Lesson 49 · 95 min · advanced

Quantum Natural Gradient & the Fubini–Study Metric

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g_{ij}=\mathrm{Re}\big[\langle\partial_i\psi\vert\partial_j\psi\rangle-\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert\partial_j\psi\rangle\big]

Prerequisites: The Parameter-Shift Rule, The Bloch Sphere

Objectives and key results

$g_{ij}=\mathrm{Re}\big[\langle\partial_i\psi\vert\partial_j\psi\rangle-\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert\partial_j\psi\rangle\big]$
$e^{-i\theta_iK_i}$
$\delta\propto-g^{-1}\nabla L$
$g=\mathrm{diag}\big(\tfrac14,\tfrac14\sin^2\theta\big)$
$F_Q=4g$
$Perform natural-gradient updates by hand and with the platform's engine, and compare them with vanilla gradient descent.$

definition · Fubini–Study metric of a parameterised state: $g_{ij}(\theta)=\mathrm{Re}\Big[\langle\partial_i\psi\vert\partial_j\psi\rangle-\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert\partial_j\psi\rangle\Big],\qquad1-\lvert\langle\psi(\theta)\vert\psi(\theta+\mathrm d\theta)\rangle\rvert^2=\sum_{ij}g_{ij}\,\mathrm d\theta_i\,\mathrm d\theta_j+O(\lvert\mathrm d\theta\rvert^3)$
theorem · Layerwise block of the metric: $i,j\in\text{layer }\ell:\qquad g_{ij}=\langle\psi_{\ell-1}\vert K_iK_j\vert\psi_{\ell-1}\rangle-\langle\psi_{\ell-1}\vert K_i\vert\psi_{\ell-1}\rangle\langle\psi_{\ell-1}\vert K_j\vert\psi_{\ell-1}\rangle,\qquad g_{ii}=\mathrm{Var}_{\psi_{\ell-1}}(K_i)$
formula · Metric of a qubit: $\lvert\psi\rangle=R_Z(\phi)R_Y(\theta)\lvert0\rangle:\qquad g=\begin{pmatrix}\frac14&0\\0&\frac14\sin^2\theta\end{pmatrix},\qquad\mathrm ds^2=\tfrac14\big(\mathrm d\theta^2+\sin^2\theta\,\mathrm d\phi^2\big)$
algorithm · Quantum natural gradient update: $\theta\leftarrow\theta-\eta\,\big(g(\theta)+\lambda I\big)^{-1}\nabla L(\theta)\qquad(\lambda\ge0;\ \lambda=0\text{ with the pseudo-inverse }g^{+})$
identity · Quantum Fisher information of a pure state: $F_{Q,ij}=\tfrac12\mathrm{Tr}\big(\rho\{L_i,L_j\}\big)=4\,g_{ij},\qquad F_C\le F_Q$

Distances between quantum states#

\nabla L(\theta)

Definition

Rays, fidelity and the Fubini–Study distance

\lvert\psi\rangle,\lvert\varphi\rangle\in\mathbb C^d

Theorem

Infinitesimal distance between rays

\lvert\psi\rangle

Proof (6 steps)

Step 1:
$a=\langle\psi\vert\delta\psi\rangle,\quad b=\langle\delta\psi\vert\delta\psi\rangle:\qquad\langle\psi\vert\psi+\delta\psi\rangle=1+a,\qquad\langle\psi+\delta\psi\vert\psi+\delta\psi\rangle=1+a+\bar a+b=1+2\,\mathrm{Re}\,a+b$
Expand the inner products by sesquilinearity, using $\langle\psi\vert\psi\rangle=1$ and $\langle\delta\psi\vert\psi\rangle=\bar a$ .
Step 2:
$\lvert\langle\psi\vert\psi+\delta\psi\rangle\rvert^2=(1+a)(1+\bar a)=1+2\,\mathrm{Re}\,a+\lvert a\rvert^2$
$\lvert z\rvert^2=z\bar z$ .
Step 3:
$F=\frac{1+2\,\mathrm{Re}\,a+\lvert a\rvert^2}{1+2\,\mathrm{Re}\,a+b}=1-\frac{b-\lvert a\rvert^2}{1+2\,\mathrm{Re}\,a+b}$
Definition of the fidelity with $\langle\psi\vert\psi\rangle=1$ ; add and subtract $b$ in the numerator.
Step 4:
$b-\lvert a\rvert^2=O(\lVert\delta\psi\rVert^2),\qquad\frac1{1+2\,\mathrm{Re}\,a+b}=1+O(\lVert\delta\psi\rVert)\ \Rightarrow\ 1-F=b-\lvert a\rvert^2+O(\lVert\delta\psi\rVert^3)$
$\lvert a\rvert\le\lVert\delta\psi\rVert$ by Cauchy–Schwarz and $b=\lVert\delta\psi\rVert^2$ ; a geometric series expands the denominator, and the product of an $O(\lVert\delta\psi\rVert^2)$ term with an $O(\lVert\delta\psi\rVert)$ correction is $O(\lVert\delta\psi\rVert^3)$ .
Step 5:
$P_\perp=I-\lvert\psi\rangle\langle\psi\rvert:\quad\lVert P_\perp\delta\psi\rVert^2=\langle\delta\psi\vert P_\perp\vert\delta\psi\rangle=\langle\delta\psi\vert\delta\psi\rangle-\langle\delta\psi\vert\psi\rangle\langle\psi\vert\delta\psi\rangle=b-\lvert a\rvert^2$
$P_\perp$ is an orthogonal projector ( $P_\perp^\dagger=P_\perp=P_\perp^2$ ), so $\lVert P_\perp v\rVert^2=\langle v\vert P_\perp\vert v\rangle$ . The component of $\delta\psi$ along $\lvert\psi\rangle$ only rescales the vector (to first order it changes the norm and the global phase) and does not move the ray.
Step 6:
$\cos^2d_{FS}=F\ \Rightarrow\ \sin^2d_{FS}=1-F,\qquad d_{FS}^2=\sin^2d_{FS}+O(d_{FS}^4)=1-F+O(\lVert\delta\psi\rVert^3)$
$\sin^2x=x^2+O(x^4)$ , and $d_{FS}^2=O(1-F)=O(\lVert\delta\psi\rVert^2)$ , so $d_{FS}^4=O(\lVert\delta\psi\rVert^4)$ .

∎

Proposition

The Fubini–Study metric of a parameterised state

\theta\mapsto\lvert\psi(\theta)\rangle\in\mathbb C^d

Proof (6 steps)

Step 1:
$\lvert\delta\psi\rangle=\lvert\psi(\theta+\mathrm d\theta)\rangle-\lvert\psi(\theta)\rangle=\sum_i\mathrm d\theta_i\,\lvert\partial_i\psi\rangle+O(\lvert\mathrm d\theta\rvert^2)$
First-order Taylor expansion of the state; the second-order remainder contributes only at order $\lvert\mathrm d\theta\rvert^3$ to the quadratic quantities below.
Step 2:
$\langle\delta\psi\vert\delta\psi\rangle=\sum_{i,j}\langle\partial_i\psi\vert\partial_j\psi\rangle\,\mathrm d\theta_i\,\mathrm d\theta_j+O(\lvert\mathrm d\theta\rvert^3)$
Insert the expansion into both slots; the parameters $\mathrm d\theta_i$ are real, so they leave the bra unconjugated.
Step 3:
$\lvert\langle\psi\vert\delta\psi\rangle\rvert^2=\overline{\langle\psi\vert\delta\psi\rangle}\,\langle\psi\vert\delta\psi\rangle=\sum_{i,j}\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert\partial_j\psi\rangle\,\mathrm d\theta_i\,\mathrm d\theta_j+O(\lvert\mathrm d\theta\rvert^3)$
$\overline{\langle\psi\vert\partial_i\psi\rangle}=\langle\partial_i\psi\vert\psi\rangle$ .
Step 4:
$1-F=\sum_{i,j}G_{ij}\,\mathrm d\theta_i\,\mathrm d\theta_j+O(\lvert\mathrm d\theta\rvert^3)$
Subtract the two lines and apply the theorem on the infinitesimal distance between rays.
Step 5:
$\overline{G_{ij}}=\langle\partial_j\psi\vert\partial_i\psi\rangle-\langle\partial_j\psi\vert\psi\rangle\langle\psi\vert\partial_i\psi\rangle=G_{ji}\ \Rightarrow\ \sum_{i,j}G_{ij}\,\mathrm d\theta_i\mathrm d\theta_j=\sum_{i,j}\mathrm{Re}\,G_{ij}\,\mathrm d\theta_i\mathrm d\theta_j$
$G$ is Hermitian, so $\mathrm{Re}\,G$ is symmetric and $\mathrm{Im}\,G$ is antisymmetric; an antisymmetric matrix contributes nothing to a quadratic form with a real vector.
Step 6:
$v\in\mathbb R^P:\quad v^\top g\,v=\sum_{i,j}v_iv_jG_{ij}=\Big\lVert P_\perp\sum_iv_i\lvert\partial_i\psi\rangle\Big\rVert^2\ge0$
The identity $\langle u\vert u\rangle-\lvert\langle\psi\vert u\rangle\rvert^2=\lVert P_\perp u\rVert^2$ of the previous proof with $u=\sum_iv_i\lvert\partial_i\psi\rangle$ : $g$ is positive semidefinite, and $v^\top g\,v=0$ exactly when moving along $v$ changes $\lvert\psi\rangle$ only along itself.

