SlowGuess/NeurIPS_2023_OCR huggingface dataset:
in which we have again ignored the dependency on  $\log (1 / \delta)$  to obtain a cleaner expression.
The error probability of  $\delta$  results from (a) conditioning on the event that  $|f(x_{s}) - y_{s}|\leq \epsilon_{s},\forall s = 1,\ldots ,m$  which holds with probability of at least  $1 - \delta /2$ , and (b) Theorem 3 which also holds with probability of at least  $1 - \delta /2$ .
Consistent with the analysis of Q-LinUCB [38], in order to apply the theoretical guarantee provided by QMC (Lemma 1) for noise with bounded variance, here we need to assume that the noise variance is not too small: We assume that  $\sigma > 1/3$ . We have chosen the value of  $1/3$  just for convenience, in fact, any  $\sigma$  larger than  $1/4$  can be used in our analysis here because these different values will only alter the constants in our regret upper bound which are absorbed by the  $\mathcal{O}$ .
Note that  $\epsilon_s = \widetilde{\sigma}_{s-1}(x_s) / \sqrt{\lambda} \leq 1$ , which follows because  $\widetilde{\sigma}_{s-1}^2(x_s) \leq 1$  (which can be easily seen from (3) and our assumption that  $k(x, x') \leq 1$ ) and  $\lambda > 1$ . As a result of the assumption of  $\sigma > 1/3$ , we have that  $\sigma > 1/3 > 2^{\sqrt{2}}/8$ . This, combined with  $\epsilon_s \leq 1$ , allows us to show that  $\log_2^{3/2}\left(\frac{8\sigma}{\epsilon_s}\right) \geq 2^{3/4}$  and that  $\log_2\left(\log_2\frac{8\sigma}{\epsilon_s}\right) \geq 1/2$ .
Next, we derive an upper bound on  $\sum_{s=1}^{m} \frac{1}{\epsilon_s^2}$  which is analogous to (8). Note that in the case of a noise with bounded variance, the number of queries to the quantum oracle in stage  $s$  is given by  $N_{\epsilon_s} = \frac{C_2 \sigma}{\epsilon_s} \log_2^{3/2} \left( \frac{8 \sigma}{\epsilon_s} \right) \log_2 \left( \log_2 \frac{8 \sigma}{\epsilon_s} \right) \log \frac{2\overline{m}}{\delta}$  with  $C_2 > 1$ . As a result, we have the following inequality which is in a similar spirit to (7)
$$
\sum_ {s = 1} ^ {m} \frac {C _ {2} \sigma}{\epsilon_ {s}} \log_ {2} ^ {3 / 2} \left(\frac {8 \sigma}{\epsilon_ {s}}\right) \log_ {2} \left(\log_ {2} \frac {8 \sigma}{\epsilon_ {s}}\right) \log \frac {2 \overline {{m}}}{\delta} \geq \sum_ {s = 1} ^ {m} \frac {1}{3} \frac {1}{\epsilon_ {s}} 2 ^ {3 / 4} 2 ^ {- 1} \geq \frac {1}{3 \times 2 ^ {1 / 4}} \sqrt {\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}} \tag {50}
$$
Now suppose that  $\sum_{s=1}^{m} \frac{1}{\epsilon_s^2} > 9\sqrt{2}T^2$ , then we have that  $\sum_{s=1}^{m} N_{\epsilon_s} > T$  which is a contradiction. Therefore, we have that
$$
\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \leq 9 \sqrt {2} T ^ {2}. \tag {51}
$$
Next, note that Theorem 1 is unaffected since its proof does not depend on the noise. Moreover, Corollary 1 also stays the same because it has only made use of (8) which gives an upper bound on  $\sum_{s=1}^{m} 1 / \epsilon_s^2$ , and compared to (8), its counterpart (51) in the analysis here only introduces an extra constant factor which is absorbed by the bit  $\mathcal{O}$  notation. Similarly, Theorem 2 is also unaltered for a similar reason, i.e., only an extra constant factor of  $3 \times 2^{1/4}$  will be introduced to the right hand side of (30) and (34), which is absorbed by the big  $\mathcal{O}$  notation.
