AVOIDING BARREN PLATEAUS VIA GAUSSIAN MIXTURE MODEL

## Abstract

Abstract Variational quantum algorithms is one of the most representative algorithms in quantum computing, which has a wide range of applications in quantum machine learning, quantum simulation and other related fields. However, they face challenges associated with the barren plateau phenomenon, especially when dealing with large numbers of qubits, deep circuit layers, or global cost functions, making them often untrainable. In this paper, we propose a novel parameter initialization strategy based on Gaussian Mixture Models. We rigorously prove that, the proposed initialization method consistently avoids the barren plateaus problem for hardware-efficient ansatz with arbitrary length and qubits and any given cost function. Specifically, we find that the gradient norm lower bound provided by the proposed method is independent of the number of qubits N 𝑁 N italic_N and increases with the circuit depth L 𝐿 L italic_L . Our results strictly highlight the significance of Gaussian Mixture model initialization strategies in determining the trainability of quantum circuits, which provides valuable guidance for future theoretical investigations and practical applications.

###### Theorem 1 .

###### Proof.

###### Theorem 2 .

###### Proof.

###### Theorem 3 .

###### Acknowledgements.

## I preliminaries

For the sake of convenience, let’s introduce some notations. If there are two observables $O_{i}=o_{1}^{i}\otimes o_{2}^{i}\otimes...\otimes o_{N}^{i}$ and $O_{j}=o_{1}^{j}\otimes o_{2}^{j}\otimes...\otimes o_{N}^{j}$, $\forall l\in[N]$, the single observables $o_{l}^{i}$ and $o_{l}^{j}$ at their corresponding positions belong to the set $\{X,X;Y,Y;Z,Z;I,Z;Z,I;I,I\}$. We define:

$$ $\displaystyle S_{1}^{ij}$ $\displaystyle:=|\{m|o^{i}_{m}=o^{j}_{m}=X,m\in[N]\}|$ (S1) $\displaystyle P_{0}^{ij}$ $\displaystyle:=\{m|o^{i}_{m}=I||o_{m}^{j}=I,m\in[N]\}$ (S2) $\displaystyle P_{1:3}^{ij}$ $\displaystyle:=\{m|o^{i}_{m}=o^{j}_{m}\neq I,m\in[N]\}$ (S3) $$

Also, the random variable $\theta$ is distributed according to $\mathcal{G}_{0}$, $\mathcal{G}_{1}(\sigma^{2})$, $\mathcal{G}_{2}(\sigma^{2})$, $\mathcal{G}_{3}(\sigma^{2})$, adhering to the same definitions as presented in the main text. Assuming $\theta$ follows the distribution $\mathcal{G}_{1}(\sigma^{2})$, we define $\alpha$, $\beta$, and $\gamma$ as follows:

$$ $\alpha=\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2})}cos^{ 2}\theta=\frac{1+e^{-2\sigma^{2}}}{2}$ (S5) $$

$$ $\beta=\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2})}sin^{2 }\theta=\frac{1-e^{-2\sigma^{2}}}{2}$ (S6) $$

$$ $\gamma=\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2})}cos \theta=e^{-\frac{\sigma^{2}}{2}}$ (S7) $$

By straightforward application of a Taylor expansion, it is evident that $\alpha\geq 1-\sigma^{2}$ and $\beta\geq\sigma^{2}(1-\sigma^{2})$.

Figure: Figure S1: When a term in the observable is $Y$, the parameters in the last block’s $R_{y}(\theta)$ in the ansatz do not contribute to the training. Moreover, when the entire observable consists of $Y$, the $\theta$ parameters in the $R_{y}$ gates of the last block have no impact on the cost function.
Refer to caption: x4.png

We will now delve into the relationship between observables and inactive parameters. Let’s assume the observable $\bm{O}$ is a global observable, i.e., $\bm{O}=o_{1}\otimes o_{2}\otimes...\otimes o_{N}$, where $\forall k\in\{1,2,...,N\},o_{k}\in\{X,Y,Z\}$. Let the density matrix of the final quantum state be $\rho_{2L}$, and the quantum state just before the final $R_{y}$ rotation gate in the last block be $\rho_{2L-1}$. We find that $f(\bm{\theta})=\text{Tr}[\bm{O}\rho_{2L}]=\text{Tr}[\bm{O}(R_{y}(\theta_{2L,1}
)\otimes R_{y}(\theta_{2L,2})\otimes...\otimes R_{y}(\theta_{2L,N}))\rho_{2L-1
}(R_{y}^{\dagger}(\theta_{2L,1})\otimes R_{y}^{\dagger}(\theta_{2L,2})\otimes.
..\otimes R_{y}^{\dagger}(\theta_{2L,N}))])$. Then, when $o_{k}=Y$, we notice that $\forall\theta_{2L,k},R_{y}(\theta_{2L,k})YR_{y}^{\dagger}(\theta_{2L,k})=Y$. Obviously, in this case, $\theta_{2L,k}$ is independent of the cost function $f(\bm{\theta})$, making it an ”inactive parameter.” When the observable $\bm{O}=Y\otimes Y\otimes...\otimes Y$, as shown in Fig. S1, all parameters in the last layer of $R_{y}$ gates are inactive parameters.

## II technical lemmas

###### Lemma 1 .

Let $\rho$ be an arbitrary linear operator, $G$ be a Hermitian unitary and $V=e^{-i\frac{\theta}{2}G}$. Consider an arbitrary Hamiltonian operator $O$ that commutes with $G$. Moreover, let $\theta$ be a random variable following an arbitrary distribution, i.e., $\theta\sim\mathcal{G}_{0}$. Then:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{0}}\text{Tr}[ OV\rho V^{\dagger}]=\text{Tr}[O\rho]$ (S8) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{0}}\text{Tr}^{ 2}[OV\rho V^{\dagger}]=\text{Tr}^{2}[O\rho]$ (S9) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{0}}\frac{ \partial}{\partial\theta}\text{Tr}[OV\rho V^{\dagger}]=0$ (S10) $$

where $\text{Tr}^{2}[\cdot]=(\text{Tr}[\cdot])^{2}$

Proof. Consider that $V=e^{-i\frac{\theta}{2}G}=I\cos\left(\frac{\theta}{2}\right)-iG\sin\left(\frac
{\theta}{2}\right)$, for any arbitrary operator $O$, we obtain:

$$ $\displaystyle\text{Tr}[OV\rho V^{\dagger}]$ $\displaystyle=\text{Tr}\left[O\left(I\cos\left(\frac{\theta}{2}\right)-iG\sin \left(\frac{\theta}{2}\right)\right)\rho\left(I\cos\left(\frac{\theta}{2} \right)+iG\sin\left(\frac{\theta}{2}\right)\right)\right]$ $\displaystyle=\frac{1+\cos\theta}{2}\text{Tr}\left[O\rho\right]+\frac{1-\cos \theta}{2}\text{Tr}[OG\rho G]+\frac{\sin\theta}{2}\left(\text{Tr}[iO\rho G]- \text{Tr}[iOG\rho]\right)$ (S11) $$

Given that $G$ is unitary and $[O,G]=0$, the above expression simplifies to:

$$ $\displaystyle\text{Tr}[OV\rho V^{\dagger}]=\text{Tr}[O\rho]$ (S12) $$

Hence, $\text{Tr}[OV\rho V^{\dagger}]$ is independent of $\theta$. Consequently, for any random variable $\theta$, we establish that $\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{0}}\text{Tr}[OV\rho V^{
\dagger}]=\text{Tr}[O\rho]$, $\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{0}}\text{Tr}^{2}[OV\rho V^{
\dagger}]=\text{Tr}^{2}[O\rho]$ and $\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{0}}\frac{\partial}{\partial
\theta}\text{Tr}[OV\rho V^{\dagger}]=0$.

###### Lemma 2 .

Let $\rho$ be an arbitrary linear operator, and let $G$ be a Hermitian unitary and $V=e^{-i\frac{\theta}{2}G}$. Consider arbitrary Hamiltonian operator $O_{1}$, $O_{2}$, $\widetilde{O}_{1}$, and $\widetilde{O}_{2}$, where $O_{1}$, $O_{2}$ anti-commute with $G$ and $\widetilde{O}_{1}$, $\widetilde{O}_{2}$ commute with $G$, implying $\{O_{1},G\}=0$, $\{O_{2},G\}=0$, $[\widetilde{O}_{1},G]=0$, and $[\widetilde{O}_{2},G]=0$. And $\theta$ is a random variable following a Gaussian distribution $\mathcal{N}(0,\sigma^{2})$, i.e., $\theta\sim\mathcal{G}_{1}(\sigma^{2})$. Then:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2}) }\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=\gamma\text{Tr}[O_{1}\rho]$ (S13) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=\gamma\text{Tr}[iGO_{1}\rho]$ (S14) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2}) }\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}]\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=\gamma\text{Tr}[\widetilde{O}_{1}\rho]\text{Tr}[O_{1}\rho]$ (S15) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}] \frac{\partial}{\partial\theta}\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2} )}\frac{\partial}{\partial\theta}\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}] \frac{\partial}{\partial\theta}\text{Tr}[\widetilde{O}_{2}V\rho V^{\dagger}]=0$ (S16) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2}) }\text{Tr}[O_{1}V\rho V^{\dagger}]\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\alpha\text{Tr}[O_{1}\rho]\text{Tr}[O_{2}\rho]+\beta\text{Tr}[ iGO_{1}\rho]\text{Tr}[iGO_{2}\rho]$ (S17) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[O_{1}V\rho V^{\dagger}]\frac{ \partial}{\partial\theta}\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\beta\text{Tr}[O_{1}\rho]\text{Tr}[O_{2}\rho]+\alpha\text{Tr}[ iGO_{1}\rho]\text{Tr}[iGO_{2}\rho]$ (S18) $$

where $i$ is the imaginary unit.

proof. According to Eq. (II), it can be see that for any operator $O$, we have

$$ $\displaystyle\text{Tr}[OV\rho V^{\dagger}]$ $\displaystyle=\frac{1+\cos\theta}{2}\text{Tr}[O\rho]+\frac{1-\cos\theta}{2} \text{Tr}[GOG\rho]+\frac{\sin\theta}{2}(\text{Tr}[iGO\rho]-\text{Tr}[iOG\rho])$ (S19) $$

Considering the unitary of $G$ and the conditions $\{O_{1},G\}=0$, as indicated in Eq. (S19), we can deduce that

$$ $\displaystyle\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=\cos\theta\text{Tr}[O_{1}\rho]+\sin\theta\text{Tr}[iGO_{1}\rho]$ (S20) $$

Based on Eq. (S20), we obtain that

$$ $\displaystyle\frac{\partial}{\partial\theta}\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=-\sin\theta\text{Tr}[O_{1}\rho]+\cos\theta\text{Tr}[iGO_{1}\rho]$ (S21) $$

Given that $\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2})}\sin\theta=
\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{1}(\sigma^{2})}\sin 2\theta=0$, and combining it with Eq. (S12), Eq. (S20) and Eq. (S21) . Therefore, we can deduce Eq. (S13) to Eq. (S18).

