition

Variational quantum model and classifier

H

Theorem

A variational model is linear in the encoded density operator

\rho(x)=U(x)\lvert0\rangle\langle0\rvert U^\dagger(x)

Proof (7 steps)

Step 1:
$f_\theta(x)=\langle0\vert U^\dagger W^\dagger HWU\vert0\rangle=\mathrm{Tr}\big(\langle0\rvert\,U^\dagger W^\dagger HWU\lvert0\rangle\big)$
A $1\times1$ matrix equals its trace. Abbreviate $U=U(x)$ and $W=W(\theta)$ .
Step 2:
$\mathrm{Tr}\big(\langle0\rvert\,\big(U^\dagger W^\dagger HWU\lvert0\rangle\big)\big)=\mathrm{Tr}\big(U^\dagger W^\dagger HWU\lvert0\rangle\langle0\rvert\big)$
Cyclicity of the trace $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$ for the row $A=\langle0\rvert\in\mathbb C^{1\times2^n}$ and the matrix-times-column $B\in\mathbb C^{2^n\times1}$ .
Step 3:
$=\mathrm{Tr}\big(W^\dagger HW\;U\lvert0\rangle\langle0\rvert U^\dagger\big)=\mathrm{Tr}\big(O_\theta\,\rho(x)\big)=\mathrm{Tr}\big(\rho(x)\,O_\theta\big)$
Cyclicity once more, moving the leftmost factor $U^\dagger$ to the right end; then the definitions of $O_\theta$ and $\rho(x)$ , and cyclicity for the last equality.
Step 4:
$O_\theta^\dagger=W^\dagger H^\dagger(W^\dagger)^\dagger=W^\dagger HW=O_\theta,\qquad O_\theta=W^\dagger\Big(\sum_k\lambda_k\lvert e_k\rangle\langle e_k\rvert\Big)W=\sum_k\lambda_k\,\lvert W^\dagger e_k\rangle\langle W^\dagger e_k\rvert$
$(ABC)^\dagger=C^\dagger B^\dagger A^\dagger$ and $H=H^\dagger$ . Inserting the spectral decomposition of $H$ shows that $O_\theta$ has the eigenvalues $\lambda_k$ with the orthonormal eigenvectors $W^\dagger\lvert e_k\rangle$ ( $W^\dagger$ is unitary).
Step 5:
$\lVert O_\theta\rVert_{HS}^2=\mathrm{Tr}\big(W^\dagger HWW^\dagger HW\big)=\mathrm{Tr}\big(W^\dagger H^2W\big)=\mathrm{Tr}\big(H^2WW^\dagger\big)=\lVert H\rVert_{HS}^2$
$WW^\dagger=I$ and cyclicity: the Hilbert–Schmidt norm is invariant under unitary conjugation.
Step 6:
$f_\theta(x)=\mathrm{Tr}\big(\rho(x)O_\theta\big)=\langle\rho(x),O_\theta\rangle_{HS},\qquad\mathrm{Tr}\big((a\rho+b\sigma)O_\theta\big)=a\,\mathrm{Tr}(\rho O_\theta)+b\,\mathrm{Tr}(\sigma O_\theta)$
For Hermitian matrices $\langle A,B\rangle_{HS}=\mathrm{Tr}(A^\dagger B)=\mathrm{Tr}(AB)$ . Linearity of the trace makes $\rho\mapsto\mathrm{Tr}(\rho O_\theta)$ a real-linear functional on $\mathrm{Herm}(2^n)$ : the model is a linear model whose feature vector is $\rho(x)$ and whose weight vector is $O_\theta$ .
Step 7:
$O_\theta=O_{\mathcal S}+O_\perp,\quad O_\perp\perp_{HS}\mathcal S=\mathrm{span}_{\mathbb R}\{\rho(x)\}\ \Rightarrow\ f_\theta=f_{O_{\mathcal S}}\in\mathcal H_K,\quad\lVert f_\theta\rVert_{\mathcal H_K}=\lVert O_{\mathcal S}\rVert_{HS}\le\lVert O_\theta\rVert_{HS}=\lVert H\rVert_{HS}$
The RKHS proposition of the kernel lesson: $\mathcal H_K=\{f_O:O\in\mathcal S\}$ with $\lVert f_O\rVert_{\mathcal H_K}=\lVert O\rVert_{HS}$ , and the component of an observable orthogonal to $\mathcal S$ does not change any expectation value $\mathrm{Tr}(O\rho(x))$ . The orthogonal projection does not increase the norm.

∎

\rho(x)

Derivation

Heisenberg picture for one qubit: the model as a dot product of Bloch vectors

Step 1:
$\rho(x)=\tfrac12\big(I+\vec r(x)\cdot\vec\sigma\big),\qquad O_\theta=W^\dagger ZW=\vec n(\theta)\cdot\vec\sigma$
Any one-qubit density matrix has a Bloch vector $\vec r(x)$ . $O_\theta$ is Hermitian and traceless with eigenvalues $\pm1$ (the spectrum of $Z$ ), so it is $\vec n\cdot\vec\sigma$ with a unit vector $\vec n(\theta)$ .
Step 2:
$f_\theta(x)=\mathrm{Tr}\big(\rho(x)\,O_\theta\big)=\tfrac12\mathrm{Tr}\big(\vec n\cdot\vec\sigma\big)+\tfrac12\sum_{a,b}r_an_b\,\mathrm{Tr}(\sigma_a\sigma_b)=\vec r(x)\cdot\vec n(\theta)$
The Pauli matrices are traceless and $\mathrm{Tr}(\sigma_a\sigma_b)=2\delta_{ab}$ : the one-qubit model is the Euclidean dot product of the Bloch vector of the encoded state with a measurement direction.
Step 3:
$R_Y(\theta)^\dagger ZR_Y(\theta)=\begin{pmatrix}c&s\\-s&c\end{pmatrix}\begin{pmatrix}c&-s\\-s&-c\end{pmatrix}=\begin{pmatrix}c^2-s^2&-2cs\\-2cs&s^2-c^2\end{pmatrix}=\cos\theta\,Z-\sin\theta\,X$
With $c=\cos\frac\theta2$ , $s=\sin\frac\theta2$ , $R_Y(\theta)=\begin{pmatrix}c&-s\\s&c\end{pmatrix}$ is real, so $R_Y^\dagger=R_Y^\top$ , and $ZR_Y(\theta)=\begin{pmatrix}c&-s\\-s&-c\end{pmatrix}$ . Double-angle formulas: $c^2-s^2=\cos\theta$ , $2cs=\sin\theta$ . Hence $\vec n(\theta)=(-\sin\theta,0,\cos\theta)$ .
Step 4:
$R_Y(x)\lvert0\rangle=\cos\tfrac x2\lvert0\rangle+\sin\tfrac x2\lvert1\rangle\ \Rightarrow\ \vec r(x)=\big(2\cos\tfrac x2\sin\tfrac x2,\ 0,\ \cos^2\tfrac x2-\sin^2\tfrac x2\big)=(\sin x,0,\cos x)$
Bloch vector $(2\,\mathrm{Re}\,\bar\alpha\beta,\,2\,\mathrm{Im}\,\bar\alpha\beta,\,\lvert\alpha\rvert^2-\lvert\beta\rvert^2)$ of a state with real amplitudes: the $R_Y$ encoding moves $\vec r$ on the great circle of the $xz$ -plane at polar angle $x$ .
Step 5:
$f_\theta(x)=(\sin x,0,\cos x)\cdot(-\sin\theta,0,\cos\theta)=\cos x\cos\theta-\sin x\sin\theta=\cos(x+\theta)$
Addition theorem of the cosine. The model of this lesson's worked example is obtained in the Heisenberg picture, where the ansatz acts on the observable instead of on the state; the Schrödinger-picture derivation follows in the section on the one-qubit classifier.
Step 6:
$\mathcal S=\mathrm{span}_{\mathbb R}\{\rho(x)\}=\mathrm{span}_{\mathbb R}\{I,X,Z\},\qquad\lVert f_\theta\rVert_{\mathcal H_K}=\lVert O_\theta\rVert_{HS}=\sqrt{\mathrm{Tr}\,Z^2}=\sqrt2$
$\rho(x)=\frac12(I+\sin x\,X+\cos x\,Z)$ ; the values $x=0,\pi,\frac\pi2$ give $\frac12(I\pm Z)$ and $\frac12(I+X)$ , which span $I,X,Z$ . Here $O_\theta\in\mathcal S$ , so the trained models form a circle of radius $\sqrt2$ in the three-dimensional RKHS spanned by the functions $1,\sin x,\cos x$ : the variational family is a one-parameter curve inside the linear model space.