Theorem 3 is also unaltered since its proof does not depend on the type of noise. In particular, the key underlying reason why it is unaffected is because for both types of noise, given a particular  $\epsilon_{s}$ , we choose the corresponding number  $N_{\epsilon_s}$  of queries (to the quantum oracle) depending on the type of noise so that the error guarantee of  $|y_{s} - f(x_{s})|\leq \epsilon_{s}$  is satisfied. This allows us to make use of the self-normalizing bound from [9] for 1-sub-Gaussian noise.
Lastly, we need to modify the proof of Theorem 4. To begin with, the total regret from stage  $s$  can now be bounded as
$$
\frac {C _ {2} \sigma}{\epsilon_ {s}} \log_ {2} ^ {3 / 2} \left(\frac {8 \sigma}{\epsilon_ {s}}\right) \log_ {2} \left(\log_ {2} \frac {8 \sigma}{\epsilon_ {s}}\right) \log \frac {2 \bar {m}}{\delta} \times 2 \beta_ {s} \epsilon_ {s} \sqrt {\lambda} = \mathcal {O} \left(\sigma \log_ {2} ^ {3 / 2} \left(\frac {\sigma}{\epsilon_ {s}}\right) \log_ {2} \left(\log_ {2} \frac {\sigma}{\epsilon_ {s}}\right) \log \frac {\bar {m}}{\delta} \beta_ {s}\right). \tag {52}
$$
Note that every  $1 / \epsilon_s$  is upper-bounded by  $1 / \epsilon_s \leq 3 \times 2^{1/4} T = \mathcal{O}(T)$  which can be easily inferred using (51). So, the total regret can be upper-bounded as
$$
\begin{array}{l} R _ {T} = \mathcal {O} \left(\sum_ {s = 1} ^ {m} \sigma \log_ {2} ^ {3 / 2} \left(\frac {\sigma}{\epsilon_ {s}}\right) \log_ {2} \left(\log_ {2} \frac {\sigma}{\epsilon_ {s}}\right) \log \frac {\bar {m}}{\delta} \beta_ {s}\right) \tag {53} \\ = \mathcal {O} \left(m \beta_ {m} \log \frac {\overline {{m}}}{\delta} \sigma \log_ {2} ^ {3 / 2} (\sigma T) \log_ {2} \left(\log_ {2} (\sigma T)\right)\right). \\ \end{array}
$$
As a result, for the linear kernel for which  $m \leq \overline{m} = \mathcal{O}(d\log T)$  (Theorem 2) and  $\beta_{m} = \sqrt{\widetilde{\gamma}_{m - 1} + \log(1 / \delta)} = \mathcal{O}(\sqrt{d\log T + \log(1 / \delta)})$  (Theorem 3 and Corollary 1),
$$
\begin{array}{l} R _ {T} = \mathcal {O} \left(d \log T \sqrt {d \log T + \log (1 / \delta)} \log \frac {d \log T}{\delta} \sigma \log_ {2} ^ {3 / 2} (\sigma T) \log_ {2} (\log_ {2} (\sigma T))\right) \tag {54} \\ = \mathcal {O} \left(\sigma d ^ {3 / 2} \log^ {3 / 2} T \log (d \log T) \log_ {2} ^ {3 / 2} (\sigma T) \log_ {2} (\log_ {2} (\sigma T))\right), \\ \end{array}
$$
in which we have ignored all  $\log (1 / \delta)$  factors for a cleaner expression.
For the SE kernel for which  $m \leq \overline{m} = \mathcal{O}(\log^{d + 1}T)$  (Theorem 2) and  $\beta_{m} = \sqrt{\widetilde{\gamma}_{m - 1} + \log(1 / \delta)} = \mathcal{O}(\sqrt{\log^{d + 1}T + \log(1 / \delta)})$  (Theorem 3 and Corollary 1),
$$
\begin{array}{l} R _ {T} = \mathcal {O} \left(\log^ {d + 1} T \sqrt {\log^ {d + 1} T + \log (1 / \delta)} \log \frac {\log^ {d + 1} T}{\delta} \sigma \log_ {2} ^ {3 / 2} (\sigma T) \log_ {2} (\log_ {2} \sigma T)\right) \tag {55} \\ = \mathcal {O} \left(\sigma (\log T) ^ {3 (d + 1) / 2} \log ((\log T) ^ {d + 1}) \log_ {2} ^ {3 / 2} (\sigma T) \log_ {2} (\log_ {2} \sigma T)\right), \\ \end{array}
$$
in which we have again ignored all  $\log (1 / \delta)$  factors for a cleaner expression.