∎

Proposition

The quantum geometric tensor is invariant under parameter-dependent phases

\alpha:\mathbb R^P\to\mathbb R

Proof (5 steps)

Step 1:
$\lvert\partial_i\tilde\psi\rangle=e^{i\alpha}\big(i\,\partial_i\alpha\,\lvert\psi\rangle+\lvert\partial_i\psi\rangle\big)$
Product rule; $\alpha$ is real, so $\overline{e^{i\alpha}}e^{i\alpha}=1$ in every inner product below.
Step 2:
$\langle\partial_i\tilde\psi\vert\partial_j\tilde\psi\rangle=\big(-i\,\partial_i\alpha\,\langle\psi\vert+\langle\partial_i\psi\vert\big)\big(i\,\partial_j\alpha\,\lvert\psi\rangle+\lvert\partial_j\psi\rangle\big)=\partial_i\alpha\,\partial_j\alpha-i\,\partial_i\alpha\,\langle\psi\vert\partial_j\psi\rangle+i\,\partial_j\alpha\,\langle\partial_i\psi\vert\psi\rangle+\langle\partial_i\psi\vert\partial_j\psi\rangle$
The bra of $i\,\partial_i\alpha\lvert\psi\rangle$ is $-i\,\partial_i\alpha\langle\psi\rvert$ ; multiply out the four terms with $\langle\psi\vert\psi\rangle=1$ .
Step 3:
$\langle\partial_i\tilde\psi\vert\tilde\psi\rangle=\langle\partial_i\psi\vert\psi\rangle-i\,\partial_i\alpha,\qquad\langle\tilde\psi\vert\partial_j\tilde\psi\rangle=\langle\psi\vert\partial_j\psi\rangle+i\,\partial_j\alpha$
The same expansion with one slot replaced by $\lvert\tilde\psi\rangle=e^{i\alpha}\lvert\psi\rangle$ .
Step 4:
$\langle\partial_i\tilde\psi\vert\tilde\psi\rangle\langle\tilde\psi\vert\partial_j\tilde\psi\rangle=\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert\partial_j\psi\rangle+i\,\partial_j\alpha\,\langle\partial_i\psi\vert\psi\rangle-i\,\partial_i\alpha\,\langle\psi\vert\partial_j\psi\rangle+\partial_i\alpha\,\partial_j\alpha$
Multiply the two factors; $(-i)(i)=1$ gives the last term.
Step 5:
$\tilde G_{ij}=\langle\partial_i\tilde\psi\vert\partial_j\tilde\psi\rangle-\langle\partial_i\tilde\psi\vert\tilde\psi\rangle\langle\tilde\psi\vert\partial_j\tilde\psi\rangle=\langle\partial_i\psi\vert\partial_j\psi\rangle-\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert\partial_j\psi\rangle=G_{ij}$
The three $\alpha$ -dependent terms of the second line and the fourth line coincide and cancel in the difference.

∎

Convention

G

Quick check

\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert\partial_j\psi\rangle

The metric of a qubit: a sphere of radius ½#

R_Y(\theta)

Circuit · 1 qubit

R_Y(\theta)

\lvert\psi\rangle = \left(0.8001 - 0.3314i\right)\lvert 0\rangle + \left(0.4619 + 0.1913i\right)\lvert 1\rangle

R_Z(\phi)R_Y(\theta)=\begin{pmatrix}e^{-i\phi/2}&0\\0&e^{i\phi/2}\end{pmatrix}\begin{pmatrix}\cos\frac\theta2&-\sin\frac\theta2\\\sin\frac\theta2&\cos\frac\theta2\end{pmatrix}=\begin{pmatrix}e^{-i\phi/2}\cos\frac\theta2&-e^{-i\phi/2}\sin\frac\theta2\\e^{i\phi/2}\sin\frac\theta2&e^{i\phi/2}\cos\frac\theta2\end{pmatrix}\overset{\theta=\pi/3,\ \phi=\pi/4}{=}\begin{pmatrix}e^{-i\pi/8}\frac{\sqrt3}2&-e^{-i\pi/8}\frac12\\e^{i\pi/8}\frac12&e^{i\pi/8}\frac{\sqrt3}2\end{pmatrix}

R_Z(π/4) R_Y(π/3)2×2

Unitary verified: $U^\dagger U = I$ (max deviation 1.1e-16)

Derivation

The metric of R_Z(φ)R_Y(θ)|0⟩ from the definition

Step 1:
$\lvert\psi\rangle=\begin{pmatrix}e^{-i\phi/2}c\\e^{i\phi/2}s\end{pmatrix},\qquad c=\cos\tfrac\theta2,\ s=\sin\tfrac\theta2$
The first column of the matrix $R_Z(\phi)R_Y(\theta)$ above.
Step 2:
$\lvert\partial_\theta\psi\rangle=\frac12\begin{pmatrix}-e^{-i\phi/2}s\\e^{i\phi/2}c\end{pmatrix},\qquad\lvert\partial_\phi\psi\rangle=\frac i2\begin{pmatrix}-e^{-i\phi/2}c\\e^{i\phi/2}s\end{pmatrix}$
$\partial_\theta c=-\tfrac12s$ , $\partial_\theta s=\tfrac12c$ and $\partial_\phi e^{\mp i\phi/2}=\mp\tfrac i2e^{\mp i\phi/2}$ .
Step 3:
$\langle\partial_\theta\psi\vert\partial_\theta\psi\rangle=\tfrac14(s^2+c^2)=\tfrac14,\qquad\langle\partial_\phi\psi\vert\partial_\phi\psi\rangle=\tfrac14(c^2+s^2)=\tfrac14$
The phases have modulus one, and $c^2+s^2=1$ .
Step 4:
$\langle\partial_\theta\psi\vert\partial_\phi\psi\rangle=\tfrac12(-e^{i\phi/2}s)\tfrac i2(-e^{-i\phi/2}c)+\tfrac12e^{-i\phi/2}c\,\tfrac i2e^{i\phi/2}s=\tfrac i2\,sc=\tfrac i4\sin\theta$
Conjugate the components of the bra; $2sc=\sin\theta$ . The result is purely imaginary.
Step 5:
$\langle\psi\vert\partial_\theta\psi\rangle=\tfrac12\big(-cs+sc\big)=0,\qquad\langle\psi\vert\partial_\phi\psi\rangle=\tfrac i2\big(-c^2+s^2\big)=-\tfrac i2\cos\theta$
Inner products with the state itself; $c^2-s^2=\cos\theta$ .
Step 6:
$G=\begin{pmatrix}\frac14-0&\frac i4\sin\theta-0\\-\frac i4\sin\theta-0&\frac14-\big\lvert-\frac i2\cos\theta\big\rvert^2\end{pmatrix}=\begin{pmatrix}\frac14&\frac i4\sin\theta\\-\frac i4\sin\theta&\frac14\sin^2\theta\end{pmatrix}$
Insert into $G_{ij}=\langle\partial_i\psi\vert\partial_j\psi\rangle-\overline{\langle\psi\vert\partial_i\psi\rangle}\langle\psi\vert\partial_j\psi\rangle$ ; the projector term vanishes except in the $\phi\phi$ entry, where it subtracts $\tfrac14\cos^2\theta$ .
Step 7:
$g=\mathrm{Re}\,G=\begin{pmatrix}\frac14&0\\0&\frac14\sin^2\theta\end{pmatrix},\qquad\mathrm ds^2=\tfrac14\big(\mathrm d\theta^2+\sin^2\theta\,\mathrm d\phi^2\big)$
The off-diagonal entries of $G$ are imaginary, so the metric is diagonal and independent of $\phi$ . At $\theta=\pi/3$ : $g=\mathrm{diag}\big(\tfrac14,\tfrac3{16}\big)$ .

