SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\sum_ {k, k ^ {\prime}} \frac {\operatorname {t r} \left(Q _ {1 , k , k ^ {\prime}} ^ {\dagger} A\right)}{\operatorname {t r} \left(Q _ {1 , k , k ^ {\prime}} ^ {\dagger} Q _ {1 , k , k ^ {\prime}}\right)} Q _ {1, k, k ^ {\prime}} = \bar {P} A \bar {P}, \tag {S11}
$$
$$
\frac {\operatorname {t r} \left(Q _ {2} ^ {\dagger} A\right)}{\operatorname {t r} \left(Q _ {2} ^ {\dagger} Q _ {2}\right)} Q _ {2} = \frac {\operatorname {t r} (P A)}{d _ {\text {s u b}}} P.
$$
Alternatively, Eq. (S10) can just be seen as a result of the block matrix multiplication, i.e.,
$$
\begin{array}{l} \mathbb {E} _ {\mathbb {U}} \left[ U ^ {\dagger} A U \right] = \mathbb {E} _ {\mathbb {U}} \left[ (\bar {P} + P U ^ {\dagger} P) A (\bar {P} + P U P) \right] \\ = \mathbb {E} _ {\mathbb {U}} \left[ \bar {P} A \bar {P} + \bar {P} A P U P + P U ^ {\dagger} P A \bar {P} + P U ^ {\dagger} P A P U P \right] \tag {S12} \\ = \bar {P} A \bar {P} + \frac {\operatorname {t r} (P A P)}{d _ {\mathrm {s u b}}} P, \\ \end{array}
$$
where  $\operatorname{tr}(PAP) = \operatorname{tr}(P^2A) = \operatorname{tr}(PA)$ .
Corollary S4 Suppose  $|\varphi\rangle$  is a Haar-random pure state in  $\mathcal{H}_{\mathrm{sub}}$ . For arbitrary linear operators  $A$  on  $\mathcal{H}$ , the following equality holds
$$
\mathbb {E} _ {\varphi} [ \langle \varphi | A | \varphi \rangle ] = \frac {\operatorname {t r} (P A)}{d _ {\text {s u b}}}, \tag {S13}
$$
where  $\mathbb{E}_{\varphi}[\cdot ]$  is the expectation with respect to the random state  $|\varphi \rangle$
Proof Suppose  $|\varphi_0\rangle$  is an arbitrary fixed state in  $\mathcal{H}_{\mathrm{sub}}$ . The random state  $|\varphi \rangle$  can be written in terms of  $U\in \mathbb{U}$  as  $|\varphi \rangle = U|\varphi_0\rangle$  such that
$$
\mathbb {E} _ {\varphi} \left[ \langle \varphi | A | \varphi \rangle \right] = \mathbb {E} _ {U \in \mathbb {U}} \left[ \langle \varphi_ {0} | U ^ {\dagger} A U | \varphi_ {0} \rangle \right]. \tag {S14}
$$
Eq. (S13) is naturally obtained from Lemma S3 by taking the expectation over  $|\varphi_0\rangle$  which satisfies  $P|\varphi_0\rangle = |\varphi_0\rangle$  and  $\tilde{P} |\varphi_0\rangle = 0$ .