###### Lemma 3 .

Let $\rho$, $G$, $V$, $O_{1}$, $O_{2}$, $\widetilde{O}_{1}$ and $\widetilde{O}_{2}$ be defined in the same manner as presented in Lemma 2. Random variable $\bm{\theta}$ follows distribution $\mathcal{G}_{2}(\sigma^{2})$. Then

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2}) }\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=0$ (S22) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=0$ (S23) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2}) }\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}]\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=0$ (S24) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2}) }\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}]\text{Tr}[\widetilde{O}_{2}V\rho V ^{\dagger}]$ $\displaystyle=\text{Tr}[\widetilde{O}_{1}\rho]\text{Tr}[\widetilde{O}_{2}\rho]$ (S25) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}] \frac{\partial}{\partial\theta}\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2} )}\frac{\partial}{\partial\theta}\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}] \frac{\partial}{\partial\theta}\text{Tr}[\widetilde{O}_{2}V\rho V^{\dagger}]=0$ (S26) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2}) }\text{Tr}[O_{1}V\rho V^{\dagger}]\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\beta\text{Tr}[O_{1}\rho]\text{Tr}[O_{2}\rho]+\alpha\text{Tr}[ iGO_{1}\rho]\text{Tr}[iGO_{2}\rho]$ (S27) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[O_{1}V\rho V^{\dagger}]\frac{ \partial}{\partial\theta}\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\alpha\text{Tr}[O_{1}\rho]\text{Tr}[O_{2}\rho]+\beta\text{Tr}[ iGO_{1}\rho]\text{Tr}[iGO_{2}\rho]$ (S28) $$

proof. Since $\theta\sim\mathcal{G}_{2}(\sigma^{2})$, we have

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2}) }cos\theta$ $\displaystyle=\frac{1}{2}\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}\sigma}e ^{-\frac{(x+\frac{\pi}{2})^{2}}{2\sigma^{2}}}cos(x)dx}+\frac{1}{2}\int_{- \infty}^{+\infty}{\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\frac{\pi}{2})^{2}}{ 2\sigma^{2}}}cos(x)dx}$ (S29) $\displaystyle=-\frac{1}{2}\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{x^{2}}{2\sigma^{2}}}sin(x)dx}+\frac{1}{2}\int_{-\infty}^{+\infty}{ \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^{2}}{2\sigma^{2}}}sin(x)dx}$ (S30) $\displaystyle=0$ (S31) $$

By following the similar calculations, we obtain $\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2})}sin(2\theta)=0$,$\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2})}cos^{2}(
\theta)=\beta$,$\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{2}(\sigma^{2})}sin^{2}(
\theta)=\alpha$.
Combining them with Eq. (S12) and Eq. (S20), it is straightforward to have Eq. (S22) to Eq. (S28).

###### Lemma 4 .

The definitions of $\rho$, $G$, $V$, $O_{1}$, $O_{2}$, $\widetilde{O}_{1}$ and $\widetilde{O}_{2}$ align with those outlined in Lemma 2. Random variable $\theta$ follows distribution $\mathcal{G}_{3}(\sigma^{2})$. Then

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2}) }\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=0$ (S32) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=0$ (S33) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2}) }\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}]\text{Tr}[O_{1}V\rho V^{\dagger}]$ $\displaystyle=0$ (S34) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2}) }\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}]\text{Tr}[\widetilde{O}_{2}V\rho V ^{\dagger}]$ $\displaystyle=\text{Tr}[\widetilde{O}_{1}\rho]\text{Tr}[\widetilde{O}_{2}\rho]$ (S35) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}] \frac{\partial}{\partial\theta}\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\mathop{\mathbb{E}}\limits_{\theta}\frac{\partial}{\partial \theta}\text{Tr}[\widetilde{O}_{1}V\rho V^{\dagger}]\frac{\partial}{\partial \theta}\text{Tr}[\widetilde{O}_{2}V\rho V^{\dagger}]=0$ (S36) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2}) }\text{Tr}[O_{1}V\rho V^{\dagger}]\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\alpha\text{Tr}[O_{1}\rho]\text{Tr}[O_{2}\rho]+\beta\text{Tr}[ iGO_{1}\rho]\text{Tr}[iGO_{2}\rho]$ (S37) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2}) }\frac{\partial}{\partial\theta}\text{Tr}[O_{1}V\rho V^{\dagger}]\frac{ \partial}{\partial\theta}\text{Tr}[O_{2}V\rho V^{\dagger}]$ $\displaystyle=\beta\text{Tr}[O_{1}\rho]\text{Tr}[O_{2}\rho]+\alpha\text{Tr}[ iGO_{1}\rho]\text{Tr}[iGO_{2}\rho]$ (S38) $$

proof. Since $\theta\sim\mathcal{G}_{3}(\sigma^{2})$, we have

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2}) }cos\theta$ $\displaystyle=\frac{1}{4}\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}\sigma}e ^{-\frac{(x+\pi)^{2}}{2\sigma^{2}}}cos(x)dx}+\frac{1}{4}\int_{-\infty}^{+ \infty}{\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\pi)^{2}}{2\sigma^{2}}}cos(x) dx}+\frac{1}{2}\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x ^{2}}{2\sigma^{2}}}cos(x)dx}$ (S39) $\displaystyle=-\frac{1}{4}\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{x^{2}}{2\sigma^{2}}}cos(x)dx}-\frac{1}{4}\int_{-\infty}^{+\infty}{ \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^{2}}{2\sigma^{2}}}cos(x)dx}+\frac{1}{2} \int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^{2}}{2\sigma^{ 2}}}cos(x)dx}$ (S40) $\displaystyle=0$ (S41) $$

By following the similar calculations, we obtain $\mathop{\mathbb{E}}\limits_{\theta}sin(2\theta)=0$, $\mathop{\mathbb{E}}\limits_{\theta}cos^{2}(\theta)=\alpha$, $\mathop{\mathbb{E}}\limits_{\theta}sin^{2}(\theta)=\beta$.
Again using Eq. (S12) and Eq. (S20), it is straightforward to have Eq. (S32) to Eq. (S38).

When $O_{1}=O_{2}$ and $\widetilde{O}_{1}=\widetilde{O}_{2}$, we can derive the following corollary:

Corollary:
Let $\rho$ be an arbitrary linear operator, and let $G$ be a Hermitian unitary and $V=e^{-i\frac{\theta}{2}G}$. Consider arbitrary quantum observables $O$, where $O$ anti-commute with $G$.

If random variable $\theta$ follows distribution $\theta\sim\mathcal{G}_{1}(\sigma^{2})$ or $\theta\sim\mathcal{G}_{3}(\sigma^{2})$ . Then

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta}\text{Tr}^{2}[OV\rho V^{ \dagger}]$ $\displaystyle=\alpha\text{Tr}^{2}[O\rho]+\beta\text{Tr}^{2}[iGO\rho],$ (S42) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta}\left(\frac{\partial}{\partial \theta}\text{Tr}[OV\rho V^{\dagger}]\right)^{2}$ $\displaystyle=\beta\text{Tr}^{2}[O\rho]+\alpha\text{Tr}^{2}[iGO\rho].$ (S43) $$

If random variable $\theta$ follows a Gaussian mixture model $\theta\sim\mathcal{G}_{2}(\sigma^{2})$. Then

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta}\text{Tr}^{2}[OV\rho V^{ \dagger}]$ $\displaystyle=\beta\text{Tr}^{2}[O\rho]+\alpha\text{Tr}^{2}[iGO\rho],$ (S44) $$

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta}\left(\frac{\partial}{\partial \theta}\text{Tr}[OV\rho V^{\dagger}]\right)^{2}$ $\displaystyle=\alpha\text{Tr}^{2}[O\rho]+\beta\text{Tr}^{2}[iGO\rho],$ (S45) $$

For clarity, we employ graphical representations to illustrate the evolution of Pauli matrices. Consider Eq. (S37):

$$ $\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2})}\text{Tr}[O_ {1}V\rho V^{\dagger}]\text{Tr}[O_{2}V\rho V^{\dagger}]=\alpha\text{Tr}[O_{1} \rho]\text{Tr}[O_{2}\rho]+\beta\text{Tr}[iGO_{1}\rho]\text{Tr}[iGO_{2}\rho]$ $$

Suppose $O_{1}=X,O_{2}=Z,G=Y$. Then, $iGO_{1}=Z$ and $iGO_{2}=-X$. Therefore, $\mathop{\mathbb{E}}\limits_{\theta\sim\mathcal{G}_{3}(\sigma^{2})}\text{Tr}[XV
\rho V^{\dagger}]\text{Tr}[ZV\rho V^{\dagger}]=\alpha\text{Tr}[X\rho]\text{Tr}
[Z\rho]-\beta\text{Tr}[Z\rho]\text{Tr}[X\rho]$. The original operators $O_{1}$ and $O_{2}$ are now split into two terms, $XZ$ and $ZX$, with coefficients $\alpha$ and $-\beta$ respectively. The corresponding graphical representation, as depicted in Fig. S2, illustrates the evolution of Pauli matrices after applying the gates, with arrows indicating the resulting Pauli matrices and lines representing their parameters.

Figure: Figure S2: In the scenario where the density matrix $\rho$ remains invariant, the Pauli matrix $XZ$ undergoes a transformation resulting in two components. One component corresponds to $\alpha XZ$, while the other corresponds to $-\beta ZX$.
Refer to caption: x5.png

The following lemma pertains to the transformations of 2-qubit Pauli tensor products after the application of a controlled-Z gate.

###### Lemma 5 .

Let CZ represent a controlled-Z gate, and $o_{i}\otimes o_{j}$ denote a 2-qubit Pauli tensor product, where $o_{i}$ and $o_{j}$ are Pauli matrices. When $o_{i^{\prime}}\otimes o_{j^{\prime}}$ is equivalent to $CZ^{\dagger}(o_{i}\otimes o_{j})CZ$, we denote this transformation as $o_{i}\otimes o_{j}\rightarrow o_{i^{\prime}}\otimes o_{j^{\prime}}$. To encapsulate all specific transformations succinctly, we present the following summary:

$$ $\displaystyle X\otimes I\leftrightarrow X\otimes Z,X\otimes X\leftrightarrow Y \otimes Y,X\otimes Y\leftrightarrow-Y\otimes X,Y\otimes I\leftrightarrow Y\otimes Z$ $$

$$ $\displaystyle Y\otimes Z\leftrightarrow Y\otimes I,Z\otimes I\leftrightarrow Z \otimes I,Z\otimes X\leftrightarrow I\otimes X,Z\otimes Y\leftrightarrow I \otimes Y,$ $$

$$ $\displaystyle Z\otimes Z\leftrightarrow Z\otimes,I\otimes I\leftrightarrow I \otimes I,I\otimes Z\leftrightarrow I\otimes Z$ $$

###### Lemma 1 .

###### Lemma 2 .

###### Lemma 3 .

###### Lemma 4 .

###### Lemma 5 .

## III Proof of theorem 1

Here, we consider an observable with only one term, i.e. $\bm{O}=o_{1}\otimes o_{2}\otimes...\otimes o_{N}$, where $o_{i}\in\{I,X,Y,Z\}$. For subsequent calculations, we establish the following notations. We define $\bm{O_{3:i;1}}$ to mean replacing all the Pauli matrices of $X$ in $O$ with $Z$, and $\bm{O_{3:i}}$ means replacing all Pauli matrices of $X$ and $Y$ in $O$ with $Z$. The parameterized quantum circuit $U(\theta)$ comprising $L$ blocks can be represented as

$$ $\displaystyle U(\boldsymbol{\theta})=U_{L}(\boldsymbol{\theta}_{2L}, \boldsymbol{\theta}_{2L-1})U_{L-1}(\boldsymbol{\theta}_{2L-2},\boldsymbol{ \theta}_{2L-3})...U_{1}(\boldsymbol{\theta}_{2},\boldsymbol{\theta}_{1})$ (S46) $$

For each block $U(\theta_{l})$, it can be represented as

$$ $\displaystyle U_{l}(\boldsymbol{\theta}_{2l},\boldsymbol{\theta}_{2l-1})=R_{2l }(\boldsymbol{\theta}_{2l})R_{2l-1}(\boldsymbol{\theta}_{2l-1})CZ_{l}$ (S47) $$

where

$$ $\displaystyle R_{2l}(\boldsymbol{\theta}_{2l})=e^{-i\frac{\theta_{2l,1}}{2}Y} \otimes e^{-i\frac{\theta_{2l,2}}{2}Y}...\otimes e^{-i\frac{\theta_{2l,N}}{2}Y}$ (S48) $$

$$ $\displaystyle R_{2l-1}(\boldsymbol{\theta}_{2l-1})=e^{-i\frac{\theta_{2l-1,1}} {2}X}\otimes e^{-i\frac{\theta_{2l-1,2}}{2}X}...\otimes e^{-i\frac{\theta_{2l- 1,N}}{2}X},$ (S49) $$

$CZ_{l}$ denotes that the circuit induces entanglement through the inclusion of multiple $CZ$ gates in the $l$-th block.

Next, we consider the intermediate state. For any $k\in\{0,1,...,2L\}$, assuming that the quantum state obtained after passing through the $k$-th block is $\rho_{k}$, we define

$$ $\displaystyle\rho_{k}:=\begin{cases}\begin{aligned} &R_{k}(\boldsymbol{\theta} _{k})\rho_{k-1}R_{k}(\boldsymbol{\theta}_{k})^{\dagger}&&\text{for }k=2l\leq 2 L\\ &R_{k}(\boldsymbol{\theta}_{k})CZ_{\frac{k+1}{2}}\rho_{k-1}CZ_{\frac{k+1}{2}}^ {\dagger}R_{k}(\boldsymbol{\theta}_{k})^{\dagger}&&\text{for }k=2l+1\leq 2L-1 \end{aligned}\end{cases}$ (S50) $$

Additionally, we define $I_{s}:=\{m|o_{m}\neq I,m\in[N]\}$ to denote the set of qubits whose observables act nontrivially. Next, we proceed to prove the content of Theorem 1.