Result

\boxed{g(\theta,\phi)=\mathrm{diag}\big(\tfrac14,\ \tfrac14\sin^2\theta\big)}

Proposition

The Fubini–Study geometry of a qubit is the Bloch sphere of radius ½

\vec r,\vec r\,'\in S^2

Proof (5 steps)

Step 1:
$\rho=\tfrac12(I+\vec r\cdot\vec\sigma),\quad\rho'=\tfrac12(I+\vec r\,'\cdot\vec\sigma)\ \Rightarrow\ F=\mathrm{Tr}(\rho\rho')=\tfrac14\Big(\mathrm{Tr}\,I+\sum_{a,b}r_ar'_b\,\mathrm{Tr}(\sigma_a\sigma_b)\Big)=\tfrac12\big(1+\vec r\cdot\vec r\,'\big)$
For pure states $\lvert\langle\psi\vert\psi'\rangle\rvert^2=\mathrm{Tr}(\rho\rho')$ ; the Pauli matrices are traceless with $\mathrm{Tr}(\sigma_a\sigma_b)=2\delta_{ab}$ .
Step 2:
$\lVert\vec r-\vec r\,'\rVert^2=2-2\,\vec r\cdot\vec r\,'\ \Rightarrow\ 1-F=\tfrac12\big(1-\vec r\cdot\vec r\,'\big)=\tfrac14\lVert\vec r-\vec r\,'\rVert^2$
Pure states have unit Bloch vectors.
Step 3:
$\vec r\cdot\vec r\,'=\cos\gamma\ \Rightarrow\ F=\tfrac12(1+\cos\gamma)=\cos^2\tfrac\gamma2\ \Rightarrow\ d_{FS}=\arccos\sqrt F=\tfrac\gamma2$
$\gamma\in[0,\pi]$ is the angle between the Bloch vectors; $\cos\frac\gamma2\ge0$ . Orthogonal states (antipodal points, $\gamma=\pi$ ) are at the maximal distance $\frac\pi2$ .
Step 4:
$\vec r=(\sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta)\ \Rightarrow\ \lVert\mathrm d\vec r\rVert^2=\mathrm d\theta^2+\sin^2\theta\,\mathrm d\phi^2$
$\partial_\theta\vec r=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta)$ and $\partial_\phi\vec r=(-\sin\theta\sin\phi,\sin\theta\cos\phi,0)$ are orthogonal with squared lengths $1$ and $\sin^2\theta$ .
Step 5:
$\mathrm ds^2=1-F\big(\vec r,\vec r+\mathrm d\vec r\big)+O(\lvert\mathrm d\theta\rvert^3)=\tfrac14\big(\mathrm d\theta^2+\sin^2\theta\,\mathrm d\phi^2\big)+O(\lvert\mathrm d\theta\rvert^3)$
Combine the exact identity $1-F=\tfrac14\lVert\Delta\vec r\rVert^2$ with $\Delta\vec r=\mathrm d\vec r+O(\lvert\mathrm d\theta\rvert^2)$ ; this agrees with the direct derivation of $g$ , as the proposition on the Fubini–Study metric requires.

∎

Quick check

\lvert\psi\rangle=R_Z(\phi)R_Y(\theta)\lvert0\rangle

The layerwise block-diagonal approximation#

P

Definition

Layered parameterised circuit

\lvert\psi(\theta)\rangle=V_L\,U_L(\theta^{(L)})\cdots V_2\,U_2(\theta^{(2)})\,V_1\,U_1(\theta^{(1)})\lvert\psi_0\rangle,\qquad U_\ell(\theta^{(\ell)})=\prod_{i\in S_\ell}e^{-i\theta_iK_i},

Theorem

In-layer blocks of the metric are generator covariances

i,j

Proof (8 steps)

Step 1:
$\lvert\psi\rangle=W_\ell\,U_\ell\,\lvert\psi_{\ell-1}\rangle,\qquad W_\ell=V_L\,U_L\cdots U_{\ell+1}V_\ell$
Split the circuit at layer $\ell$ . Neither the unitary $W_\ell$ (the gates after the layer) nor $\lvert\psi_{\ell-1}\rangle$ (the gates before it) depends on a parameter $\theta_i$ with $i\in S_\ell$ .
Step 2:
$U_\ell=\prod_{k\in S_\ell}e^{-i\theta_kK_k}=\exp\Big(-i\sum_{k\in S_\ell}\theta_kK_k\Big),\qquad\frac{\partial U_\ell}{\partial\theta_i}=-iK_iU_\ell=-iU_\ell K_i$
Commuting generators: $e^Ae^B=e^{A+B}$ for $[A,B]=0$ . Differentiating the factor $e^{-i\theta_iK_i}$ gives $-iK_ie^{-i\theta_iK_i}$ , and $K_i$ commutes with every other factor, so it can be placed on either side of $U_\ell$ .
Step 3:
$\lvert\partial_i\psi\rangle=-i\,W_\ell U_\ell K_i\lvert\psi_{\ell-1}\rangle\qquad(i\in S_\ell)$
Product rule: only $U_\ell$ depends on $\theta_i$ .
Step 4:
$\langle\partial_i\psi\vert\partial_j\psi\rangle=\langle\psi_{\ell-1}\vert K_iU_\ell^\dagger W_\ell^\dagger W_\ell U_\ell K_j\vert\psi_{\ell-1}\rangle=\langle\psi_{\ell-1}\vert K_iK_j\vert\psi_{\ell-1}\rangle$
The bra of $-iW_\ell U_\ell K_i\lvert\psi_{\ell-1}\rangle$ is $i\langle\psi_{\ell-1}\rvert K_iU_\ell^\dagger W_\ell^\dagger$ ( $K_i$ Hermitian), and $(i)(-i)=1$ . Unitary invariance: $W_\ell^\dagger W_\ell=I$ and $U_\ell^\dagger U_\ell=I$ , so all gates after the layer drop out.
Step 5:
$\langle\psi\vert\partial_j\psi\rangle=-i\,\langle\psi_{\ell-1}\vert U_\ell^\dagger W_\ell^\dagger W_\ell U_\ell K_j\vert\psi_{\ell-1}\rangle=-i\,\langle K_j\rangle_{\ell-1},\qquad\langle\partial_i\psi\vert\psi\rangle=i\,\langle K_i\rangle_{\ell-1}$
The same cancellation with one derivative replaced by the state; the second identity is the complex conjugate of the first with $j\to i$ (expectation values of Hermitian operators are real).
Step 6:
$G_{ij}=\langle K_iK_j\rangle_{\ell-1}-\big(i\langle K_i\rangle_{\ell-1}\big)\big(-i\langle K_j\rangle_{\ell-1}\big)=\langle K_iK_j\rangle_{\ell-1}-\langle K_i\rangle_{\ell-1}\langle K_j\rangle_{\ell-1}$
Definition of the quantum geometric tensor.
Step 7:
$(K_iK_j)^\dagger=K_jK_i=K_iK_j\ \Rightarrow\ \langle K_iK_j\rangle_{\ell-1}\in\mathbb R\ \Rightarrow\ g_{ij}=\mathrm{Re}\,G_{ij}=G_{ij},\qquad g_{ii}=\langle K_i^2\rangle-\langle K_i\rangle^2=\mathrm{Var}(K_i)$
The product of two commuting Hermitian operators is Hermitian, so the in-layer block of $G$ is already real.
Step 8:
$i\in S_\ell,\ j\in S_m,\ \ell<m:\quad G_{ij}=\langle\varphi\vert K_i\,M^\dagger K_jM\vert\varphi\rangle-\langle\varphi\vert K_i\vert\varphi\rangle\langle\varphi\vert M^\dagger K_jM\vert\varphi\rangle,\qquad\lvert\varphi\rangle=U_\ell\lvert\psi_{\ell-1}\rangle,\quad M=V_{m-1}U_{m-1}\cdots U_{\ell+1}V_\ell$
The same computation with $\lvert\partial_j\psi\rangle=-iW_mU_mK_jM\lvert\varphi\rangle$ and $W_\ell=W_mU_mM$ : the gates after layer $m$ cancel, but the gates $M$ between the layers do not. The entry is a covariance of $K_i$ with the generator $M^\dagger K_jM$ transported back through $M$ , which in general does not commute with $K_i$ ; it depends on the intermediate gates and is complex. The approximation sets these blocks to zero: it is exact when they vanish (as for the qubit model below) and an approximation otherwise.

∎

Derivation

The qubit metric from the layerwise formula

Step 1:
$S_1=\{\theta\},\ K_\theta=\tfrac Y2,\ \lvert\psi_0\rangle=\lvert0\rangle;\qquad S_2=\{\phi\},\ K_\phi=\tfrac Z2,\ \lvert\psi_1\rangle=R_Y(\theta)\lvert0\rangle$
The circuit $R_Y(\theta)$ then $R_Z(\phi)$ consists of two one-gate layers with $V_1=V_2=I$ ; $R_P(t)=e^{-itP/2}$ has generator $P/2$ .
Step 2:
$g_{\theta\theta}=\mathrm{Var}_{\lvert0\rangle}\big(\tfrac Y2\big)=\tfrac14\Big(\langle0\vert Y^2\vert0\rangle-\langle0\vert Y\vert0\rangle^2\Big)=\tfrac14(1-0)=\tfrac14$
$Y^2=I$ and $\langle0\vert Y\vert0\rangle=0$ .
Step 3:
$\langle Z\rangle_1=\cos^2\tfrac\theta2-\sin^2\tfrac\theta2=\cos\theta\ \Rightarrow\ g_{\phi\phi}=\mathrm{Var}_{\lvert\psi_1\rangle}\big(\tfrac Z2\big)=\tfrac14\big(1-\cos^2\theta\big)=\tfrac14\sin^2\theta$
$R_Y(\theta)\lvert0\rangle=\cos\frac\theta2\lvert0\rangle+\sin\frac\theta2\lvert1\rangle$ and $Z^2=I$ . The state before the $R_Z$ gate matters: on $\lvert0\rangle$ the variance of $Z/2$ would be $0$ .
Step 4:
$g^{\text{bd}}=\begin{pmatrix}\frac14&0\\0&\frac14\sin^2\theta\end{pmatrix}=g$
The block-diagonal approximation drops the cross-layer entry $g_{\theta\phi}$ ; the direct derivation showed that its exact value is $\mathrm{Re}\big(\tfrac i4\sin\theta\big)=0$ . For this model the approximation is exact: block-diagonal $=$ full metric.