Result

\boxed{f_\theta(x)=\vec r(x)\cdot\vec n(\theta)=(\sin x,0,\cos x)\cdot(-\sin\theta,0,\cos\theta)=\cos(x+\theta)}

Quick check

\theta

Output range, decision rule and losses#

Theorem

The model output lies between the extreme eigenvalues of the observable

H

Proof (6 steps)

Step 1:
$H=H^\dagger\ \Rightarrow\ H=\sum_{k}\lambda_k\lvert e_k\rangle\langle e_k\rvert,\quad\lambda_k\in\mathbb R,\quad\langle e_k\vert e_l\rangle=\delta_{kl}$
Spectral theorem for Hermitian matrices: real eigenvalues and an orthonormal eigenbasis.
Step 2:
$f_\theta(x)=\langle\psi\vert H\vert\psi\rangle=\sum_k\lambda_k\,\langle\psi\vert e_k\rangle\langle e_k\vert\psi\rangle=\sum_k\lambda_k\,p_k,\qquad p_k=\lvert\langle e_k\vert\psi\rangle\rvert^2$
Abbreviate $\psi=\psi_\theta(x)$ and insert the decomposition; $\langle\psi\vert e_k\rangle=\overline{\langle e_k\vert\psi\rangle}$ .
Step 3:
$p_k\ge0,\qquad\sum_kp_k=\sum_k\langle\psi\vert e_k\rangle\langle e_k\vert\psi\rangle=\langle\psi\vert\Big(\sum_k\lvert e_k\rangle\langle e_k\rvert\Big)\vert\psi\rangle=\langle\psi\vert\psi\rangle=1$
Completeness of the orthonormal basis and normalisation of $\psi$ ( $W$ and $U$ are unitary). By the Born rule $p_k$ is the probability of measuring the eigenvalue $\lambda_k$ .
Step 4:
$\lambda_{\min}=\sum_kp_k\lambda_{\min}\le\sum_kp_k\lambda_k\le\sum_kp_k\lambda_{\max}=\lambda_{\max}$
$f$ is a convex combination of the eigenvalues; replace every $\lambda_k$ by the smallest or the largest one, using $p_k\ge0$ and $\sum_kp_k=1$ .
Step 5:
$\lambda_{\max}-f=\sum_kp_k(\lambda_{\max}-\lambda_k)=0\iff p_k=0\ \text{whenever}\ \lambda_k<\lambda_{\max}$
A sum of nonnegative terms vanishes only if every term vanishes: equality holds exactly when $\psi$ has no component outside the top eigenspace (and likewise for $\lambda_{\min}$ ).
Step 6:
$P^2=I\ \Rightarrow\ \lambda\in\{-1,+1\};\qquad\Big\lvert\sum_kc_k\langle\psi\vert P_k\vert\psi\rangle\Big\rvert\le\sum_k\lvert c_k\rvert\,\big\lvert\langle\psi\vert P_k\vert\psi\rangle\big\rvert\le\sum_k\lvert c_k\rvert$
A Pauli string is Hermitian and squares to the identity, so its eigenvalues satisfy $\lambda^2=1$ ; it has both eigenvalues because $\mathrm{Tr}\,P=0$ for $P\ne I$ . For a sum, the triangle inequality and the Pauli bound give the stated interval (a bound, not always attained).

∎

Derivation

MSE and hinge losses and their gradients

Step 1:
$L(\theta)=\frac1M\sum_{m=1}^M\ell\big(f_\theta(x_m),y_m\big),\qquad\frac{\partial L}{\partial\theta_j}=\frac1M\sum_{m=1}^M\frac{\partial\ell}{\partial f}\big(f_\theta(x_m),y_m\big)\,\frac{\partial f_\theta(x_m)}{\partial\theta_j}$
Empirical risk on the training set $\{(x_m,y_m)\}_{m=1}^M$ with a per-sample loss $\ell(f,y)$ ; the chain rule separates a classical factor $\partial\ell/\partial f$ from a quantum factor $\partial f/\partial\theta_j$ , which the circuit must supply.
Step 2:
$\ell_{\text{MSE}}(f,y)=(f-y)^2,\qquad\frac{\partial\ell_{\text{MSE}}}{\partial f}=2(f-y)$
The squared error treats the labels $\pm1$ as regression targets. Because $f\in[-1,1]$ for a Pauli observable, the targets lie at the end points of the attainable range.
Step 3:
$\frac{\partial L_{\text{MSE}}}{\partial\theta_j}=\frac2M\sum_{m=1}^M\big(f_\theta(x_m)-y_m\big)\,\frac{\partial f_\theta(x_m)}{\partial\theta_j}$
Insert the derivative into the chain rule: each sample needs its output $f_\theta(x_m)$ (for the residual) and its gradient $\nabla_\theta f_\theta(x_m)$ .
Step 4:
$\ell_{\text{hinge}}(f,y)=\max(0,1-yf),\qquad\frac{\partial\ell_{\text{hinge}}}{\partial f}=\begin{cases}-y,&yf<1\\0,&yf>1\end{cases}$
The hinge loss of the support vector machine. At the kink $yf=1$ it is not differentiable; every number between $-y$ and $0$ is a subgradient, and the platform uses $0$ .
Step 5:
$\lvert f\rvert\le1\ \Rightarrow\ yf\le1\ \Rightarrow\ \ell_{\text{hinge}}(f,y)=1-yf,\qquad L_{\text{hinge}}(\theta)=1-\frac1M\sum_{m=1}^My_m\,f_\theta(x_m)$
For a Pauli observable the output never exceeds $1$ in modulus, so the margin $yf$ never exceeds $1$ and the hinge is always on its linear branch: the hinge loss becomes $1$ minus the average signed output, a linear function of the outputs.
Step 6:
$\frac{\partial L_{\text{hinge}}}{\partial\theta_j}=-\frac1M\sum_{m:\,y_mf_\theta(x_m)<1}y_m\,\frac{\partial f_\theta(x_m)}{\partial\theta_j}$
The chain rule with the hinge derivative; for a Pauli observable the sum runs over all samples except those with $y_mf_\theta(x_m)=1$ exactly.
Step 7:
$\mathrm{acc}(\theta)=\frac1M\#\big\{m:\ y_mf_\theta(x_m)>0\big\}$
The accuracy of the classifier $\hat y=\mathrm{sign}\,f$ is piecewise constant in $\theta$ , with zero gradient almost everywhere; it is used to evaluate a model, while the differentiable losses above are used to train it.