Lemma S5 For arbitrary linear operators  $A, B, C$  on  $\mathcal{H}$  and  $U \in \mathbb{U}$ , the following equality holds
$$
\begin{array}{l} \mathbb {E} _ {\mathbb {U}} \left[ U ^ {\dagger} A U B U ^ {\dagger} C U \right] \\ = \bar {P} A \bar {P} B \bar {P} C \bar {P} + \frac {\operatorname {t r} (P B)}{d _ {\mathrm {s u b}}} \bar {P} A P C \bar {P} + \frac {\operatorname {t r} (P C)}{d _ {\mathrm {s u b}}} \bar {P} A \bar {P} B P + \frac {\operatorname {t r} (P A)}{d _ {\mathrm {s u b}}} P B \bar {P} C \bar {P} \\ + \frac {\operatorname {t r} (P A \bar {P} B \bar {P} C)}{d _ {\text {s u b}}} P + \frac {\operatorname {t r} (P A P C) \operatorname {t r} (P B)}{d _ {\text {s u b}} ^ {2}} P \tag {S15} \\ + \frac {d _ {\mathrm {s u b}} \operatorname {t r} (P A) \operatorname {t r} (P C) - \operatorname {t r} (P A P C)}{d _ {\mathrm {s u b}} (d _ {\mathrm {s u b}} ^ {2} - 1)} \left(P B P - \frac {\operatorname {t r} (P B)}{d _ {\mathrm {s u b}}} P\right). \\ \end{array}
$$
Proof Here we simply employ the block matrix multiplication to prove this equality. We denote the  $2 \times 2$  blocks with indices  $\begin{pmatrix} 11 & 12 \\ 21 & 22 \end{pmatrix}$  respectively where the index 2 corresponds to  $\mathcal{H}_{\mathrm{sub}}$ . Thus the random unitary  $U$  can be written as  $U = \begin{pmatrix} I_{11} & 0 \\ 0 & U_{22} \end{pmatrix}$  where  $I_{11}$  is the identity matrix on the orthogonal complement of  $\mathcal{H}_{\mathrm{sub}}$  and  $U_{22}$  is a Haar-random unitary on  $\mathcal{H}_{\mathrm{sub}}$ . The integrand becomes
$$
U ^ {\dagger} A U B U ^ {\dagger} C U = \left( \begin{array}{c c} A _ {1 1} & A _ {1 2} U _ {2 2} \\ U _ {2 2} ^ {\dagger} A _ {2 1} & U _ {2 2} ^ {\dagger} A _ {2 2} U _ {2 2} \end{array} \right) \left( \begin{array}{c c} B _ {1 1} & B _ {1 2} \\ B _ {2 1} & B _ {2 2} \end{array} \right) \left( \begin{array}{c c} C _ {1 1} & C _ {1 2} U _ {2 2} \\ U _ {2 2} ^ {\dagger} C _ {2 1} & U _ {2 2} ^ {\dagger} C _ {2 2} U _ {2 2} \end{array} \right). \tag {S16}
$$
The four matrix elements of the multiplication results are
11:  $A_{11}B_{11}C_{11} + A_{12}U_{22}B_{21}C_{11} + A_{11}B_{12}U_{22}^{\dagger}C_{11} + A_{12}U_{22}B_{22}U_{22}^{\dagger}C_{21}$
12:  $A_{11}B_{11}C_{12}U_{22} + A_{12}U_{22}B_{21}C_{12}U_{22} + A_{11}B_{12}U_{22}^{\dagger}C_{22}U_{22} + A_{12}U_{22}B_{22}U_{22}^{\dagger}C_{22}U_{22},$
21:  $U_{22}^{\dagger}A_{21}B_{11}C_{11} + U_{22}^{\dagger}A_{22}U_{22}B_{21}C_{11} + U_{22}^{\dagger}A_{21}B_{12}U_{22}^{\dagger}C_{21} + U_{22}^{\dagger}A_{22}U_{22}B_{22}U_{22}^{\dagger}C_{21}$
22:  $U_{22}^{\dagger}A_{21}B_{11}C_{12}U_{22} + U_{22}^{\dagger}A_{22}U_{22}B_{21}C_{12}U_{22} + U_{22}^{\dagger}A_{21}B_{12}U_{22}^{\dagger}C_{22}U_{22}$
$$
+ U _ {2 2} ^ {\dagger} A _ {2 2} U _ {2 2} B _ {2 2} U _ {2 2} ^ {\dagger} C _ {2 2} U _ {2 2}. \tag {S17}
$$
Since inhomogeneous Haar integrals always vanish on  $\mathcal{H}_{\mathrm{sub}}$ , the elements above can be reduced to
11:  $A_{11}B_{11}C_{11} + A_{12}U_{22}B_{22}U_{22}^{\dagger}C_{21}$
12:  $A_{11}B_{12}U_{22}^{\dagger}C_{22}U_{22},\quad 21: U_{22}^{\dagger}A_{22}U_{22}B_{21}C_{11},$  (S18)
22:  $U_{22}^{\dagger}A_{21}B_{11}C_{12}U_{22} + U_{22}^{\dagger}A_{22}U_{22}B_{22}U_{22}^{\dagger}C_{22}U_{22}$
Let  $d_{2} = d_{\mathrm{sub}} = \dim \mathcal{H}_{\mathrm{sub}}$  and  $I_{22}$  be the identity matrix in  $\mathcal{H}_{\mathrm{sub}}$ . Utilizing Eqs. (S4) and (S5), the expectation of each block becomes