Result

\boxed{g_{\theta\theta}=\mathrm{Var}_{\lvert0\rangle}\big(\tfrac Y2\big)=\tfrac14,\qquad g_{\phi\phi}=\mathrm{Var}_{R_Y(\theta)\lvert0\rangle}\big(\tfrac Z2\big)=\tfrac14\sin^2\theta}

On hardware

K_i=P_i/2

Quick check

\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)

The natural gradient as steepest descent#

\delta

Theorem

Steepest descent under a metric constraint

g\in\mathbb R^{P\times P}

Proof (6 steps)

Step 1:
$\mathcal L(\delta,\mu)=\nabla L^\top\delta+\frac\mu2\big(\delta^\top g\,\delta-\epsilon^2\big)$
Lagrangian of the equality-constrained problem with multiplier $\mu\in\mathbb R$ ; the factor $\frac12$ is a convenient normalisation. The constraint set is a compact ellipsoid and the objective is continuous, so a minimiser exists and satisfies the Lagrange conditions (the constraint gradient $2g\delta$ is nonzero on the ellipsoid).
Step 2:
$\nabla_\delta\mathcal L=\nabla L+\mu\,g\,\delta=0\ \Rightarrow\ \delta=-\frac1\mu\,g^{-1}\nabla L\qquad(\mu\ne0)$
$\nabla_\delta(\delta^\top g\delta)=(g+g^\top)\delta=2g\delta$ for symmetric $g$ . $\mu=0$ would force $\nabla L=0$ , which is excluded, and $g$ is invertible.
Step 3:
$\delta^\top g\,\delta=\frac1{\mu^2}\nabla L^\top g^{-1}g\,g^{-1}\nabla L=\frac{\nabla L^\top g^{-1}\nabla L}{\mu^2}=\epsilon^2\ \Rightarrow\ \mu=\pm\frac{\sqrt{\nabla L^\top g^{-1}\nabla L}}{\epsilon}$
Insert into the constraint; $g^{-1}$ is symmetric, and $\nabla L^\top g^{-1}\nabla L>0$ because $g^{-1}$ is positive definite.
Step 4:
$\nabla L^\top\delta=-\frac1\mu\,\nabla L^\top g^{-1}\nabla L=\mp\,\epsilon\sqrt{\nabla L^\top g^{-1}\nabla L}$
The two stationary points are the steepest descent ( $\mu>0$ , negative value) and the steepest ascent ( $\mu<0$ ); the minimum is attained at $\mu>0$ , which gives $\delta^\ast$ .
Step 5:
$\langle u,v\rangle_g=u^\top g\,v:\qquad\nabla L^\top\delta=\langle g^{-1}\nabla L,\delta\rangle_g\ge-\lVert g^{-1}\nabla L\rVert_g\,\lVert\delta\rVert_g=-\sqrt{\nabla L^\top g^{-1}\nabla L}\;\epsilon$
An independent proof of global optimality and uniqueness: $g$ defines an inner product, and the Cauchy–Schwarz inequality holds with equality only if $\delta$ is a negative multiple of $g^{-1}\nabla L$ ; the constraint fixes the multiple.
Step 6:
$L(\theta+\delta)=L(\theta)+\nabla L^\top\delta+O(\lVert\delta\rVert^2),\qquad d_{FS}\big(\psi(\theta),\psi(\theta+\delta)\big)^2=\delta^\top g\,\delta+O(\lVert\delta\rVert^3)$
The objective is the first-order change of the loss and the constraint fixes the first-order Fubini–Study length of the step. Absorbing $\epsilon/\sqrt{\nabla L^\top g^{-1}\nabla L}$ into a learning rate $\eta>0$ gives the quantum natural gradient update $\theta\leftarrow\theta-\eta\,g^{-1}\nabla L$ .

∎

Proposition

Singular metrics: the gradient lies in the range of g

L(\theta)=\langle\psi(\theta)\vert H\vert\psi(\theta)\rangle

Proof (6 steps)

Step 1:
$g\,v=0\ \Rightarrow\ v^\top g\,v=\big\lVert P_\perp\lvert\partial_v\psi\rangle\big\rVert^2=0\ \Rightarrow\ \lvert\partial_v\psi\rangle=c\,\lvert\psi\rangle,\quad c=\langle\psi\vert\partial_v\psi\rangle$
$\lvert\partial_v\psi\rangle=\sum_iv_i\lvert\partial_i\psi\rangle$ is the directional derivative; the last step of the proposition on the Fubini–Study metric shows $v^\top gv=\lVert P_\perp\partial_v\psi\rVert^2$ , and a vector with no component orthogonal to $\lvert\psi\rangle$ is a multiple of it.
Step 2:
$0=\partial_v\langle\psi\vert\psi\rangle=\langle\partial_v\psi\vert\psi\rangle+\langle\psi\vert\partial_v\psi\rangle=2\,\mathrm{Re}\,c\ \Rightarrow\ c=i\beta,\ \beta\in\mathbb R$
The state is normalised for every $\theta$ .
Step 3:
$v^\top\nabla L=\partial_vL=\langle\partial_v\psi\vert H\vert\psi\rangle+\langle\psi\vert H\vert\partial_v\psi\rangle=2\,\mathrm{Re}\big(c\,\langle\psi\vert H\vert\psi\rangle\big)=2\,\mathrm{Re}\big(i\beta\langle H\rangle\big)=0$
$\langle H\rangle$ is real, so $i\beta\langle H\rangle$ is purely imaginary. Moving along $v$ only changes the global phase, which no expectation value detects.
Step 4:
$g=\sum_k\mu_ku_ku_k^\top,\quad g^{+}=\sum_{k:\,\mu_k>0}\mu_k^{-1}u_ku_k^\top,\quad g\,g^{+}=\sum_{k:\,\mu_k>0}u_ku_k^\top=\Pi_{\mathrm{range}(g)}$
Spectral theorem for the real symmetric positive semidefinite $g$ with orthonormal eigenvectors $u_k$ ; $gg^{+}$ is the orthogonal projector onto $\mathrm{range}(g)=(\ker g)^\perp$ .
Step 5:
$\nabla L\perp\ker g\ \Rightarrow\ \nabla L\in\mathrm{range}(g)\ \Rightarrow\ g\big(-\eta\,g^{+}\nabla L\big)=-\eta\,\nabla L$
The first three steps show $\nabla L\perp\ker g$ ; for symmetric $g$ , $(\ker g)^\perp=\mathrm{range}(g)$ . Every solution of $g\delta=-\eta\nabla L$ is $-\eta g^{+}\nabla L+w$ with $w\in\ker g$ , and $-\eta g^{+}\nabla L\in\mathrm{range}(g)\perp w$ has the smallest norm.
Step 6:
$\theta=0:\quad g=\mathrm{diag}\big(\tfrac14,0\big),\quad\nabla L=(\partial_\theta L,0),\quad g^{+}\nabla L=\big(4\,\partial_\theta L,\ 0\big)$
At the north pole of the qubit model $\phi$ is a pure phase direction, so $\partial_\phi L=0$ for every observable, and the pseudo-inverse leaves $\phi$ unchanged.

∎

Derivation

Tikhonov regularisation and the proximal form of the update

Step 1:
$\delta_\lambda=\arg\min_\delta\Big[\nabla L^\top\delta+\frac1{2\eta}\,\delta^\top(g+\lambda I)\,\delta\Big],\qquad\lambda>0$
Replace the hard constraint by a quadratic penalty on the step length in the regularised metric $g+\lambda I$ , which is positive definite even when $g$ is singular.
Step 2:
$\nabla L+\frac1\eta(g+\lambda I)\,\delta_\lambda=0\ \Rightarrow\ \delta_\lambda=-\eta\,(g+\lambda I)^{-1}\nabla L$
The objective is strictly convex (positive definite Hessian $\frac1\eta(g+\lambda I)$ ), so the unique minimiser is its stationary point.
Step 3:
$(g+\lambda I)^{-1}=\sum_k\frac{1}{\mu_k+\lambda}\,u_ku_k^\top\ \xrightarrow{\ \lambda\to0^+\ }\ g^{+}\ \text{on }\mathrm{range}(g)$
Eigen-decomposition of $g$ . Directions with tiny eigenvalues $\mu_k\ll\lambda$ are damped to $1/\lambda$ instead of being amplified by $1/\mu_k$ , which keeps the step finite near singular points; since $\nabla L\in\mathrm{range}(g)$ , the limit is the pseudo-inverse step.
Step 4:
$\lambda\to\infty:\quad\delta_\lambda=-\frac\eta\lambda\,\nabla L+O(\lambda^{-2})$
A large regulariser turns the natural gradient back into vanilla gradient descent with learning rate $\eta/\lambda$ . The platform's engine uses $\lambda=\varepsilon=10^{-8}$ (a numerical safeguard for a Cholesky solve) and the QNGvsGD widget uses $\lambda=0.01$ by default.