Result

\boxed{\nabla_\theta L_{\text{MSE}}=\frac2M\sum_m\big(f_\theta(x_m)-y_m\big)\nabla_\theta f_\theta(x_m),\qquad\nabla_\theta L_{\text{hinge}}=-\frac1M\sum_{m:\,y_mf_m<1}y_m\nabla_\theta f_\theta(x_m)}

Quick check

H=\tfrac12\,Z\otimes I+Z\otimes Z

A one-qubit classifier worked out exactly#

U(x)=R_Y(x)

Lemma

Rotations about the same axis add their angles

a,b

Proof (5 steps)

Step 1:
$AB=BA\ \Rightarrow\ (A+B)^k=\sum_{l=0}^k\binom klA^lB^{k-l}$
For commuting matrices the binomial theorem holds, by induction on $k$ : commutativity is exactly what is needed to collect the terms $A^lB^{k-l}$ .
Step 2:
$e^Ae^B=\sum_{l\ge0}\frac{A^l}{l!}\sum_{r\ge0}\frac{B^r}{r!}=\sum_{k\ge0}\frac1{k!}\sum_{l=0}^k\binom klA^lB^{k-l}=\sum_{k\ge0}\frac{(A+B)^k}{k!}=e^{A+B}$
Both exponential series converge absolutely, so their Cauchy product may be regrouped by total degree $k=l+r$ , using $\frac1{l!\,(k-l)!}=\frac1{k!}\binom kl$ .
Step 3:
$A=-\tfrac{ia}2P,\quad B=-\tfrac{ib}2P,\quad AB=-\tfrac{ab}4P^2=BA\ \Rightarrow\ R_P(a)R_P(b)=e^{-iaP/2}e^{-ibP/2}=e^{-i(a+b)P/2}=R_P(a+b)$
Two multiples of the same matrix commute, so the exponents add.
Step 4:
$R_Y(a)=\begin{pmatrix}\cos\frac a2&-\sin\frac a2\\\sin\frac a2&\cos\frac a2\end{pmatrix},\qquad R_Y(a)R_Y(b)=\begin{pmatrix}c_ac_b-s_as_b&-(c_as_b+s_ac_b)\\s_ac_b+c_as_b&c_ac_b-s_as_b\end{pmatrix}$
An independent check by matrix multiplication, with $c_a=\cos\frac a2$ , $s_a=\sin\frac a2$ and the same for $b$ ; the matrix form of $R_Y$ is $\cos\frac a2\,I-i\sin\frac a2\,Y$ with $-iY=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ .
Step 5:
$c_ac_b-s_as_b=\cos\tfrac{a+b}2,\quad s_ac_b+c_as_b=\sin\tfrac{a+b}2\ \Rightarrow\ R_Y(a)R_Y(b)=R_Y(a+b)$
Addition theorems of cosine and sine at the half angles. The product is again a rotation about $y$ , by the summed angle.

∎

Derivation

State and output of U(x) = R_Y(x), W(θ) = R_Y(θ), H = Z

Step 1:
$\lvert\psi_\theta(x)\rangle=R_Y(\theta)R_Y(x)\lvert0\rangle=R_Y(x+\theta)\lvert0\rangle$
The circuit applies $R_Y(x)$ first and $R_Y(\theta)$ second; the lemma combines them.
Step 2:
$R_Y(x+\theta)\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}\cos\frac{x+\theta}2&-\sin\frac{x+\theta}2\\\sin\frac{x+\theta}2&\cos\frac{x+\theta}2\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}\cos\frac{x+\theta}2\\\sin\frac{x+\theta}2\end{pmatrix}$
Matrix times the first basis vector is the first column.
Step 3:
$\lvert\psi_\theta(x)\rangle=\cos\tfrac{x+\theta}2\lvert0\rangle+\sin\tfrac{x+\theta}2\lvert1\rangle$
A real state at polar angle $x+\theta$ on the $xz$ -great circle of the Bloch sphere.
Step 4:
$f_\theta(x)=\langle\psi\vert Z\vert\psi\rangle=\begin{pmatrix}\cos\frac{x+\theta}2&\sin\frac{x+\theta}2\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}\cos\frac{x+\theta}2\\\sin\frac{x+\theta}2\end{pmatrix}=\cos^2\tfrac{x+\theta}2-\sin^2\tfrac{x+\theta}2$
The amplitudes are real, so the bra is the transposed column; $Z$ flips the sign of the second component.
Step 5:
$f_\theta(x)=\cos(x+\theta)$
Double-angle formula $\cos^2u-\sin^2u=\cos2u$ with $u=\frac{x+\theta}2$ . The value lies in $[-1,1]=[\lambda_{\min}(Z),\lambda_{\max}(Z)]$ , as the output-range theorem requires.
Step 6:
$x=0.3,\ \theta=\pi:\quad\lvert\psi\rangle=\cos\tfrac{0.3+\pi}2\lvert0\rangle+\sin\tfrac{0.3+\pi}2\lvert1\rangle=-\sin0.15\,\lvert0\rangle+\cos0.15\,\lvert1\rangle\approx-0.149438\lvert0\rangle+0.988771\lvert1\rangle$
$\cos(u+\frac\pi2)=-\sin u$ and $\sin(u+\frac\pi2)=\cos u$ . Then $f=\cos(0.3+\pi)=-\cos0.3\approx-0.955336$ , so $\hat y=-1$ , the label of $x=0.3$ in the lab. The circuit below shows this state.

Result

\boxed{f_\theta(x)=\langle0\vert R_Y(x)^\dagger R_Y(\theta)^\dagger ZR_Y(\theta)R_Y(x)\vert0\rangle=\cos(x+\theta)}

Circuit · 1 qubit

R_Y(x)

\lvert\psi\rangle = -0.1494\lvert 0\rangle + 0.9888\lvert 1\rangle

R_Y(\pi)R_Y(0.3)=R_Y(0.3+\pi)=\begin{pmatrix}\cos\frac{0.3+\pi}2&-\sin\frac{0.3+\pi}2\\\sin\frac{0.3+\pi}2&\cos\frac{0.3+\pi}2\end{pmatrix}=\begin{pmatrix}-\sin0.15&-\cos0.15\\\cos0.15&-\sin0.15\end{pmatrix}\approx\begin{pmatrix}-0.149438&-0.988771\\0.988771&-0.149438\end{pmatrix}

W(π)U(0.3) = R_Y(0.3 + π)2×2

Unitary verified: $U^\dagger U = I$ (max deviation 0.0e+0)

Derivation

Decision boundary and decision regions

Step 1:
$f_\theta(x)=0\iff\cos(x+\theta)=0\iff x+\theta=\pm\tfrac\pi2\pmod{2\pi}$
The zeros of the cosine are the odd multiples of $\frac\pi2$ ; modulo $2\pi$ these are $\frac\pi2$ and $-\frac\pi2$ .
Step 2:
$\hat y(x)=+1\iff x+\theta\in\big(-\tfrac\pi2,\tfrac\pi2\big)\pmod{2\pi},\qquad\hat y(x)=-1\iff x+\theta\in\big(\tfrac\pi2,\tfrac{3\pi}2\big)\pmod{2\pi}$
The cosine is positive on $(-\frac\pi2,\frac\pi2)$ and negative on $(\frac\pi2,\frac{3\pi}2)$ , each modulo $2\pi$ .
Step 3:
$x^\ast_\uparrow=-\tfrac\pi2-\theta,\qquad x^\ast_\downarrow=\tfrac\pi2-\theta\pmod{2\pi}$
On the input axis the classifier has a rising boundary $x^\ast_\uparrow$ , where the prediction changes from $-1$ to $+1$ as $x$ increases, and a falling boundary $x^\ast_\downarrow=x^\ast_\uparrow+\pi$ , where it changes back.
Step 4:
$\theta\mapsto\theta+\Delta\ \Rightarrow\ x^\ast\mapsto x^\ast-\Delta,\qquad\theta\mapsto\theta+\pi\ \Rightarrow\ f\mapsto-f$
The weight translates both boundaries rigidly; the two regions always have length $\pi$ . Shifting by $\pi$ exchanges the regions, since $\cos(u+\pi)=-\cos u$ : every prediction is negated.