Considering the case where $o_{i}=Z/I$ has two possible choices, namely $\mathcal{G}_{1}(\sigma^{2})$ and $\mathcal{G}_{3}(\sigma^{2})$, we choose $\mathcal{G}_{1}(\sigma^{2})$ without loss of generality. A similar proof can be conducted for the other distribution. First, we consider the expectation of $f(\theta)$. Suppose there exists an index $i$ such that $o_{i}=X$ or $o_{i}=Y$. According to Eq.(S22) and Eq. (S23), it is evident that for all $q,n$, $\mathop{\mathbb{E}}\limits_{\theta}\partial_{\theta_{q,n}}f=0$.

If the Hamiltonian comprises only Pauli $Z$ and the identity matrix $I$, and $q=2L-1$ or $2L$ with $o_{n}=I$, Eq. (S8) and Eq. (S10) imply $\mathop{\mathbb{E}}\limits_{\theta}\partial_{\theta_{q,n}}f=0$. When $o_{n}=Z$, using Eq. (S13) and Eq. (S14), it inevitably transforms into the Pauli matrix $X$ or $Y$. Combining this with Eq. (S13) and Lemma 5, in the final $\text{Tr}[O^{\prime}\rho]$, the Pauli matrix at the $n$-th position of $O^{\prime}$ must be $X$ or $Y$. Furthermore, due to $\langle 0|X|0\rangle=\langle 1|X|1\rangle=\langle 0|Y|0\rangle=\langle 1|Y|1
\rangle=0$, we have $\mathop{\mathbb{E}}\limits_{\theta_{[2L]}}\partial_{\theta_{q,n}}\text{Tr}[O
\rho_{2L-2}]=0$. If $q\in\{1,...,2L-2\}$, then

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta}\partial_{\theta_{q,n}}f$ $\displaystyle=\mathop{\mathbb{E}}\limits_{\theta_{[2L]}}\partial_{\theta_{q,n} }\text{Tr}[\bm{O}\rho_{2L}]$ (S51) $\displaystyle=\gamma^{S_{3}}\mathop{\mathbb{E}}\limits_{\theta_{[2L-1]}} \partial_{\theta_{q,n}}\text{Tr}[\bm{O}\rho_{2L-1}]$ (S52) $\displaystyle=\gamma^{2S_{3}}\mathop{\mathbb{E}}\limits_{\theta_{[2L-2]}} \partial_{\theta_{q,n}}\text{Tr}[CZ_{l}^{\dagger}\bm{O}CZ_{l}\rho_{2L-2}]$ (S53) $\displaystyle=\gamma^{2S_{3}}\mathop{\mathbb{E}}\limits_{\theta_{[2L-2]}} \partial_{\theta_{q,n}}\text{Tr}[\bm{O}\rho_{2L-2}]$ (S54) $\displaystyle=\gamma^{(2L-q-1)S_{3}}\mathop{\mathbb{E}}\limits_{\theta_{[q]}} \partial_{\theta_{q,n}}\text{Tr}[\bm{O}\rho_{q}]$ (S55) $$

According to Eq. (S8) and (S13), we can infer that when $n\in I_{s}$, the expectation of $\theta_{n}$ yields $\gamma$, and when $n\notin I_{s}$, the expectation of $\theta_{n}$ results in a constant 1. Thus, we obtain Eq. (S52). Similarly, we can derive Eq. (S53). Eq. (S54) is derived from Lemma 5. By repeating this process, we arrive at Eq. (S55).

We are currently directing our attention to the subscript $n$. If $n\notin I_{S}$, then, based on Eq. (S10), we can obtain,

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta_{[q]}}\partial_{\theta_{q,n}} \text{Tr}[O\rho_{q}]=0$ (S56) $$

which means

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta_{[2L]}}\partial_{\theta_{q,n}} f=0.$ (S57) $$

When $n\in I_{S}$, according to Eq. (S14), we have

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta_{[q]}}\partial_{\theta_{q,n}} \text{Tr}[O\rho_{2L-2}]=\gamma^{S_{3}}\mathop{\mathbb{E}}\limits_{\theta_{[q-1 ]}}\partial_{\theta_{q,n}}\text{Tr}[O^{\prime}\rho_{2L-2}]$ (S58) $$

Among these, $O^{\prime}$ entails transforming the Pauli $Z$ matrix at the $n$th position of the Hamiltonian $O$ into $Y$ or $-X$. Subsequently, Eq. (S8) and Eq. (S13) elucidate that applying an expectation to $\bm{\theta_{1}},\bm{\theta_{2}},...,\bm{\theta_{q}}$ does not alter the form of the observable but merely augments certain coefficients from the previous state. Additionally, considering that the observable at this juncture comprises only $Y$ or $-X$ at the $n$th position, with the remaining positions being $Z$ or $I$, Lemma 5 implies that we have

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta_{[2L]}}\partial_{\theta_{q,n}} \text{Tr}[O\rho_{2L-2}]=\gamma^{c}\text{Tr}[O^{\prime\prime}\rho_{0}]$ (S59) $$

Here, $c$ is a constant greater than or equal to $L$ and less than or equal to $2L$. Considering that the observable $O^{\prime\prime}$ involves the Pauli operators $X$ or $Y$ at position $n$, and $\langle 0|X|0\rangle=0,\langle 0|Y|0\rangle=0$, we obtain

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta_{[2L]}}\partial_{\theta_{q,n}} \text{Tr}[O\rho_{2L-2}]=0$ (S60) $$

Thus far, we have successfully demonstrated that its expectation is equal to 0. Next, we will establish the lower bound of its gardient norm. Note that

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \bm{\theta})||^{2}$ $\displaystyle=\sum_{q=1}^{2L}\sum_{n=1}^{N}\mathop{\mathbb{E}}\limits_{\bm{ \theta}}\left(\frac{\partial f(\bm{\theta})}{\partial\theta_{q,n}}\right)^{2}$ $\displaystyle=\sum_{q=1}^{2L}\sum_{n\in I_{S}}\mathop{\mathbb{E}}\limits_{\bm{ \theta}}\left(\frac{\partial f(\bm{\theta})}{\partial\theta_{q,n}}\right)^{2}+ \sum_{q=1}^{2L}\sum_{n\notin I_{S}}\mathop{\mathbb{E}}\limits_{\bm{\theta}} \left(\frac{\partial f(\bm{\theta})}{\partial\theta_{q,n}}\right)^{2}$ $\displaystyle\geq\sum_{q=1}^{2L}\sum_{n\in I_{S}}\mathop{\mathbb{E}}\limits_{ \bm{\theta}}\left(\frac{\partial f(\bm{\theta})}{\partial\theta_{q,n}}\right)^ {2}$ (S61) $$

For each term within the first $L-1$ blocks of $\mathop{\mathbb{E}}\limits_{\bm{\theta}}(\frac{\partial f(\theta)}{\partial
\theta_{q,n}})^{2}$, it follows that

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f( \theta)}{\partial\theta_{q,n}}\right)^{2}$ $\displaystyle=\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}}\left(\frac{ \partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}\rho_{2L}]\right)^{2}$ (S62) $\displaystyle=\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta_{1}}}...\mathop{ \mathbb{E}}\limits_{\boldsymbol{\theta_{2L}}}\left(\frac{\partial}{\partial \theta_{q,n}}\text{Tr}[\boldsymbol{O}R_{2L}(\boldsymbol{\theta}_{2L})\rho_{2L- 1}R^{\dagger}_{2L}(\boldsymbol{\theta}_{2L})]\right)^{2}$ (S63) $\displaystyle\geq\alpha^{S_{1}+S_{3}}\mathop{\mathbb{E}}\limits_{\boldsymbol{ \theta_{1}}}...\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta_{2L-1}}}\left( \frac{\partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}_{3:i;1}\rho_{2L- 1}]\right)^{2}$ (S64) $\displaystyle\geq\alpha^{S_{1}+S_{3}}\mathop{\mathbb{E}}\limits_{\boldsymbol{ \theta_{1}}}...\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta_{2L-1}}}\left( \frac{\partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}_{3:i;1}R_{2L-1}( \boldsymbol{\theta}_{2L-1})CZ_{L}\rho_{2L-2}CZ_{L}^{\dagger}R^{\dagger}_{2L}( \boldsymbol{\theta}_{2L-1})\right)^{2}$ (S65) $\displaystyle\geq\alpha^{S_{1}+S_{3}}\alpha^{S_{1}+S_{3}+S_{2}}\mathop{\mathbb {E}}\limits_{\boldsymbol{\theta_{1}}}...\mathop{\mathbb{E}}\limits_{ \boldsymbol{\theta_{2L-2}}}\left(\frac{\partial}{\partial\theta_{q,n}}\text{Tr }[\boldsymbol{O}_{3:i}CZ_{L}\rho_{2L-2}CZ_{L}^{\dagger}]\right)^{2}$ (S66) $\displaystyle=\alpha^{S_{1}+S_{3}}\alpha^{S_{1}+S_{3}+S_{2}}\mathop{\mathbb{E} }\limits_{\boldsymbol{\theta_{1}}}...\mathop{\mathbb{E}}\limits_{\boldsymbol{ \theta_{2L-2}}}\left(\frac{\partial}{\partial\theta_{q,n}}\text{Tr}[ \boldsymbol{O}_{3:i}\rho_{2L-2}]\right)^{2}$ (S67) $\displaystyle\geq\alpha^{S_{1}+S_{3}}\alpha^{S(2L-1-q)}\mathop{\mathbb{E}} \limits_{\boldsymbol{\theta_{1}}}...\mathop{\mathbb{E}}\limits_{\boldsymbol{ \theta_{q}}}\left(\frac{\partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O }_{3:i}\rho_{q}]\right)^{2}$ (S68) $$

In Eq. (S64), the formulation arises from the utilization of Eq. (S42) when $n$ is in $I_{s_{3}}$ and Eq. (S44) when $n$ is in $I_{s_{1}}$, contributing a parameter $\alpha$ for each term. Conversely, when $n$ is in either $I_{s_{0}}$ or $I_{s_{2}}$, Eq. (S9) is employed without altering the preceding coefficients. Through analogous analysis, Eq. (S66) is derived. Eq. (S67) is a consequence of the deductions stemming from Lemma 5. By iterating through these steps, we arrive at Eq. (S68).

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}}(\frac{\partial f (\bm{\theta})}{\partial\theta_{q,n}})^{2}$ $\displaystyle\geq\alpha^{S_{1}+S_{3}}\alpha^{S(2L-1-q)}\alpha^{S-1}\beta \mathop{\mathbb{E}}\limits_{\boldsymbol{\theta_{1}}}...\mathop{\mathbb{E}} \limits_{\boldsymbol{\theta_{q-1}}}(\text{Tr}[\boldsymbol{O}_{3:i}\rho_{q-1}]) ^{2}$ (S69) $\displaystyle\geq\alpha^{S_{1}+S_{3}}\alpha^{S(2L-1-q)}\alpha^{S-1}\beta\alpha ^{S(q-1)}\text{Tr}^{2}[\boldsymbol{O}_{3:i}\rho_{0}]$ (S70) $\displaystyle\geq\alpha^{2LS-1}\beta$ (S71) $\displaystyle\geq(1-\sigma^{2})^{2LS-1}\sigma^{2}(1-\sigma^{2})$ (S72) $\displaystyle=\frac{1}{2LS}(1-\frac{1}{2LS})^{2LS}$ (S73) $\displaystyle\geq\frac{1}{8LS}$ (S74) $$

In Eq. (S69), the coefficient $\beta$ is determined by taking the expectation with respect to $\theta_{q,n}$ based on Eq. (S43). Here, we retain the terms with the coefficient $\beta$ instead of $\alpha$. The remaining $\alpha^{S-1}$ terms remain consistent with Eq. (S42). Eq. (S70) follows a process similar to Eq. (S68), obtained by taking the expectation over the remaining $\theta$. Considering $\text{Tr}[\bm{O_{3:i}\rho_{0}}]=1$, $S_{1}+S_{3}\leq S$, and $\alpha<1$, we arrive at Eq. (S71). Eq. (S72) is derived from a Taylor expansion. Taking into account $h(x)=(1-\frac{1}{x})^{x}$ being monotonically increasing when $x\geq 2$, Eq. (S74) is thus proven.

