FERMILAB-PUB-23-387-SQMS current page.north east -3.5cm -1.5cm

# Investigating Parameter Trainability in the SNAP-Displacement Protocol of a
Qudit system

Oluwadara Ogunkoya  Superconducting and Quantum Materials System Center
(SQMS), Batavia, IL, USA.  Fermi National Accelerator Laboratory, Batavia, IL,
USA.  Kirsten Morris  Department of Mathematics, University of Nebraska-
Lincoln, USA.  Dog̃a Murat Kürkçüog̃lu  Superconducting and Quantum Materials
System Center (SQMS), Batavia, IL, USA.  Fermi National Accelerator
Laboratory, Batavia, IL, USA.

(Received: date / Accepted: date)

###### Abstract

In this study, we explore the universality of Selective Number-dependent
Arbitrary Phase (SNAP) and Displacement gates for quantum control in qudit-
based systems. However, optimizing the parameters of these gates poses a
challenging task. Our main focus is to investigate the sensitivity of training
any of the SNAP parameters in the SNAP-Displacement protocol. We analyze
conditions that could potentially lead to the Barren Plateau problem in a
qudit system and draw comparisons with multi-qubit systems. The parameterized
ansatz we consider consists of blocks, where each block is composed of
hardware operations, namely SNAP and Displacement gates [9]. Applying
Variational Quantum Algorithm (VQA) with observable and gate cost functions,
we utilize techniques similar to those in [14] and [4] along with the concept
of $t-$design. Through this analysis, we make the following key observations:
(a) The trainability of a SNAP-parameter does not exhibit a preference for any
particular direction within our cost function landscape, (b) By leveraging the
first and second moments properties of Haar measures, we establish new lemmas
concerning the expectation of certain polynomial functions, and (c) utilizing
these new lemmas, we identify a general condition that indicates an expected
trainability advantage in a qudit system when compared to multi-qubit systems.

##  1 Introduction

Variational Quantum Algorithms (VQAs) have emerged as promising candidates for
the implementation on Noisy Intermediate Scale Quantum (NISQ) computers.
However, the scalability of VQAs faces a significant obstacle due to the
presence of the barren plateau phenomenon. This phenomenon was first
introduced by McClean et al. in 2018, where they demonstrated its existence in
a wide range of random quantum circuits [14]. Furthermore, Cerezo et al.
investigated the presence of the barren plateau phenomenon even in shallow
circuit depths when using global observables, although the effect can be
mitigated when employing local observables [4]. Despite these studies, the
literature lacks a comprehensive analysis regarding the potential avoidance of
the barren plateau phenomenon by utilizing qudits instead of qubits. In this
context, our research investigates the scaling behavior of VQAs concerning the
qudit dimension rather than the number of qubits. Rather than focusing on the
gradient scaling with respect to the number of qubits, we direct our attention
to understanding the impact of increasing the qudit dimension on the VQA’s
performance. Through our analysis, we aim to provide insights into the
trainability and scalability of VQAs using qudits, potentially offering a
solution to overcome the barren plateau phenomenon. By exploring this
unexplored avenue, we hope to uncover novel strategies for enhancing the
efficiency and applicability of VQAs in practical quantum computing tasks.
Understanding the implications of qudit-based VQAs can lead to advancements in
quantum algorithm design, optimization techniques, and pave the way for the
utilization of more powerful quantum systems in various real-world
applications.

The objective of this research is to examine how the parameter trainability
scales with the increasing qudit dimension in a variational quantum algorithm,
focusing on two distinct cost functions. In the context of qubits, certain
cost functions have exhibited the barren plateau phenomenon, leading to a
significant reduction in trainability, particularly for parameters initialized
randomly. This trainability challenge has been observed in both deep and
shallow Parametrized Quantum Circuits (PQCs) [14], [4]. Therefore, our
investigation aims to ascertain whether similar trainability issues arise when
a qudit is utilized instead of multiple qubits. By analyzing the scaling
behavior, we aim to gain valuable insights into how the qudit dimension
impacts trainability of parameters in the specified cost functions. This
exploration is pivotal in understanding the practical advantages and
limitations of qudits over qubits within the domain of variational quantum
computing. Through our study, we seek to shed light on the comparative
efficiency and effectiveness of qudits in quantum algorithms, with a
particular focus on trainability. By identifying and understanding potential
trainability challenges and limitations specific to qudits, we can devise
tailored strategies to overcome these hurdles and enhance the overall
performance of variational quantum algorithms, thereby widening their
applicability in various quantum computing tasks. Our investigation holds
significant implications for the quantum computing community, offering
insights into the suitability of qudits in different applications and shedding
light on the general challenges associated with trainability in variational
quantum algorithms. The outcomes of our research could drive further
advancements in the field, aiding the development of robust quantum algorithms
and contributing to the realization of practical quantum computing solutions.

To achieve universal cavity control, a combination of the Selective Number-
dependent Arbitrary Phase (SNAP) gate and Displacement gate set is utilized.
According to [12], a significant improvement in the scaling of the ansatz
length is achieved by alternating the gates in layers of size
$\mathcal{O}(d)$, where $d$ represents the Hilbert space dimension, as opposed
to the $\mathcal{O}(d^{2})$ scaling discussed in [9]. For a SNAP gate, the
execution time is approximately $3-5;\mu$s, while for a Displacement gate, it
takes about $0.1;\mu$s [2]. Consequently, for $d=10$, the total time required
to execute the ansatz is approximately $1.1;\mu$s (accounting for $d+1$
Displacement gates) and $30-50;\mu$s (for $d$ SNAP gates). The overall time
needed for the ansatz ranges from $31.1$ to $51.1;\mu$s, which is significant
compared to the cavity’s lifetime of $100;\mu$s, considering the additional
operations such as readout that need to be performed. While ongoing research
aims to further reduce the ansatz length, the focus of this paper is to work
with the $\mathcal{O}(d)$-length ansatz. The efficiency of the proposed
universal cavity control scheme using SNAP and Displacement gates showcases
its potential for various quantum operations. However, continued efforts to
optimize the ansatz duration are crucial to accommodate additional tasks and
maximize the utility of the cavity within its limited lifetime. By refining
the control techniques, this approach can pave the way for enhanced quantum
processing and information tasks, contributing to the advancement of quantum
technology.

$t$-designs are important tools in the field of science, ranging from
classical statistics to quantum information theory. A whole theory of
combinatorial $t$-designs can be found in [5] and [11]. Quantum $t$-designs
have been studied in [1] which are analogous to spherical $2t$-design in
$\mathbb{R}^{d}$. In [7] and [10], the extension of spherical $t$-designs to
quantum $t$-designs, i.e. averaging polynomials over $d$\- dimensional spheres
as compared to averaging polynomial functions of matrices over the set of
unitaries (with respect to the Haar measure) was formally introduced. Over a
decade before now, interesting results concerning quantum designs have been
shown. In [17], the existence of quantum $t$-designs for every possible
combinations of $t$ and $d$ (where $t$ is the polynomial degree and $d$ is the
dimension of the underlying space) was proven. For fixed $t$ and $d$, it is
quite difficult to explicitly find the designs but there are computational
results about the least number of unitaries that make up each quantum
$t$-designs: for example, [10] and [17] proved that for a quantum 2-design in
and dimension $d$, there are at least $d^{4}-2d^{2}+2$ unitaries. The
uniqueness of these sets are not guaranteed and finding the quantum
$t$-designs with minimal number of unitaries is still a difficult problem [3]
even though much new methods are being suggested [15]. Another approach is
finding how close (up to an error $\epsilon>0$) an ensemble is to a quantum
$t$-design. Much of this can be found in [7], [8] and [1].

In section 2, we present a comprehensive collection of precise symbols and
definitions. Our proposed ansatz is formulated using hardware blocks, each
housing SNAPS and Displacements gates. Additionally, we introduce the concept
of trainability, which revolves around assessing the gradient variance of a
cost function along specific directions. By delving into this notion, we aim
to understand the system’s adaptability and its ability to be effectively
trained through gradient-based optimization techniques. The combination of
detailed notations, hardware-based ansatz, and the exploration of trainability
enhances the foundation of our research and opens new possibilities for
advancing the field. In section 3, we delve into the study of parameter
trainability for diverse cost functions, presenting both analytical and
numerical findings. Leveraging the theory of quantum $2-$designs in $1D$, we
provide analytical results and bounds for the variance of observable and gate
cost function gradients, which are directly linked to the traces of the
corresponding observables and gates. Through our analysis, we establish that
the trainability condition is intricately governed by these trace functions.
To validate our theoretical claims, we offer numerical examples that serve as
concrete evidence of the principles put forth in our analytical proofs. These
numerical illustrations provide tangible support for the trainability outcomes
we have identified, further reinforcing the robustness of our research. A
noteworthy highlight of our investigation is the identification of a general
condition that leads to achieving trainability advantage specifically in the
SNAP-Displacement protocol when compared to the multi-qubit scenario. We
elaborate on this condition, demonstrating its potential implications and
significance in gradient-based optimization domains. Furthermore, we furnish a
specific example that satisfies this condition, showcasing its real-world
applicability and relevance. Our approach combines rigorous analytical
derivations and practical numerical demonstrations, yielding a comprehensive
understanding of parameter trainability in quantum systems. By unveiling the
relationships between cost function variance, trace functions, and
trainability, our research contributes valuable insights to the field. The
implications of our findings can potentially advance quantum computation
techniques, paving the way for more efficient and effective quantum algorithms
and protocols. Rigorous proofs for the results in section 3 are found in the
appendix.