Result

\boxed{\{x:\hat y(x)=+1\}=\big(-\tfrac\pi2-\theta,\ \tfrac\pi2-\theta\big)\pmod{2\pi}}

Bloch sphereBlochSphere

Shows a single-qubit state as the vector (sin θ cos φ, sin θ sin φ, cos θ) on the unit sphere; mixed states appear inside the ball with length |r| < 1.

3D view unavailable (WebGL is not supported in this browser). The Bloch coordinates of the state are listed in the readout.

θ (polar angle)1.800 (1.80)

φ (azimuth)0.000 (0)

\lvert\psi\rangle = \cos\tfrac{1.800}{2}\,\lvert 0\rangle + e^{i\,0.000}\sin\tfrac{1.800}{2}\,\lvert 1\rangle = 0.622\,\lvert 0\rangle + 0.783\,\lvert 1\rangle

Bloch vector r = (0.974, 0.000, -0.227), |r| = 1.000; P(0) = 0.386, P(1) = 0.614

\alpha=\cos\tfrac\theta2=0.6216,\qquad \beta=e^{i\phi}\sin\tfrac\theta2=0.7833

P(0)=\lvert\alpha\rvert^2=0.3864,\qquad P(1)=\lvert\beta\rvert^2=0.6136,\qquad P(0)+P(1)=1

R_Y(1.8)\lvert0\rangle

Variational circuit pipelineVQCPipeline

Three panes — feature map U(x), ansatz W(θ), measurement ⟨H⟩ — of a one-qubit variational classifier, with a 1-D dataset strip coloured by the decision regions sign f_θ(x) as θ changes.

Weight θθ = 0.000

Feature map

Loss

1. Feature map U(x)

U(x)|0\rangle = R_Y(x)|0\rangle = \cos\tfrac{x}{2}|0\rangle + \sin\tfrac{x}{2}|1\rangle

2. Ansatz W(θ)

W(\theta) = R_Y(\theta),\quad \theta = 0.000

3. Measurement ⟨Z⟩

f_\theta(x) = \langle Z\rangle = \cos(x + \theta)

Dataset and decision regions (shaded: f_θ(x) > 0, class +1)

● label +1 · ■ label −1 · ◯ misclassified. Dashed lines are the decision boundary f_θ(x) = 0: x + θ = ±π/2 (mod 2π).

Model output f_θ(x)

Loss L(θ) on [−π, π]

L(\theta) = \frac{1}{M}\sum_{m=1}^{M}\big(f_\theta(x_m) - y_m\big)^2

accuracy: 0/4 (0%)
MSE loss: 2.6206

Per-point outputs
x	y	f_θ(x)	ŷ = sign f
0.300	−1	0.9553	+1
0.900	−1	0.6216	+1
1.800	+1	-0.2272	−1
2.200	+1	-0.5885	−1

\theta=0

Derivation

Changing the encoding: U(x) = R_X(x) gives f = cos x cos θ

Step 1:
$R_X(x)\lvert0\rangle=\begin{pmatrix}\cos\frac x2&-i\sin\frac x2\\-i\sin\frac x2&\cos\frac x2\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}c_x\\-is_x\end{pmatrix},\qquad c_x=\cos\tfrac x2,\ s_x=\sin\tfrac x2$
$R_X(x)=\cos\frac x2\,I-i\sin\frac x2\,X$ ; its first column is the encoded state.
Step 2:
$R_Y(\theta)R_X(x)\lvert0\rangle=\begin{pmatrix}c&-s\\s&c\end{pmatrix}\begin{pmatrix}c_x\\-is_x\end{pmatrix}=\begin{pmatrix}c\,c_x+i\,s\,s_x\\s\,c_x-i\,c\,s_x\end{pmatrix},\qquad c=\cos\tfrac\theta2,\ s=\sin\tfrac\theta2$
Matrix–vector product with the real matrix $R_Y(\theta)$ . The two rotations are about different axes, so the lemma does not apply and the angles do not add.
Step 3:
$\lvert c\,c_x+i\,s\,s_x\rvert^2=c^2c_x^2+s^2s_x^2,\qquad\lvert s\,c_x-i\,c\,s_x\rvert^2=s^2c_x^2+c^2s_x^2$
$\lvert a+ib\rvert^2=a^2+b^2$ for real $a,b$ ; the cross terms cancel because one part is real and the other imaginary.
Step 4:
$f_\theta(x)=c^2c_x^2+s^2s_x^2-s^2c_x^2-c^2s_x^2=(c^2-s^2)(c_x^2-s_x^2)=\cos\theta\,\cos x$
$\langle Z\rangle=\lvert a_0\rvert^2-\lvert a_1\rvert^2$ ; factor and use the double-angle formula twice.
Step 5:
$\cos(x+\theta)-\cos x\cos\theta=-\sin x\sin\theta\ \not\equiv0,\qquad x=\theta=\tfrac\pi2:\ \cos\pi-0=-1$
The $R_X$ model is not $\cos(x+\theta)=\cos x\cos\theta-\sin x\sin\theta$ : the term $-\sin x\sin\theta$ is missing, and the two models differ by up to $1$ (the platform's simulator confirms the maximum difference $1$ on a grid). They agree only where $\sin x\sin\theta=0$ .
Step 6:
$\vec r_X(x)=(0,-\sin x,\cos x),\quad\vec n(\theta)=(-\sin\theta,0,\cos\theta)\ \Rightarrow\ f_\theta(x)=\vec r_X(x)\cdot\vec n(\theta)=\cos x\cos\theta$
The Heisenberg picture explains why: $R_X$ places the data on the $yz$ -great circle, the ansatz $R_Y$ only tilts the measurement direction within the $xz$ -plane, and the $x$ -component of the encoded Bloch vector, which would carry $\sin x$ , is identically $0$ .
Step 7:
$\cos\theta\ne0:\quad f_\theta(x)=0\iff\cos x=0\iff x=\tfrac\pi2\pmod\pi$
The weight only rescales the output by $\cos\theta$ and possibly flips its sign; the decision boundary is fixed at $x=\frac\pi2$ for every $\theta$ (for $\cos\theta=0$ the model is identically $0$ ).

Result

\boxed{U(x)=R_X(x),\ W(\theta)=R_Y(\theta),\ H=Z:\qquad f_\theta(x)=\cos x\cos\theta\ne\cos(x+\theta)}

Variational circuit pipelineVQCPipeline

Three panes — feature map U(x), ansatz W(θ), measurement ⟨H⟩ — of a one-qubit variational classifier, with a 1-D dataset strip coloured by the decision regions sign f_θ(x) as θ changes.

Weight θθ = 3.142

Feature map

Loss

1. Feature map U(x)

U(x)|0\rangle = R_X(x)|0\rangle = \cos\tfrac{x}{2}|0\rangle - i\sin\tfrac{x}{2}|1\rangle

2. Ansatz W(θ)

W(\theta) = R_Y(\theta),\quad \theta = 3.142

3. Measurement ⟨Z⟩

f_\theta(x) = \langle Z\rangle = \cos x\,\cos\theta

Dataset and decision regions (shaded: f_θ(x) > 0, class +1)

● label +1 · ■ label −1 · ◯ misclassified. Dashed lines are the decision boundary f_θ(x) = 0: x = π/2 and 3π/2 for every θ — R_X encoding cannot move it.