Applying the identical methodology for analysis, we can similarly derive the same results for the $R_{X}$ rotation layer in the final block. Thus, we can conclude that

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \theta)||^{2}$ $\displaystyle\geq\sum_{q=1}^{2L-1}\sum_{n\in I_{S}}\mathop{\mathbb{E}}\limits_ {\bm{\theta}}(\frac{\partial f(\bm{\theta})}{\partial\theta_{q,n}})^{2}$ $\displaystyle\geq\sum_{q=1}^{2L-1}\sum_{n\in I_{S}}\frac{1}{8LS}$ $\displaystyle=(2L-1)\times S\times\frac{1}{8LS}$ $\displaystyle=\frac{1}{4}-\frac{1}{8L}$ (S75) $$

## IV Proof of theorem 2

Before proving Theorem 2, let’s first consider a special case where both $O_{i}$ and $O_{j}$ are global. We can provide the following lemma:

###### Lemma 6 .

Considering a quantum circuit $U(\bm{\theta})$ with $N$ qubits, initialized with $\rho_{0}$ as a pure state, and employing a hardware-efficient ansatz with $L$ blocks, as depicted in Fig. 1, the cost function is defined as $f(\bm{\theta})=\text{Tr}[(\sum_{i}\boldsymbol{O}_{i}-\sum_{j}\boldsymbol{O}_{j
})U(\bm{\theta})\rho_{o}U(\bm{\theta})^{\dagger}]$, where observable $\boldsymbol{O}_{i},\boldsymbol{O}_{j}$ are global observables, denoted as $o_{i},o_{j}\in\{X,Y,Z\}$. Randomly choose either $\bm{O}_{i}$ or $\bm{O}_{j}$ and initialize it in accordance with the procedure outlined in Theorem 2. Consequently, we obtain:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta}||\nabla_{\bm{\theta}}f(\bm{ \theta})||^{2}_{2}\geq\frac{1}{4}-\frac{1}{8L}$ (S76) $$

proof: Without loss of generality, let us opt to specify $C_{1}$ and initialize the parameters within $U(\theta)$ following the methodology expounded in Theorem 1. Subsequently, we have

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \theta)||^{2}$ $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{ \partial f(\theta)}{\partial\theta_{q,n}}\right)^{2}$ (S77) $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i} \frac{\partial f_{i}(\theta)}{\partial\theta_{q,n}}-\sum_{j}\frac{\partial f_{ j}(\theta)}{\partial\theta_{q,n}}\right)^{2}$ (S78) $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i} \frac{\partial f_{i}(\theta)}{\partial\theta_{q,n}}\right)^{2}-2\sum_{q,n} \mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i,j}\frac{\partial f_{i}( \theta)}{\partial\theta_{q,n}}\cdot\frac{\partial f_{j}(\theta)}{\partial \theta_{q,n}}\right)+\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left( \sum_{j}\frac{\partial f_{j}(\theta)}{\partial\theta_{q,n}}\right)^{2}$ (S79) $\displaystyle=\sum_{q,n,i}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{ \partial f_{i}(\theta)}{\partial\theta_{q,n}}\right)^{2}+\sum_{q,n,i_{1}\neq i _{2}}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{i_{1}}( \theta)}{\partial\theta_{q,n}}\cdot\frac{\partial f_{i_{2}}(\theta)}{\partial \theta_{q,n}}\right)-2\sum_{q,n,i,j}\mathop{\mathbb{E}}\limits_{\bm{\theta}} \left(\frac{\partial f_{i}(\theta)}{\partial\theta_{q,n}}\cdot\frac{\partial f _{j}(\theta)}{\partial\theta_{q,n}}\right)$ $\displaystyle+\sum_{q,n,j}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{ \partial f_{j}(\theta)}{\partial\theta_{q,n}}\right)^{2}+\sum_{q,n,j_{1}\neq j _{2}}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{j_{1}}( \theta)}{\partial\theta_{q,n}}\cdot\frac{\partial f_{j_{2}}(\theta)}{\partial \theta_{q,n}}\right)$ (S80) $$

We expand the function $f(\theta)$, resulting in Eq. (S80). Here, $f_{i}(\theta)=\text{Tr}[\bm{O}_{i}U(\theta)\rho_{0}U(\theta)^{\dagger}],f_{j}(
\theta)=\text{Tr}[\bm{O}_{j}U(\theta)\rho_{0}U(\theta)^{\dagger}]$. Moving forward, let’s consider the cross terms. Without loss of generality, let’s examine each element in the third term. Let’s denote $O_{i}=\vec{\sigma}_{i,2L}=\sigma_{1,i,2L}\otimes\sigma_{2,i,2L}\otimes...
\otimes\sigma_{N,i,2L}$ and $O_{j}=\vec{\sigma}_{j,2L}=\widetilde{\sigma}_{1,j,2L}\otimes\widetilde{\sigma}
_{2,j,2L}\otimes...\otimes\widetilde{\sigma}_{N,j,2L}$. Next, we focus on the evolution of these Pauli matrices throughout the process, we have:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{i }(\theta)}{\partial\theta}\frac{\partial f_{j}(\theta)}{\partial\theta}\right)$ $\displaystyle=\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial}{ \partial\theta}\mathrm{Tr}[\vec{\sigma}_{i,2L}\rho_{2L}]\frac{\partial}{ \partial\theta}\mathrm{Tr}[\vec{\sigma}_{j,2L}\rho_{2L}]\right)$ (S81) $\displaystyle=\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial}{ \partial\theta}\mathrm{Tr}[\vec{\sigma}_{i,2L}R_{2L}(\theta)\frac{\partial\rho _{2L-1}}{\partial\theta}R^{\dagger}_{2L}(\theta)]\frac{\partial}{\partial \theta}\mathrm{Tr}[\vec{\sigma}_{j,2L}R_{2L}(\theta)\frac{\partial\rho_{2L-1}} {\partial\theta}R^{\dagger}_{2L}(\theta)]\right)$ (S82) $\displaystyle=\sum_{k_{1}}h_{k_{1}}\mathop{\mathbb{E}}\limits_{\bm{\theta}} \left(\mathrm{Tr}[\vec{\sigma}_{i,2L-1}^{k_{1}}\frac{\partial\rho_{2L-1}}{ \partial\theta}]\mathrm{Tr}[\vec{\sigma}_{j,2L-1}^{k_{1}}\frac{\partial\rho_{2 L-1}}{\partial\theta}]\right)$ (S83) $\displaystyle=\sum_{k_{2}}h_{k_{2}}\mathop{\mathbb{E}}\limits_{\bm{\theta}} \left(\mathrm{Tr}[CZ^{\dagger}\vec{\sigma}_{i,2L-2}^{k_{2}}CZ\frac{\partial \rho_{2L-2}}{\partial\theta}]\mathrm{Tr}[CZ^{\dagger}\vec{\sigma}_{j,2L-2}^{k_ {2}}CZ\frac{\partial\rho_{2L-2}}{\partial\theta}]\right)$ (S84) $\displaystyle=\sum_{k_{2}^{\prime}}h_{k_{2}^{\prime}}\mathop{\mathbb{E}} \limits_{\bm{\theta}}\left(\mathrm{Tr}[\vec{\sigma}_{i,2L-2}^{k_{2}^{\prime}} \frac{\partial\rho_{2L-2}}{\partial\theta}]\mathrm{Tr}[\vec{\sigma}_{j,2L-2}^{ k_{2}^{\prime}}\frac{\partial\rho_{2L-2}}{\partial\theta}]\right)$ (S85) $\displaystyle\ldots$ (S86) $\displaystyle=\sum_{k_{2L}^{\prime}}h_{k_{2L}^{\prime}}\mathrm{Tr}[\vec{\sigma }_{i,0}^{k_{2L}^{\prime}}\rho_{0}]\mathrm{Tr}[\vec{\sigma}_{j,0}^{k_{2L}^{ \prime}}\rho_{0}]$ (S87) $$

Among these, the coefficients $h_{k_{1}},h_{k_{2}},h_{k_{2}^{\prime}},\ldots,h_{k_{2L}},h_{k_{2L}^{\prime}}$ take the form $\pm\alpha^{g_{1}}\beta^{g_{2}}\gamma^{g_{3}}$, where $g_{1},g_{2},g_{3}\in\mathbb{N}$.
$\vec{\sigma}_{i,0}^{k_{2L}^{\prime}},\vec{\sigma}_{i,0}^{k_{2L}},\ldots,\vec{
\sigma}_{i,2L-1}^{k_{1}},\vec{\sigma}_{i,2L},\vec{\sigma}_{j,0}^{k_{2L}^{
\prime}},\vec{\sigma}_{j,0}^{k_{2L}},\ldots,\vec{\sigma}_{j,2L-1}^{k_{1}},\vec
{\sigma}_{j,2L}$ are all in the form of Pauli matrix tensor product. Furthermore, since $\bm{O_{i}}$ and $\bm{O_{j}}$ are both globally observable operators, and $\bm{O_{i}}\neq\bm{O_{j}}$, there exists $k\in[N]$ such that the Pauli matrix on the $k$-th qubit of $\sigma_{k,i,2L}$ and $\widetilde{\sigma}_{k,j,2L}$ is one of the cases $\{X,Y;Y,X;X,Z;Z,X;Y,Z;Z,Y\}$. Next, we will prove that for all these combinations, $\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{i}(\theta)}{
\partial\theta}\frac{\partial f_{j}(\theta)}{\partial\theta}\right)=0$. Without loss of generality, let’s assume that there exists $k$ such that the $k$-th position of $\sigma_{k,i,2L}$ is $X$ and the $k$-th position of $\widetilde{\sigma}_{k,j,2L}$ is $Z$.

Next, let’s consider the changes in observables. According to Lemma 1, 2, 3, and 4, after the last block’s $R_{y}$ rotation gate, regardless of the distribution followed by $\theta$ in $R_{x}(\theta)$, based on Eq. (S17), Eq. (S27) and Eq. (S37), the value at position $k$ becomes $\{X,Z\}$ or $\{Z,X\}$, the coefficients for the other terms are zero. However, different distributions will result in varying coefficients in front of $\{X,Z\}$ or $\{Z,X\}$. $\{X,Z\},\{Z,X\}$ remains $\{X,Z\},\{Z,X\}$ or 0 after the $R_{x}$ rotation gate, according to Eq. (S15), Eq. (S24) and Eq. (S34). If it’s non-zero, according to Lemma 5, the $CZ$ operation can transform the original $X$ or $Y$ into $X$ or $Y$, without changing them into $Z$ or $I$. Similarly, it cannot transform $Z$ and $I$ into $X$ or $Y$. If, after the application of $CZ$, the original Pauli matrix undergoes a change, such as turning $X$ into $Y$ or $Z$ into $I$, we refer to this process as a ”flip.” Clearly, for any observable $C=c_{1}\otimes c_{2}\otimes...\otimes c_{n}$, if it aims to achieve a ”flip” operation at its $k$-th position, it must satisfy the condition that the Pauli matrix at the $(k-1)$-th position belongs to ${X,Y}$, the Pauli matrix at the $(k+1)$-th position belongs to ${I,Z}$, or the Pauli matrix at the $(k-1)$-th position belongs to ${I,Z}$, and the Pauli matrix at the $(k+1)$-th position belongs to ${Z,I}$. Therefore, after the CZ entanglement gate, its situation becomes one of $\{X,Z;Z,X;Y,Z;Z,Y;X,I;I,X;Y,I;I,Y\}$. Furthermore, taking partial derivatives with respect to any position $\theta_{q,n}$ only alters the coefficients in front, and it does not lead to the appearance of the four possible combinations $\{I,I;Z,Z;I,Z;Z,I\}$ for Pauli matrices.