##  2 Preliminaries and Notation

In this section, we draw inspiration from [13] and shift our focus to qudits,
moving away from qubits. Specifically, our investigation revolves around a $d$
dimensional qudit space, with parameters $\alpha\in\mathbb{C}$ and
$\vec{\theta}=\\{\theta_{n}\\}_{n=0}^{d}$. Each phase $\theta_{n}$ is chosen
uniformly at random from the interval $[0,2\pi)$. To harness their
universality [12], we employ the Displacement operator $\hat{D}(\alpha)$ and
the Selective Number-dependent Arbitrary Phase (SNAP) gate
$\hat{S}(\vec{\theta})$. The displacement operator $\hat{D}(\alpha)$ is
crucial in our study. Its definition is as follows:

$\hat{D}(\alpha)=e^{\alpha a-\alpha^{\ast}a^{\dagger}},$ (1)

where $a$ is the lowering operator and $a^{\dagger}$ is the raising operator
such that $a|n\rangle=\sqrt{n}|n-1\rangle$,
$a^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle$. While the SNAP gate is defined
as

$\hat{S}(\vec{\theta})=\sum_{n=0}^{\infty}e^{i\theta_{n}}\ket{n}\bra{n}.$ (2)

In practice, given the finite dimensional qudit space context we use a
truncated SNAP gate defined as

$\hat{S}_{d}(\vec{\theta})=\sum_{n=0}^{d}e^{i\theta_{n}}\ket{n}\bra{n}$ (3)

for $\vec{\theta}=\\{\theta_{n}\\}_{n=0}^{d}$.

By utilizing these powerful and versatile operators, we venture into the realm
of qudits, expanding the potential for quantum computation and control. The
adoption of parameters randomly chosen from a uniform distribution in the
range $[0,2\pi)$ adds an element of quantum randomness to the setup, which can
yield interesting insights into the behavior and performance of qudit-based
quantum algorithms. Our approach holds the promise of enhancing the scope and
applicability of quantum information processing using qudits. By leveraging
the universality of the displacement operator and the SNAP gate, we pave the
way for novel quantum computing strategies and explore new possibilities for
quantum advantage in various quantum computing tasks. The use of qudits offers
a broader state space and additional computational resources, which could be
harnessed to tackle more complex problems and drive innovations in quantum
technology.

Throughout this paper we consider a particular ansatz as given by [9]:

$U(\vec{\theta},\vec{\alpha})=\hat{B}(\alpha_{T},\vec{\theta}_{T})\cdots\hat{B}(\alpha_{2},\vec{\theta}_{2})\hat{B}(\alpha_{1},\vec{\theta}_{1}).$
(4)

In this ansatz, $T$ is an integer representing the number of SNAP gates,
$\;\vec{\alpha}=\\{\alpha_{i}\\}_{i=1}^{T}\in\mathbb{R}^{T},\;\;\vec{\theta}_{j}=\\{\theta_{kj}\\}_{k=1}^{d}$
where $j=1,\ldots,T$ and $\theta_{kj}\in[0,2\pi)$. Each term
$\hat{B}(\alpha,\vec{\theta})$ is a block of the form

$\hat{B}(\alpha,\vec{\theta})=\hat{D}^{\dagger}(\alpha)\hat{S}_{d}(\vec{\theta})\hat{D}(\alpha).$

Consider a PQC $V(\vec{\theta})$ where $C$ is any cost function, e.g. the
state/observable cost function

$C_{s}=1-\operatorname{Tr}(\mathcal{\hat{O}}U(\vec{\theta},\vec{\alpha})\rho_{{}_{0}}U^{\dagger}(\vec{\theta},\vec{\alpha})),\;\;\;\rho_{0}=\ket{0}\bra{0}$
(5)

where $\hat{O}$ is the target observable, or the gate cost function

$C_{g}=1-|\tfrac{1}{d}\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}(\vec{\theta},\vec{\alpha})]|^{2}$
(6)

where $\hat{U}_{t}$ is the target unitary.

In general, a quantum circuit experiences a barren plateau if for any
parameter $\theta_{\nu}$, there exists some positive real numbers $A,b$ and
$n$ such that

$\operatorname{Var}\left[\frac{\partial
C}{\partial\theta_{\nu}}\right]=\mathcal{O}(A^{-bn})$ (7)

and mean value

$\left<\frac{\partial C}{\partial\theta_{\nu}}\right>=0$ (8)

For the multi-qubit case, $A=2$ and $n$ is the number of qubits, while in the
single qudit case, $A$ is the dimension of the qudit, and $n=1$.

Both McClean et al. and Cerezo et al. and noted the relevance of the Haar
measure and $t$-designs in investigating the possible existence of a barren
plateau in Parametrized Quantum Circuits (PQC). These notions are also
relevant in this paper so we include their definitions below.

######  Definition 1.

The Haar measure $\mu$ over the unitary group $\mathcal{U}(d)$ is the uniquely
defined left and right invariant measure such that for any unitary matrix
$U\in\mathcal{U}(d)$, and $f:\mathcal{U}(d)\rightarrow\mathcal{U}(d)$ we have
that

$\displaystyle\int_{\mathcal{U}(d)}f(W)\;d\mu(W)$
$\displaystyle=\int_{\mathcal{U}(d)}f(UW)\;d\mu(W)$ (9)
$\displaystyle=\int_{\mathcal{U}(d)}f(WU)\;d\mu(W)$ (10)

######  Definition 2.

Let $\mathcal{S}$ be an ensemble of unitaries from $\mathcal{U}(d)$ with a
uniform distribution $\nu,$ and $P_{(t,t)}(U)$ be a polynomial of degree at
most $t$ in the matrix elements of both $U$ and $U^{\dagger}$. Then
$\mathcal{S}$ is called a unitary $t$-design if

$\displaystyle\int_{\mathcal{S}}P_{(t,t)}(S)\;d\nu(S)=\int_{\mathcal{U}(d)}P_{(t,t)}(U)\;d\mu(U).$
(11)

##  3 Trainability of Parameters for different Cost functions

The following lemmas ( with proofs presented in the appendix section) hold
significant importance for evaluating the expectation of operators over a
specific set of unitaries in $\mathcal{U}(d)$. Rather than computing complex
integrals over this particular set, we transform the problem into evaluating
integrals (with respect to the Haar measure) over the entire set of unitaries
$\mathcal{U}(d)$. This approach offers two key advantages:

Firstly, the subset of $\mathcal{U}(d)$ over which the integral is to be
performed is sometimes vaguely defined, which can introduce uncertainty and
challenges in the calculations. By employing the concept of unitary
$t-$designs, we can circumvent this ambiguity and achieve a well-defined
integration procedure, simplifying the overall evaluation.

Secondly, even when the subset is well-defined, traditional mathematical tools
may not be sufficient for performing such complex calculations. The
utilization of unitary $t-$designs provides us with a powerful and efficient
method to tackle these integrals, facilitating more tractable and manageable
evaluations.

By leveraging the concept of unitary $t-$designs, we can avoid complications
and streamline the evaluation process for these integrals. This approach
offers a valuable toolset for handling expectations of operators over unitary
sets, enabling more straightforward and effective calculations.

######  Lemma 1.

Let an ensemble $\mathcal{S}$ (subset of $\mathcal{U}(d)$) with a uniform
distribution $\nu$ be a unitary $t-$design with $t\geq 1$, and let
$C,D:\mathbb{C}^{d}\rightarrow\mathbb{C}^{d}$ be linear operators. Then

$\int_{\mathcal{S}}\operatorname{Tr}[WC]\operatorname{Tr}[W^{\dagger}D]\;d\nu(W)=\int_{\mathcal{U}(d)}\operatorname{Tr}[WC]\operatorname{Tr}[WD]\;d\mu(W)=\tfrac{1}{d}\operatorname{Tr}[CD]$
(12)

######  Lemma 2.

Let an ensemble $\mathcal{S}$ (subset of $\mathcal{U}(d)$) with a uniform
distribution $\nu$ be a unitary $t-$design with $t\geq 2$, and let
$C,D,E,F:\mathbb{C}^{d}\rightarrow\mathbb{C}^{d}$ be linear operators. Then

$\displaystyle\int_{\mathcal{S}}\operatorname{Tr}[WC]$
$\displaystyle\operatorname{Tr}[W^{\dagger}D]\operatorname{Tr}[WE]\operatorname{Tr}[W^{\dagger}F]\;d\nu(W)=\int_{\mathcal{U}(d)}\operatorname{Tr}[WC]\operatorname{Tr}[W^{\dagger}D]\operatorname{Tr}[WE]\operatorname{Tr}[W^{\dagger}F]\;d\mu(W)$
$\displaystyle=\tfrac{1}{d^{2}-1}\left(\operatorname{Tr}[CD]\operatorname{Tr}[EF]+\operatorname{Tr}[CF]\operatorname{Tr}[ED]\right)-\tfrac{1}{d^{2}-1}\left(\operatorname{Tr}[CDEF]+\operatorname{Tr}[CFED]\right)$
(13)

Moreover,

$\displaystyle\int\left(\operatorname{Tr}[WC]\operatorname{Tr}[W^{\dagger}D]\right)^{2}\;d\mu(W)=\frac{2}{d^{2}-1}\left(\operatorname{Tr}[CD]\right)^{2}-\frac{1}{d(d^{2}-1)}\left(\operatorname{Tr}[(CD)^{2}]\right)^{2}$
(14)

The proofs for the second equality of the above lemmas is seen in Appendix A
(The first equality follows from the assumption). They are essential to
compute the expectation of the cost functions in the later sections.