$$
\begin{array}{l} 1 1: A _ {1 1} B _ {1 1} C _ {1 1} + \frac {\operatorname {t r} B _ {2 2}}{d _ {2}} A _ {1 2} C _ {2 1}, \\ 1 2: \frac {\operatorname {t r} C _ {2 2}}{d _ {2}} A _ {1 1} B _ {1 2}, \quad 2 1: \frac {\operatorname {t r} A _ {2 2}}{d _ {2}} B _ {2 1} C _ {1 1}, \\ 2 2: \frac {\operatorname {t r} \left(A _ {2 1} B _ {1 1} C _ {1 2}\right)}{d _ {2}} I _ {2 2} + \frac {\operatorname {t r} \left(A _ {2 2} C _ {2 2}\right) \operatorname {t r} \left(B _ {2 2}\right)}{d _ {2} ^ {2}} I _ {2 2} \tag {S19} \\ + \frac {d _ {2} \operatorname {t r} \left(A _ {2 2}\right) \operatorname {t r} \left(C _ {2 2}\right) - \operatorname {t r} \left(A _ {2 2} C _ {2 2}\right)}{d _ {2} \left(d _ {2} ^ {2} - 1\right)} \left(B _ {2 2} - \frac {\operatorname {t r} \left(B _ {2 2}\right)}{d _ {2}} I _ {2 2}\right). \\ \end{array}
$$
Written in terms of subspace projectors  $P$  and  $\bar{P}$ , the results become exactly as Eq. (S15).
Corollary S6 Suppose  $|\varphi\rangle$  is a Haar-random pure state in  $\mathcal{H}_{\mathrm{sub}}$ . For arbitrary linear operators  $A$  on  $\mathcal{H}$ , the following equality holds
$$
\mathbb {E} _ {\varphi} \left[ \langle \varphi | A | \varphi \rangle^ {2} \right] = \frac {\operatorname {t r} ((P A) ^ {2}) + (\operatorname {t r} (P A)) ^ {2}}{d _ {\mathrm {s u b}} (d _ {\mathrm {s u b}} + 1)}, \tag {S20}
$$
$$
\mathbb {E} _ {\varphi} \left[ \left(\langle \varphi | A | \varphi \rangle\right) ^ {2} \right] = \mathbb {E} _ {U \in \mathbb {U}} \left[ \langle \varphi_ {0} | U ^ {\dagger} A U | \varphi_ {0} \rangle \langle \varphi_ {0} | U ^ {\dagger} A U | \varphi_ {0} \rangle \right]. \tag {S21}
$$
Eq. (S20) is naturally obtained from Lemma S5 by taking  $C = A$  and  $B = |\varphi_0\rangle \langle \varphi_0|$  which satisfies  $\bar{P}B = B\bar{P} = 0, P\bar{B}P = B$  and  $\operatorname{tr} B = 1$ .
Lemma S7 For arbitrary linear operators  $A, B, C, D$  on  $\mathcal{H}$  and  $U \in \mathbb{U}$ , the following equality holds
$$
\begin{array}{l} \mathbb {E} _ {\mathbb {U}} \left[ \operatorname {t r} \left(U ^ {\dagger} A U B\right) \operatorname {t r} \left(U ^ {\dagger} C U D\right) \right] = \operatorname {t r} (\bar {P} A \bar {P} B) \operatorname {t r} (\bar {P} C \bar {P} D) \\ + \frac {\operatorname {t r} (\bar {P} A \bar {P} B) \operatorname {t r} (P C) \operatorname {t r} (P D)}{d _ {\mathrm {s u b}}} + \frac {\operatorname {t r} (\bar {P} C \bar {P} D) \operatorname {t r} (P A) \operatorname {t r} (P B)}{d _ {\mathrm {s u b}}} \\ + \frac {\operatorname {t r} (P B \bar {P} A P C \bar {P} D)}{d _ {\text {s u b}}} + \frac {\operatorname {t r} (P A \bar {P} B P D \bar {P} C)}{d _ {\text {s u b}}} \tag {S22} \\ + \frac {\operatorname {t r} (P A) \operatorname {t r} (P B) \operatorname {t r} (P C) \operatorname {t r} (P D) + \operatorname {t r} (P A P C) \operatorname {t r} (P B P D)}{d _ {\mathrm {s u b}} ^ {2} - 1} \\ - \frac {\mathrm {t r} (P A P C) \mathrm {t r} (P B) \mathrm {t r} (P D) + \mathrm {t r} (P A) \mathrm {t r} (P C) \mathrm {t r} (P B P D)}{d _ {\mathrm {s u b}} (d _ {\mathrm {s u b}} ^ {2} - 1)}. \\ \end{array}
$$
Proof Similarly with the proof of Lemma S5, the block matrix multiplication gives
$$