Result

\boxed{\theta\leftarrow\theta-\eta\,\big(g(\theta)+\lambda I\big)^{-1}\nabla L(\theta)}

Proposition

The natural gradient is invariant under reparametrisation

\theta=\Phi(\xi)

Proof (5 steps)

Step 1:
$\frac{\partial L}{\partial\xi_a}=\sum_i\frac{\partial\theta_i}{\partial\xi_a}\frac{\partial L}{\partial\theta_i}=\sum_iJ_{ia}\frac{\partial L}{\partial\theta_i}\ \Rightarrow\ \nabla_\xi L=J^\top\nabla_\theta L$
Chain rule.
Step 2:
$\lvert\partial_{\xi_a}\psi\rangle=\sum_iJ_{ia}\lvert\partial_{\theta_i}\psi\rangle\ \Rightarrow\ G^{(\xi)}_{ab}=\sum_{i,j}J_{ia}J_{jb}\,G^{(\theta)}_{ij}\ \Rightarrow\ g^{(\xi)}=J^\top g^{(\theta)}J$
Both terms of $G$ are linear in each tangent vector and $J$ is real; take real parts. Equivalently, $\mathrm ds^2=\mathrm d\theta^\top g^{(\theta)}\mathrm d\theta$ with $\mathrm d\theta=J\,\mathrm d\xi$ is a property of the states, not of the coordinates.
Step 3:
$\delta\xi=-\eta\,\big(J^\top g^{(\theta)}J\big)^{-1}J^\top\nabla_\theta L=-\eta\,J^{-1}\big(g^{(\theta)}\big)^{-1}J^{-\top}J^\top\nabla_\theta L=-\eta\,J^{-1}\big(g^{(\theta)}\big)^{-1}\nabla_\theta L$
$(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$ for invertible square matrices.
Step 4:
$\Phi(\xi+\delta\xi)-\Phi(\xi)=J\,\delta\xi+O(\lVert\delta\xi\rVert^2)=-\eta\,\big(g^{(\theta)}\big)^{-1}\nabla_\theta L+O(\eta^2)$
First-order Taylor expansion of $\Phi$ ; $\delta\xi=O(\eta)$ . For an affine $\Phi$ the remainder vanishes and the two steps agree exactly.
Step 5:
$\delta\xi_{\text{GD}}=-\eta\,\nabla_\xi L=-\eta\,J^\top\nabla_\theta L\ \Rightarrow\ \Phi(\xi+\delta\xi_{\text{GD}})-\Phi(\xi)=-\eta\,JJ^\top\nabla_\theta L+O(\eta^2)$
The vanilla step depends on the coordinates through $JJ^\top$ .

∎

Derivation

Example: rescaling the rotation angle, θ = 3ξ

Step 1:
$R_Y(3\xi)\ \text{then}\ R_Z(\phi):\quad J=\frac{\partial\theta}{\partial\xi}=3,\qquad g^{(\xi)}_{\xi\xi}=9\cdot\tfrac14=\tfrac94$
The same states, parameterised by $\xi=\theta/3$ ; the metric transforms with $J^2$ .
Step 2:
$L=\cos\theta=\cos3\xi,\quad\theta=\tfrac\pi6:\qquad\partial_\theta L=-\tfrac12,\qquad\partial_\xi L=-3\sin3\xi=-\tfrac32$
Chain rule at $\xi=\pi/18$ .
Step 3:
$\eta=0.1:\quad\Delta\theta_{\text{GD},\theta}=0.1\cdot\tfrac12=0.05,\qquad\Delta\theta_{\text{GD},\xi}=3\cdot0.1\cdot\tfrac32=0.45$
Vanilla gradient descent: the same learning rate produces a nine times longer step after the rescaling, $JJ^\top=9$ .
Step 4:
$\Delta\theta_{\text{QNG},\theta}=0.1\cdot\frac{1/2}{1/4}=0.2,\qquad\Delta\theta_{\text{QNG},\xi}=3\cdot0.1\cdot\frac{3/2}{9/4}=0.2$
The natural-gradient steps coincide, as the proposition predicts; the TypeScript cell below reproduces both with the engine (an affine scale of a gate angle is a valid circuit parameter).

Result

\boxed{\theta=\tfrac\pi6\ \to\ \tfrac\pi6+0.2\ \text{(QNG in either coordinate)},\qquad\tfrac\pi6+0.05\ \text{or}\ \tfrac\pi6+0.45\ \text{(GD)}}

Derivation

Natural-gradient updates for the qubit model

Step 1:
$L=\langle Z\rangle=\cos\theta:\quad\nabla L=(-\sin\theta,\,0),\quad g^{-1}\nabla L=\big(-4\sin\theta,\,0\big)\quad(0<\theta<\pi)$
The $\phi$ -component is $0/g_{\phi\phi}=0$ ; at the poles the pseudo-inverse gives the same $0$ .
Step 2:
$\text{QNG: }\theta\leftarrow\theta+4\eta\sin\theta,\qquad\text{GD: }\theta\leftarrow\theta+\eta\sin\theta$
For this loss the natural gradient is vanilla gradient descent with a four times larger learning rate, because $g_{\theta\theta}=\frac14$ is constant.
Step 3:
$L=\langle X\rangle=\sin\theta\cos\phi:\quad\nabla L=(\cos\theta\cos\phi,\,-\sin\theta\sin\phi),\quad g^{-1}\nabla L=\Big(4\cos\theta\cos\phi,\ -\frac{4\sin\phi}{\sin\theta}\Big)$
Divide each gradient component by the corresponding diagonal metric entry.
Step 4:
$\Delta s_\phi=\tfrac12\sin\theta\,\lvert\Delta\phi\rvert:\qquad\text{GD: }\tfrac12\eta\sin^2\theta\,\lvert\sin\phi\rvert,\qquad\text{QNG: }2\eta\,\lvert\sin\phi\rvert$
The Fubini–Study length of the azimuthal part of a step is the arc on the circle of latitude of radius $\frac12\sin\theta$ . Near the pole the vanilla step barely moves the state, while the natural step moves it by a distance independent of $\theta$ .

Result

\boxed{\theta_{t+1}=\theta_t-\eta\,\big(g^{-1}\nabla L\big)_\theta,\qquad L=\langle Z\rangle:\ \theta_{t+1}=\theta_t+4\eta\sin\theta_t}

Natural gradient vs gradient descentQNGvsGD

Optimisation trajectories on the Bloch sphere for L = ⟨Z⟩ under vanilla gradient descent and the quantum natural gradient with the Fubini–Study metric, with a step-size slider.

Gradient descent (■)
Quantum natural gradient (●)
○ start · ✕ minimiser n* = −o/‖o‖

L(\theta,\phi) = 1.000\,\cos\theta,\qquad H = 1.000\,Z

g(\theta,\phi) = \operatorname{diag}\!\big(\tfrac14,\ \tfrac14\sin^2\theta\big)

\text{GD: } \boldsymbol\theta \leftarrow \boldsymbol\theta - \eta\,\nabla L

\text{QNG: } \boldsymbol\theta \leftarrow \boldsymbol\theta - \eta\,(g + \lambda I)^{-1}\nabla L,\quad \eta = 0.050,\ \lambda = 0.010

Trajectories on the Bloch sphere

View azimuth25°

Loss L against step

Computing the trajectories…

Stepk = 60 / 60

Optimiser and start point

Learning rate ηη = 0.050

Regularisation λλ = 0.010

Steps60

Start θ₀θ₀ = 0.400

Start φ₀φ₀ = 0.300

Observable H = xX + yY + zZ

‖o‖ = 1.0000; the minimum of L over the sphere is −‖o‖, reached at n* = −o/‖o‖. The metric shrinks the φ direction by sin²θ near the poles, so QNG takes larger coordinate steps in φ there; λ bounds that step by 1/λ.

L=\langle Z\rangle

Natural gradient vs gradient descentQNGvsGD

Optimisation trajectories on the Bloch sphere for L = ⟨Z⟩ under vanilla gradient descent and the quantum natural gradient with the Fubini–Study metric, with a step-size slider.