###  3.1 Trainability of Parameters for State/Observable Cost function

In this section we investigate the trainability of a parameter within the
context of a State/Observable cost function. In order to compare the
‘distance’ between the targeted state and the engineered state, we consider
the State/Observable cost function $C_{s}$ as defined in the Preliminaries
Section, and an observable $\mathcal{\hat{O}}$,

$C_{s}=1-\operatorname{Tr}(\mathcal{\hat{O}}U(\vec{\theta},\vec{\alpha})\rho_{{}_{0}}U^{\dagger}(\vec{\theta},\vec{\alpha})),\;\;\;\rho_{0}=\ket{0}\bra{0}$
(15)

######  Proposition 1.

The average of the partial derivatives of the cost function $C_{s}$ with
respect to $\theta_{\nu}$ in a block $\hat{B}(\alpha,\vec{\theta})$ of the
ansatz $U(\vec{\theta},\vec{\alpha})$

$\braket{\partial_{\nu}C_{s}}_{\hat{U}}=0$ (16)

Moreover,

$\braket{(\partial_{\mu}C_{s})^{2}}_{\hat{W}_{A},\hat{W}_{B}}=\frac{2}{(d-1)(d+1)^{2}}\left((\operatorname{Tr}\hat{\mathcal{O}}^{2})-\frac{(\operatorname{Tr}\hat{\mathcal{O}})^{2}}{d}\right)$
(17)

provided $W_{A}$ or $W_{B}$ forms a 2-design.

A zero expectation for the directional derivative of the cost function implies
that no particular direction in the landscape of the cost function is
function. i.e., in the landscape of the cost function, the chances of
encountering a local minimum or maximum are the same irrespective of the
direction. Hence, we can choose any particular direction for a start. Having a
zero expectation also reduces the computations to evaluate the variance. In
order analyze the chances of a trainability problem in a particular direction,
it suffices to evaluate the variance (in the same direction) by Chebyshev
inequality. This shows that the above proposition is a vital result in
exploring the possible problems in parameter trainability. The second term on
the right hand side of (17) hence controls the trainabilty of the parameter
$\theta_{\mu}$. We give a similar result for a gate cost function later in
this paper.

For numerical examples, we consider the particular ansatz given in (4) to
analyze the trainability of a parameter while considering the fock state
$\ket{0}$ as a particular example.

Fig. 1: Plots showing the mean and the variance of the partial derivative of
the State/Observable cost function (with final state $\ket{0}\bra{0}$) with
respect to the parameter $\theta_{k,\nu}$ (where $\theta_{k,\nu}$ is the phase
associated with $\ket{\nu}\bra{\nu}$ in the $k-$th block.) For the above
plots, $k=3$, $\nu=1$, and the total number of blocks $T=5,10,15,20$

The computation of these variances was carried out with the assumption that
the means of the partial derivative of the cost function is zero. It is safe
to do this since the analytical results in (16) is already validated by Fig.
1(a). Suppose $\mathcal{O}=\ket{0}\bra{0}$ in (17). With increase in the
dimension of the qudit system, the variance of the partial derivative of the
cost function decays quadratically, since the $\mathcal{O}$ is a projection
and
$\operatorname{Tr}(\mathcal{O}^{2})-(\operatorname{Tr}\mathcal{O})^{2}/d\leq
d$. As seen in the Fig. 1(b), the variance is independent of the number of
blocks present in the ansatz (i.e. for different block sizes $T=5,10,15$), and
the polynomial decay is confirmed by numerical results.

For this case $\mathcal{O}=\ket{0}\bra{0}$, we do not have any advantage of
trainability as compared to the the results in [4]. In fact the quadratic
decay rate according to (17) is much worse compared to the decay rate of
approximately $0.21$ as seen in [Cor. 1(i), [4]]. Therefore, the vital
question remains: Are there cases where a qudit system poses trainability
advantages over multi-qubits systems?

We will give some general case where we expect qudit trainability advantage
over the multi-qubit case. Consider the cases where $\hat{\mathcal{O}}$
satisfy conditions in part (ii), Theorem 1 of [4] (i.e. $\hat{\mathcal{O}}$
can only be written as an sum of operator tensors:
$\hat{\mathcal{O}}=c_{0}{\mathchoice{\mathrm{1\mskip-4.0mul}}{\mathrm{1\mskip-4.0mul}}{\mathrm{1\mskip-4.5mul}}{\mathrm{1\mskip-5.0mul}}}+\sum^{N}_{i=1}c_{i}\hat{\mathcal{O}}_{i1}\otimes\hat{\mathcal{O}}_{i2}\otimes\cdots\otimes\hat{\mathcal{O}}_{i\xi}$,
with
$\operatorname{Tr}[\hat{\mathcal{O}}_{ik}]=0,\;\;\operatorname{Tr}[(\hat{\mathcal{O}}_{ik})^{2}]\leq
2^{m}$, $m\geq 2$), then Var$(\partial_{\mu}C_{s})\leq 2^{-an}$ where
$a=(1-1/m)$. If $d=2^{n}$ in (17), and
$\operatorname{Tr}[\hat{\mathcal{O}}^{2}]-2^{-n}[\operatorname{Tr}(\hat{\mathcal{O}})]^{2})>2^{(3-a)n+1}$
(where $n$ is the number of qubits, $\;a\in(0.5,1))$, then by the trivial
bound $2^{n}>1]$, we get that Var$(\partial_{\mu}C_{s})>2^{-an}$. Thus, an
advantage over the multi-qubit case.

To better comprehend the aforementioned analytical argument, consider
introducing the Particle Number operator denoted by $\hat{\mathcal{O}}$. This
operator exhibits non-zero entries exclusively on its diagonal, forming the
sequence $(d-1,d-2,\ldots,0)$. The trace of $\hat{\mathcal{O}}$, represented
as $\operatorname{Tr}[\hat{\mathcal{O}}]$, is given by $(d/2)(d-1)$, while the
trace of $(\hat{\mathcal{O}})^{2}$ is
$\operatorname{Tr}[(\hat{\mathcal{O}})^{2}]=(d/6)(d-1)(2d-1)$. By leveraging
equation (17), we deduce that the variance of $\partial_{\mu}C_{s}$, denoted
as Var$(\partial_{\mu}C_{s})$, equals $(2d-3)/(3d+3)$. Notably, this function
is both convergent and monotonically increasing with respect to the variable
$d$. As a result, we can conclude that Var$(\partial_{\mu}C_{s})$ remains
bounded as $d$ varies, with $1/9\leq$Var$(\partial_{\mu}C_{s})\leq 2/3$. This
implies that the variability in $\partial_{\mu}C_{s}$ does not grow without
bound and instead lies within a specific range. Understanding the behavior of
Var$(\partial_{\mu}C_{s})$ is crucial in the context of the analytical
argument presented. It allows us to gain insights into the particle number
distribution and the associated state cost function $C_{s}$ for different
values of $d$.

Fig. 2: Plots showing the mean and the variance of
$\partial_{\theta_{k,\nu}}C_{s}$ where the observable is the Particle Number
Operator. 3(b) combines plots representing the decay bound as provided in [4]
(with different decay rates $a=0.5$ and $a=0.67$) and the experimental plots
from the qudit case.

Observe that in the above plots, trainability of a parameter in the few qubits
case (1-2 qubits) is much faster (depending on the decay rate $a$) than
training in a qudit case. But as qudit system dimension $d$ gets larger,
trainability becomes better in the qudit case since blue plot is increasing
(not too fast though) and the red and black plots decrease. Evidently for
observables requiring large number of qubits, training on a qudit system is of
advantage over multi-qubits systems.

###  3.2 Trainability of Parameters for Gate Cost function

In this context, our focus is on examining the trainability concerning the
gate cost function. The preparation of a unitary gate using the SNAP-
Displacement protocol holds a significance akin to that of preparing a state.
Remarkably, if we possess accurate gates, it alleviates the complexities
surrounding the preparation of highly excited states. Thus, the quality of
unitary gates is crucial for efficient quantum operations. Similar to our
approach for the state/observable cost function, we adopt a comparable
argument for the gate cost function. This means employing the same partition
for the unitary operator $\hat{U}(\vec{\theta},\vec{\alpha})$, as presented in
(34). The gate cost function, denoted as $C_{g}$, captures the essence of the
cost associated with generating the desired unitary gate and is defined in the
preliminaries section 2. Examining the trainability within the realm of the
gate cost function allows us to assess the feasibility and efficiency of
obtaining the desired unitary transformations through the SNAP-Displacement
protocol. By understanding the factors influencing the trainability of gates,
we can devise strategies to improve the protocol’s performance and enhance the
accuracy of unitary operations. This, in turn, contributes to the overall
effectiveness of quantum computations and tasks. In summary, the trainability
analysis of the gate cost function emphasizes the importance of precise
unitary gates in quantum computing applications. It underscores the role of
the SNAP-Displacement protocol in generating reliable unitary operations and
highlights the potential benefits of having accurate gates, leading to
simplified state preparations and improved quantum computation capabilities.