\begin{array}{l} \operatorname {t r} \left(U ^ {\dagger} A U B\right) = \operatorname {t r} \left[ \left( \begin{array}{c c} A _ {1 1} & A _ {1 2} U _ {2 2} \\ U _ {2 2} ^ {\dagger} A _ {2 1} & U _ {2 2} ^ {\dagger} A _ {2 2} U _ {2 2} \end{array} \right) \left( \begin{array}{l l} B _ {1 1} & B _ {1 2} \\ B _ {2 1} & B _ {2 2} \end{array} \right) \right] \tag {S23} \\ = \mathrm {t r} (A _ {1 1} B _ {1 1}) + \mathrm {t r} (A _ {1 2} U _ {2 2} B _ {2 1}) + \mathrm {t r} (U _ {2 2} ^ {\dagger} A _ {2 1} B _ {1 2}) + \mathrm {t r} (U _ {2 2} ^ {\dagger} A _ {2 2} U _ {2 2} B _ {2 2}). \\ \end{array}
$$
$$
\begin{array}{l} \mathbb {E} _ {\mathbb {U}} \left[ \operatorname {t r} \left(U ^ {\dagger} A U B\right) \operatorname {t r} \left(U ^ {\dagger} C U D\right) \right] = \mathbb {E} _ {\mathbb {U}} \left[ \operatorname {t r} \left(A _ {1 1} B _ {1 1}\right) \operatorname {t r} \left(C _ {1 1} D _ {1 1}\right) \right. \\ + \operatorname {t r} \left(A _ {1 1} B _ {1 1}\right) \operatorname {t r} \left(U _ {2 2} ^ {\dagger} C _ {2 2} U _ {2 2} D _ {2 2}\right) + \operatorname {t r} \left(C _ {1 1} D _ {1 1}\right) \operatorname {t r} \left(U _ {2 2} ^ {\dagger} A _ {2 2} U _ {2 2} B _ {2 2}\right) \tag {S24} \\ + \operatorname {t r} \left(A _ {1 2} U _ {2 2} B _ {2 1}\right) \operatorname {t r} \left(U _ {2 2} ^ {\dagger} C _ {2 1} D _ {1 2}\right) + \operatorname {t r} \left(U _ {2 2} ^ {\dagger} A _ {2 1} B _ {1 2}\right) \operatorname {t r} \left(C _ {1 2} U _ {2 2} D _ {2 1}\right) \\ \left. + \operatorname {t r} \left(U _ {2 2} ^ {\dagger} A _ {2 2} U _ {2 2} B _ {2 2}\right) \operatorname {t r} \left(U _ {2 2} ^ {\dagger} C _ {2 2} U _ {2 2} D _ {2 2}\right) \right]. \\ \end{array}
$$
where all inhomogeneous terms have been ignored. Utilizing Eqs. (S4), (S6) and (S7), the expectation becomes
$$
\begin{array}{l} \mathbb {E} _ {\mathbb {U}} \left[ \operatorname {t r} \left(U ^ {\dagger} A U B\right) \operatorname {t r} \left(U ^ {\dagger} C U D\right) \right] = \operatorname {t r} \left(A _ {1 1} B _ {1 1}\right) \operatorname {t r} \left(C _ {1 1} D _ {1 1}\right) \\ + \frac {\operatorname {t r} \left(A _ {1 1} B _ {1 1}\right) \operatorname {t r} \left(C _ {2 2}\right) \operatorname {t r} \left(D _ {2 2}\right)}{d _ {2}} + \frac {\operatorname {t r} \left(C _ {1 1} D _ {1 1}\right) \operatorname {t r} \left(A _ {2 2}\right) \operatorname {t r} \left(B _ {2 2}\right)}{d _ {2}} \\ + \frac {\operatorname {t r} \left(B _ {2 1} A _ {1 2} C _ {2 1} D _ {1 2}\right)}{d _ {2}} + \frac {\operatorname {t r} \left(A _ {2 1} B _ {1 2} D _ {2 1} C _ {1 2}\right)}{d _ {2}} \tag {S25} \\ + \frac {1}{d _ {2} ^ {2} - 1} \left(\operatorname {t r} \left(A _ {2 2}\right) \operatorname {t r} \left(B _ {2 2}\right) \operatorname {t r} \left(C _ {2 2}\right) \operatorname {t r} \left(D _ {2 2}\right) + \operatorname {t r} \left(A _ {2 2} C _ {2 2}\right) \operatorname {t r} \left(B _ {2 2} D _ {2 2}\right)\right) \\ - \frac {1}{d _ {2} \left(d _ {2} ^ {2} - 1\right)} \left(\operatorname {t r} \left(A _ {2 2} C _ {2 2}\right) \operatorname {t r} \left(B _ {2 2}\right) \operatorname {t r} \left(D _ {2 2}\right) + \operatorname {t r} \left(A _ {2 2}\right) \operatorname {t r} \left(C _ {2 2}\right) \operatorname {t r} \left(B _ {2 2} D _ {2 2}\right)\right) \\ \end{array}
$$
Written in terms of subspace projectors  $P$  and  $\bar{P}$ , the results become exactly as Eq. (S22).