Gradient descent (■)
Quantum natural gradient (●)
○ start · ✕ minimiser n* = −o/‖o‖

L(\theta,\phi) = 1.000\,\sin\theta\cos\phi,\qquad H = 1.000\,X

g(\theta,\phi) = \operatorname{diag}\!\big(\tfrac14,\ \tfrac14\sin^2\theta\big)

\text{GD: } \boldsymbol\theta \leftarrow \boldsymbol\theta - \eta\,\nabla L

\text{QNG: } \boldsymbol\theta \leftarrow \boldsymbol\theta - \eta\,(g + \lambda I)^{-1}\nabla L,\quad \eta = 0.050,\ \lambda = 0.010

Trajectories on the Bloch sphere

View azimuth25°

Loss L against step

Computing the trajectories…

Stepk = 60 / 60

Optimiser and start point

Learning rate ηη = 0.050

Regularisation λλ = 0.010

Steps60

Start θ₀θ₀ = 0.300

Start φ₀φ₀ = 2.500

Observable H = xX + yY + zZ

\langle Z\rangle

TypeScriptrunnable

import { fubiniStudyMetric, parameterShiftGradient, qngStep } from '@/lib/quantum-simulator';

// The model |psi(theta, phi)> = RZ(phi) RY(theta)|0>: RY acts first, then RZ, with R_P(t) = exp(-i t P / 2).
const model = {
  version: 1, numQubits: 1, numClbits: 0,
  parameters: [{ name: 'theta', value: 0 }, { name: 'phi', value: 0 }],
  ops: [
    { kind: 'gate', gate: 'RY', targets: [0], params: [{ param: 'theta' }] },
    { kind: 'gate', gate: 'RZ', targets: [0], params: [{ param: 'phi' }] },
  ],
};
const Z = [{ coeff: 1, ops: 'Z' }];
const X = [{ coeff: 1, ops: 'X' }];
const fmt = (v: number) => (Math.abs(v) < 5e-13 ? 0 : v).toFixed(6);

// (1) The exact Fubini-Study metric is diag(1/4, sin^2(theta)/4), whatever the value of phi.
for (const theta of [Math.PI / 6, Math.PI / 2, 2.5]) {
  const g = fubiniStudyMetric(model, { bindings: { theta, phi: 0.7 } }).metric;
  console.log('theta = ' + fmt(theta) + ': g = [[' + fmt(g[0][0]) + ', ' + fmt(g[0][1]) + '], [' + fmt(g[1][0]) + ', ' + fmt(g[1][1]) + ']]  sin^2(theta)/4 = ' + fmt(Math.sin(theta) ** 2 / 4));
}

// (2) One step on L = <Z> = cos(theta) from (pi/6, 0) with eta = 0.1: the natural step is 4 times longer in theta.
const start = { theta: Math.PI / 6, phi: 0 };
const grad = parameterShiftGradient(model, Z, { bindings: start });
const natural = qngStep(model, Z, 0.1, { bindings: start });
console.log('gradient (' + fmt(grad.gradient.theta) + ', ' + fmt(grad.gradient.phi) + ')  GD theta = ' + fmt(start.theta - 0.1 * grad.gradient.theta) + '  QNG theta = ' + fmt(natural.bindings.theta) + ', phi = ' + fmt(natural.bindings.phi));

// (3) Minimise L = <X> = sin(theta) cos(phi) from (0.3, 2.5), close to the pole where the phi-gradient is small.
const lr = 0.05;
let gd = { theta: 0.3, phi: 2.5 };
let ng = { theta: 0.3, phi: 2.5 };
for (let k = 0; k <= 40; k++) {
  const a = parameterShiftGradient(model, X, { bindings: gd });
  const b = qngStep(model, X, lr, { bindings: ng });
  if (k % 10 === 0) console.log('step ' + k + ': GD L = ' + fmt(a.value) + ' at (' + fmt(gd.theta) + ', ' + fmt(gd.phi) + ')   QNG L = ' + fmt(b.value) + ' at (' + fmt(ng.theta) + ', ' + fmt(ng.phi) + ')');
  gd = { theta: gd.theta - lr * a.gradient.theta, phi: gd.phi - lr * a.gradient.phi };
  ng = { theta: b.bindings.theta, phi: b.bindings.phi };
}

// (4) Reparametrise theta = 3 xi. Gradient descent changes, the natural-gradient step does not.
const scaled = { ...model, parameters: [{ name: 'xi', value: 0 }, { name: 'phi', value: 0 }], ops: [{ kind: 'gate', gate: 'RY', targets: [0], params: [{ param: 'xi', scale: 3 }] }, model.ops[1]] };
const at = { xi: Math.PI / 18, phi: 0 };
const gXi = parameterShiftGradient(scaled, Z, { bindings: at });
const nXi = qngStep(scaled, Z, 0.1, { bindings: at });
console.log('in xi: g_xixi = ' + fmt(nXi.metric[0][0]) + ', dL/dxi = ' + fmt(gXi.gradient.xi) + '  GD theta = ' + fmt(3 * (at.xi - 0.1 * gXi.gradient.xi)) + '  QNG theta = ' + fmt(3 * nXi.bindings.xi));

Expected output

theta = 0.523599: g = [[0.250000, 0.000000], [0.000000, 0.062500]]  sin^2(theta)/4 = 0.062500
theta = 1.570796: g = [[0.250000, 0.000000], [0.000000, 0.250000]]  sin^2(theta)/4 = 0.250000
theta = 2.500000: g = [[0.250000, 0.000000], [0.000000, 0.089542]]  sin^2(theta)/4 = 0.089542
gradient (-0.500000, 0.000000)  GD theta = 0.573599  QNG theta = 0.723599, phi = 0.000000
step 0: GD L = -0.236754 at (0.300000, 2.500000)   QNG L = -0.236754 at (0.300000, 2.500000)
step 10: GD L = -0.537879 at (0.667508, 2.623782)   QNG L = -0.984420 at (1.394575, 3.127851)
step 20: GD L = -0.774000 at (0.975289, 2.778876)   QNG L = -0.999819 at (1.551806, 3.140133)
step 30: GD L = -0.906291 at (1.198246, 2.908859)   QNG L = -0.999998 at (1.568757, 3.141436)
step 40: GD L = -0.964375 at (1.343799, 2.998404)   QNG L = -1.000000 at (1.570577, 3.141576)
in xi: g_xixi = 2.250000, dL/dxi = -1.500000  GD theta = 0.973599  QNG theta = 0.723599

Quick check

L=\langle Z\rangle

Fisher information and imaginary-time evolution#

4

Theorem

The quantum Fisher information of a pure state is 4g

\rho(\theta)=\lvert\psi(\theta)\rangle\langle\psi(\theta)\rvert

Proof (6 steps)

Step 1:
$\rho^2=\rho\ \Rightarrow\ \partial_i\rho=(\partial_i\rho)\rho+\rho(\partial_i\rho)=\tfrac12\big((2\partial_i\rho)\rho+\rho(2\partial_i\rho)\big)$
Differentiate the projector identity $\rho^2=\rho$ of a pure state with the product rule: $L_i=2\partial_i\rho$ solves the defining equation.
Step 2:
$F_{Q,ij}=\tfrac12\mathrm{Tr}\big(\rho L_iL_j\big)+\tfrac12\mathrm{Tr}\big(\rho L_jL_i\big)=\mathrm{Re}\,\mathrm{Tr}\big(\rho L_iL_j\big)=4\,\mathrm{Re}\,\mathrm{Tr}\big(\rho\,\partial_i\rho\,\partial_j\rho\big)$
$\overline{\mathrm{Tr}(\rho L_iL_j)}=\mathrm{Tr}\big((\rho L_iL_j)^\dagger\big)=\mathrm{Tr}(L_jL_i\rho)=\mathrm{Tr}(\rho L_jL_i)$ for Hermitian $\rho,L_i,L_j$ , so the two traces are complex conjugates.
Step 3:
$\partial_i\rho=\lvert\partial_i\psi\rangle\langle\psi\rvert+\lvert\psi\rangle\langle\partial_i\psi\rvert,\qquad a_i=\langle\psi\vert\partial_i\psi\rangle,\qquad a_i+\bar a_i=\partial_i\langle\psi\vert\psi\rangle=0$
Product rule for the outer product; normalisation makes $a_i$ purely imaginary.
Step 4:
$\mathrm{Tr}(\rho\,\partial_i\rho\,\partial_j\rho)=\langle\psi\vert\partial_i\rho\,\partial_j\rho\vert\psi\rangle=\big(a_i\langle\psi\rvert+\langle\partial_i\psi\rvert\big)\big(\lvert\partial_j\psi\rangle+\bar a_j\lvert\psi\rangle\big)$
$\mathrm{Tr}(\lvert\psi\rangle\langle\psi\rvert A)=\langle\psi\vert A\vert\psi\rangle$ ; $\langle\psi\rvert\partial_i\rho=a_i\langle\psi\rvert+\langle\partial_i\psi\rvert$ and $\partial_j\rho\lvert\psi\rangle=\lvert\partial_j\psi\rangle+\bar a_j\lvert\psi\rangle$ .
Step 5:
$=a_ia_j+a_i\bar a_j+\langle\partial_i\psi\vert\partial_j\psi\rangle+\bar a_i\bar a_j=\langle\partial_i\psi\vert\partial_j\psi\rangle+a_ia_j$
Multiply out, using $\langle\psi\vert\psi\rangle=1$ and $\langle\partial_i\psi\vert\psi\rangle=\bar a_i$ ; with $\bar a=-a$ the terms $a_i\bar a_j=-a_ia_j$ and $\bar a_i\bar a_j=a_ia_j$ .
Step 6:
$a_ia_j=-\bar a_i\,a_j=-\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert\partial_j\psi\rangle\ \Rightarrow\ \mathrm{Tr}(\rho\,\partial_i\rho\,\partial_j\rho)=G_{ij},\qquad F_{Q,ij}=4\,\mathrm{Re}\,G_{ij}=4\,g_{ij}$
Again $a_i=-\bar a_i$ . The trace is the quantum geometric tensor, and its real part is the metric.