For simplicity, we write $\hat{U}$ in place of
$\hat{U}(\vec{\theta},\vec{\alpha})$ henceforth. A similar result as in (1) is
provided.

######  Proposition 2.

The average of the partial derivatives of the cost function $C_{g}$ with
respect to $\theta_{\nu}$ in a block $\hat{B}(\alpha,\vec{\theta})$ of the
ansatz $U(\vec{\theta},\vec{\alpha})$

$\braket{\partial_{\nu}C_{g}}_{\hat{U}}=0$ (18)

Moreover,

$\displaystyle\braket{(\partial_{\nu}C_{g})^{2}}_{\hat{U}}=\frac{2(d^{6}+d^{5}-4d^{4}-3d^{3}+5d^{2}+2d-1)}{d^{4}(d^{2}-1)^{4}}+\frac{2}{d^{4}(d^{2}-1)^{4}}(|\operatorname{Tr}[\hat{U}_{t}]|^{4}-2|\operatorname{Tr}[\hat{U}_{t}]|^{2})$
(19)

$\braket{(\partial_{\nu}C_{g})^{2}}_{\hat{U}}=\mathcal{O}(\mathrm{poly}(1/d^{6}))$
(20)

provided $W_{A}$ or $W_{B}$ form a 2-design.

We further explore what happens when we try to train parameters in trying to
obtain certain gates. A particular example is shown in the figure below for
the Identity gate. Analytical results in (18) and (20) are validated.

Fig. 3: Plots showing the mean and the variance of the partial derivative of
the Gate cost function (with target gate being the identity map $I_{d}$) with
respect to the parameter $\theta_{k,\nu}$ (where $\theta_{k,\nu}$ is the phase
associated with $\ket{\nu}\bra{\nu}$ in the $k-$th block.) For the above
plots, $k=3$, $\nu=1$, and the total number of blocks $T=5,10,15,20,25$.

The above plots show that the leading order decay is independent of the number
of blocks in the ansatz. Unlike the plots for the state cost function,
trainability of parameters in the ansatz for the gate cost function is not
dependent on any particular gate. According to equation (20), the variance
leading order decay is of degree 6, hence, confirmed by the plots of uniformly
distributed variables in Fig. 3(b).

To justify the 2-design assumptions in the above propositions, notice that the
plots in Fig.(3) are plots from means over uniform distribution of unitaries
in the sets $\\{\hat{W}_{A}|\;\alpha,\theta_{j}\in[0,2\pi),\;j=1,\ldots,d\\}$
and $\\{\hat{W}_{B}|\;\alpha,\theta_{j}\in[0,2\pi),\;j=1,\ldots,d\\}$. This
indicates these sets form 2-designs (and hence 1-designs by definition).

###### Acknowledgements.

##  4 Acknowledgments

This material is based upon work supported by the U.S. Department of Energy,
Office of Science, National Quantum Information Science Research Centers,
Superconducting Quantum Materials and Systems Center (SQMS) under contract
number DE-AC02-07CH11359. This research was supported in part by an
appointment with the National Science Foundation (NSF) Mathematical Sciences
Graduate Internship (MSGI) Program sponsored by the NSF Division of
Mathematical Sciences. This program is administered by the Oak Ridge Institute
for Science and Education (ORISE) through an interagency agreement between the
U.S. Department of Energy (DOE) and NSF. ORISE is managed for DOE by ORAU. All
opinions expressed in this paper are the author’s and do not necessarily
reflect the policies and views of NSF, ORAU/ORISE, or DOE.

##  Appendix A Appendix

The first and second moments of Haar measures are used to analyze expectations
of trace operators [6, 16]. These calculations would have been very difficult
since it entails numerous integrations of polynomial functions over the set of
unitaries $\mathcal{U}(d)$, where $d$ is the dimension of the unitary group.
The first two moment results are listed below and some of the lemmas from [4]
(based on the these moments) for parameterized circuits with $t$-designs are
also applied.

$\displaystyle\int_{\mathcal{U}(d)}w_{\bf i_{1},j_{1}}\;w^{*}_{\bf
i_{2},j_{2}}\;d\mu(W)$ $\displaystyle=\frac{\delta_{\bf
i_{1},i_{2}}\delta_{\bf j_{1},j_{2}}}{d},$ (21)
$\displaystyle\int_{\mathcal{U}(d)}w_{\bf i_{1},j_{1}}w_{\bf
i_{2},j_{2}}w^{*}_{\bf p_{1},k_{1}}w^{*}_{\bf p_{2},k_{2}}\;d\mu(W)$
$\displaystyle=\frac{1}{d^{2}-1}(\delta_{\bf i_{1},p_{1}}\delta_{\bf
i_{2},p_{2}}\delta_{\bf j_{1},k_{1}}\delta_{\bf j_{2},k_{2}}+\delta_{\bf
i_{i},p_{2}}\delta_{\bf i_{2},p_{1}}\delta_{\bf j_{1},k_{2}}\delta_{\bf
j_{2},k_{1}})$ (22) $\displaystyle-\frac{1}{d(d^{2}-1)}(\delta_{\bf
i_{1},p_{1}}\delta_{\bf i_{2},p_{2}}\delta_{\bf j_{1},k_{2}}\delta_{\bf
j_{2},k_{1}}+\delta_{\bf i_{i},p_{2}}\delta_{\bf i_{2},p_{1}}\delta_{\bf
j_{1},k_{1}}\delta_{\bf j_{2},k_{2}})$

The proofs to Lemmas 3 and 4 below can be found in [4] while we give the
proofs of some new lemmas.

######  Lemma 3.

Let an ensemble $\mathcal{S}$ (subset of $\mathcal{U}(d)$) with a uniform
distribution $\nu$ be a unitary $t-$design with $t\geq 1$, and let
$A,B:\mathbb{C}^{d}\rightarrow\mathbb{C}^{d}$ be arbitrary linear operators.
Then

$\int_{\mathcal{S}}\operatorname{Tr}[WAW^{\dagger}B]\;d\nu(W)=\int_{\mathcal{U}(d)}\operatorname{Tr}[WAW^{\dagger}B]\;d\mu(W)=\frac{\operatorname{Tr}[A]\operatorname{Tr}[B]}{d}$
(23)

######  Lemma 4.

Let an ensemble $\mathcal{S}$ (subset of $\mathcal{U}(d)$) with a uniform
distribution $\nu$ be a unitary $t-$design with $t\geq 2$, and let
$A,B,C,D:\mathbb{C}^{d}\rightarrow\mathbb{C}^{d}$ be linear operators. Then

$\displaystyle\int_{\mathcal{S}}\operatorname{Tr}[WAW^{\dagger}BWCW^{\dagger}D]\;d\nu(W)$
$\displaystyle=\int_{\mathcal{U}(d)}\operatorname{Tr}[WAW^{\dagger}BWCW^{\dagger}D]\;d\mu(W)$
$\displaystyle=\frac{1}{d^{2}-1}\left(\operatorname{Tr}[A]\operatorname{Tr}[C]\operatorname{Tr}[BD]+\operatorname{Tr}[AC]\operatorname{Tr}[B]\operatorname{Tr}[D]\right)$
(24)
$\displaystyle-\frac{1}{d(d^{2}-1)}\left(\operatorname{Tr}[AC]\operatorname{Tr}[BD]+\operatorname{Tr}[A]\operatorname{Tr}[B]\operatorname{Tr}[C]\operatorname{Tr}[D]\right)$

######  Lemma 5.

Let an ensemble $\mathcal{S}$ (subset of $\mathcal{U}(d)$) with a uniform
distribution $\nu$ be a unitary $t-$design with $t\geq 2$ and let
$A,B,C,D:\mathbb{C}^{d}\rightarrow\mathbb{C}^{d}$ be linear operators. Then

$\displaystyle\int_{\mathcal{S}}\operatorname{Tr}[WAW^{\dagger}B]\operatorname{Tr}[WCW^{\dagger}D]\;d\nu(W)$
$\displaystyle=\int_{\mathcal{U}(d)}\operatorname{Tr}[WAW^{\dagger}B]\operatorname{Tr}[WCW^{\dagger}D]\;d\mu(W)$
$\displaystyle=\frac{1}{d^{2}-1}\left(\operatorname{Tr}[A]\operatorname{Tr}[B]\operatorname{Tr}[C]\operatorname{Tr}[D]+\operatorname{Tr}[AC]\operatorname{Tr}[BD]\right)$
(25)
$\displaystyle-\frac{1}{d(d^{2}-1)}\left(\operatorname{Tr}[AC]\operatorname{Tr}[B]\operatorname{Tr}[D]+\operatorname{Tr}[A]\operatorname{Tr}[C]\operatorname{Tr}[BD]\right)$