This analysis applies to each block similarly. Consequently, it generates numerous terms, but in each term, on the $k$-th qubit, all possible situations that eventually arise are $\{X,Z;Z,X;Y,Z;Z,Y;X,I;I,X;Y,I;I,Y\}$. This implies that in $\vec{\sigma}_{i,0}^{k_{2L}^{\prime}},\vec{\sigma}_{j,0}^{k_{2L}^{\prime}}$, there is at least one term with $X$ or $Y$. Additionally, since $\langle 0|X|0\rangle=\langle 0|Y|0\rangle=\langle 1|X|1\rangle=\langle 1|Y|1
\rangle=0$, it follows that $\mathrm{Tr}[\vec{\sigma}_{i,0}^{k_{2L}^{\prime}}\rho_{0}]\mathrm{Tr}[\vec{
\sigma}_{j,0}^{k_{2L}^{\prime}}\rho_{0}]=0$. Therefore, we conclude that when $\sigma_{k,i,2L}=X$ and $\widetilde{\sigma}_{k,j,2L}=Z$, Eq. (S87) equals 0.

In an analogous manner, when the initial Pauli matrix of the $k$-th qubit is $\{X,Y;Y,X;Y,Z;Z,X;Z,Y\}$, we can still obtain $\mathrm{Tr}[\vec{\sigma}_{i,0}^{k_{2L}^{\prime}}\rho_{0}]\mathrm{Tr}[\vec{
\sigma}_{j,0}^{k_{2L}^{\prime}}\rho_{0}]=0$. Only when the initial state is one of $\{X,X;Y,Y;Z,Z;Z,I;I,Z;I,I\}$, $\mathrm{Tr}[\vec{\sigma}_{i,0}^{k_{2L}^{\prime}}\rho_{0}]\mathrm{Tr}[\vec{
\sigma}_{j,0}^{k_{2L}^{\prime}}\rho_{0}]\neq 0$. In light of the fact that both $\bm{O_{i}}$ and $\bm{O_{j}}$ are global observables, and $\bm{O_{i}}\neq\bm{O_{j}}$, it follows that there exists at least one position, such that the Pauli matrices at the $k$-th position of $\bm{O_{i}}$ and $\bm{O_{j}}$ belong to the set $\{X,Y;Y,X;Y,Z;Z,X;Z,Y\}$. Thus, for global observable operators $\bm{O}_{i}$ and $\bm{O}_{j}$, $\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{i}(\theta)}{
\partial\theta}\frac{\partial f_{j}(\theta)}{\partial\theta}\right)=0$.

Following a similar analysis, we obtain $\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{i_{1}}(\theta)
}{\partial\theta_{q,n}}\cdot\frac{\partial f_{i_{2}}(\theta)}{\partial\theta_{
q,n}}\right)=\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{j
_{1}}(\theta)}{\partial\theta_{q,n}}\cdot\frac{\partial f_{j_{2}}(\theta)}{
\partial\theta_{q,n}}\right)=0$. Thus, Eq. (S80) can be simplified to:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \theta)||^{2}$ $\displaystyle=\sum_{q,n,i}E\left(\frac{\partial f_{i}(\theta)}{\partial\theta_ {q,n}}\right)^{2}+\sum_{q,n,j}E\left(\frac{\partial f_{j}(\theta)}{\partial \theta_{q,n}}\right)^{2}$ (S88) $\displaystyle\geq\sum_{q,n}E\left(\frac{\partial f_{1}(\theta)}{\partial\theta _{q,n}}\right)^{2}$ (S89) $\displaystyle\geq\frac{1}{4}-\frac{1}{8L}$ (S90) $$

Thus, we have completed the proof of the lemma.

Next, let’s proceed with the proof of Theorem 2. Without loss of generality, we select $\bm{O}_{1}$ and initialize the parameters of the quantum circuit according to it. Next, we will expand $f(\theta)$ to obtain its expression:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \theta)||^{2}$ $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{ \partial f(\theta)}{\partial\theta_{q,n}}\right)^{2}$ (S91) $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i_ {1}}\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}-\sum_{j_{ 1}}\frac{\partial f_{j_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}+\sum_{i_{2 }}\frac{\partial f_{i_{2}}^{\prime\prime}(\theta)}{\partial\theta_{q,n}}-\sum_ {j_{2}}\frac{\partial f_{j_{2}}^{\prime\prime}(\theta)}{\partial\theta_{q,n}} \right)^{2}$ (S92) $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i_ {1}}\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}-\sum_{j_{ 1}}\frac{\partial f_{j_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}$ $\displaystyle+2\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i _{1}}\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}-\sum_{j_ {1}}\frac{\partial f_{j_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right) \left(\sum_{i_{2}}\frac{\partial f_{i_{2}}^{\prime\prime}(\theta)}{\partial \theta_{q,n}}-\sum_{j_{2}}\frac{\partial f_{j_{2}}^{\prime\prime}(\theta)}{ \partial\theta_{q,n}}\right)$ $\displaystyle+\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i_ {2}}\frac{\partial f_{i_{2}}^{\prime\prime}(\theta)}{\partial\theta_{q,n}}- \sum_{j_{2}}\frac{\partial f_{j_{2}}^{\prime\prime}(\theta)}{\partial\theta_{q ,n}}\right)^{2},$ (S93) $$

where $f_{i_{1}}^{\prime}(\theta)=\text{Tr}[\bm{O}_{i_{1}}^{\prime}U(\theta)\rho_{0}U
(\theta)^{\dagger}]$, $f_{j_{1}}^{\prime}(\theta)=\text{Tr}[\bm{O}_{j_{1}}^{\prime}U(\theta)\rho_{0}U
(\theta)^{\dagger}]$, $f_{i_{2}}^{\prime\prime}(\theta)=\text{Tr}[\bm{O}_{i_{2}}^{\prime\prime}U(
\theta)\rho_{0}U(\theta)^{\dagger}]$, $f_{j_{2}}^{\prime\prime}(\theta)=\text{Tr}[\bm{O}_{j_{2}}^{\prime\prime}U(
\theta)\rho_{0}U(\theta)^{\dagger}]$. The notations $\bm{O}_{i_{1}}^{\prime}$ and $\bm{O}_{j_{1}}^{\prime}$ suggest that, in comparison to $\bm{O}_{1}$, they simply involve replacing some Pauli matrices Z with I or vice versa. For instance, consider $X\otimes Y\otimes Z\otimes I$ and $X\otimes Y\otimes I\otimes Z$. On the other hand, $\bm{O}_{i_{2}}^{\prime\prime}$, $\bm{O}_{j_{2}}^{\prime\prime}$ represent other observables.

Following similar analyses from Lemma 6, we determine that the second term in Eq. S93 is equal to 0. Now, let’s expand the remaining terms. Therefore:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \theta)||^{2}$ $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left((\sum_{i _{1}}\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}-\sum_{j_ {1}}\frac{\partial f_{j_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}})^{2}+( \sum_{i_{2}}\frac{\partial f_{i_{2}}^{\prime\prime}(\theta)}{\partial\theta_{q ,n}}-\sum_{j_{2}}\frac{\partial f_{j_{2}}^{\prime\prime}(\theta)}{\partial \theta_{q,n}})^{2}\right)$ (S94) $\displaystyle\geq\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_ {i_{1}}\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}-\sum_{ j_{1}}\frac{\partial f_{j_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^ {2}$ (S95) $\displaystyle=\sum_{q,n,i_{1}}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left( \frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}+ \sum_{q,n,i_{1}^{\prime}\neq i_{1}^{\prime\prime}}\mathop{\mathbb{E}}\limits_{ \bm{\theta}}\left(\frac{\partial f_{i_{1}^{\prime}}^{\prime}(\theta)}{\partial \theta_{q,n}}\cdot\frac{\partial f_{i_{1}^{\prime\prime}}^{\prime}(\theta)}{ \partial\theta_{q,n}}\right)-2\sum_{q,n,i_{1},j_{1}}\mathop{\mathbb{E}}\limits _{\bm{\theta}}\left(\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_ {q,n}}\cdot\frac{\partial f_{j_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)$ $\displaystyle+\sum_{q,n,j_{1}}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left( \frac{\partial f_{j_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}+ \sum_{q,n,j_{1}^{\prime}\neq j_{1}^{\prime\prime}}\mathop{\mathbb{E}}\limits_{ \bm{\theta}}\left(\frac{\partial f_{j_{1}^{\prime}}^{\prime}(\theta)}{\partial \theta_{q,n}}\cdot\frac{\partial f_{j_{1}^{\prime\prime}}^{\prime}(\theta)}{ \partial\theta_{q,n}}\right)$ (S96) $$

It is easy to see that all the cross terms in this expression differ in the positions where $I$ and $Z$ occur. Therefore, there exists a $k$ such that the $k$-th position in $f_{i_{1}}^{\prime}(\theta)$ and $f_{j_{1}}^{\prime}(\theta)$ is either ${I,Z}$ or $Z,I$. According to Eq. (S34) and Eq. (S36), we know that the third term in Eq. (S96) is equal to 0. Similarly, we can analyze the other cross terms in Eq. (S96) and conclude that they are all equal to 0. Therefore, we have:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \theta)||^{2}$ $\displaystyle\geq\sum_{q,n,i_{1}}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left (\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}+ \sum_{q,n,j_{1}}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f _{j_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}$ (S97) $$

Given that $\bm{O}_{i_{1}}^{\prime}$ and $\bm{O}_{1}$ differ only in certain terms that flip $I$ to $Z$ or $Z$ to $I$, and during the initialization of quantum circuit parameters, the $k$-th position in $\bm{O_{1}}$ follows $\mathcal{G}_{3}(\sigma^{2})$ if it is $I$ or $Z$. Therefore, for all $i_{1}$, $\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{i_{1
}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}$ are all equal. According to Eq. (S42) and Eq. (S43), and employing a similar analysis to Theorem 1, we obtain:

$$ $\displaystyle\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{ \partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}\geq\frac {1}{4}-\frac{1}{8L}$ (S98) $$

Thus, we have:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \theta)||^{2}\geq M\left(\frac{1}{4}-\frac{1}{8L}\right)$ (S99) $$

###### Lemma 6 .

## V Proof of theorem 3

Without loss of generality, we select $\bm{O}_{1}$ and initialize according to $\bm{O}_{1}$. Let $\bm{O}_{1}=o_{1}^{1}\otimes o_{2}^{1}\otimes...\otimes o_{N}^{1}$. We expand $f(\theta)$ to obtain:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f( \theta)||^{2}$ $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{ \partial f(\theta)}{\partial\theta_{q,n}}\right)^{2}$ (S100) $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i} \frac{\partial f_{i}^{\prime}(\theta)}{\partial\theta_{q,n}}+\sum_{j}\frac{ \partial f_{j}^{\prime\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}$ (S101) $\displaystyle=\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i} \frac{\partial f_{i}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}+2\sum_ {q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\sum_{i,j}\frac{\partial f_ {i}^{\prime}(\theta)}{\partial\theta_{q,n}}\cdot\frac{\partial f_{j}^{\prime \prime}(\theta)}{\partial\theta_{q,n}}\right)+\sum_{q,n}\mathop{\mathbb{E}} \limits_{\bm{\theta}}\left(\sum_{j}\frac{\partial f_{j}^{\prime\prime}(\theta) }{\partial\theta_{q,n}}\right)^{2}$ (S102) $\displaystyle\geq\sum_{q,n,i}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left( \frac{\partial f_{i}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}+\sum_{ q,n,i_{1}\neq i_{2}}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{ \partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{q,n}}\cdot\frac{\partial f _{i_{2}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)+2\sum_{q,n,i,j}\mathop {\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{i}^{\prime}(\theta)}{ \partial\theta_{q,n}}\cdot\frac{\partial f_{j}^{\prime\prime}(\theta)}{ \partial\theta_{q,n}}\right),$ (S103) $$

where $f_{i}^{\prime}(\theta)=\text{Tr}[\bm{O}_{i}^{\prime}U(\theta)\rho_{0}U(\theta)
^{\dagger}]$ and $f_{j}^{\prime\prime}(\theta)=\text{Tr}[\bm{O}_{j}^{\prime}U(\theta)\rho_{0}U(
\theta)^{\dagger}]$. $\bm{O}_{i}^{\prime}$ implies that, compared to $\bm{O}_{1}$, they might have operations that flip some $I$ to $Z$ or $Z$ to $I$, while the rest of the Pauli matrices are the same. $\bm{O}_{j}^{\prime}$ represents observables that do not satisfy these conditions.