H Proof of Theorem 6
Our proof here follows closely the proof of Theorem 2 from the work of [2]. For consistency, in our proof here, we follow the notations from [38] and [2].
Since we focus on linear bandits here, we assume that the reward function is linear with a groundtruth  $d$ -dimensional parameter vector  $\theta^{*}$ :  $f(x) = x^{\top}\theta^{*},\forall x$ . Again following [2] and [38] as well as many other works on linear bandits, we assume that  $\| \theta^{*}\|_{2}\leq S$ , and  $\| x\| _2\leq L,\forall x$ . We use  $X_{s}$  to denote  $X_{s}\triangleq [x_{1},\ldots ,x_{s}]^{\top}$  which is an  $s\times d$  matrix, and denote  $Y_{s}\triangleq [y_{1},\dots,y_{s}]^{\top}$  which is an  $s\times 1$ -dimensional vector of observations. We use  $\zeta_{1:s}\triangleq [\zeta_1,\dots,\zeta_s]^{\top}$  to denote the  $s\times 1$ -dimensional vector of noise. These notations allow us to write  $Y_{s} = X_{s}\theta^{*} + \zeta_{1:s}$ . We denote  $V_{s}\triangleq X_{s}^{\top}W_{s}X_{s} + \lambda I$ . Following [38], we use  $\widehat{\theta}_s$  to denote the MLE estimate of the parameter  $\theta^{*}$  given the observations after the first  $s$  stages:  $\widehat{\theta}_{s} = V_{s}^{-1}X_{s}^{\top}W_{s}Y_{s}$ . Given these notations, we firstly have that
$$
\begin{array}{l} \widehat {\theta} _ {s} = \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) ^ {- 1} X _ {s} ^ {\top} W _ {s} \left(X _ {s} \theta^ {*} + \zeta_ {1: s}\right) \\ = \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) ^ {- 1} X _ {s} ^ {\top} W _ {s} \zeta_ {1: s} + \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) ^ {- 1} \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) \theta^ {*} \tag {56} \\ - \lambda \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) ^ {- 1} \theta^ {*} \\ = \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) ^ {- 1} X _ {s} ^ {\top} W _ {s} \zeta_ {1: s} + \theta^ {*} - \lambda \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) ^ {- 1} \theta^ {*}. \\ \end{array}
$$
$$
\begin{array}{l} x ^ {\top} \widehat {\theta} _ {s} - x ^ {\top} \theta^ {*} = x ^ {\top} \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) ^ {- 1} X _ {s} ^ {\top} W _ {s} \zeta_ {1: s} - \lambda x ^ {\top} \left(X _ {s} ^ {\top} W _ {s} X _ {s} + \lambda I\right) ^ {- 1} \theta^ {*} \tag {57} \\ = \left\langle x, X _ {s} ^ {\top} W _ {s} \zeta_ {1: s} \right\rangle_ {V _ {s} ^ {- 1}} - \lambda \left\langle x, \theta^ {*} \right\rangle_ {V _ {s} ^ {- 1}}. \\ \end{array}
$$
This allows us to use the Cauchy-Schwarz inequality to show that
$$
\begin{array}{l} \left| x ^ {\top} \widehat {\theta} _ {s} - x ^ {\top} \theta^ {*} \right| \leq \| x \| _ {V _ {s} ^ {- 1}} \left(\left\| X _ {s} ^ {\top} W _ {s} \zeta_ {1: s} \right\| _ {V _ {s} ^ {- 1}} + \lambda \| \theta^ {*} \| _ {V _ {s} ^ {- 1}}\right) \tag {58} \\ \leq \| x \| _ {V _ {s} ^ {- 1}} \left(\left\| X _ {s} ^ {\top} W _ {s} \zeta_ {1: s} \right\| _ {V _ {s} ^ {- 1}} + \sqrt {\lambda} \| \theta^ {*} \| _ {2}\right). \\ \end{array}
$$
The last inequality follows because  $\| \theta^{*}\|_{V_{s}^{-1}}^{2}\leq \frac{1}{\lambda_{\min}(V_{s})}\| \theta^{*}\|_{2}^{2}\leq \frac{1}{\lambda}\| \theta^{*}\|_{2}^{2}$ . Note that  $X_{s}^{\top}W_{s}\zeta_{1:s} = \sum_{k = 1}^{s}\frac{1}{\epsilon_{k}^{2}}\zeta_{k}x_{k}$ .