Finally, similar to the unitary  $t$ -design, we introduce the concept of "subspace  $t$ -design". If an ensemble  $\mathbb{W}$  of unitaries  $V$  matches the ensemble  $\mathbb{U}$  rotating the subspace  $\mathcal{H}_{\mathrm{sub}}$  up to the  $t$ -degree moment, then  $\mathbb{W}$  is called a subspace unitary  $t$ -design with respect to  $\mathcal{H}_{\mathrm{sub}}$ . In the main text, the ensemble comes from the unknown target state. Alternatively, if a random QNN  $\mathbf{U}(\boldsymbol{\theta})$  with some constraints such as keeping the loss function constant  $\mathcal{L}(\boldsymbol{\theta}) = \mathcal{L}_0$ , i.e.,
$$
\mathbb {W} = \mathbf {U} (\Theta), \quad \Theta = \left\{\boldsymbol {\theta} \mid \mathcal {L} (\boldsymbol {\theta}) = \mathcal {L} _ {0} \right\}, \tag {S26}
$$
forms a approximate subspace 2-design, then similar results as in the main text can be established yet with a different interpretation: there is an exponentially large proportion of local minima on a constant-loss-section of the training landscape.
A.2 Perturbation on positive definite matrices
To identify whether a parameter point is a local minimum, we need to check whether the Hessian matrix is positive definite, where the following sufficient condition is used in the proof of our main theorem in the next section.
Lemma S8 Suppose  $X$  is a positive definite matrix and  $Y$  is a Hermitian matrix. If the distance between  $Y$  and  $X$  is smaller than the minimal eigenvalue of  $X$ , i.e.,  $\| Y - X \|_{\infty} < \| X^{-1} \|_{\infty}^{-1}$ , then  $Y$  is positive definite. Here  $\| \cdot \|_{\infty}$  denotes the Schatten- $\infty$  norm.
Proof For an arbitrary vector  $|v\rangle$ , we have
$$
\langle v | Y | v \rangle = \langle v | X | v \rangle + \langle v | Y - X | v \rangle \geq \| X ^ {- 1} \| _ {\infty} ^ {- 1} - \| Y - X \| _ {\infty} > 0. \tag {S27}
$$
Note that  $\| X^{-1}\|_{\infty}^{-1}$  just represents the minimal eigenvalue of the positive matrix  $X$ . Thus,  $Y$  is positive definite.
A.3 Tail inequalities
In order to bound the probability of avoiding local minima, we need to use some "tail inequalities" in probability theory, especially the generalized Chebyshev's inequality for matrices, which we summarize below for clarity.