According to a similar analysis as in Lemma 6, we can see that the third term in Eq. (S103) is equal to 0. In the context of the final block, where the positions of $I$ and $Z$ in $\bm{O}_{1}$ follow Gaussian distributions $\mathcal{N}(0,\sigma^{2})$, and considering that $\bm{O}_{i}^{\prime}$, compared to $\bm{O}_{1}$, only involves flipping Pauli I to Pauli Z or Pauli Z to Pauli I, we can apply a similar analysis as in Theorem 1. As a result, in the first term of Eq. (S103), for each $\bm{O}_{i}^{\prime}$, we find that $\sum_{q,n}\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial f_{i}^{
\prime}(\theta)}{\partial\theta_{q,n}}\right)^{2}\geq\frac{1}{4}-\frac{1}{8L}$. For the second term in Eq. (S103), when $n\in P_{1:3}^{ij}$ and $q\in[2L-2]$, note that:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}}$ $\displaystyle\left(\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{ q,n}}\cdot\frac{\partial f_{i_{2}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)$ $\displaystyle=\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}}\left(\frac{ \partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}_{i_{1}}^{\prime}\rho_{ 2L}]\cdot\frac{\partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}_{i_{2}} ^{\prime}\rho_{2L}]\right)$ (S104) $\displaystyle=\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}_{1}}...\mathop{ \mathbb{E}}\limits_{\boldsymbol{\theta}_{2L}}\left(\frac{\partial}{\partial \theta_{q,n}}\text{Tr}[\boldsymbol{O}_{i_{1}}^{\prime}R_{2L}(\boldsymbol{ \theta}_{2L})\rho_{2L-1}R^{\dagger}_{2L}(\boldsymbol{\theta}_{2L})]\cdot\frac{ \partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}_{i_{2}}^{\prime}R_{2L} (\boldsymbol{\theta}_{2L})\rho_{2L-1}R^{\dagger}_{2L}(\boldsymbol{\theta}_{2L} )]\right)$ (S105) $\displaystyle\geq\alpha^{S_{1}^{i_{1}i_{2}}+S_{3}^{i_{1}i_{2}}}\gamma^{S_{0,3} ^{i_{1}i_{2}}}\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}_{1}}...\mathop{ \mathbb{E}}\limits_{\boldsymbol{\theta}_{2L-1}}\left(\frac{\partial}{\partial \theta_{q,n}}\text{Tr}[\boldsymbol{O}_{3:i_{1};1}^{\prime}\rho_{2L-1}]\cdot \frac{\partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}_{3:i_{2};1}^{ \prime}\rho_{2L-1}]\right)$ (S106) $\displaystyle\geq\alpha^{S_{1}^{i_{1}i_{2}}+S_{3}^{i_{1}i_{2}}}\gamma^{S_{0,3} ^{i_{1}i_{2}}}\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}_{1}}...\mathop{ \mathbb{E}}\limits_{\boldsymbol{\theta}_{2L-1}}\left(\frac{\partial}{\partial \theta_{q,n}}\text{Tr}[\boldsymbol{O}_{3:i_{1};1}^{\prime}R_{2L-1}(\boldsymbol {\theta}_{2L-1})CZ_{L}\rho_{2L-2}CZ_{L}^{\dagger}R^{\dagger}_{2L}(\boldsymbol{ \theta}_{2L-1})]\right.$ $\displaystyle\left.\cdot\frac{\partial}{\partial\theta_{q,n}}\text{Tr}[ \boldsymbol{O}_{3:i_{2};1}^{\prime}R_{2L-1}(\boldsymbol{\theta}_{2L-1})CZ_{L} \rho_{2L-2}CZ_{L}^{\dagger}R^{\dagger}_{2L}(\boldsymbol{\theta}_{2L-1})]\right)$ (S107) $\displaystyle\geq\alpha^{S_{1}^{i_{1}i_{2}}+S_{3}^{i_{1}i_{2}}}\alpha^{S_{1:3} ^{i_{1}i_{2}}}\gamma^{2S_{0,3}^{i_{1}i_{2}}}\mathop{\mathbb{E}}\limits_{ \boldsymbol{\theta}_{1}}...\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}_{2L -2}}\left(\frac{\partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}_{3:i_{ 1}}^{\prime}CZ_{L}\rho_{2L-2}CZ_{L}^{\dagger}]\cdot\frac{\partial}{\partial \theta_{q,n}}\text{Tr}[\boldsymbol{O}_{3:i_{2}}^{\prime}CZ_{L}\rho_{2L-2}CZ_{L }^{\dagger}]\right)$ (S108) $\displaystyle=\alpha^{S_{1}^{i_{1}i_{2}}+S_{3}^{i_{1}i_{2}}}\alpha^{S_{1:3}^{i _{1}i_{2}}}\gamma^{2S_{0,3}^{i_{1}i_{2}}}\mathop{\mathbb{E}}\limits_{ \boldsymbol{\theta}_{1}}...\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}_{2L -2}}\left(\frac{\partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O}_{3:i_{ 1}}^{\prime}\rho_{2L-2}]\cdot\frac{\partial}{\partial\theta_{q,n}}\text{Tr}[ \boldsymbol{O}_{3:i_{2}}^{\prime}\rho_{2L-2}]\right)$ (S109) $\displaystyle\geq\alpha^{S_{1}^{i_{1}i_{2}}+S_{3}^{i_{1}i_{2}}}\alpha^{(2L-q-1 )S_{1:3}^{i_{1}i_{2}}}\gamma^{(2L-q)S_{0,3}^{i_{1}i_{2}}}\mathop{\mathbb{E}} \limits_{\boldsymbol{\theta}_{1}}...\mathop{\mathbb{E}}\limits_{\boldsymbol{ \theta}_{q}}\left(\frac{\partial}{\partial\theta_{q,n}}\text{Tr}[\boldsymbol{O }_{3:i_{1}}^{\prime}\rho_{q}]\cdot\frac{\partial}{\partial\theta_{q,n}}\text{ Tr}[\boldsymbol{O}_{3:i_{2}}^{\prime}\rho_{q}]\right)$ (S110) $$

Similar to Eq. (S64), Eq. (S106) is derived from Eq. (S9), (S15), (S42) and (S44). Similarly, we obtain Eq. (S108). Eq. (S109) is simplified through Lemma 5. Continuing this analysis up to layer $q$, we arrive at Eq. (S110).

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\bm{\theta}}$ $\displaystyle\left(\frac{\partial f_{i_{1}}^{\prime}(\theta)}{\partial\theta_{ q,n}}\cdot\frac{\partial f_{i_{2}}^{\prime}(\theta)}{\partial\theta_{q,n}}\right)$ $\displaystyle=\mathop{\mathbb{E}}\limits_{\bm{\theta}}\left(\frac{\partial}{ \partial\theta_{q,n}}\text{Tr}[\bm{O}_{i_{1}}^{\prime}\rho_{2L}]\cdot\frac{ \partial}{\partial\theta_{q,n}}\text{Tr}[\bm{O}_{i_{2}}^{\prime}\rho_{2L}]\right)$ (S111) $\displaystyle\geq\alpha^{S_{1}^{i_{1}i_{2}}+S_{3}^{i_{1}i_{2}}}\alpha^{(2L-q-1 )S_{1:3}^{i_{1}i_{2}}}\gamma^{(2L-q+1)S_{0,3}^{i_{1}i_{2}}}\alpha^{S_{1:3}^{i_ {1}i_{2}}-1}\beta\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta_{1}}}... \mathop{\mathbb{E}}\limits_{\boldsymbol{\theta_{q-1}}}(\text{Tr}[\boldsymbol{O }_{3:i_{1}}^{\prime}\rho_{q-1}]\text{Tr}[\boldsymbol{O}_{3:i_{2}}^{\prime}\rho _{q-1}])$ (S112) $\displaystyle\geq\alpha^{S_{1}^{i_{1}i_{2}}+S_{3}^{i_{1}i_{2}}}\alpha^{(2L-1)S _{1:3}^{i_{1}i_{2}}}\gamma^{2LS_{0,3}^{i_{1}i_{2}}}\beta\text{Tr}[\boldsymbol{ O}_{3:i_{1}}^{\prime}\rho_{0}]\text{Tr}[\boldsymbol{O}_{3:i_{2}}^{\prime}\rho_ {0}]$ (S113) $\displaystyle\geq\alpha^{2LS_{1:3}^{i_{1}i_{2}}-1}\gamma^{2LS_{0,3}^{i_{1}i_{2 }}}\beta$ (S114) $\displaystyle\geq(1-\sigma^{2})^{2LS_{1:3}^{i_{1}i_{2}}-1}e^{-L\sigma^{2}S_{0, 3}^{i_{1}i_{2}}}\sigma^{2}(1-\sigma^{2})$ (S115) $\displaystyle=\frac{1}{2LS}(1-\frac{1}{2LS})^{2LS_{1:3}^{i_{1}i_{2}}}e^{-\frac {S_{0,3}^{i_{1}i_{2}}}{2S}},$ (S116) $$

Eq. (S111) to Eq. (S116) follow a similar analysis to Eq. (S69) and Eq. (S73). When $n\in P_{1:3}^{ij}$, a similar analysis reveals that when $q=2L-1$,

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}}(\frac{\partial f (\theta)}{\partial\theta_{q,n}})^{2}\geq\frac{1}{2LS}(1-\frac{1}{2LS})^{2LS_{1 :3}^{i_{1}i_{2}}}e^{-\frac{S_{0,3}^{i_{1}i_{2}}}{2S}},$ (S117) $$

Figure: Figure S3: At each position, depending on the different initial Pauli matrices, various terms are generated. This indicates that when the initial Pauli matrix at any position belongs to $\{XX,YY,ZZ,II,IZ,ZI\}$, it shows the transformation of the Pauli matrix and the corresponding coefficients. When the Pauli matrix undergoes a CZ gate, according to Lemma 5, it may involve flipping operations. Here, it illustrates the scenario when no flips exist, showcasing the changes in the Pauli matrix.
Refer to caption: x6.png

Figure: Figure S4: As before, it illustrates changes in the Pauli matrix. However, in this case, we assume that the CZ gate introduces flip operations.
Refer to caption: x7.png

and when $q=2L$, $\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}}(\frac{\partial f(\theta)}{
\partial\theta_{q,n}})^{2}\geq 0$. Fig. S3 and S4 illustrate the evolution of the first cross-terms in Eq. S103 for different configurations of Pauli matrices at each position. According to Lemma 5, $CZ$ may execute a flip operation. Therefore, we discuss two scenarios: one where no flip occurs, as shown in Fig. S3, and another where $CZ$ causes a flip of Pauli matrices, as depicted in Fig S4. As mentioned earlier, we find that if the k-th Pauli matrix is to undergo a flip operation, we require the (k-1)-th position to have a Pauli matrix of $X$ or $Y$, and the (k+1)-th position to have a Pauli matrix of $Z$ or $I$, or vice versa. Taking into account that some terms in the evolution of $iGO\rho$ may yield coefficients with negative signs, our specific setup ensures that when the coefficient for the preceding Pauli matrix becomes negative, the succeeding Pauli matrix will also inevitably have a negative coefficient. Consequently, the final coefficients are positive. When $n\notin P_{1:3}^{ij}$, i.e., $n\in P_{0}^{i,j}$, we can easily deduce that $\mathop{\mathbb{E}}\limits_{\boldsymbol{\theta}}(\frac{\partial f(\theta)}{
\partial\theta_{q,n}})^{2}\geq 0$.
In conclusion, we can draw the following conclusions:

$$ $\displaystyle\mathop{\mathbb{E}}\limits_{\theta}||\nabla_{\bm{\theta}}f(\theta )||^{2}_{2}$ $\displaystyle\geq M(\frac{1}{4}-\frac{1}{8L})+\sum_{i\neq j=1}^{M}\frac{(2L-1) S_{3}^{ij}}{2LS}(1-\frac{1}{2LS})^{2LS_{1:3}^{ij}}e^{-\frac{S_{0,3}^{ij}}{2S}} +\sum_{q,n,j}E\left(\frac{\partial f_{j}^{\prime\prime}(\theta)}{\partial \theta_{q,n}}\right)^{2}$ (S118) $\displaystyle\geq M(\frac{1}{4}-\frac{1}{8L})+\sum_{i\neq j=1}^{M}\frac{(2L-1) S_{3}^{ij}}{2LS}(1-\frac{1}{2LS})^{2LS_{1:3}^{ij}}e^{-\frac{S_{0,3}^{ij}}{2S}}$ (S119) $$

## VI Additional numerical experiments and details

### VI.1 Experiments with arbitrary global cost functions

Finally, we randomly generate some global observables to calculate their initial gradients. In this case, the cost function is given by $f(\bm{\theta})=\text{Tr}[(\sum_{i=1}^{10}\bm{O}_{i}-\sum_{j=1}^{10}\bm{O}_{j})
U(\theta)\rho_{in}U^{\dagger}(\theta)]$, where the Pauli matrices in $\bm{O}_{i}$ and $\bm{O}_{j}$ are randomly selected from $\{X,Y,Z\}$. We set $L$ to be 2 and computed $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||_{2}^
{2}$ for different numbers of qubits $N$. The results are presented in Table S4. Given that each term is global and excludes Pauli $I$, in this case, $M=1$. Consequently, according to Theorem 2, our lower bound on $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||_{2}^
{2}$ is 0.25. From the results, it is evident that with an increase in the number of qubits, the $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||_{2}^
{2}$ for Gaussian, uniform, and reduced-domain distributions undergoes a sharp reduction. While our method also exhibits a decreasing trend in $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||_{2}^
{2}$, it aligns closely with the outcome predicted by Theorem 2 and significantly surpasses other methods by several orders of magnitude.