Next, we make use of the self-normalized concentration inequality from Theorem 1 of [2]. Specifically, we will use  $\{\frac{\zeta_s}{\epsilon_s}\}_{s=1}^{\infty}$  as to replace the stochastic process  $\{\zeta_s\}_{s=1}^{\infty}$  (i.e., representing the noise,  $\{\eta_s\}_{s=1}^{\infty}$  according to the notations from [2]), and use  $\{\frac{1}{\epsilon_s}x_s\}_{s=1}^{\infty}$  to replace the vector-valued stochastic process  $\{x_s\}_{s=1}^{\infty}$  (i.e., representing the sequence of inputs). Now we explain why this is feasible.
Similar to the proof of Theorem 3, we again denote the observation noise as  $\zeta_s$ :  $y_s = f(x_s) + \zeta_s$  and define  $\mathcal{F}_s$  as the  $\sigma$ -algebra  $\mathcal{F}_s = \sigma(x_1, x_2, \ldots, x_{s+1}, \zeta_1, \zeta_2, \ldots, \zeta_s)$ . Similarly, an immediate consequence is that  $x_s$  is  $\mathcal{F}_{s-1}$ -measurable, and  $\zeta_s$  is  $\mathcal{F}_s$ -measurable. Following a similar analysis to that used in the proof of Theorem 3 allows us to show that  $\epsilon_k$  is  $\mathcal{F}_{k-1}$ -measurable (i.e.,  $\epsilon_k$  can be predicted based on  $\mathcal{F}_{k-1}$ ). Therefore, we have that  $\frac{1}{\epsilon_k} x_k$  is  $\mathcal{F}_{k-1}$ -measurable (i.e.,  $\frac{1}{\epsilon_k} x_k$  can be predicted based on  $\mathcal{F}_{k-1}$ ) and that  $\frac{\zeta_k}{\epsilon_k}$  is  $\mathcal{F}_k$ -measurable. Moreover, conditioned on  $\mathcal{F}_{k-1}$ ,  $\frac{\zeta_k}{\epsilon_k}$  is 1-sub-Gaussian. This is because the QMC subroutine guarantees that  $|y_k - f(x_k)| \leq \epsilon_k$ ,  $\forall k = 1, \ldots, m$  (with high probability), which means that the absolute value of the noise  $\zeta_k = y_k - f(x_k)$  is upper-bounded by  $\epsilon_k$ :  $|\zeta_k| \leq \epsilon_k$ . Therefore, conditioned on  $\mathcal{F}_{k-1}$ ,  $\frac{\zeta_k}{\epsilon_k}$  is 1-sub-Gaussian.