######  Corollary 1.

Suppose $A,B$ satisfy the assumptions in the above lemma,

$\displaystyle\int_{\mathcal{S}}(\operatorname{Tr}[WAW^{\dagger}B])^{2}\;d\nu(W)$
$\displaystyle=\int_{\mathcal{U}(d)}(\operatorname{Tr}[WAW^{\dagger}B])^{2}\;d\mu(W)$
(26)
$\displaystyle=\frac{1}{d^{2}-1}\left[(\operatorname{Tr}[A])^{2}\left((\operatorname{Tr}[B])^{2}-\frac{\operatorname{Tr}[B^{2}]}{d}\right)+\operatorname{Tr}[A^{2}]\left(\operatorname{Tr}[B^{2}]-\frac{1}{d}(\operatorname{Tr}[B])^{2}\right)\right]$

###### Proof.

Setting $A=C$ and $B=D$ in the above lemma and combining similar terms, we get
the desired result. ∎

###### Proof of Lemma 1:.

By expressing the trace and by applying (21), we have

$\displaystyle\int_{\mathcal{U}(d)}\operatorname{Tr}[WC]\operatorname{Tr}[W^{\dagger}D]\;d\mu(W)$
$\displaystyle=\sum_{i,j,k,m}^{d}c_{ji}d_{mk}\int_{\mathcal{U}(d)}w_{ij}w^{*}_{mk}\;d\mu(W)$
(27)
$\displaystyle=\frac{1}{d}\sum_{i,j}^{d}c_{ji}d_{ij}=\frac{1}{d}\operatorname{Tr}[CD]$
(28)

∎

###### Proof of Lemma 2:.

By expressing the trace and applying (22),

$\displaystyle\int_{\mathcal{U}(d)}\operatorname{Tr}[WC]\operatorname{Tr}[W^{\dagger}D]$
$\displaystyle\operatorname{Tr}[WE]\operatorname{Tr}[W^{\dagger}F]\;d\mu(W)=\int_{\mathcal{U}(d)}\sum_{\begin{subarray}{c}i,j,k,m\\\
i^{\prime},j^{\prime},k^{\prime},m^{\prime}\end{subarray}}^{d}c_{ji}d_{mk}e_{j^{\prime}i^{\prime}}f_{m^{\prime}k^{\prime}}w_{ij}w_{mk}^{*}w_{i^{\prime}j^{\prime}}w^{*}_{m^{\prime}k^{\prime}}\;d\mu(W)$
(29) $\displaystyle=\sum_{\begin{subarray}{c}i,j,k,m\\\
i^{\prime},j^{\prime},k^{\prime},m^{\prime}\end{subarray}}^{d}c_{ji}d_{mk}e_{j^{\prime}i^{\prime}}f_{m^{\prime}k^{\prime}}\left[\frac{1}{d^{2}-1}(\delta_{im}\delta_{i^{\prime}m^{\prime}}\delta_{jk}\delta_{j^{\prime}k^{\prime}}+\delta_{im^{\prime}}\delta_{i^{\prime}m}\delta_{jk^{\prime}}\delta_{j^{\prime}k})\right.$
(30) $\displaystyle\hskip
128.0374pt\left.-\frac{1}{d(d^{2}-1)}(\delta_{im}\delta_{i^{\prime}m^{\prime}}\delta_{jk^{\prime}}\delta_{j^{\prime}k}+\delta_{im^{\prime}}\delta_{i^{\prime}m}\delta_{jk}\delta_{j^{\prime}k^{\prime}})\right]$

By summing over the indices $k,m,k^{\prime},m^{\prime}$, the last equality
yields

$\displaystyle\sum_{\begin{subarray}{c}i,j\\\
i^{\prime},j^{\prime}\end{subarray}}^{d}c_{ji}e_{j^{\prime}i^{\prime}}\left[\frac{1}{d^{2}-1}(d_{ij}f_{i^{\prime}j^{\prime}}\right.$
$\displaystyle\left.+d_{i^{\prime}j^{\prime}}f_{ij})-\frac{1}{d(d^{2}-1)}(f_{i^{\prime}j}d_{ij^{\prime}}+d_{i^{\prime}j}f_{ij^{\prime}})\right]$
(32)
$\displaystyle=\frac{1}{d^{2}-1}(\operatorname{Tr}[CD\operatorname{Tr}[EF]+\operatorname{Tr}[CF]\operatorname{Tr}[ED])-\frac{1}{d(d^{2}-1)}(\operatorname{Tr}[CDEF]+\operatorname{Tr}[CFED])$

Setting $C=E$ and $D=F$ proves the last part of the lemma. ∎∎

Following the notations in section 2, where the parameterized ansatz
$U(\vec{\theta},\vec{\alpha})$ given by

$U(\vec{\theta},\vec{\alpha})=\hat{B}(\alpha_{T},\vec{\theta}_{T})\cdots\hat{B}(\alpha_{2},\vec{\theta}_{2})\hat{B}(\alpha_{1},\vec{\theta}_{1})$
(33)

and
$\hat{B}(\alpha,\vec{\theta})=\hat{D}^{\dagger}(\alpha)\hat{S}_{N}(\vec{\theta})\hat{D}(\alpha)$,
we can then write
$\hat{U}(\vec{\theta},\vec{\alpha})=\hat{U_{R}}\hat{B}\hat{U_{L}}$ where

$\displaystyle\hat{U_{R}}=$
$\displaystyle\hat{B}(\alpha_{T},\vec{\theta}_{T})\cdots\hat{B}(\alpha_{\nu-1},\vec{\theta}_{\nu-1}),$
(34) $\displaystyle\hat{U_{L}}$
$\displaystyle=\hat{B}(\alpha_{\nu+1},\vec{\theta}_{\nu+1})\cdots\hat{B}(\alpha_{1},\vec{\theta}_{1})\;\;\;\;\text{and}\;\;\hat{B}(\alpha_{\nu},\vec{\theta}_{\nu})=\hat{B}$

From the definition of the SNAP gate,

$\hat{B}(\vec{\theta},\vec{\alpha})=\underset{\hat{W}_{A}}{\underbrace{\hat{D^{\dagger}}(\alpha)\prod_{j=0}^{\nu-1}e^{i\theta_{j}\ket{j}\bra{j}}e^{i\theta_{\nu}\ket{\nu}\bra{\nu}}}}\;\underset{\hat{W}_{B}}{\underbrace{\prod_{j=\nu+1}^{d}e^{i\theta_{j}\ket{j}\bra{j}}D(\alpha)}}$
(35)

Hence, $\hat{B}(\vec{\theta},\vec{\alpha})=\hat{W}_{A}\hat{W}_{B}$. For ease
of writing we write $B$ in place of the block
$\hat{B}(\vec{\theta},\vec{\alpha})$. Taking the partial derivative of $B$
with respect to $\nu$, i.e.,

$\partial_{\nu}B\equiv\frac{\partial
B}{\partial\theta_{\nu}}=i\hat{W}_{A}\rho_{\nu}\hat{W}_{B}\;\;\;\;\text{where}\;\;\;\rho_{\nu}=\ket{\nu}\bra{\nu}.$
(36)

###### Proof of Proposition 1:.

$\displaystyle\partial_{\nu}C_{s}$
$\displaystyle=-\operatorname{Tr}[\hat{\mathcal{O}}(\partial_{\nu}\hat{U}\rho_{0}\hat{U}^{\dagger}+\hat{U}\rho_{0}\partial_{\nu}\hat{U}^{\dagger})]$
(37)
$\displaystyle=-i\operatorname{Tr}(\hat{\mathcal{O}}\hat{U}_{R}\hat{W}_{A}\rho_{\nu}\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\hat{W}^{\dagger}_{A}\hat{U}^{\dagger}_{R}-\hat{\mathcal{O}}\hat{U}_{R}\hat{W}_{A}\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\rho_{\nu}\hat{W}^{\dagger}_{A}\hat{U}^{\dagger}_{R})$
(38)
$\displaystyle=-i\operatorname{Tr}[\hat{W}^{\dagger}_{A}\hat{U}^{\dagger}_{R}\hat{\mathcal{O}}\hat{U}_{R}\hat{W}_{A}(\rho_{\nu}\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}-\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\rho_{\nu})]$
(39)
$\displaystyle=-i\operatorname{Tr}(\hat{W}^{\dagger}_{A}\hat{U}^{\dagger}_{R}\hat{\mathcal{O}}\hat{U}_{R}\hat{W}_{A}[\rho_{\nu},\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}])$
(40)

Similarly,

$\partial_{\nu}C_{s}=-i\operatorname{Tr}([\hat{W}^{\dagger}_{A}\hat{U}^{\dagger}_{R}\hat{\mathcal{O}}\hat{U}_{R}\hat{W}_{A},\rho_{\nu}]\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B})$
(41)

To compute $\braket{\partial_{\mu}C_{s}}$, there is need to compute
expectation over $\hat{W}_{A},\hat{W}_{B},\hat{U}_{L},\hat{U}_{R}$. We then
consider the three cases (1) $\hat{W}_{A}$ is a 1-design, (2) $\hat{W}_{B}$ is
a 1-design, (3) $\hat{W}_{A}$ and $\hat{W}_{B}$ are both 1-designs.