∎

Derivation

Fidelity susceptibility and the Bures distance

Step 1:
$F(\theta,\theta+\mathrm d\theta)=\lvert\langle\psi(\theta)\vert\psi(\theta+\mathrm d\theta)\rangle\rvert^2=1-\sum_{i,j}g_{ij}\,\mathrm d\theta_i\,\mathrm d\theta_j+O(\lvert\mathrm d\theta\rvert^3)$
The proposition on the Fubini–Study metric.
Step 2:
$-\frac{\partial^2F(\theta,\theta+\mathrm d\theta)}{\partial(\mathrm d\theta_i)\,\partial(\mathrm d\theta_j)}\bigg\rvert_{\mathrm d\theta=0}=2\,g_{ij}=\tfrac12F_{Q,ij}$
Differentiate the quadratic form twice; $g$ is symmetric. The QFI is minus twice the Hessian of the fidelity, the fidelity susceptibility; the tests of this lesson compute it by finite differences and compare it with $4g$ .
Step 3:
$d_B^2=2\big(1-\sqrt F\big)=\sum_{i,j}g_{ij}\,\mathrm d\theta_i\,\mathrm d\theta_j+O(\lvert\mathrm d\theta\rvert^3)=\tfrac14\sum_{i,j}F_{Q,ij}\,\mathrm d\theta_i\,\mathrm d\theta_j+O(\lvert\mathrm d\theta\rvert^3)$
$\sqrt{1-x}=1-\frac x2+O(x^2)$ . This is the Bures distance, whose infinitesimal form Braunstein and Caves identified with $\frac14F_Q$ ; for pure states it reduces to the Fubini–Study metric.

Result

\boxed{F_Q=4g,\qquad\mathrm ds^2_{\text{Bures}}=\tfrac14\sum_{i,j}F_{Q,ij}\,\mathrm d\theta_i\,\mathrm d\theta_j}

\{E_k\}

Derivation

Classical versus quantum Fisher information for a Z measurement on the qubit model

Step 1:
$p_0=\lvert\langle0\vert\psi\rangle\rvert^2=\cos^2\tfrac\theta2,\qquad p_1=\sin^2\tfrac\theta2$
The phases $e^{\mp i\phi/2}$ drop out of the moduli: the $Z$ statistics do not depend on $\phi$ .
Step 2:
$\partial_\theta p_0=-\tfrac12\sin\theta,\quad\partial_\theta p_1=\tfrac12\sin\theta\ \Rightarrow\ F_{C,\theta\theta}=\frac{\frac14\sin^2\theta}{\cos^2\frac\theta2}+\frac{\frac14\sin^2\theta}{\sin^2\frac\theta2}=\frac{\frac14\sin^2\theta}{\cos^2\frac\theta2\sin^2\frac\theta2}=1$
$2\sin\frac\theta2\cos\frac\theta2=\sin\theta$ for $0<\theta<\pi$ ; common denominator with $\cos^2+\sin^2=1$ .
Step 3:
$F_{Q}=4g=\begin{pmatrix}1&0\\0&\sin^2\theta\end{pmatrix},\qquad F_C=\begin{pmatrix}1&0\\0&0\end{pmatrix}\le F_Q$
$\partial_\phi p_k=0$ , so all $\phi$ -entries of $F_C$ vanish. The $Z$ measurement is optimal for $\theta$ (equality) and blind to $\phi$ ; estimating $\phi$ requires measuring in the $xy$ -plane of the Bloch sphere.

Result

\boxed{F_{C,\theta\theta}=F_{Q,\theta\theta}=1,\qquad F_{C,\phi\phi}=0<F_{Q,\phi\phi}=\sin^2\theta}

Derivation

Variational imaginary-time evolution is a natural-gradient flow

Step 1:
$\lvert\psi(\tau)\rangle=\frac{e^{-\tau H}\lvert\psi_0\rangle}{\lVert e^{-\tau H}\lvert\psi_0\rangle\rVert}\ \Rightarrow\ \partial_\tau\lvert\psi\rangle=-(H-E)\lvert\psi\rangle,\qquad E=\langle\psi\vert H\vert\psi\rangle$
Differentiate the normalised state: $\partial_\tau e^{-\tau H}=-He^{-\tau H}$ and $\partial_\tau\lVert e^{-\tau H}\psi_0\rVert^{-1}=E\,\lVert e^{-\tau H}\psi_0\rVert^{-1}$ . For large $\tau$ the state converges to a ground state of $H$ (when $\lvert\psi_0\rangle$ overlaps it).
Step 2:
$\min_{\dot\theta}\Big\lVert\sum_j\dot\theta_j\lvert\tilde\partial_j\psi\rangle+(H-E)\lvert\psi\rangle\Big\rVert^2,\qquad\lvert\tilde\partial_j\psi\rangle=P_\perp\lvert\partial_j\psi\rangle$
McLachlan's variational principle: choose the parameter velocity whose tangent vector is closest to the exact one. The projected tangent vectors remove the global-phase direction, which does not change the ray.
Step 3:
$\sum_j\mathrm{Re}\langle\tilde\partial_i\psi\vert\tilde\partial_j\psi\rangle\,\dot\theta_j=-\mathrm{Re}\langle\tilde\partial_i\psi\vert(H-E)\vert\psi\rangle$
Normal equations of the real least-squares problem: the derivative of $\lVert\sum_j\dot\theta_jv_j+w\rVert^2$ with respect to $\dot\theta_i$ is $2\,\mathrm{Re}\langle v_i\vert\sum_j\dot\theta_jv_j+w\rangle$ .
Step 4:
$\mathrm{Re}\langle\tilde\partial_i\psi\vert\tilde\partial_j\psi\rangle=\mathrm{Re}\langle\partial_i\psi\vert P_\perp\vert\partial_j\psi\rangle=g_{ij}$
$P_\perp^\dagger P_\perp=P_\perp$ and the definition of the metric.
Step 5:
$\mathrm{Re}\langle\tilde\partial_i\psi\vert(H-E)\vert\psi\rangle=\mathrm{Re}\langle\partial_i\psi\vert H\vert\psi\rangle-E\,\mathrm{Re}\langle\partial_i\psi\vert\psi\rangle-\mathrm{Re}\big(\langle\partial_i\psi\vert\psi\rangle\langle\psi\vert(H-E)\vert\psi\rangle\big)=\tfrac12\,\partial_iE$
$\langle\psi\vert(H-E)\vert\psi\rangle=0$ , $\mathrm{Re}\langle\partial_i\psi\vert\psi\rangle=0$ by normalisation, and $\partial_iE=2\,\mathrm{Re}\langle\partial_i\psi\vert H\vert\psi\rangle$ .
Step 6:
$g\,\dot\theta=-\tfrac12\nabla E\ \Rightarrow\ \dot\theta=-\tfrac12\,g^{-1}\nabla E\ \Rightarrow\ \theta(\tau+\Delta\tau)\approx\theta(\tau)-\tfrac{\Delta\tau}2\,g^{-1}\nabla E$
An explicit Euler step of the variational imaginary-time flow is a quantum natural-gradient step on the energy with learning rate $\eta=\Delta\tau/2$ (McArdle et al. 2019).
Step 7:
$H=Z:\quad\dot\theta=-\tfrac12\cdot4\cdot(-\sin\theta)=2\sin\theta;\qquad e^{-\tau Z}R_Y(\theta_0)\lvert0\rangle\propto e^{-\tau}\cos\tfrac{\theta_0}2\lvert0\rangle+e^{\tau}\sin\tfrac{\theta_0}2\lvert1\rangle$
For the qubit model with $E=\cos\theta$ and $\phi=0$ . The exact state stays real, so it is $R_Y(\theta(\tau))\lvert0\rangle$ with $\tan\frac{\theta(\tau)}2=e^{2\tau}\tan\frac{\theta_0}2$ .
Step 8:
$\tfrac12\sec^2\tfrac\theta2\,\dot\theta=2\tan\tfrac\theta2\ \Rightarrow\ \dot\theta=4\sin\tfrac\theta2\cos\tfrac\theta2=2\sin\theta$
Differentiate $\tan\frac\theta2=e^{2\tau}\tan\frac{\theta_0}2$ in $\tau$ . The variational flow coincides with the exact evolution, because the ansatz reaches every state that the evolution visits; the NumPy cell confirms $\theta(0.5)=0.779612$ for $\theta_0=0.3$ .