Given that these conditions are satisfied, we can apply the self-normalized concentration inequality from Theorem 1 of [2] to  $\{\frac{\zeta_s}{\epsilon_s}\}_{s=1}^{\infty}$  and  $\{\frac{1}{\epsilon_s}x_s\}_{s=1}^{\infty}$  with (following the notations from Theorem 1 of [2])
$$
\begin{array}{l} V _ {s} = V + \sum_ {k = 1} ^ {s} \left(\frac {1}{\epsilon_ {k}} x _ {k}\right) \left(\frac {1}{\epsilon_ {k}} x _ {k} ^ {\top}\right) = \lambda I + \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}} x _ {k} x _ {k} ^ {\top}, \tag {59} \\ S _ {s} = \sum_ {k = 1} ^ {s} \frac {\zeta_ {k}}{\epsilon_ {k}} \left(\frac {1}{\epsilon_ {k}} x _ {k}\right) = \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}} \zeta_ {k} x _ {k}, \\ \end{array}
$$
in which we have used  $V = \lambda I$ . This allows us to show that
$$
\left\| S _ {s} \right\| _ {V _ {s} ^ {- 1}} = \left\| X _ {s} ^ {\top} W _ {s} \zeta_ {1: s} \right\| _ {V _ {s} ^ {- 1}} \leq \sqrt {2 \log \left(\frac {\det (V _ {s}) ^ {1 / 2} \det (\lambda I) ^ {- 1 / 2}}{\delta}\right)} \tag {60}
$$
which holds with probability of at least  $1 - \delta$ . Next, again making use of the trace-determinant inequality from (16) and using the aforementioned assumption that  $\|x\|_2 \leq L$ , we can show that  $\operatorname{det}(V_s) \leq (\lambda + \frac{L^2}{d} \sum_{k=1}^{s} \frac{1}{\epsilon_k^2})^d$ . Also note that  $\operatorname{det}(\lambda I) = \lambda^d$ . As a result, we can further upper-bound the equation above by:
$$
\begin{array}{l} \left\| X _ {s} ^ {\top} W _ {s} \zeta_ {1: s} \right\| _ {V _ {s} ^ {- 1}} \leq \sqrt {2 \log \left(\sqrt {\frac {\left(\lambda + \frac {L ^ {2}}{d} \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}}\right) ^ {d}}{\lambda^ {d}}} \frac {1}{\delta}\right)} \\ = \sqrt {2 \log \left(\sqrt {\left(1 + \frac {L ^ {2}}{d \lambda} \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}}\right) ^ {d}} \frac {1}{\delta}\right)} \tag {61} \\ \leq \sqrt {2 d \log \frac {1 + \frac {L ^ {2}}{\lambda} \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}}}{\delta}}, \\ \end{array}
$$
in which we have used  $d \geq 1$  in the last step to get a cleaner expression.
Therefore, we can re-write (58) as
$$
\left| x ^ {\top} \widehat {\theta} _ {s} - x ^ {\top} \theta^ {*} \right| \leq \| x \| _ {V _ {s} ^ {- 1}} \left(\sqrt {2 d \log \frac {1 + \frac {L ^ {2}}{\lambda} \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}}}{\delta}} + \sqrt {\lambda} S\right), \quad \forall x, s, \tag {62}
$$
in which we have also used the above-mentioned assumption of  $\| \theta^{*}\|_{2}\leq S$ . Now again following [2], we plug in  $x = V_{s}(\widehat{\theta}_{s} - \theta^{*})$ , which gives us
$$
\begin{array}{l} \left\| \widehat {\theta} _ {s} - \theta^ {*} \right\| _ {V _ {s}} ^ {2} \leq \left\| V _ {s} \left(\widehat {\theta} _ {s} - \theta^ {*}\right) \right\| _ {V _ {s} ^ {- 1}} \left(\sqrt {2 d \log \frac {1 + \frac {L ^ {2}}{\lambda} \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}}}{\delta}} + \sqrt {\lambda} S\right) \tag {63} \\ = \left\| \widehat {\theta} _ {s} - \theta^ {*} \right\| _ {V _ {s}} \left(\sqrt {2 d \log \frac {1 + \frac {L ^ {2}}{\lambda} \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}}}{\delta}} + \sqrt {\lambda} S\right). \\ \end{array}
$$
$$
\left\| \widehat {\theta} _ {s} - \theta^ {*} \right\| _ {V _ {s}} \leq \left(\sqrt {2 d \log \frac {1 + \frac {L ^ {2}}{\lambda} \sum_ {k = 1} ^ {s} \frac {1}{\epsilon_ {k} ^ {2}}}{\delta}} + \sqrt {\lambda} S\right). \tag {64}
$$
This completes the proof of Theorem 6.