Model output f_θ(x)

Loss L(θ) on [−π, π]

L(\theta) = \frac{1}{M}\sum_{m=1}^{M}\big(f_\theta(x_m) - y_m\big)^2

accuracy: 4/4 (100%)
MSE loss: 0.2279

Per-point outputs
x	y	f_θ(x)	ŷ = sign f
0.300	−1	-0.9553	−1
0.900	−1	-0.6216	−1
1.800	+1	0.2272	+1
2.200	+1	0.5885	+1

R_X

Quick check

U(x)=R_Y(x)

Quick check

U(x)=R_X(x)

Estimating the output with finitely many shots#

Theorem

Shot noise of a Pauli expectation value

H=P

Proof (7 steps)

Step 1:
$P^2=I\ \Rightarrow\ P=\Pi_+-\Pi_-,\qquad\Pi_\pm=\tfrac12(I\pm P),\qquad\Pi_\pm^2=\Pi_\pm,\ \Pi_++\Pi_-=I$
The eigenprojectors of an operator with eigenvalues $\pm1$ : $\Pi_\pm^2=\frac14(I\pm2P+P^2)=\frac12(I\pm P)$ , and they sum to the identity.
Step 2:
$\Pr(z_s=\pm1)=\mathrm{Tr}\big(\Pi_\pm\,\rho_\theta(x)\big)=\tfrac12\big(1\pm\mathrm{Tr}(P\rho_\theta(x))\big)=\frac{1\pm f}2$
Born rule for projective measurements with $\rho_\theta(x)=\lvert\psi_\theta(x)\rangle\langle\psi_\theta(x)\rvert$ , $\mathrm{Tr}\,\rho=1$ and $f=\mathrm{Tr}(P\rho)$ .
Step 3:
$\mathbb E[z_s]=\frac{1+f}2-\frac{1-f}2=f,\qquad\mathbb E[z_s^2]=1,\qquad\mathrm{Var}[z_s]=1-f^2$
The outcome is $\pm1$ , so its square is always $1$ .
Step 4:
$\mathbb E[\hat f]=\frac1N\sum_s\mathbb E[z_s]=f,\qquad\mathrm{Var}[\hat f]=\frac1{N^2}\sum_s\mathrm{Var}[z_s]=\frac{1-f^2}N$
Each shot re-prepares the state, so the outcomes are independent and identically distributed: linearity of expectation, and variances of independent variables add.
Step 5:
$\Pr\big(\lvert\hat f-f\rvert\ge\varepsilon\big)\le2\exp\Big(-\frac{2N^2\varepsilon^2}{\sum_{s=1}^N(1-(-1))^2}\Big)=2e^{-N\varepsilon^2/2}$
Hoeffding's inequality for the mean of $N$ independent variables with values in $[a,b]=[-1,1]$ : the exponent is $-2N^2\varepsilon^2/\big(N(b-a)^2\big)=-N\varepsilon^2/2$ .
Step 6:
$2e^{-N\varepsilon^2/2}\le\delta\iff N\ge\frac{2\ln(2/\delta)}{\varepsilon^2}$
Solve for $N$ by taking logarithms. The number of shots grows like $\varepsilon^{-2}$ and only logarithmically in the confidence.
Step 7:
$f>0:\quad\Pr(\hat f\le0)=\Pr\big(\hat f-f\le-f\big)\le e^{-Nf^2/2}$
The one-sided Hoeffding bound $\Pr(\hat f-f\le-t)\le e^{-Nt^2/2}$ with $t=f$ ; the case $f<0$ is symmetric. A point far from the decision boundary (large $\lvert f\rvert$ ) is classified correctly with few shots; a point near it needs about $f^{-2}$ shots.

∎

Derivation

Shot budget for the lab point x = 1.8 at θ = −2.9

Step 1:
$f=\cos(1.8-2.9)=\cos1.1\approx0.453596,\qquad \Pr(z_s=+1)=\frac{1+f}2\approx0.726798$
Exact output of the worked model and the Born probability of the outcome $+1$ .
Step 2:
$\mathrm{Var}[\hat f]=\frac{1-f^2}N\approx\frac{0.794251}{100}\approx0.00794,\qquad\sqrt{\mathrm{Var}[\hat f]}\approx0.0891\quad(N=100)$
With $100$ shots the estimate scatters by about $0.09$ around $0.4536$ . The exact binomial computation in the NumPy cell reproduces $N\,\mathrm{Var}[\hat f]=0.794251$ .
Step 3:
$\Pr(\hat f\le0)\le e^{-100\cdot0.205750/2}=e^{-10.2875}\approx3.41\times10^{-5}$
Part (d) of the theorem. The exact binomial probability is $1.24\times10^{-6}$ : Hoeffding's bound is valid but loose.
Step 4:
$\varepsilon=0.01,\ \delta=0.05:\quad N\ge\frac{2\ln40}{10^{-4}}=73777.6\ \Rightarrow\ N=73778$
Estimating the output itself to two decimals with $95\%$ confidence costs tens of thousands of shots, far more than deciding its sign.

Result

\boxed{N=100:\ \sqrt{\mathrm{Var}[\hat f]}\approx0.089,\ P(\text{sign error})\le3.4\times10^{-5};\qquad\varepsilon=0.01,\ \delta=0.05:\ N=73778}

On hardware

H=\sum_kc_kP_k

Quick check

\varepsilon=0.01

The hybrid quantum–classical training loop#

\theta

Derivation

The parameter-shift rule on the worked model

Step 1:
$\frac{\partial}{\partial\theta}\cos(x+\theta)=-\sin(x+\theta)$
The exact derivative of the closed-form output.
Step 2:
$\tfrac12\big[\cos\big(x+\theta+\tfrac\pi2\big)-\cos\big(x+\theta-\tfrac\pi2\big)\big]=\tfrac12\big[-\sin(x+\theta)-\sin(x+\theta)\big]=-\sin(x+\theta)$
$\cos(u\pm\frac\pi2)=\mp\sin u$ . The two shifted circuits reproduce the derivative exactly, not approximately: there is no step size and no truncation error, unlike a finite difference.
Step 3:
$\frac{\partial L_{\text{MSE}}}{\partial\theta}=-\frac2M\sum_{m=1}^M\big(\cos(x_m+\theta)-y_m\big)\sin(x_m+\theta)$
Insert into the MSE gradient of the previous section; at $\theta=0$ on the lab data this is $\approx0.316>0$ , so gradient descent decreases $\theta$ , towards the feasible interval.

Result

\boxed{\partial_\theta f_\theta(x)=\tfrac12\big[f_{\theta+\pi/2}(x)-f_{\theta-\pi/2}(x)\big]=-\sin(x+\theta)}

Proposition

Cost of one epoch of gradient-based training

M

Proof (5 steps)

Step 1:
$\nabla_\theta L=\frac1M\sum_{m=1}^M\ell'\big(f_\theta(x_m),y_m\big)\,\nabla_\theta f_\theta(x_m)$
The chain-rule form of the loss gradient: each sample contributes separately.
Step 2:
$\ell'_{\text{MSE}}=2\big(f_\theta(x_m)-y_m\big):\quad1\ \text{circuit per sample for }f_\theta(x_m)$
The classical factor needs the unshifted output (for the hinge loss it needs the sign of $1-y_mf$ , which also requires $f$ ).
Step 3:
$\frac{\partial f_\theta(x_m)}{\partial\theta_j}=\tfrac12\big[f_{\theta+\frac\pi2e_j}(x_m)-f_{\theta-\frac\pi2e_j}(x_m)\big]:\quad2\ \text{circuits per weight and sample}$
Parameter-shift rule with the shift applied only to weight $j$ ( $e_j$ is the $j$ -th unit vector); for $P$ weights $2P$ circuits.
Step 4:
$\#\text{circuits}=M\,(1+2P),\qquad\#\text{shots}=M\,(2P+1)\,N_{\text{shots}}=O\big(M\,P\,N_{\text{shots}}\big)$
Sum over the $M$ samples and multiply by the shots per circuit. For the lab model ( $M=4$ , $P=1$ ) this is $12$ circuits per epoch.
Step 5:
$\hat g_{\text{SPSA}}=\frac{L(\theta+c\Delta)-L(\theta-c\Delta)}{2c}\,\Delta^{-1},\qquad\Delta\in\{\pm1\}^P:\quad2M\ \text{circuits}$
Simultaneous-perturbation stochastic approximation perturbs all weights at once along a random sign vector, so a step costs two loss evaluations, i.e. $2M$ circuits, at the price of a noisy, only asymptotically unbiased gradient estimate.