Result

\boxed{\dot\theta=-\tfrac12\,g^{-1}\nabla E\qquad(\text{McLachlan; QNG step with }\eta=\Delta\tau/2)}

Quick check

F_{Q,\phi\phi}

Numerical checks and the lab#

g=\mathrm{diag}\big(\tfrac14,\tfrac3{16}\big)

Pythonrunnable

import numpy as np

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

def rz(t):
    return np.diag([np.exp(-0.5j * t), np.exp(0.5j * t)])

ket0 = np.array([1, 0], dtype=complex)

def psi(th, ph, alpha=lambda th, ph: 0.0):
    # |psi> = RZ(phi) RY(theta)|0>, optionally multiplied by a parameter-dependent global phase e^{i alpha}
    return np.exp(1j * alpha(th, ph)) * (rz(ph) @ ry(th) @ ket0)

def tangents(th, ph, alpha=lambda th, ph: 0.0, h=1e-6):
    d_th = (psi(th + h, ph, alpha) - psi(th - h, ph, alpha)) / (2 * h)
    d_ph = (psi(th, ph + h, alpha) - psi(th, ph - h, alpha)) / (2 * h)
    return psi(th, ph, alpha), [d_th, d_ph]

def qgt(th, ph, alpha=lambda th, ph: 0.0):
    # quantum geometric tensor G_ij = <d_i psi|d_j psi> - <d_i psi|psi><psi|d_j psi>
    p, d = tangents(th, ph, alpha)
    return np.array([[np.vdot(a, b) - np.vdot(a, p) * np.vdot(p, b) for b in d] for a in d])

def naive(th, ph, alpha=lambda th, ph: 0.0):
    _, d = tangents(th, ph, alpha)
    return np.array([[np.vdot(a, b) for b in d] for a in d])

def show(m):
    m = np.where(np.abs(m) < 5e-9, 0.0, m)
    return "[[" + "], [".join(", ".join(f"{v:.6f}" for v in row) for row in m) + "]]"

th, ph = np.pi / 3, 0.4
G = qgt(th, ph)
print("g = Re G at (pi/3, 0.4):", show(G.real), "| diag(1/4, sin^2(theta)/4) =", show(np.diag([0.25, np.sin(th) ** 2 / 4])))
print(f"Im G_theta_phi = {G[0, 1].imag:.6f} = sin(theta)/4 = {np.sin(th) / 4:.6f}")

# Phase invariance: a parameter-dependent global phase leaves G unchanged but changes <d_i psi|d_j psi>.
alpha = lambda th, ph: th ** 2 + 3 * ph
print("max |G' - G| < 1e-8 under psi -> e^{i(theta^2 + 3 phi)} psi:", bool(np.abs(qgt(th, ph, alpha) - G).max() < 1e-8),
      "| naive <d psi|d psi> changes from", show(naive(th, ph).real), "to", show(naive(th, ph, alpha).real))

# Projective distance: 1 - |<psi(theta)|psi(theta + delta)>|^2 = g(delta, delta) + O(delta^3).
v = np.array([0.6, -0.8])
for delta in (1e-1, 1e-2, 1e-3):
    infid = 1 - abs(np.vdot(psi(th, ph), psi(th + delta * v[0], ph + delta * v[1]))) ** 2
    print(f"delta = {delta:.0e}: (1 - F)/delta^2 = {infid / delta ** 2:.8f}, v^T g v = {v @ G.real @ v:.8f}")

# Quantum Fisher information 4g versus the classical Fisher information of a Z measurement.
def fisher_z(th, ph, i, h=1e-6):
    p = lambda a, b: np.abs(psi(a, b)) ** 2
    dp = (p(th + h, ph) - p(th - h, ph)) / (2 * h) if i == 0 else (p(th, ph + h) - p(th, ph - h)) / (2 * h)
    return float(np.sum(dp ** 2 / p(th, ph)))
print(f"F_Q = 4g = {show(4 * G.real)}; classical Fisher of a Z measurement: F_theta = {fisher_z(th, ph, 0):.6f}, F_phi = {fisher_z(th, ph, 1):.6f}")

# One QNG step and one vanilla step on L = <Z> = cos(theta) from (pi/3, 0) with eta = 0.1.
eta, t0 = 0.1, np.pi / 3
grad = np.array([-np.sin(t0), 0.0])
g = np.diag([0.25, np.sin(t0) ** 2 / 4])
print(f"QNG: theta' = {t0 - eta * np.linalg.solve(g, grad)[0]:.10f}  (pi/3 + 0.2 sqrt 3 = {t0 + 0.2 * np.sqrt(3):.10f});  GD: theta' = {t0 - eta * grad[0]:.10f}")

# Imaginary-time evolution under H = Z: the variational flow theta' = -1/2 g^{-1} dE/dtheta = 2 sin(theta) is exact.
tau, n = 0.5, 1000
flow = lambda t: 2 * np.sin(t)
t, dt = 0.3, tau / n
for _ in range(n):  # classical Runge-Kutta 4 integration of the parameter flow
    k1 = flow(t); k2 = flow(t + dt * k1 / 2); k3 = flow(t + dt * k2 / 2); k4 = flow(t + dt * k3)
    t += dt * (k1 + 2 * k2 + 2 * k3 + k4) / 6
exact = 2 * np.arctan(np.exp(2 * tau) * np.tan(0.3 / 2))
evolved = np.diag([np.exp(-tau), np.exp(tau)]) @ ry(0.3) @ ket0
evolved /= np.linalg.norm(evolved)
print(f"imaginary time tau = {tau}: flow theta = {t:.6f}, 2 atan(e^(2 tau) tan(0.15)) = {exact:.6f}, from e^(-tau Z) = {2 * np.arctan2(evolved[1].real, evolved[0].real):.6f}")

Expected output

g = Re G at (pi/3, 0.4): [[0.250000, 0.000000], [0.000000, 0.187500]] | diag(1/4, sin^2(theta)/4) = [[0.250000, 0.000000], [0.000000, 0.187500]]
Im G_theta_phi = 0.216506 = sin(theta)/4 = 0.216506
max |G' - G| < 1e-8 under psi -> e^{i(theta^2 + 3 phi)} psi: True | naive <d psi|d psi> changes from [[0.250000, 0.000000], [0.000000, 0.250000]] to [[4.636491, 5.759587], [5.759587, 7.750000]]
delta = 1e-01: (1 - F)/delta^2 = 0.21384541, v^T g v = 0.21000000
delta = 1e-02: (1 - F)/delta^2 = 0.21041262, v^T g v = 0.21000000
delta = 1e-03: (1 - F)/delta^2 = 0.21004154, v^T g v = 0.21000000
F_Q = 4g = [[1.000000, 0.000000], [0.000000, 0.750000]]; classical Fisher of a Z measurement: F_theta = 1.000000, F_phi = 0.000000
QNG: theta' = 1.3936077127  (pi/3 + 0.2 sqrt 3 = 1.3936077127);  GD: theta' = 1.1338000916
imaginary time tau = 0.5: flow theta = 0.779612, 2 atan(e^(2 tau) tan(0.15)) = 0.779612, from e^(-tau Z) = 0.779612

Natural gradient vs gradient descentQNGvsGD

Optimisation trajectories on the Bloch sphere for L = ⟨Z⟩ under vanilla gradient descent and the quantum natural gradient with the Fubini–Study metric, with a step-size slider.

Gradient descent (■)
Quantum natural gradient (●)
○ start · ✕ minimiser n* = −o/‖o‖

L(\theta,\phi) = 1.000\,\cos\theta,\qquad H = 1.000\,Z

g(\theta,\phi) = \operatorname{diag}\!\big(\tfrac14,\ \tfrac14\sin^2\theta\big)

\text{GD: } \boldsymbol\theta \leftarrow \boldsymbol\theta - \eta\,\nabla L

\text{QNG: } \boldsymbol\theta \leftarrow \boldsymbol\theta - \eta\,(g + \lambda I)^{-1}\nabla L,\quad \eta = 0.200,\ \lambda = 0.010

Trajectories on the Bloch sphere

View azimuth25°

Loss L against step

Computing the trajectories…

Stepk = 60 / 60

Optimiser and start point

Learning rate ηη = 0.200

Regularisation λλ = 0.010

Steps60

Start θ₀θ₀ = 0.400

Start φ₀φ₀ = 0.300

Observable H = xX + yY + zZ

\eta=0.2

References

J. Stokes, J. Izaac, N. Killoran and G. Carleo, Quantum natural gradient, Quantum 4, 269 (2020).doi:10.22331/q-2020-05-25-269
S. Amari, Natural gradient works efficiently in learning, Neural Computation 10(2), 251–276 (1998).doi:10.1162/089976698300017746
S. L. Braunstein and C. M. Caves, Statistical distance and the geometry of quantum states, Physical Review Letters 72, 3439–3443 (1994).doi:10.1103/PhysRevLett.72.3439
S. McArdle, T. Jones, S. Endo, Y. Li, S. C. Benjamin and X. Yuan, Variational ansatz-based quantum simulation of imaginary time evolution, npj Quantum Information 5, 75 (2019).doi:10.1038/s41534-019-0187-2
J. J. Meyer, Fisher information in noisy intermediate-scale quantum applications, Quantum 5, 539 (2021).doi:10.22331/q-2021-09-09-539
J. P. Provost and G. Vallee, Riemannian structure on manifolds of quantum states, Communications in Mathematical Physics 76, 289–301 (1980).doi:10.1007/BF02193559

Hands-on labs

The Fubini–Study metric and one natural-gradient step
2 steps · auto-graded
$\lvert\psi(\theta,\phi)\rangle=R_Z(\phi)R_Y(\theta)\lvert0\rangle$
Open lab

Distances between quantum states#

Rays, fidelity and the Fubini–Study distance

The metric of a qubit: a sphere of radius ½#

The layerwise block-diagonal approximation#

Layered parameterised circuit

The natural gradient as steepest descent#

Fisher information and imaginary-time evolution#

Numerical checks and the lab#

References

Hands-on labs

The Fubini–Study metric and one natural-gradient step