Now we briefly show how the improved confidence ellipsoid from our Theorem 6 leads to an improved regret upper bound for Q-LinUCB. Here we follow the proof of Theorem 3 from [38], and hence defer the detailed explanations of some of the steps to the proof there. To begin with, we have that
$$
\begin{array}{l} f (x ^ {*}) - f (x _ {s}) = \left(x ^ {*} - x _ {s}\right) ^ {\top} \theta^ {*} \\ \leq \epsilon_ {s} \left(\left\| \widetilde {\theta} _ {s} - \widehat {\theta} _ {s - 1} \right\| _ {V _ {s - 1}} + \left\| \widehat {\theta} _ {s - 1} - \theta^ {*} \right\| _ {V _ {s - 1}}\right) \\ \leq 2 \epsilon_ {s} \times \left(\sqrt {2 d \log \frac {1 + L ^ {2} T ^ {2} / \lambda}{\delta}} + \sqrt {\lambda} S\right) \tag {65} \\ = \mathcal {O} \left(\epsilon_ {s} \sqrt {d \log T}\right). \\ \end{array}
$$
Here for simplicity, we have upper-bounded  $\sum_{k=1}^{s} \left( \frac{1}{\epsilon_k^2} \right)$  by  $T^2$  (Lemma 2) in the second inequality, and have omitted the dependency on  $\log(1/\delta)$ .
Then the total regrets in stage  $s$  can be upper-bounded by
$$
\frac {C _ {1}}{\epsilon_ {s}} \log \frac {2 \bar {m}}{\delta} \times \mathcal {O} \left(\epsilon_ {s} \sqrt {d \log T}\right) = \mathcal {O} \left(\sqrt {d \log T} \log \frac {\bar {m}}{\delta}\right) \tag {66}
$$
Plugging in our tighter confidence ellipsoid (Theorem 6) into the regret analysis of QLinUCB, we have that the regret upper bound of QLinUCB can be analyzed as
$$
\begin{array}{l} R _ {T} = \mathcal {O} (\sum_ {s = 1} ^ {m} \sqrt {d \log T} \log (\frac {\overline {{m}}}{\delta})) \\ = \mathcal {O} \left(m \sqrt {d \log T} \log \left(\frac {\bar {m}}{\delta}\right)\right) \tag {67} \\ = \mathcal {O} (d \log T \sqrt {d \log (T)} \log (d \log T)) \\ = \mathcal {O} \left(d ^ {3 / 2} (\log T) ^ {3 / 2} \log (d \log T)\right), \\ \end{array}
$$
which gives the tighter regret upper bound for Q-LinUCB that we have discussed in the main text (Sec. 5.6).
I Regret Upper Bound for the Matérn Kernel (Proof of Theorem 7)
In this section, we modify our analysis to derive a regret upper bound for the Matérn kernel. We focus on bounded noise here (i.e., the first scenario in Lemma 1), since the analysis for noise with bounded variance (i.e., the second scenario in Lemma 1) is similar and follows from the analysis of App. G. In our analysis here, for simplicity, we ignore the logarithmic factors.
To begin with, note that Theorem 1 is not kernel-specific and hence still holds in our analysis here. It has been shown that for the Matérn kernel, the upper bound on the standard maximum information gain is  $\gamma_{T} = \widetilde{\mathcal{O}}(T^{\frac{d}{2\nu + d}})$  [35]. Therefore, similar to our Corollary 1, for the Matérn kernel, we can use Theorem 1 to show that
$$
\widetilde {\gamma} _ {m} = \widetilde {\mathcal {O}} \left(\left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right) ^ {\frac {d}{2 \nu + d}}\right) = \widetilde {\mathcal {O}} \left(T ^ {\frac {2 d}{2 \nu + d}}\right). \tag {68}
$$
Next, we modify the proof of our Theorem 2 to derive the corresponding upper bound on the total number of stages for the Matérn kernel, i.e., we discuss here how we modify the proof of Theorem 2 in App. D. To begin with, (68) suggests that there exists a  $C > 0$  such that  $\widetilde{\gamma}_m \leq C\left(\sum_{s=1}^{m} \frac{1}{\epsilon_s^2}\right)^{\frac{d}{2\nu + d}}$ . Different from our analysis for the linear and SE kernels in App. D (i.e., (27) and (32)), the absolute constant  $C$  here also contains logarithmic factors. Similar to (27), we can show that
$$
m \log 2 = \log \frac {\det  \left(V _ {m}\right)}{\det  \left(V _ {0}\right)} = 2 \widetilde {\gamma} _ {m} \leq 2 C \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right) ^ {\frac {d}{2 \nu + d}}. \tag {69}