**Table S4: Comparison of initial gradients norm $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||_{2}^ {2}$ for different methods at various numbers of qubits.**
| $N$ | GMM | Gaussian | Uniform | Reduced-domain |
| --- | --- | --- | --- | --- |
| 5 | 1.26 | 0.99 | 2.02 | 1.21 |
| 10 | 0.75 | $2.86\times 10^{-2}$ | 0.41 | $6.22\times 10^{-2}$ |
| 15 | 0.73 | $1.92\times 10^{-7}$ | $6.65\times 10^{-2}$ | $8.56\times 10^{-4}$ |
| 20 | 0.74 | $3.47\times 10^{-16}$ | $8.78\times 10^{-3}$ | $4.61\times 10^{-6}$ |
| 25 | 0.74 | $2.55\times 10^{-23}$ | $1.37\times 10^{-3}$ | $6.87\times 10^{-8}$ |

### VI.2 Simulated experiments in quantum chemistry

In the following, we explore the application of our initialization method to compute the ground-state energy of the LiH molecule, a benchmark in quantum chemistry. For an electronic system with $N$ electrons distributed over $M$ spin molecular orbitals, the initial state is the Hartree-Fock (HF) state:

$$ $|\Phi\rangle_{HF}=|\underbrace{\overbrace{11...11}^{N}00...00}_{M}\rangle.$ $$

In the LiH molecule, with an electron count of $N=2$ and $M=10$ free spin orbitals, simulating electronic structure problems on a quantum computer requires establishing a mapping that transforms fermionic operators of electrons into Pauli operators. Common mappings include the Jordan-Wigner (JW) transformation, Bravyi-Kitaev (BK) transformation, and Parity transformation. Here, we adopt the JW mapping to compute its ground-state energy.

Figure: Figure S5: When $L=10$, we examine the variation of the cost function and $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||^{2}$ under noisy and noise-free conditions, using both uniform distribution ($\mathcal{U}[-\pi,\pi]$) and GMM-initialized parameters. Where (a) and (b) represent the noise-free scenario, while (c) and (d) represent the case with noise.
Refer to caption: x8.png

Figure: Figure S6: When $L=20$, (a) and (c) depict the loss function under noise-free and noisy conditions, respectively, with a uniform distribution ($\mathcal{U}[-\pi,\pi]$) and GMM-initialized parameters. On the other hand, (b) and (d) illustrate the changes in $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||^{2}$ under noise-free and noisy conditions, respectively.
Refer to caption: x9.png

Figure: Figure S7: When $L=30$, (a) illustrates the variation of the loss under noise-free conditions; (b) depicts $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||^{2}$ under noise-free conditions; (c) shows the change in loss under noisy conditions; and (d) displays $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||^{2}$ under noisy conditions.
Refer to caption: x10.png

We set the number of layers ($L$) to 10, 20, and 30, using a gradient descent optimizer with a learning rate of 0.01. Additionally, we consider the impact of the noise on the barren plateau problem by introducing a moderate amount of noise during training to simulate real-world quantum computer operation. We compare the evolution of the cost function and $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||^{2}$ during training when initializing parameters using GMM and uniform distribution $\mathcal{U}[-\pi,\pi]$. The results are shown in Fig. S5, S6, and S7. In each figure, (a) and (b) represent the condition without noise, while (c) and (d) represent the noisy condition. From the results, we observe that regardless of the value of $L$ or the presence of noise, initializing parameters using the GMM method consistently provides a larger $\mathop{\mathbb{E}}\limits_{\bm{\theta}}||\nabla_{\bm{\theta}}f(\theta)||^{2}$ at the beginning of training and it consistently stays much higher than the lower bound we have provided. This value remains relatively high before the convergence of the cost function, therefore, the GMM initialization ensures a rapid convergence. On the other hand, the uniform distribution $\mathcal{U}[-\pi,\pi]$ maintains a consistently lower level of gradient norm, resulting in a significantly slower convergence process.

Next, let’s consider the impact of the parameter $\sigma^{2}$ in the GMM. In the main text, we set $\sigma^{2}$ to be $\frac{1}{2LS}$. We compare the training scenarios with different $\sigma^{2}$ values under noisy and noise-free conditions when $L=10,20,30$. Here, $\sigma^{2}$ is chosen as $0.1\times\frac{1}{2LS}$, $\frac{1}{2LS}$, and $10\times\frac{1}{2LS}$. The results are shown in Fig. S8, S9, and S10.

As before, (a) and (b) represent noise-free conditions, while (c) and (d) represent scenarios with noise. The results in the figures indicate that when $\sigma^{2}=10\times\frac{1}{2LS}$, the convergence of the cost function is significantly slower. On the other hand, when $\sigma^{2}=0.1\times\frac{1}{2LS}$, although the cost function converges, its results are often inferior to the original case, especially in the presence of noise. We believe that as $\sigma^{2}$ increases, the peaks of the probability density function in the GMM become lower, and its distribution becomes closer to the uniform distribution, leading to a smaller KL divergence between them. Conversely, when $\sigma^{2}$ decreases, the peaks of the GMM’s probability density function become higher. Therefore, the data becomes more concentrated around the peaks, making it less dispersed. This may be the reason why the convergence results are not as good as when $\sigma^{2}=\frac{1}{2LS}$.

Figure: Figure S8: In the configuration with $L=10$, the impact of different $\sigma^{2}$ on training under noisy and noise-free conditions is depicted. Here, (a) and (b) represent the noise-free scenario, while (c) and (d) represent the noisy scenario.
Refer to caption: x11.png

Figure: Figure S9: For a 20-layer configuration, the impact of different $\sigma^{2}$ on training under noisy and noise-free conditions is depicted. Here, (a) and (b) represent the noise-free scenario, while (c) and (d) represent the noisy situation.
Refer to caption: x12.png

Figure: Figure S10: In the $L=30$ configuration, (a) and (b) illustrate the impact of different $\sigma^{2}$ on training under noise-free conditions, while (c) and (d) depict the influence of various $\sigma^{2}$ under noisy conditions.
Refer to caption: x13.png

## VII Discussion reamrk

According to Lemma 5, we observe that when Pauli matrices are limited to $I$ and $Z$, the CZ gate does not alter their forms. In other words, $CZ^{\dagger}(o_{i}\otimes o_{j})CZ=o_{i}\otimes o_{j}$ for all $o_{i},o_{j}\in{I,Z}$. Therefore, $CZ_{l}$ can be any combination of CZ gates, and it only changes the conditions for ’flip,’ which does not affect our results. Also, although our method is specifically effective for the $R_{x}-R_{y}$ gate structure, it can be readily extended to other combinations of rotation gates. For instance, as shown in Theorem 2, if we interchange the positions of $R_{x}$ and $R_{y}$ in the arrangement of rotation gates, i.e., the arrangement is $R_{y}-R_{x}$, then we initialize the parameters of the last block according to Table S5, and the initialization of parameters in other layers follows the distribution $\mathcal{G}_{1}(\sigma^{2})$. Alternatively, when the rotation gates consist of three $R_{x}-R_{y}-R_{x}$ gates, under the same conditions as in Theorem 1, we initialize the parameters of the last block as shown in Table S6, and the initialization of parameters in other layers follows the distribution $\mathcal{G}_{1}(\sigma^{2})$. In both cases, the results are consistent with those of Theorem 1. Certainly, our analysis method remains applicable when using CNOT to provide entanglement.

**Table S5: For the $R_{y}-R_{x}$ gate structure, we initialize the parameters $\theta$ in both $R_{y}(\theta)$ and $R_{x}(\theta)$ gates using a Gaussian distribution $\mathcal{G}_{1}(\sigma^{2})$.**
| $o_{i}$ | X | Y | Z | I |
| --- | --- | --- | --- | --- |
| Init method of $R_{x}(\theta)$ | $\mathcal{G}_{2}(\sigma^{2})$ | $\mathcal{G}_{1}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ |
| Init method of $R_{y}(\theta)$ | $\mathcal{G}_{1}(\sigma^{2})$ | $\mathcal{G}_{2}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ |

**Table S6: For the $R_{x}-R_{y}-R_{x}$ gate structure, we initialize the parameters $\theta$ in both $R_{y}(\theta)$ and $R_{x}(\theta)$ gates using a Gaussian distribution $\mathcal{G}_{1}(\sigma^{2})$.**
| $o_{i}$ | X | Y | Z | I |
| --- | --- | --- | --- | --- |
| Init method of first $R_{x}(\theta)$ | $\mathcal{G}_{3}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ |
| Init method of $R_{y}(\theta)$ | $\mathcal{G}_{1}(\sigma^{2})$ | $\mathcal{G}_{2}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ |
| Init method of second $R_{x}(\theta)$ | $\mathcal{G}_{1}(\sigma^{2})$ | $\mathcal{G}_{2}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ | $\mathcal{G}_{3}(\sigma^{2})$ |

## Peer Reviews

confidence: 3
contribution: 2
initial_rating: 5
presentation: 2
questions: 1 In Figure2, how to determine the inactive parameters
2 in theorem1, what’s the impact of the partial of f(\theta)
3 in theorem 2, there is no definition of M, how to determine the value of M for the different number of layers.
4 what’s more, there is no comparison with other approaches,
soundness: 2
strengths: The novel parameter initialization strategy is provided for the barren plateau phenomenon. And the prove is given,
summary: This paper considers the variational quantum algorithms and deal with the barren plateau phenomenon. The new parameter initialization strategy is proposed combing with gaussian mixture models. The prove is provided that the initialization could avoids the barren plateaus problem.
weaknesses: The provided parameter initialization strategy is not clear in the figure 1. What’s more, the comparison with other methods is not given. Furthermore, the induced Gaussian mixture models is not the firstly introduced.