Lemma S9 (Markov's inequality) For a non-negative random variable  $X$  and  $a > 0$ , the probability that  $X$  is at least a is upper bounded by the expectation of  $X$  divided by  $a$ , i.e.,
$$
\Pr [ X \geq a ] \leq \frac {\mathbb {E} [ X ]}{a}. \tag {S28}
$$
Proof The expectation can be rewritten and bounded as
$$
\begin{array}{l} \mathbb {E} [ X ] = \Pr [ X <   a ] \cdot \mathbb {E} [ X \mid X <   a ] + \Pr [ X \geq a ] \cdot \mathbb {E} [ X \mid X \geq a ] \tag {S29} \\ \geq \Pr [ X \geq a ] \cdot \mathbb {E} [ X \mid X \geq a ] \geq \Pr [ X \geq a ] \cdot a. \\ \end{array}
$$
Thus we have  $\operatorname*{Pr}[X\geq a]\leq \mathbb{E}[X] / a$
Lemma S10 (Chebyshev's inequality) For a real random variable  $X$  and  $\varepsilon > 0$ , the probability that  $X$  deviates from the expectation  $\mathbb{E}[X]$  by  $\varepsilon$  is upper bounded by the variance of  $X$  divided by  $\varepsilon^2$ , i.e.,
$$
\Pr [ | X - \mathbb {E} [ X ] | \geq \varepsilon ] \leq \frac {\operatorname {V a r} [ X ]}{\varepsilon^ {2}}. \tag {S30}
$$
Proof Applying Markov's inequality in Lemma S9 to the random variable  $(X - \mathbb{E}[X])^2$  gives
$$
\Pr \left[ | X - \mathbb {E} [ X ] | \geq \varepsilon \right] = \Pr \left[ (X - \mathbb {E} [ X ]) ^ {2} \geq \varepsilon^ {2} \right] \leq \frac {\mathbb {E} \left[ (X - \mathbb {E} [ X ]) ^ {2} \right]}{\varepsilon^ {2}} = \frac {\operatorname {V a r} [ X ]}{\varepsilon^ {2}}. \tag {S31}
$$
Alternatively, the proof can be carried out similarly as in Eq. (S29) with respect to  $(X - \mathbb{E}[X])^2$ .
Lemma S11 (Chebyshev's inequality for matrices) For a random matrix  $X$  and  $\varepsilon > 0$ , the probability that  $X$  deviates from the expectation  $\mathbb{E}[X]$  by  $\varepsilon$  in terms of the norm  $\| \cdot \|_{\alpha}$  satisfies
$$
\Pr \left[ \| X - \mathbb {E} [ X ] \| _ {\alpha} \geq \varepsilon \right] \leq \frac {\sigma_ {\alpha} ^ {2}}{\varepsilon^ {2}} \tag {S32}
$$
where  $\sigma_{\alpha}^{2} = \mathbb{E}[\| X - \mathbb{E}[X]\|_{\alpha}^{2}]$  denotes the variance of  $X$  in terms of the norm  $\| \cdot \|_{\alpha}$ .
Proof Applying Markov's inequality in Lemma S9 to the random variable  $\| X - \mathbb{E}[X]\|_{\alpha}^{2}$  gives
$$
\Pr [ \| X - \mathbb {E} [ X ] \| _ {\alpha} \geq \varepsilon ] = \Pr [ \| X - \mathbb {E} [ X ] \| _ {\alpha} ^ {2} \geq \varepsilon^ {2} ] \leq \frac {\mathbb {E} [ \| X - \mathbb {E} [ X ] \| _ {\alpha} ^ {2} ]}{\varepsilon^ {2}} = \frac {\sigma_ {\alpha} ^ {2}}{\varepsilon^ {2}}. \tag {S33}
$$
Note that here the expectation  $\mathbb{E}[X]$  is still a matrix while the "variance"  $\sigma_{\alpha}^{2}$  is a real number.