For the case (1) and using (41),

$\displaystyle\braket{\partial_{\mu}C_{g}}_{\hat{W}_{A}}$
$\displaystyle=-i\operatorname{Tr}\left(\left[\int\hat{W}^{\dagger}_{A}\hat{U}^{\dagger}_{R}\hat{\mathcal{O}}\hat{U}_{R}\hat{W}_{A}\;\;d\mu(\hat{W}_{A}),\rho_{\nu}\right]\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\right)$
(42)
$\displaystyle=-i\operatorname{Tr}\left(\left[\frac{\operatorname{Tr}(\hat{U}^{\dagger}_{R}\hat{\mathcal{O}}\hat{U}_{R})\operatorname{Tr}({\mathchoice{\mathrm{1\mskip-4.0mul}}{\mathrm{1\mskip-4.0mul}}{\mathrm{1\mskip-4.5mul}}{\mathrm{1\mskip-5.0mul}}})}{d},\rho_{\nu}\right]\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\right)$
(43) $\displaystyle=0$ (44)

where we have applied Lemma 3. Similarly,
$\braket{\partial_{\mu}C_{s}}_{\hat{W}_{B}}=0$. Hence for the three cases,

$\braket{\partial_{\mu}C_{s}}_{\hat{U}}=0$ (45)

To evaluate the variance, it suffices to only consider the expectation of the
squared of the gradient.

$\displaystyle(\partial_{\mu}C_{s})^{2}=-\operatorname{Tr}\left[\hat{W}^{\dagger}_{A}\hat{U}^{\dagger}_{R}\hat{\mathcal{O}}\hat{U}_{R}\hat{W}_{A}[\rho_{\nu},\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}]\right]^{2}$
(46)

Therefore by setting
$\hat{A}=[\rho_{\nu},\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}]$,
and $\hat{B}=\hat{U}^{\dagger}_{R}\hat{\mathcal{O}}\hat{U}_{R}$, and applying
Corollary 1,

$\displaystyle\braket{(\partial_{\mu}C_{s})^{2}}_{\hat{W}_{A}}$
$\displaystyle=\frac{1}{d^{2}-1}\left[(\operatorname{Tr}\hat{A})^{2}\left((\operatorname{Tr}\hat{B})^{2}-\frac{\operatorname{Tr}\hat{B}^{2}}{d}\right)\right]+(\operatorname{Tr}\hat{A}^{2})\left((\operatorname{Tr}\hat{B}^{2})-\frac{(\operatorname{Tr}\hat{B})^{2}}{d}\right)$
(47)
$\displaystyle=\frac{1}{d^{2}-1}\left[(\operatorname{Tr}\hat{A}^{2})\left((\operatorname{Tr}\hat{B}^{2})-\frac{(\operatorname{Tr}\hat{B})^{2}}{d}\right)\right]$
(48)
$\displaystyle=\frac{1}{d^{2}-1}\left[(\operatorname{Tr}\hat{A}^{2})\left((\operatorname{Tr}\hat{\mathcal{O}}^{2})-\frac{(\operatorname{Tr}\hat{\mathcal{O}})^{2}}{d}\right)\right]$
(49)

where we have used the fact that $\operatorname{Tr}[\hat{A}]=0$ in the second
in the second equality.

We simplify $\operatorname{Tr}[\hat{B}^{2}]$ to have

$\operatorname{Tr}\hat{B}^{2}=2\left(\operatorname{Tr}[\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\rho_{\nu}]\right)^{2}-2\operatorname{Tr}[\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\rho_{\nu}]$
(50)

Applying Corollary 1,

$\int\left(\operatorname{Tr}[\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\rho_{\nu}]\right)^{2}\;d\mu(\hat{W}_{B})=\frac{1}{d(d+1)}$
(51)

and by Lemma 3, we have that

$\int\operatorname{Tr}\left(\hat{W}_{B}\hat{U}_{L}\rho_{0}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\rho_{\nu}\right)\;d\mu(\hat{W}_{B})=\frac{1}{d}$
(52)

Hence,

$\braket{(\partial_{\mu}C_{s})^{2}}_{\hat{W}_{A},\hat{W}_{B}}=\frac{2}{(d-1)(d+1)^{2}}\left((\operatorname{Tr}\hat{\mathcal{O}}^{2})-\frac{(\operatorname{Tr}\hat{\mathcal{O}})^{2}}{d}\right)$
(53)

∎

###### Proof of Proposition 2:.

$\displaystyle\partial_{\nu}C_{g}$
$\displaystyle=-\tfrac{1}{d^{2}}(\operatorname{Tr}[\hat{U}_{t}\hat{U}^{\dagger}_{f}]\partial_{\nu}(\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{f}])+\text{c.c})$
(54)
$\displaystyle=-\frac{1}{d^{2}}(\operatorname{Tr}[\hat{U}_{t}\hat{U}^{\dagger}_{f}]\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{R}(\partial_{\nu}B)\hat{U}_{L}]+\text{c.c})$
(55)
$\displaystyle=-\frac{i}{d^{2}}(\operatorname{Tr}[\hat{U}_{t}\hat{U}^{\dagger}_{f}]\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{R}\hat{W}_{A}\rho_{\nu}\hat{W}_{B}\hat{U}_{L}]+\text{c.c})$
(56)
$\displaystyle=-\frac{i}{d^{2}}(\operatorname{Tr}[\hat{U}_{t}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\hat{W}^{\dagger}_{A}\hat{U}^{\dagger}_{R}]\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{R}\hat{W}_{A}\rho_{\nu}\hat{W}_{B}\hat{U}_{L}]-\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{L}\hat{W}_{B}\hat{W}_{A}\hat{U}_{R}]\operatorname{Tr}[\hat{U}_{t}\hat{U}^{\dagger}_{R}\hat{W}^{\dagger}_{A}\rho_{\nu}\hat{W}^{\dagger}_{B}\hat{U}^{\dagger}_{L}])$
(57)

where we have used (36) in third equation.

By setting
$C_{1}=\rho_{\nu}\hat{W}_{B}\hat{U}_{L}\hat{U}^{\dagger}_{t}\hat{U}_{R},\;\;D_{1}=\hat{U}^{\dagger}_{R}\hat{U}_{t}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{b},\;\;C_{2}=\hat{U}_{R}\hat{U}^{\dagger}_{t}\hat{U}_{L}\hat{W}_{B},\;\;D_{2}=\rho_{\nu}\hat{W}^{\dagger}_{B}\hat{U}^{\dagger}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{R}$,
it is obvious that $C_{1}=\rho_{\nu}D_{1}^{\dagger}$ and
$D_{2}=\rho_{\nu}D^{\dagger}_{2}$. By applying lemma 1, we have that

$\displaystyle\braket{\partial_{\nu}C_{g}}_{\hat{W}_{A}}$
$\displaystyle=-\frac{i}{d^{3}}(\operatorname{Tr}[C_{1}D_{1}]-\operatorname{Tr}[D_{2}^{\dagger}C_{2}^{\dagger}])$
(58)
$\displaystyle=-\frac{i}{d^{3}}(\operatorname{Tr}[\rho_{\nu}D^{\dagger}_{1}D_{1}]-\operatorname{Tr}[D^{\dagger}_{2}D\rho_{\nu}])=0$
(59)

A similar result holds for $\braket{\partial_{\nu}C_{g}}_{\hat{W}_{B}}$.
Therefore, we have a zero expectation if either $\hat{W}_{A}$ or
$\hat{W}_{B}$, (or both) is a 1-design.

The variance is hence calculated directly as the second moment of the gradient
since we have zero expectation.

$\displaystyle(\partial_{\nu}C_{g})^{2}=-\frac{1}{d^{4}}\\{(\operatorname{Tr}[\hat{W}_{A}C_{1}]\operatorname{Tr}[\hat{W}^{\dagger}_{A}D_{1}])^{2}-2\operatorname{Tr}[\hat{W}_{A}C_{1}]\operatorname{Tr}[\hat{W}^{\dagger}_{A}D_{1}]\operatorname{Tr}[\hat{W}_{A}C_{2}]\operatorname{Tr}[\hat{W}^{\dagger}_{A}D_{2}]+(\operatorname{Tr}[\hat{W}_{A}C_{2}]\operatorname{Tr}[\hat{W}^{\dagger}_{A}D_{2}])^{2}\\}$
(60)

We then consider each terms separately and apply Lemma 2.