---

confidence: 3
contribution: 2
initial_rating: 3
presentation: 1
questions: - If I understand the result correctly, this only prevents “barren plateau” at initialization. I wonder if there is any comment the authors can make about the optimization trajectory (can you say anything about the norm of the gradient other than initialization?)
- I understand the above point is empirically argued e.g., in Figure 4. But the interpretation starting from line 413: “Moreover, the gradient norm remains within a relatively large range throughout the entire training process. This enables our approach to escape … vanishing gradient problem … . *These observations are entirely consistent with the conclusions drawn in Theorem 1*.” But isn’t Theorem 1 ONLY about initialization?
- In Theorem 1, the assumption of the parameters $\theta$ is that it follows $\mathcal{G}_1(\sigma^2)$ ? But this is just the Gaussian distribution $\mathcal{N}(0, \sigma^2)$. Could you explain how to interpret this? Why is this not compared to [1]? (other than one line sentence in line 228, “This is in stark contrast to the exponential lower bound $O(1/L^N)$ found in previous works for global cost functions Zhang et al. (2022a); Wang et al. (2023).)
- How is the experiment set up? Are these results of actually applying VQA to a quantum computer? Or are these some numerical simulations?
- Line 370: “…we compare our proposed method with… , Gaussian distribution $\mathcal{N} (0, \frac{1}{4S(L+2)})$ : is this variance taken from [1, Theorem 4.1]? It should really be cited clearly…
- Can you say anything about the solution quality? (converged $\theta$ after some iterations).
soundness: 1
strengths: This paper proposes and proves that the mixture of gaussians as an initialization scheme avoids barren plateau (at initialization) even when the cost function is global. Experiments, even though settings are a bit unclear, seem to support their claim.
summary: This paper proposes using mixture of gaussians as initialization for variational quantum algorithms using mixture of gaussians. Theoretical results show the expectation of the norm of the gradient following the assumed distribution is lower bounded.
weaknesses: - My biggest concern is that the relation to the previous work [1] is hardly discussed. The authors should at least mention that [1] proposed for the first time the gaussian initialization precisely to prevent barren plateau at initialization—the exact setting this paper is addresses. I understand there are some differences discussed briefly starting from line 258, but even there the authors do not mention [1] uses Gaussian initialization. Such writing gives me the impression that the authors intentionally hide due to significant similarity with [1].
- I believe some people refer to “barren plateau” not only at the initialization but also more generally, i.e., vanish exponentially with the size of the system, c.f. [2, 3]. The authors should clearly state the “barren plateau” that they mention is only with regards to initialization; this is only mentioned in line 52 as: “the phenomenon of the barren plateau is characterized by the *randomized initialization* of parameters $\theta$ in VQAs,”…
- Continuing the above point, in my opinion it is misleading for instance to write in line 87 as: 
$\theta_{k+1} = \theta_k - \alpha \nabla_\alpha f(\theta_k)$ … Therefore, typically $|| \nabla_\theta f (\theta_k) ||^2$ is used to determine whether the cost function can be updated.” 
Given that this paper is only about initialization, what it really shows is that $|| \nabla_\theta f (\theta_0) \|^2||$ has significant magnitude, but the result does not say *anything* about $||\nabla_\theta f (\theta_k) \|^2||$ for $k > 1$.
- Please use parenthesized citations correctly; it’s very hard to read especially since the citation text colors are the same as the main text color
- Typos:
    - line 214: “Then We expand…”
    - line 231: wrong quotation marks for “inactive parameters”, etc

[1] Zhang, et al. (2022) “Escaping from the Barren Plateau via Gaussian Initializations in Deep Variational Quantum Circuits” 

[2] Fontana et al. (2024) “Characterizing barren plateaus in quantum ansätze with the adjoint representation”

[3] Loracca et al. (2024) “A Review of Barren Plateaus in Variational Quantum Computing”

---

confidence: 4
contribution: 2
initial_rating: 5
presentation: 2
questions: * You claim your method "avoids barren plateau". However, the theoretical results only guarantee barren plateau is avoided at initialization. Do you have any insight on how this method may help avoid this phenomenon during training? 
* Where does the eq for $f(\theta_{k+1})$ on line 87 come from?
* What is the importance of the result given by eq. (5)?
* How do you achieve a bound that does not depend on the number of qubits $N$ for Theorem 1? This seems surprising to me.
* You interchangeably use $O$ and $\mathbf{O}$ to describe observables. You also index this $O$ sometimes. Is this notation you did not define or simply a typo?
* In Figure 4, your method seems to reach the desired solution, but then as iterations continue, it diverges away before coming back. What explains this phenomenon you think?
soundness: 3
strengths: 1. As far as I am concerned, this is a novel contribution in the field of quantum machine learning.
2. Given it concerns only parameter initialization, the proposed initialization scheme is simple and easy to implement and compute efficient.
3. The method is backed up by solid theoretical guarantees which are also validated empirically through experiments. Also, the theoretical guarantees hold for rather practical situations, not just an idealized case.
4. The proofs in the appendix seem correct and are easy to follow.
summary: The paper presents a novel initialization scheme for parameterized quantum circuits optimized with variation quantum algorithms (VQA). The method employs a strategy based on Gaussian Mixture Models (GMMs) to initialize the parameters of an $L$ block $N$ qubit ansatz. The paper claims that for the considered ansatz, the initialization scheme avoids the barren plateau phenomenon (BP).

Theoretically, the authors prove lower bounds on the expectation of the gradient norm under three different assumptions for the observable in the loss function. By showing the expected gradient norm is nonzero, they are able to theoretically guarantee absence of the barren plateau at initialization

The authors also validate their initialization strategy on synthetic experiments. These experiments confirm that this initialization scheme indeed avoids the BP in practice as well.
weaknesses: 1. Although your method is applied to a popular ansatz, it does not seem to generalize to other ansatz structures. Could you discuss potential for generalizing to different ansatz?
2. I found the paper to be written in a style that hinders understanding. There are various errors/inconsistencies in the notation and long inline math sections (around lines 158, 236, 250 for instance) that I personally found difficult to read. Please see the minor comments section for examples. It would be constructive to break up long inline math sections and give more context around complex mathematical equations.
3. It seems as though you did not validate your experiments using multiple runs/seeds. Number of runs/standard error/variance is not reported. If you indeed ran your experiments only once, this would be a major weakness of your experimental section; given your method is based on pseudorandom initialization, this would hinder the statistical validity of the results. Would it be possible for you to provide results from multiple runs along with error bars and confidence intervals?

Also here are some minor comments you may want to address for the final version:
* Generally, it would be clearer if you define all variables present in a theorem in the theorem statement for clarity
* In lines 86-100, you introduce the VQA problem and define the cost function. This should go in the notation/background section.
* Line 54 typo BP is underlined for citation
* Line 73 typo "expressibilityRagone et al."
* Line 227 typo, you have a citation in your big O
* Line 1131 "Theorem 1" should be "Lemma 1" I believe
* Generally inconsistent use of $cos$ and $\cos$
* In the proofs, inconsistent use of $I_S$ vs $I_s$

---

confidence: 3
contribution: 1
initial_rating: 3
presentation: 1
questions: Please refer to the section above as often question associates to the weaknesses of the paper.
soundness: 1
strengths: - The paper adresses a fundamental problem in the context of variational quantum algoritms.
summary: Variational Quantum Algorithms (VQAs) are important tools to exploit the capabilities of Noisy Intermediate Scale Quantum devices. The basic idea is to leverage parametrized quantum circuits (PQCs), whose parameters are often angles of rotational gates, to find the ground state of a given Hamiltonian, which plays the role of the cost function. 

VQAs, however, suffer from several shortcomings. One of the most sever one is known as Barren Plateaus (BPs). When dealing with many qubits and layers in the quantum circuit, the number of variational parameters rapidly increases thus making the optmization problem more challenging. Furthermore, the optimization landscape becomes extrmely difficult to navigate and this results in very small gradient signals (during the optimization of the parametrized quantum circuits) which prevents to converge to global minima and, therefore, to the desired ground state wave function. 

In this paper the authors propose to use Gaussian Mixture Models (GMMs) for the initialization of the parameters in PQCs.  The main contribution of the paper claims to **rigorously** solve the problem of **barren plateaus** by initializing the parameters of PQCs which in turn leads to higher gradient signals throughout the optimization of the parameters.
weaknesses: - The major weakness of this paper lies in its claim. In the abstract the authors claim that "*rigorously prove that, the proposed initialization method consistently avoids the barren plateaus problem for hardware-efficient ansatz*". While this may even be true in theory, I do find their numerical experiments and the PQCs considered in the paper not to be general such that this claim can hold with the current phrasing. This very **strong** claim is reiterated several times throughout the paper. I strongly advise to motigate such claim to something more aligned with the results of the manuscript.  
- Another important weakness of the paper is that the code for reproducing the experiments is not provided thus preventing reproducibility of the experiments and further investigation of the implementation. 
- Furthermore, I find the presentation of the paper hard to follow at times. The notation is often confusing cluttered. I believe that recalling what each variable refers to, use more display math (instead of inline equations) and provide some intuitive sketches may help.
- The ansatzes used in the paper are not general and do not expliclty account for entanglement. As it is known in the field of quantum computing, a set of universal gates, e.g., a collection of gates able to represent any possible unitary transformation, consist of the rotational gates and entangling gates such as CNOT. I believe this aspect substantially limits the generizability of the proposed claims and theorems. Does the set of gates considered in this work represents a set of universal gates? If yes that should be made clear. 
- I find the numerical results shown in the main paper to be somewhat inconclusive. In particular, the authors omit many fundamental details such as the type of algorithm used to optimize the PQC, if it consists of global or local optimizations at each iteration, if measurement noise and/or hardware noise are taken into account for these experiments, whether the results change when with different algorithms etc. I believe that a thorough ablation study is necessary in order to support the strong claims of this work. 
- I furthermore find puzzling that other results on quantum chemistry simulation are limited to the appendix as I believe those may arguably be even more relevant (or at least complementary) compared to the Ising model.
- About the layout of the paper, it looks to me as if Table 2 and Table 3, as well as Table 5 and Table 6 are duplicates of each other. Could it be, or am I missing something?
- Another concern about the layout of the paper is about the citation style. I strongly recommend to adopt fix the citation styles, e.g., substituting \cite with \citep where needed in order to wrap refs around parenthesis where suitable. 
- The notion of "observable" may not be immediate to grasp for the general audience, I believe it would be useful to provide more concrete examples of real physical examples which could be mapped onto these generalized $\mathbf{\mathcal{O}}$ discussed in the paper. 
- The writing and the clarity of the paper I think can in general be improved. 
- Table 4: I think it would be good to mention that higher gradients at initialization is better. This may not be intuitive at first sight. Perhaps making the best results in bold without  editing the caption would already help. 
- Line 32: I find the claim "*[...] VQAs provides a feasible approach to solving complex problems [...]*" to be far too general. I strongly encourage to be more specific about what VQA can be good at. 
- Line 83: I recommend to change the wording *complex distribution* to complicated/non-trivial as the former may be confused to have different meaning, e.g., distribution of complex values.
- I storngly recommend to add a "Related work" section to give more structure to the paper and streamline the reading. Furthermore, I'd recommend a more thorough review citing also other work using ML methods to enhance optimization of PQCs such as Refs. [1-3] below. Similarly, I recommend to provide the list of contributions at the bottom of page 2 in bulletpoints so that they end up being more accessible and more evident to readers. 
- For the people non familiar with optimization of PQCs, I'd briefly introduce the parameter shift rule and how to compute gradients on quantum computers at the begining of the last paragraph before section 2. I believe this would be useful to make the paper self-contained.   
- Line 158: Something seems wrong with the parenthesis in the last equation of the sentence. As mentioned above, I strongly recommend to use more dispay math instead of inline equations which are often cluttered and hard to parse. If the authors need space I'd recommend removing one of the two figures Figure 1 and Figure 2 as I think the key messages therein can be merged into one figure. 
- On the other hand I'd find beneficial to have some intuitive sketch about the main results/theorem the authors claim in the paper. That might make the paper more accessible also to a general audience (more from the ML community). At the moment the paper seems very much suited to an audience of physicists. For instance the authors never define what's a pure state. This cannot be assumed as common knowledge in the broader audience targeting this conference. 
- To streamline the reading of the paper I'd find it useful to often recall what $q,n$ and $L,N,M$ are. That would help a lot to navigate both theorems and follow the sketch of proofs. 
- Why does the gradient norm shown in figure 4 (right panel) has this double peak structure? Does this has any physical meaning? Is it intuitive why someone should expect such a steep increase in gradient norm during optimization? I believe this associates to the capability of the proposed algorithms to overcome barren plateaus but this is not discussed explicitly neither in the caption nor in the text. This might make it hard to the reader not familiar with the problem of Barren Plateaus to immediately grasp this. 
- In line 470: "*We validate our algorithm for diverse problems, [...]*" I think this is not entirely correct. The paper only tackles (in the main part) the Transverse Field Ising Model with different setup. I think the authors should be clearer and more explicit here. This comment often applies to other parts in the paper where it would be useful to revisit the paper in ordare to ensure more precise claims. 
- In line 483: what does the HEA acronym mean?

### References

- [1] [Tamiya, Shiro, and Hayata Yamasaki. "Stochastic gradient line Bayesian optimization for efficient noise-robust optimization of parameterized quantum circuits." npj Quantum Information 8.1 (2022): 90.](https://www.nature.com/articles/s41534-022-00592-6)
- [2] [Nicoli, Kim, et al. "Physics-informed bayesian optimization of variational quantum circuits." Advances in Neural Information Processing Systems 36 (2024).](https://proceedings.neurips.cc/paper_files/paper/2023/file/3adb85a348a18cdd74ce99fbbab20301-Paper-Conference.pdf)
- [3] [Anders, Christopher J., et al. "Adaptive Observation Cost Control for Variational Quantum Eigensolvers." Forty-first International Conference on Machine Learning.](https://openreview.net/pdf?id=dSrdnhLS2h)