A.4 Quantum Fisher information matrix
Given a parameterized pure quantum state  $|\psi(\theta)\rangle$ , the quantum Fisher information (QFI) matrix  $\mathcal{F}_{\mu \nu}$  [64] is defined as the Riemannian metric induced from the Bures fidelity distance  $d_{\mathrm{f}}(\pmb{\theta},\pmb{\theta}^{\prime}) = 1 - |\langle \psi(\pmb{\theta})|\psi(\pmb{\theta}^{\prime})\rangle|^{2}$  (up to a factor 2 depending on convention), i.e.,
$$
\mathcal {F} _ {\mu \nu} (\boldsymbol {\theta}) = \left. \frac {\partial^ {2}}{\partial \delta_ {\mu} \partial \delta_ {\nu}} d _ {\mathrm {f}} (\boldsymbol {\theta}, \boldsymbol {\theta} + \boldsymbol {\delta}) \right| _ {\boldsymbol {\delta} = 0} = - 2 \operatorname {R e} \left[ \langle \partial_ {\mu} \partial_ {\nu} \psi | \psi \rangle + \langle \partial_ {\mu} \psi | \psi \rangle \langle \psi | \partial_ {\nu} \psi \rangle \right]. \tag {S34}
$$
Note that  $|\partial_{\mu}\psi \rangle$  actually refers to  $\frac{\partial}{\partial\theta_{\mu}} |\psi (\pmb {\theta})\rangle$  . Using the normalization condition
$$
\langle \psi | \psi \rangle = 1,
$$
$$
\partial_ {\mu} \left(\langle \psi | \psi \rangle\right) = \left\langle \partial_ {\mu} \psi | \psi \right\rangle + \left\langle \psi | \partial_ {\mu} \psi \right\rangle = 2 \operatorname {R e} \left[ \left\langle \partial_ {\mu} \psi | \psi \right\rangle \right] = 0, \tag {S35}
$$
$$
\partial_ {\mu} \partial_ {\nu} (\langle \psi | \psi \rangle) = 2 \operatorname {R e} [ \langle \partial_ {\mu} \partial_ {\nu} \psi | \psi \rangle + \langle \partial_ {\mu} \psi | \partial_ {\nu} \psi \rangle ] = 0,
$$
the QFI can be rewritten as
$$
\mathcal {F} _ {\mu \nu} = 2 \operatorname {R e} \left[ \langle \partial_ {\mu} \psi | \partial_ {\nu} \psi \rangle - \langle \partial_ {\mu} \psi | \psi \rangle \langle \psi | \partial_ {\nu} \psi \rangle \right]. \tag {S36}
$$
The QFI characterizes the sensibility of a parameterized quantum state to a small change of parameters, and can be viewed as the real part of the quantum geometric tensor.
B Detailed proofs
In this section, we provide the detailed proofs of Lemma 1, Theorem 2 and Proposition 3 in the main text. Here we use  $d$  to denote the dimension of the Hilbert space. For a qubit system with  $N$  qubits, we have  $d = 2^{N}$ . As in the main text, we represent the value of a certain function at  $\pmb{\theta} = \pmb{\theta}^*$  by appending the superscript "  $\ast$  " for simplicity of notation, e.g.,  $\nabla \mathcal{L}|_{\pmb{\theta} = \pmb{\theta}^*}$  as  $\nabla \mathcal{L}^*$  and  $H_{\mathcal{L}}|_{\pmb{\theta} = \pmb{\theta}^*}$  as  $H_{\mathcal{L}}^*$ . In addition, for a parameterized quantum circuit  $\mathbf{U}(\pmb{\theta}) = \prod_{\mu=1}^{M} U_{\mu}(\theta_{\mu}) W_{\mu}$ , we introduce the notation  $V_{\alpha \rightarrow \beta} = \prod_{\mu=\alpha}^{\beta} U_{\mu} W_{\mu}$  if  $\alpha \leq \beta$  and  $V_{\alpha \rightarrow \beta} = I$  if  $\alpha > \beta$ . Note that the product  $\prod_{\mu}$  is by default in the increasing order from the right to the left. The derivative with respect to the parameter  $\theta_{\mu}$  is simply denoted as  $\partial_{\mu} = \frac{\partial}{\partial \theta_{\mu}}$ . We remark that our results hold for all kinds of input states into QNNs in spite that we use  $|0\rangle^{\otimes N}$  in the definition of  $|\psi(\pmb{\theta})\rangle$  for simplicity.