$$
This naturally leads to
$$
\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \geq \left(m \frac {\log 2}{2 C}\right) ^ {\frac {2 \nu + d}{d}}, \tag {70}
$$
which is analogous to (28). Next, similar to (29), the total number of iterations can be analyzed as
$$
\sum_ {s = 1} ^ {m} \frac {C _ {1}}{\epsilon_ {s}} \log \left(\frac {2 \overline {{m}}}{\delta}\right) \geq \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s}} \geq \sqrt {\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}} \geq \sqrt {\left(m \frac {\log 2}{2 C}\right) ^ {\frac {2 \nu + d}{d}}}. \tag {71}
$$
Now suppose  $m > T^{\frac{2d}{2\nu + d}}\frac{2C}{\log 2}$ , then this implies that
$$
\sqrt {\left(m \frac {\log 2}{2 C}\right) ^ {\frac {2 \nu + d}{d}}} > T. \tag {72}
$$
This equation, combined with (71), suggests that  $\sum_{s=1}^{m} \frac{C_1}{\epsilon_s} \log \left( \frac{2\overline{m}}{\delta} \right) > T$ , which is a contradiction. Therefore, we have that
$$
m \leq T ^ {\frac {2 d}{2 \nu + d}} \frac {2 C}{\log 2} = \widetilde {\mathcal {O}} \left(T ^ {\frac {2 d}{2 \nu + d}}\right). \tag {73}
$$
Next, note that our confidence ellipsoid in Theorem 3 does not depend on the choice of the kernel and hence also holds for the Matérn kernel.
Lastly, we can modify the proof of Theorem 4 (App. F). Specifically, we can adopt the result from (47):  $R_{T} \leq \mathcal{O}\left(m\beta_{m}\log \frac{\overline{m}}{\delta}\right)$ . For the Matérn kernel, Theorem 3 and (68) allow us to show that  $\beta_{m} = \widetilde{\mathcal{O}} (\sqrt{\widetilde{\gamma}_{m - 1}}) = \widetilde{\mathcal{O}} (\sqrt{T^{\frac{2d}{2\nu + d}}}) = \widetilde{\mathcal{O}} (T^{\frac{d}{2\nu + d}})$ . Combining this with (73) allows us to show that for the Matérn kernel,
$$
R _ {T} = \widetilde {\mathcal {O}} \left(T ^ {\frac {2 d}{2 \nu + d}} T ^ {\frac {d}{2 \nu + d}}\right) = \widetilde {\mathcal {O}} \left(T ^ {\frac {3 d}{2 \nu + d}}\right). \tag {74}
$$
This completes the proof of Theorem 7.
J More Experimental Details, and Experiment on Real Quantum Computer
In our experiments, when implementing both the classical GP-UCB and our Q-GP-UCB algorithms, we use random Fourier features (RFF) approximation to approximate the kernel:  $k(x,x^{\prime})\approx \widetilde{\phi} (x)^{\top}\widetilde{\phi} (x^{\prime})$ . As a result, we can implement the weighted GP posterior following (6) as discussed in App. A, in which the  $M$ -dimensional ( $M < \infty$ ) random features  $\widetilde{\phi} (\cdot)$  are used to replace the infinite-dimensional RKHS features  $\phi (\cdot)$ . We have used  $M = 100$  random features in the synthetic experiment and  $M = 200$  in the AutoML experiment. As we have mentioned in the main paper (Sec. 6), we have implemented the QMC algorithm from the recent work of [20] using the Qiskit python package. We have used 6 qubits for the Gaussian noise and 1 qubits for the Bernoulli noise. For each experiment, we perform multiple independent trials (10 trials for the synthetic experiment and 5 trials for the AutoML experiment) and have plotted the resulting mean and standard error as the solid lines and error bars in each figure, respectively. For our Q-GP-UCB, we choose  $\beta_{s} = 1 + \log s,\forall s\geq 1$  following the order given by the theoretical value of  $\beta_{s}$  (Theorem 3). For classical GP-UCB, we tried different approaches to setting  $\beta_{s}\colon \beta_{s} = 1 + \log s$ ,  $\beta_{s} = \sqrt{2}$  and  $\beta_{s} = 1$ , all of which are commonly adopted practices in BO; we found that  $\beta_{s} = \sqrt{2}$  and  $\beta_{s} = 1$  have led to the best performances for GP-UCB in, respectively, the synthetic and AutoML experiments.