∎

\big[f(\theta_j+h)-f(\theta_j-h)\big]/(2h)

\cos(x+\theta)

TypeScriptrunnable

import { singleQubitThresholdModel, trainVQC, vqcAccuracy, vqcGradient, vqcPredict } from '@/lib/qml';

// The model of this lesson: U(x) = RY(x), W(theta) = RY(theta), H = Z, so f(x) = cos(x + theta).
const model = singleQubitThresholdModel();
const X = [[0.3], [0.9], [1.8], [2.2]];
const y = [-1, -1, 1, 1];
const M = X.length;

// (1) Quantum evaluation: f and its parameter-shift derivative equal cos(x + theta) and -sin(x + theta).
for (const theta of [0, -2.9, Math.PI]) {
  const f = X.map((x) => vqcPredict(model, x, { theta }));
  const df = X.map((x) => vqcGradient(model, x, { theta }).theta);
  const exact = X.every((x, m) => Math.abs(f[m] - Math.cos(x[0] + theta)) < 1e-12 && Math.abs(df[m] + Math.sin(x[0] + theta)) < 1e-12);
  console.log('theta = ' + theta.toFixed(4) + '  f = [' + f.map((v) => v.toFixed(4)).join(', ') + ']  accuracy = ' + vqcAccuracy(model, X, y, { theta }) + '  closed form: ' + exact);
}

// (2) The hybrid loop written out: circuit evaluations inside, a classical gradient step outside.
const lr = 0.3;
let theta = 0;
for (let epoch = 0; epoch < 30; epoch++) {
  let loss = 0;
  let grad = 0;
  for (let m = 0; m < M; m++) {
    const f = vqcPredict(model, X[m], { theta }); // 1 circuit (N_shots shots on hardware)
    const df = vqcGradient(model, X[m], { theta }).theta; // 2P = 2 shifted circuits
    loss += (f - y[m]) ** 2 / M;
    grad += (2 / M) * (f - y[m]) * df; // dL/df times df/dtheta
  }
  if (epoch % 5 === 0) console.log('epoch ' + epoch + ': theta = ' + theta.toFixed(4) + '  MSE = ' + loss.toFixed(4) + '  accuracy = ' + vqcAccuracy(model, X, y, { theta }));
  theta -= lr * grad; // classical optimiser step
}

// (3) The library routine trainVQC runs the same loop.
const run = trainVQC(model, X, y, { init: { theta: 0 }, epochs: 30, optimizer: 'gd', lr, loss: 'mse' });
console.log('hand-written loop: theta = ' + theta.toFixed(6) + ', trainVQC: theta = ' + run.weights.theta.toFixed(6) + ', final accuracy = ' + run.history[30].accuracy);
console.log('circuit evaluations per epoch: M (2P + 1) = ' + M * (2 * model.weightParams.length + 1));

Expected output

theta = 0.0000  f = [0.9553, 0.6216, -0.2272, -0.5885]  accuracy = 0  closed form: true
theta = -2.9000  f = [-0.8569, -0.4161, 0.4536, 0.7648]  accuracy = 1  closed form: true
theta = 3.1416  f = [-0.9553, -0.6216, 0.2272, 0.5885]  accuracy = 1  closed form: true
epoch 0: theta = 0.0000  MSE = 2.6206  accuracy = 0
epoch 5: theta = -0.7866  MSE = 2.1141  accuracy = 0.5
epoch 10: theta = -2.4224  MSE = 0.3403  accuracy = 0.75
epoch 15: theta = -2.8662  MSE = 0.1791  accuracy = 1
epoch 20: theta = -2.8881  MSE = 0.1787  accuracy = 1
epoch 25: theta = -2.8890  MSE = 0.1787  accuracy = 1
hand-written loop: theta = -2.889079, trainVQC: theta = -2.889079, final accuracy = 1
circuit evaluations per epoch: M (2P + 1) = 12

Quick check

P=6

The lab data set: the exact feasible interval#

x\in\{0.3,0.9\}

Proposition

Feasible weights for the lab data

f_\theta(x)=\cos(x+\theta)

Proof (7 steps)

Step 1:
$\mathrm{acc}(\theta)=1\iff\cos(0.3+\theta)<0,\ \cos(0.9+\theta)<0,\ \cos(1.8+\theta)>0,\ \cos(2.2+\theta)>0$
Every prediction $\mathrm{sign}\cos(x_m+\theta)$ must equal its label, and an output $0$ counts as wrong, so the inequalities are strict.
Step 2:
$\cos(1.8+\theta)>0,\ \cos(2.2+\theta)>0\iff\theta\in\big(-\tfrac\pi2-1.8,\ \tfrac\pi2-2.2\big)+2\pi k$
$1.8+\theta$ and $2.2+\theta$ must lie in a window $(-\frac\pi2,\frac\pi2)+2\pi k$ ; both lie in the same window, because they are only $0.4<\pi$ apart and two consecutive positive windows are separated by a gap of length $\pi$ . So $1.8+\theta>-\frac\pi2+2\pi k$ and $2.2+\theta<\frac\pi2+2\pi k$ .
Step 3:
$\cos(0.3+\theta)<0,\ \cos(0.9+\theta)<0\iff\theta\in\big(\tfrac\pi2-0.3,\ \tfrac{3\pi}2-0.9\big)+2\pi k'$
The same argument with the negative windows $(\frac\pi2,\frac{3\pi}2)+2\pi k'$ for the two points $0.6$ apart.
Step 4:
$k'=k-1:\quad\big(\tfrac\pi2-0.3-2\pi,\ \tfrac{3\pi}2-0.9-2\pi\big)=\big(-\tfrac{3\pi}2-0.3,\ -\tfrac\pi2-0.9\big)\approx(-5.0124,\ -2.4708)$
Among the translates of the second interval only $k'=k-1$ meets the first one, $(-\frac\pi2-1.8,\frac\pi2-2.2)\approx(-3.3708,-0.6292)$ for $k=0$ : the translate $k'=k$ starts at $1.2708>-0.6292$ , and $k'=k-2$ ends at $-8.7540<-3.3708$ .
Step 5:
$\Big(-\tfrac\pi2-1.8,\ \tfrac\pi2-2.2\Big)\cap\Big(-\tfrac{3\pi}2-0.3,\ -\tfrac\pi2-0.9\Big)=\Big(-\tfrac\pi2-1.8,\ -\tfrac\pi2-0.9\Big)\approx(-3.370796,\ -2.470796)$
The larger lower end point is $-\frac\pi2-1.8$ (since $-\frac{3\pi}2-0.3<-\frac\pi2-1.8$ ) and the smaller upper end point is $-\frac\pi2-0.9$ (since $-\frac\pi2-0.9<\frac\pi2-2.2$ ). Length $0.9$ . Equivalently, the rising boundary $x^\ast_\uparrow=-\frac\pi2-\theta$ must lie in the gap $(0.9,1.8)$ between the classes.
Step 6:
$+2\pi:\ \Big(\tfrac{3\pi}2-1.8,\ \tfrac{3\pi}2-0.9\Big)\approx(2.912389,\ 3.812389),\qquad\pi\approx3.141593\in(2.912389,\ 3.812389)$
Adding $2\pi$ gives the representative in $[0,2\pi)$ . It contains $\pi$ , so in $(-\pi,\pi]$ the set splits into $(2.912389,\pi]$ and $(-\pi,3.812389-2\pi)=(-\pi,-2.470796)$ .
Step 7:
$\mathrm{acc}(\theta+\pi)=1-\mathrm{acc}(\theta),\qquad\mathrm{acc}(\theta)=0\iff\theta\in\Big(\tfrac\pi2-1.8,\ \tfrac\pi2-0.9\Big)\approx(-0.2292,\ 0.6708)\ni0$
$\cos(x+\theta+\pi)=-\cos(x+\theta)$ negates every output, so a correct prediction becomes wrong and vice versa (no output is $0$ unless one already was). The all-wrong set is therefore the feasible set shifted by $\pi$ .