$\displaystyle\braket{(\operatorname{Tr}[\hat{W}_{A}C_{1}]\operatorname{Tr}[\hat{W}^{\dagger}_{A}D_{1}])^{2}}_{\hat{W}_{A}}$
$\displaystyle=\frac{2}{d^{2}-1}(\operatorname{Tr}[C_{1}D_{1}])^{2}-\frac{1}{d(d^{2}-1)}(\operatorname{Tr}[(C_{1}D_{1})^{2}])^{2}$
(61)
$\displaystyle=\frac{2}{d^{2}-1}(\operatorname{Tr}[\rho_{\nu}])^{2}-\frac{1}{d(d^{2}-1)}(\operatorname{Tr}[\rho_{\nu}])^{2}$
(62) $\displaystyle=\frac{2d-1}{d(d^{2}-1)}$ (63)

$\displaystyle\braket{\operatorname{Tr}[\hat{W}_{A}C_{1}]\operatorname{Tr}[\hat{W}^{\dagger}_{A}D_{1}]\operatorname{Tr}[\hat{W}_{A}C_{2}]\operatorname{Tr}[\hat{W}^{\dagger}_{A}D_{2}]}_{\hat{W}_{A}}=\frac{1}{d^{2}-1}(\operatorname{Tr}[C_{1}D_{1}]\operatorname{Tr}[C_{2}D_{2}]+\operatorname{Tr}[C_{2}D_{1}]\operatorname{Tr}[C_{1}D_{2}])$
$\displaystyle\hskip
241.84842pt-\frac{1}{d(d^{2}-1)}(\operatorname{Tr}[C_{1}D_{1}C_{2}D_{2}]+\operatorname{Tr}[C_{1}D_{2}C_{2}D_{1}])$
(64) $\displaystyle\hskip
71.13188pt=\frac{1}{d^{2}-1}((\operatorname{Tr}[\rho_{\nu}])^{2}+\operatorname{Tr}[C_{2}D_{1}]\operatorname{Tr}[C_{1}D_{2}])-\frac{1}{d(d^{2}-1)}(\operatorname{Tr}[\rho_{\nu}C_{2}\rho_{\nu}C^{\dagger}_{2}]+\operatorname{Tr}[C_{1}D_{2}C_{2}D_{1}])$
(65) $\displaystyle\hskip
71.13188pt=\frac{1}{d^{2}-1}+\frac{1}{d^{2}-1}\operatorname{Tr}[C_{2}D_{1}]\operatorname{Tr}[C_{1}D_{2}]-\frac{1}{d(d^{2}-1)}(\operatorname{Tr}[\rho_{\nu}C_{2}\rho_{\nu}C^{\dagger}_{2}]+\operatorname{Tr}[C_{1}D_{2}C_{2}D_{1}])$
(66)

$\displaystyle\braket{(\operatorname{Tr}[\hat{W}_{A}C_{2}]\operatorname{Tr}[\hat{W}^{\dagger}_{A}D_{2}])^{2}}_{\hat{W}_{A}}$
$\displaystyle=\frac{2}{d^{2}-1}(\operatorname{Tr}[C_{2}D_{2}])^{2}-\frac{1}{d(d^{2}-1)}(\operatorname{Tr}[(C_{2}D_{2})^{2}])^{2}$
(67)
$\displaystyle=\frac{2}{d^{2}-1}(\operatorname{Tr}[\rho_{\nu}])^{2}-\frac{1}{d(d^{2}-1)}(\operatorname{Tr}[\rho_{\nu}])^{2}$
(68) $\displaystyle=\frac{2d-1}{d(d^{2}-1)}$ (69)

Substituting equations (63), (66) and (69) into (60), we have that

$\displaystyle{\braket{(\partial_{\nu}C_{g})^{2}}}_{\hat{W}_{A}}=-\frac{2}{d^{5}(d+1)}+\frac{2}{d^{4}(d^{2}-1)}\operatorname{Tr}[C_{2}D_{1}]\operatorname{Tr}[C_{1}D_{2}]-\frac{2}{d^{5}(d^{2}-1)}(\operatorname{Tr}[\rho_{\nu}C_{2}\rho_{\nu}C^{\dagger}_{2}]+\operatorname{Tr}[C_{1}D_{2}C_{2}D_{1}])$
(70)

Next we take expectation of the non-constant terms with respect to
$\hat{W}_{B}$. Starting with the last two terms,

$\displaystyle\braket{\operatorname{Tr}[\rho_{\nu}C_{2}\rho_{\nu}C^{\dagger}_{2}]}_{\hat{W}_{B}}$
$\displaystyle=\braket{\operatorname{Tr}[\hat{W}_{B}\rho_{\nu}\hat{W}^{\dagger}_{B}\hat{U}^{\dagger}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{R}\rho_{\nu}\hat{U}_{R}\hat{U}^{\dagger}_{t}\hat{U}_{L}]}_{\hat{W}_{B}}$
(71)
$\displaystyle=\tfrac{1}{d}\operatorname{Tr}[\rho_{\nu}]\operatorname{Tr}[\hat{U}^{\dagger}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{R}\rho_{\nu}\hat{U}_{R}\hat{U}^{\dagger}_{t}\hat{U}_{L}]$
(72) $\displaystyle=\tfrac{1}{d}$ (73)

$\displaystyle\braket{\operatorname{Tr}[C_{1}D_{2}C_{2}D_{1}]}_{\hat{W}_{B}}$
$\displaystyle=\braket{\hat{W}_{B}\hat{U}_{L}\hat{U}^{\dagger}_{t}\hat{U}_{R}\rho_{\nu}\hat{U}^{\dagger}_{R}\hat{U}_{t}\hat{U}^{\dagger}_{L}\hat{W}^{\dagger}_{B}\rho_{\nu}}_{\hat{W}_{B}}$
(74)
$\displaystyle=\tfrac{1}{d}\operatorname{Tr}[\hat{U}_{L}\hat{U}^{\dagger}_{t}\hat{U}_{R}\rho_{\nu}\hat{U}^{\dagger}_{R}\hat{U}_{t}\hat{U}^{\dagger}_{L}]$
(75) $\displaystyle=\tfrac{1}{d}$ (76)

From the definition of $C_{1},\;C_{2},\;D_{1}$ and $D_{2}$, we can re-write

$\operatorname{Tr}[C_{2}D_{1}]=\operatorname{Tr}[\hat{W}_{B}C_{3}\hat{W}^{\dagger}_{B}D_{3}]\qquad\text{and}\qquad\operatorname{Tr}[C_{1}D_{2}]=\operatorname{Tr}[\hat{W}_{B}C_{3}^{\dagger}\hat{W}^{\dagger}_{B}D^{\dagger}_{3}]$
(77)

where

$C_{3}=\hat{U}^{\dagger}_{R}\hat{U}_{t}\hat{U}^{\dagger}_{L}\qquad\text{and}\qquad
D_{3}=\hat{U}_{R}\hat{U}^{\dagger}_{t}\hat{U}_{L}\qquad$ (78)

Therefore, by lemma 4,

$\displaystyle\braket{\operatorname{Tr}[\hat{W}_{B}C_{3}\hat{W}^{\dagger}_{B}D_{3}]\operatorname{Tr}[\hat{W}_{B}C_{4}\hat{W}^{\dagger}_{B}D_{4}]}_{\hat{W}_{B}}=\frac{1}{d^{2}-1}(\operatorname{Tr}[C_{3}]\operatorname{Tr}[D_{3}]\operatorname{Tr}[C^{\dagger}_{3}]\operatorname{Tr}[D^{\dagger}_{3}]+d^{2})$
$\displaystyle\hskip
227.62204pt-\frac{1}{d(d^{2}-1)}(d\operatorname{Tr}[D_{3}]\operatorname{Tr}[D^{\dagger}_{3}]+d\operatorname{Tr}[C_{3}]\operatorname{Tr}[C^{\dagger}_{3}])$
(79) $\displaystyle\hskip
56.9055pt=\frac{1}{d^{2}-1}\operatorname{Tr}[C_{3}]\operatorname{Tr}[C^{\dagger}_{3}]\operatorname{Tr}[D_{3}]\operatorname{Tr}[D^{\dagger}_{3}]+\frac{d^{2}}{d^{2}-1}-\frac{1}{d^{2}-1}(\operatorname{Tr}[D_{3}]\operatorname{Tr}[D^{\dagger}_{3}]+\operatorname{Tr}[C_{3}]\operatorname{Tr}[C^{\dagger}_{3}])$
(80)

Hence,

$\displaystyle\braket{(\partial_{\nu}C_{g})^{2}}_{\hat{W}_{A},\hat{W}_{B}}$
$\displaystyle=\frac{2(d^{3}-d^{2}-d+2)}{d^{6}(d^{2}-1)^{2}}+\frac{2}{d^{4}(d^{2}-1)^{2}}(\operatorname{Tr}[C_{3}]\operatorname{Tr}[C^{\dagger}_{3}]\operatorname{Tr}[D_{3}]\operatorname{Tr}[D^{\dagger}_{3}]-\operatorname{Tr}[D_{3}]\operatorname{Tr}[D^{\dagger}_{3}]-\operatorname{Tr}[C_{3}]\operatorname{Tr}[C^{\dagger}_{3}])$
(81)

From the construction of $\hat{U}(\vec{\theta},\vec{\alpha})$ as an
alternating layer in equation (33), we find the expectation of
$(\partial_{\nu}C_{g})^{2}$ with respect to $\hat{U}_{R}$ ( and in fact
$\hat{U}_{L}$) by considering the each of the blocks composed in $\hat{U}_{R}$
(respectively in $\hat{U}_{L}$) since each of the blocks are independent.
Recall from (35) that each block is of the form
$\hat{B}=\hat{W}_{A}\hat{W}_{B}$.