$$
\mathbb {E} _ {\mathbb {T}} \left[ \nabla \mathcal {L} ^ {*} \right] = 0, \quad \operatorname {V a r} _ {\mathbb {T}} \left[ \partial_ {\mu} \mathcal {L} ^ {*} \right] = f _ {1} (p, d) \mathcal {F} _ {\mu \mu} ^ {*}. \tag {S37}
$$
$$
\mathbb {E} _ {\mathbb {T}} \left[ H _ {\mathcal {L}} ^ {*} \right] = \frac {d p ^ {2} - 1}{d - 1} \mathcal {F} ^ {*}, \quad \operatorname {V a r} _ {\mathbb {T}} \left[ \partial_ {\mu} \partial_ {\nu} \mathcal {L} ^ {*} \right] \leq f _ {2} (p, d) \| \Omega_ {\mu} \| _ {\infty} ^ {2} \| \Omega_ {\nu} \| _ {\infty} ^ {2}. \tag {S38}
$$
where  $\mathcal{F}$  denote the QFI matrix.  $f_{1}$  and  $f_{2}$  are functions of the overlap  $p$  and the Hilbert space dimension  $d$ , i.e.,
$$
f _ {1} (p, d) = \frac {p ^ {2} \left(1 - p ^ {2}\right)}{d - 1}, \quad f _ {2} (p, d) = \frac {3 2 \left(1 - p ^ {2}\right)}{d - 1} \left[ p ^ {2} + \frac {2 \left(1 - p ^ {2}\right)}{d} \right]. \tag {S39}
$$
Proof Using the decomposition in Eq. (4), the loss function can be expressed by
$$
\mathcal {L} = 1 - \langle \phi | \varrho | \phi \rangle = 1 - p ^ {2} \langle \psi^ {*} | \varrho | \psi^ {*} \rangle - (1 - p ^ {2}) \langle \psi^ {\perp} | \varrho | \psi^ {\perp} \rangle - 2 p \sqrt {1 - p ^ {2}} \operatorname {R e} \left(\langle \psi^ {\perp} | \varrho | \psi^ {*} \rangle\right), \tag {S40}
$$
where  $\varrho (\pmb {\theta}) = |\psi (\pmb {\theta})\rangle \langle \psi (\pmb {\theta})|$  denotes the density matrix of the output state from the QNN. According to Lemma S2 and Corollary S4, the expectation of the loss function with respect to the ensemble  $\mathbb{T}$  can be calculated as
$$
\begin{array}{l} \mathbb {E} _ {\mathbb {T}} [ \mathcal {L} (\boldsymbol {\theta}) ] = 1 - p ^ {2} \left\langle \psi^ {*} \mid \varrho \mid \psi^ {*} \right\rangle - (1 - p ^ {2}) \frac {\operatorname {t r} [ (I - \left| \psi^ {*} \right\rangle \left\langle \psi^ {*} \right|) \varrho ]}{d - 1} \tag {S41} \\ = 1 - p ^ {2} + \frac {d p ^ {2} - 1}{d - 1} g (\boldsymbol {\theta}), \\ \end{array}
$$
where  $g(\pmb{\theta}) = 1 - \langle \psi^{*}|\varrho (\pmb{\theta})|\psi^{*}\rangle$  denotes the fidelity distance between the output states at  $\pmb{\theta}$  and  $\pmb{\theta}^*$ . By definition,  $g(\pmb{\theta})$  takes the global minimum at  $\pmb{\theta} = \pmb{\theta}^{*}$ , i.e., at  $\varrho = |\psi^{*}\rangle \langle \psi^{*}|$ . Thus the commutation between the expectation and differentiation gives
$$
\left. \mathbb {E} _ {\mathbb {T}} [ \nabla \mathcal {L} ^ {*} ] = \nabla \left(\mathbb {E} _ {\mathbb {T}} [ C ]\right) \right| _ {\boldsymbol {\theta} = \boldsymbol {\theta} ^ {*}} = \frac {d p ^ {2} - 1}{d - 1} \left. \nabla g (\boldsymbol {\theta}) \right| _ {\boldsymbol {\theta} = \boldsymbol {\theta} ^ {*}} = 0, \tag {S42}
$$
$$
\mathbb {E} _ {\mathbb {T}} \left[ H _ {\mathcal {L}} ^ {*} \right] = \frac {d p ^ {2} - 1}{d - 1} \left. H _ {g} (\boldsymbol {\theta}) \right| _ {\boldsymbol {\theta} = \boldsymbol {\theta} ^ {*}} = \frac {d p ^ {2} - 1}{d - 1} \left. H _ {g} (\boldsymbol {\theta}) \right| _ {\boldsymbol {\theta} = \boldsymbol {\theta} ^ {*}} = \frac {d p ^ {2} - 1}{d - 1} \mathcal {F} ^ {*}.
$$
Note that  $H_{g}(\theta)|_{\theta = \theta^{*}}$  is actually the QFI matrix  $\mathcal{F}$  of  $|\psi (\theta)\rangle$  at  $\theta = \theta^{*}$  (see Appendix A.4), which is always positive semidefinite. To estimate the variance, we need to calculate the expression of derivatives first due to the non-linearity of the variance, unlike the case of Eq. (S42) where the operations of taking the expectation and derivative is exchanged. The first order derivative can be expressed by