∎

Derivation

Checking particular weights

Step 1:
$\theta=0:\quad f=(\cos0.3,\cos0.9,\cos1.8,\cos2.2)\approx(0.9553,\ 0.6216,\ -0.2272,\ -0.5885),\quad\mathrm{acc}=0$
Every sign is opposite to the label $(-1,-1,+1,+1)$ : the initial circuit of the lab misclassifies all points, which triggers the lab's diagnostic.
Step 2:
$\theta=-2.9\in(-3.3708,-2.4708):\quad f\approx(-0.8569,\ -0.4161,\ 0.4536,\ 0.7648),\quad\mathrm{acc}=1$
The lab's reference solution; the smallest margin is $\lvert f(0.9)\rvert\approx0.416$ .
Step 3:
$\theta=\pi:\quad f=-\cos x\approx(-0.9553,\ -0.6216,\ 0.2272,\ 0.5885),\quad\mathrm{acc}=1$
The flipped initial model: shifting $\theta=0$ by $\pi$ negates every output.
Step 4:
$\theta=2.5\notin(2.9124,3.8124):\quad f\approx(-0.9422,\ -0.9668,\ -0.4008,\ -0.0124),\quad\mathrm{acc}=\tfrac12$
The rising boundary $x^\ast_\uparrow=\frac{3\pi}2-2.5\approx2.212$ lies to the right of both positive points, so both are predicted $-1$ ; note how close $f(2.2)$ is to $0$ .
Step 5:
$\theta=2.8:\quad f\approx(-0.9991,\ -0.8481,\ -0.1122,\ 0.2837),\quad\mathrm{acc}=\tfrac34$
Now $x^\ast_\uparrow\approx1.912\in(1.8,2.2)$ : the boundary separates the two positive points, and only $x=1.8$ is misclassified.
Step 6:
$\theta_{\text{mid}}=-\tfrac\pi2-1.35\approx-2.920796,\qquad\min_m\lvert f_{\theta_{\text{mid}}}(x_m)\rvert=\sin0.45\approx0.434966$
The midpoint of the interval puts the boundary at $x^\ast_\uparrow=1.35$ , halfway between $0.9$ and $1.8$ , which maximises the smallest margin: the two nearest points have $\lvert f\rvert=\lvert\cos(\mp0.45-\frac\pi2)\rvert=\sin0.45$ .

Result

\boxed{\mathrm{acc}(\theta)=1\iff\theta\in\big(\tfrac{3\pi}2-1.8,\ \tfrac{3\pi}2-0.9\big)=(2.912389,\ 3.812389)\pmod{2\pi}}

Derivation

Closed-form minimiser of the hinge loss

Step 1:
$L_{\text{hinge}}(\theta)=1-\frac14\sum_{m=1}^4y_m\cos(x_m+\theta)=1-\frac14\,\mathrm{Re}\Big(e^{i\theta}\sum_my_me^{ix_m}\Big)$
The hinge loss is linear in the outputs for a Pauli observable (loss section), and $\cos(x+\theta)=\mathrm{Re}\,e^{i(x+\theta)}$ .
Step 2:
$z=\sum_my_me^{ix_m}=-e^{0.3i}-e^{0.9i}+e^{1.8i}+e^{2.2i}\approx-2.392650+0.703497\,i=Re^{i\varphi}$
Sum of the four unit phasors weighted by the labels.
Step 3:
$R=\lvert z\rvert\approx2.493929,\qquad\varphi=\arg z\approx2.855627$
Polar form; $z$ lies in the second quadrant.
Step 4:
$L_{\text{hinge}}(\theta)=1-\frac R4\cos(\theta+\varphi)\ \ge\ 1-\frac R4\approx0.376518,\qquad\theta^\ast=-\varphi\approx-2.855627$
$\mathrm{Re}(e^{i\theta}Re^{i\varphi})=R\cos(\theta+\varphi)$ is maximal at $\theta=-\varphi$ . The hinge minimiser lies inside the feasible interval $(-3.3708,-2.4708)$ , so minimising the hinge loss solves the lab.
Step 5:
$\theta^\ast_{\text{MSE}}\approx-2.889081,\qquad L_{\text{MSE}}(\theta^\ast_{\text{MSE}})\approx0.178713$
The MSE loss contains the extra $\theta$ -dependent term $\frac14\sum_m\cos^2(x_m+\theta)$ and has no equally simple closed form; its minimiser, found numerically and by the training loop of the code cell, is also feasible. The three natural choices (hinge, MSE, maximal margin) differ but all classify the data correctly.

Result

\boxed{\theta^\ast_{\text{hinge}}=-\arg\Big(\sum_my_me^{ix_m}\Big)\approx-2.855627\in\big(-\tfrac\pi2-1.8,\,-\tfrac\pi2-0.9\big)}

Variational circuit pipelineVQCPipeline

Three panes — feature map U(x), ansatz W(θ), measurement ⟨H⟩ — of a one-qubit variational classifier, with a 1-D dataset strip coloured by the decision regions sign f_θ(x) as θ changes.

Weight θθ = 3.142

Feature map

Loss

1. Feature map U(x)

U(x)|0\rangle = R_Y(x)|0\rangle = \cos\tfrac{x}{2}|0\rangle + \sin\tfrac{x}{2}|1\rangle

2. Ansatz W(θ)

W(\theta) = R_Y(\theta),\quad \theta = 3.142

3. Measurement ⟨Z⟩

f_\theta(x) = \langle Z\rangle = \cos(x + \theta)

Dataset and decision regions (shaded: f_θ(x) > 0, class +1)

● label +1 · ■ label −1 · ◯ misclassified. Dashed lines are the decision boundary f_θ(x) = 0: x + θ = ±π/2 (mod 2π).

Model output f_θ(x)

Loss L(θ) on [−π, π]

L(\theta) = \frac{1}{M}\sum_{m=1}^{M}\big(f_\theta(x_m) - y_m\big)^2

accuracy: 4/4 (100%)
MSE loss: 0.2279

Per-point outputs
x	y	f_θ(x)	ŷ = sign f
0.300	−1	-0.9553	−1
0.900	−1	-0.6216	−1
1.800	+1	0.2272	+1
2.200	+1	0.5885	+1

\theta=\pi

Pythonrunnable

import numpy as np
from math import comb, ceil, exp, log

def ry(t):
    c, s = np.cos(t / 2), np.sin(t / 2)
    return np.array([[c, -s], [s, c]], dtype=complex)

def rx(t):
    c, s = np.cos(t / 2), np.sin(t / 2)
    return np.array([[c, -1j * s], [-1j * s, c]])

Z = np.diag([1.0, -1.0]).astype(complex)
Xp = np.array([[0, 1], [1, 0]], dtype=complex)
ket0 = np.array([1, 0], dtype=complex)

def f(theta, x, enc=ry):
    psi = ry(theta) @ enc(x) @ ket0
    return float(np.vdot(psi, Z @ psi).real)