Without loss of generality, set the first block in $\hat{U}_{R}$ as
$\hat{B}_{1}=\hat{W}_{A_{1}}\hat{W}_{B_{1}}$ such that
$\hat{U}_{R}=\hat{B_{1}}\hat{U}_{R_{1}}=\hat{W}_{A_{1}}\hat{W}_{B_{1}}\hat{U}_{R_{1}}$.
Note that $\hat{U}_{R_{1}}$ denotes $\hat{U}_{R}$ without the first
(parameterized) block. Hence

$C_{3}=\hat{U}^{\dagger}_{R_{1}}\hat{W}^{\dagger}_{B_{1}}\hat{W}^{\dagger}_{A_{1}}\hat{U}_{t}\hat{U}^{\dagger}_{L}\qquad\text{and}\qquad
D_{3}=\hat{W}_{A_{1}}\hat{W}_{B_{1}}\hat{U}_{R_{1}}\hat{U}^{\dagger}_{t}\hat{U}_{L}$
(82)

Now we take expectation (with respect to $\hat{W}_{A_{1}}$ and
$\hat{W}_{B_{1}}$) of the last three terms of (81).

$\displaystyle\braket{\operatorname{Tr}[D_{3}]\operatorname{Tr}[D^{\dagger}_{3}]}_{\hat{W}_{A_{1}}}$
$\displaystyle=\braket{\operatorname{Tr}[\hat{W}_{A_{1}}\hat{W}_{B_{1}}\hat{U}_{R_{1}}\hat{U}^{\dagger}_{t}\hat{U}_{L}]\operatorname{Tr}[\hat{W}^{\dagger}_{A_{1}}\hat{U}^{\dagger}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{R_{1}}\hat{W}^{\dagger}_{B_{1}}]}_{\hat{W}_{A_{1}}}$
(83)
$\displaystyle=\frac{1}{d}\operatorname{Tr}[\hat{W}_{B_{1}}\hat{U}_{R_{1}}\hat{U}^{\dagger}_{t}\hat{U}_{L}\hat{U}^{\dagger}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{R_{1}}\hat{W}^{\dagger}_{B_{1}}]=1$
(84)

where we have applied property of trace in the first equality and lemma 1 in
the second equation.

Similarly,

$\braket{\operatorname{Tr}[C_{3}]\operatorname{Tr}[C^{\dagger}_{3}]}_{\hat{W}_{A_{1}}}=\braket{\operatorname{Tr}[\hat{W}^{\dagger}_{A_{1}}\hat{U}_{t}\hat{U}^{\dagger}_{L}\hat{U}^{\dagger}_{R_{1}}\hat{W}^{\dagger}_{B_{1}}]\operatorname{Tr}[\hat{W}_{A_{1}}\hat{W}_{B_{1}}\hat{U}_{R_{1}}\hat{U}_{L}\hat{U}^{\dagger}_{t}]}_{\hat{W}_{A_{1}}}=1$
(85)

$\displaystyle\braket{\operatorname{Tr}[C_{3}]\operatorname{Tr}[C^{\dagger}_{3}]\operatorname{Tr}[D_{3}]\operatorname{Tr}[D^{\dagger}_{3}]}_{\hat{W}_{A_{1}}}$
$\displaystyle=\langle\operatorname{Tr}[\hat{W}_{A_{1}}\hat{W}_{B_{1}}\hat{U}_{R_{1}}\hat{U}_{L}\hat{U}^{\dagger}_{t}]\operatorname{Tr}[\hat{W}^{\dagger}_{A_{1}}\hat{U}_{t}\hat{U}^{\dagger}_{L}\hat{U}^{\dagger}_{R_{1}}\hat{W}^{\dagger}_{B_{1}}]$
$\displaystyle\hskip
85.35826pt\times\operatorname{Tr}[\hat{W}_{A_{1}}\hat{W}_{B_{1}}\hat{U}_{R_{1}}\hat{U}^{\dagger}_{t}\hat{U}_{L}]\operatorname{Tr}[\hat{W}^{\dagger}_{A_{1}}\hat{U}^{\dagger}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{R_{1}}\hat{W}^{\dagger}_{B_{1}}]\rangle_{\hat{W}_{A_{1}}}$
(86)
$\displaystyle=\frac{1}{d^{2}-1}(d^{2}+\operatorname{Tr}[\hat{W}_{B_{1}}\hat{U}_{R_{1}}\hat{U}^{\dagger}_{t}\hat{U}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{L}\hat{U}^{\dagger}_{R_{1}}\hat{W}^{\dagger}_{B_{1}}]\operatorname{Tr}[\hat{W}_{B_{1}}\hat{U}_{R_{1}}\hat{U}_{L}\hat{U}^{\dagger}_{t}\hat{U}^{\dagger}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{R_{1}}\hat{W}^{\dagger}_{B_{1}}])$
$\displaystyle\hskip 85.35826pt-\frac{1}{d(d^{2}-1)}(2d)$ (87)
$\displaystyle=\frac{d^{2}-2}{d^{2}-1}+\frac{1}{d^{2}-1}\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{L}]\operatorname{Tr}[\hat{U}_{L}\hat{U}^{\dagger}_{t}\hat{U}^{\dagger}_{L}\hat{U}_{t}]$
(88)

the first equation is gotten by re-arranging the traces and applying the trace
property while the second equation is as a result of lemma 2.

Equation (81) becomes

$\displaystyle\braket{(\partial_{\nu}C_{g})^{2}}_{\hat{W}_{A},\hat{W}_{B},\hat{W}_{A_{1}}}=\frac{2(d^{4}+d^{3}-3d^{2}-2d+1)}{d^{4}(d^{2}-1)^{3}}+\frac{2}{d^{4}(d^{2}-1)^{3}}\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{L}]\operatorname{Tr}[\hat{U}_{L}\hat{U}^{\dagger}_{t}\hat{U}^{\dagger}_{L}\hat{U}_{t}]$
(89)

Notice that taking expectation of the last equation with respect to
$\hat{W}_{B_{1}}$ does not change anything since the equation is independent
on $\hat{W}_{B_{1}}$.

By a similar construction made on $\hat{U}_{R}$, consider writing
$\hat{U}_{L}$ in terms of its last block i.e., let
$\hat{U}_{L}=\hat{U}_{L_{0}}\hat{B_{0}}=\hat{U}_{L_{0}}\hat{W}_{A_{0}}\hat{W}_{B_{0}}$.
By using trace property,

$\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{L}]=\operatorname{Tr}[\hat{W}_{A_{0}}\hat{W}_{B_{0}}\hat{U}_{t}\hat{W}^{\dagger}_{B_{0}}\hat{W}^{\dagger}_{A_{0}}\hat{U}^{\dagger}_{L_{0}}\hat{U}^{\dagger}_{t}\hat{U}_{L_{0}}],\;\;\;\;\operatorname{Tr}[\hat{U}_{L}\hat{U}^{\dagger}_{t}\hat{U}^{\dagger}_{L}\hat{U}_{t}]=\operatorname{Tr}[\hat{W}_{A_{0}}\hat{W}_{B_{0}}\hat{U}^{\dagger}_{t}\hat{W}^{\dagger}_{B_{0}}\hat{W}^{\dagger}_{A_{0}}\hat{U}^{\dagger}_{L_{0}}\hat{U}_{t}\hat{U}_{L_{0}}]$
(90)

Taking expectation of the above with respect to $\hat{W}_{A_{0}}$ and applying
lemma 4, we have

$\displaystyle\braket{\operatorname{Tr}[\hat{U}^{\dagger}_{t}\hat{U}_{L}\hat{U}_{t}\hat{U}^{\dagger}_{L}]\operatorname{Tr}[\hat{U}_{L}\hat{U}^{\dagger}_{t}\hat{U}^{\dagger}_{L}\hat{U}_{t}]}_{\hat{W}_{A_{0}}}$
$\displaystyle=\frac{1}{d^{2}-1}(\operatorname{Tr}[\hat{U}_{t}]\operatorname{Tr}[\hat{U}^{\dagger}_{t}]\operatorname{Tr}[\hat{U}^{\dagger}_{t}]\operatorname{Tr}[\hat{U}_{t}]+d^{2})$
$\displaystyle\hskip
56.9055pt-\frac{1}{d(d^{2}-1)}(d\operatorname{Tr}[\hat{U}^{\dagger}_{t}]\operatorname{Tr}[\hat{U}_{t}]+d\operatorname{Tr}[\hat{U}_{t}]\operatorname{Tr}[\hat{U}^{\dagger}_{t}])$
(91)
$\displaystyle=\frac{1}{d^{2}-1}(|\operatorname{Tr}[\hat{U}_{t}]|^{4}-2|\operatorname{Tr}[\hat{U}_{t}]|^{2})+\frac{d^{2}}{d^{2}-1}$
(92)

Therefore,

$\displaystyle\braket{(\partial_{\nu}C_{g})^{2}}_{\hat{U}}=\frac{2(d^{6}+d^{5}-4d^{4}-3d^{3}+5d^{2}+2d-1)}{d^{4}(d^{2}-1)^{4}}+\frac{2}{d^{4}(d^{2}-1)^{4}}(|\operatorname{Tr}[\hat{U}_{t}]|^{4}-2|\operatorname{Tr}[\hat{U}_{t}]|^{2})$
(93)

$\braket{(\partial_{\nu}C_{g})^{2}}_{\hat{U}}=\mathcal{O}(\mathrm{poly}(1/d^{6}))$
(94)

∎

Since all eigenvalues of unitaries are on the unit complex disk then the
absolute value of the trace is bounded by the dimension $d$ of the qudit
system; hence the equality.

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