# (1) R_Y(t)R_Y(u) = R_Y(t + u), f = cos(x + theta) for R_Y encoding, f = cos(x)cos(theta) for R_X encoding.
grid = np.linspace(-np.pi, np.pi, 41)
add_ok = all(np.allclose(ry(t) @ ry(u), ry(t + u), atol=1e-12) for t in grid for u in grid)
ry_ok = all(abs(f(t, u) - np.cos(u + t)) < 1e-12 for t in grid for u in grid)
rx_ok = all(abs(f(t, u, rx) - np.cos(u) * np.cos(t)) < 1e-12 for t in grid for u in grid)
rx_gap = max(abs(f(t, u, rx) - np.cos(u + t)) for t in grid for u in grid)
print("RY(t) RY(u) = RY(t+u):", add_ok, "| RY encoding f = cos(x+theta):", ry_ok, "| RX encoding f = cos(x)cos(theta):", rx_ok)
print(f"RX encoding: max |f - cos(x+theta)| on the grid = {rx_gap:.6f}")

# (2) Heisenberg picture: W^dagger Z W = cos(theta) Z - sin(theta) X, so f = r(x) . n(theta).
heis_ok = all(np.allclose(ry(t).conj().T @ Z @ ry(t), np.cos(t) * Z - np.sin(t) * Xp, atol=1e-12) for t in grid)
print("W^dagger Z W = cos(theta) Z - sin(theta) X:", heis_ok)

# (3) The lab data: accuracy over a fine theta grid and the exact feasible interval.
xs = np.array([0.3, 0.9, 1.8, 2.2])
ys = np.array([-1, -1, 1, 1])
thetas = np.round(np.arange(-3.5, -2.0, 1e-4), 4)
acc = np.array([np.mean(np.cos(xs + t) * ys > 0) for t in thetas])
ok = thetas[acc == 1.0]
print(f"grid points with accuracy 1: from {ok.min():.4f} to {ok.max():.4f}")
print(f"exact interval: (-pi/2 - 1.8, -pi/2 - 0.9) = ({-np.pi / 2 - 1.8:.6f}, {-np.pi / 2 - 0.9:.6f})")
print(f"plus 2 pi: (3 pi/2 - 1.8, 3 pi/2 - 0.9) = ({1.5 * np.pi - 1.8:.6f}, {1.5 * np.pi - 0.9:.6f})")
for t in (0.0, 2.5, 2.8, -2.9, np.pi):
    print(f"theta = {t:7.4f}: f = {np.array2string(np.cos(xs + t), precision=4, floatmode='fixed')}, accuracy = {np.mean(np.cos(xs + t) * ys > 0):.2f}")

# (4) Shot noise of f_hat = (N_plus - N_minus)/N at x = 1.8, theta = -2.9 with N = 100 shots (exact binomial sums).
N, fx = 100, float(np.cos(1.8 - 2.9))
p = (1 + fx) / 2
pmf = [comb(N, k) * p ** k * (1 - p) ** (N - k) for k in range(N + 1)]
vals = [2 * k / N - 1 for k in range(N + 1)]
mean = sum(w * v for w, v in zip(pmf, vals))
var = sum(w * (v - mean) ** 2 for w, v in zip(pmf, vals))
p_wrong = sum(w for w, v in zip(pmf, vals) if v <= 0)
print(f"f = {fx:.6f}: E[f_hat] = {mean:.6f}, N Var[f_hat] = {N * var:.6f}, 1 - f^2 = {1 - fx ** 2:.6f}")
print(f"P(sign error) = {p_wrong:.3e} <= Hoeffding bound exp(-N f^2 / 2) = {exp(-N * fx ** 2 / 2):.3e}")
print(f"Hoeffding shots for eps = 0.01, delta = 0.05: N = ceil(2 ln(2/delta) / eps^2) = {ceil(2 * log(2 / 0.05) / 0.01 ** 2)}")

Expected output

RY(t) RY(u) = RY(t+u): True | RY encoding f = cos(x+theta): True | RX encoding f = cos(x)cos(theta): True
RX encoding: max |f - cos(x+theta)| on the grid = 1.000000
W^dagger Z W = cos(theta) Z - sin(theta) X: True
grid points with accuracy 1: from -3.3707 to -2.4708
exact interval: (-pi/2 - 1.8, -pi/2 - 0.9) = (-3.370796, -2.470796)
plus 2 pi: (3 pi/2 - 1.8, 3 pi/2 - 0.9) = (2.912389, 3.812389)
theta =  0.0000: f = [ 0.9553  0.6216 -0.2272 -0.5885], accuracy = 0.00
theta =  2.5000: f = [-0.9422 -0.9668 -0.4008 -0.0124], accuracy = 0.50
theta =  2.8000: f = [-0.9991 -0.8481 -0.1122  0.2837], accuracy = 0.75
theta = -2.9000: f = [-0.8569 -0.4161  0.4536  0.7648], accuracy = 1.00
theta =  3.1416: f = [-0.9553 -0.6216  0.2272  0.5885], accuracy = 1.00
f = 0.453596: E[f_hat] = 0.453596, N Var[f_hat] = 0.794251, 1 - f^2 = 0.794251
P(sign error) = 1.239e-06 <= Hoeffding bound exp(-N f^2 / 2) = 3.406e-05
Hoeffding shots for eps = 0.01, delta = 0.05: N = ceil(2 ln(2/delta) / eps^2) = 73778

Quick check

\theta

References

K. Mitarai, M. Negoro, M. Kitagawa and K. Fujii, Quantum circuit learning, Physical Review A 98, 032309 (2018).doi:10.1103/PhysRevA.98.032309
E. Farhi and H. Neven, Classification with quantum neural networks on near term processors, arXiv:1802.06002 (2018).arxiv.org
M. Schuld, A. Bocharov, K. M. Svore and N. Wiebe, Circuit-centric quantum classifiers, Physical Review A 101, 032308 (2020).doi:10.1103/PhysRevA.101.032308
V. Havlíček, A. D. Córcoles, K. Temme, A. W. Harrow, A. Kandala, J. M. Chow and J. M. Gambetta, Supervised learning with quantum-enhanced feature spaces, Nature 567, 209–212 (2019).doi:10.1038/s41586-019-0980-2
M. Benedetti, E. Lloyd, S. Sack and M. Fiorentini, Parameterized quantum circuits as machine learning models, Quantum Science and Technology 4, 043001 (2019).doi:10.1088/2058-9565/ab4eb5
M. Cerezo, A. Arrasmith, R. Babbush, S. C. Benjamin, S. Endo, K. Fujii, J. R. McClean, K. Mitarai, X. Yuan, L. Cincio and P. J. Coles, Variational quantum algorithms, Nature Reviews Physics 3, 625–644 (2021).doi:10.1038/s42254-021-00348-9
M. Schuld, Supervised quantum machine learning models are kernel methods, arXiv:2101.11020 (2021).arxiv.org
M. Schuld, R. Sweke and J. J. Meyer, Effect of data encoding on the expressive power of variational quantum-machine-learning models, Physical Review A 103, 032430 (2021).doi:10.1103/PhysRevA.103.032430
J. C. Spall, Multivariate stochastic approximation using a simultaneous perturbation gradient approximation, IEEE Transactions on Automatic Control 37(3), 332–341 (1992).doi:10.1109/9.119632
W. Hoeffding, Probability inequalities for sums of bounded random variables, Journal of the American Statistical Association 58(301), 13–30 (1963).doi:10.1080/01621459.1963.10500830

Hands-on labs

Train a one-qubit variational classifier
1 step · auto-graded
$R_Y(\theta)R_Y(x)\lvert0\rangle$
Open lab

Feature map, ansatz and observable#

Feature map, ansatz and observable

Variational quantum model and classifier

Output range, decision rule and losses#

A one-qubit classifier worked out exactly#

Estimating the output with finitely many shots#

The hybrid quantum–classical training loop#

The lab data set: the exact feasible interval#

References

Hands-on labs

Train a one-qubit variational classifier