Entanglement-assisted Reed–Solomon codes over qudits: theory and architecture

Abstract

We develop a systematic theory for the construction of quantum codes from classical Reed–Solomon over ${\mathbb {F}}_{p^k}$, where p is prime and $k \in {\mathbb {Z}}^{+}$. Based on two classical $[n,n-d_m+1,d_m]$ Reed–Solomon codes over ${\mathbb {F}}_{p^k}$, we provide the construction of an $[[n,n+n_e-2(d_m-1),d_m]]$ entanglement-assisted Reed–Solomon code over qudits of dimension $d=p^k$ that saturates the quantum Singleton bound and needs $n_e$ entangled qudits, which involves obtaining the explicit form of the stabilizers for the code. Our contributions towards the entanglement-assisted Reed–Solomon code are multi-fold as follows: (a) Based on the parity check matrices of the classical Reed–Solomon codes, the explicit form of the stabilizers of the entanglement-assisted Reed–Solomon code are provided and the rate-optimal code is obtained. (b) Based on the entanglement-assisted Reed–Solomon code, a burst error correcting code for qudits of dimension p that corrects a burst of $k(t_m-1)$ qudits or less, where $t_m = \lfloor (d_m-1)/2\rfloor$ is provided. (c) Finally, we provide amenable circuits for encoding and error correction with quantum circuit complexity of ${\mathcal {O}}(n^2)$, useful for practical implementations.

Appendices

A Proof of Lemma 1

Proof

When $\beta =0$, for $i \in \{0,1,\dots ,k-1\}$, $\mathrm {Tr}_{p^k/p}(\alpha ^i \beta ) = \mathrm {Tr}_{p^k/p}(0) = 0$.

For some $\beta \in {\mathbb {F}}_{p^k}$, let $\mathrm {Tr}_{p^k/p}(\alpha ^i \beta ) = 0$ $\forall i \in \{0,1,\dots ,k-1\}$. For every $\kappa \in {\mathbb {F}}_{p^k}$, there exists $\kappa _0, \dots , \kappa _{k-1} \in {\mathbb {F}}_p$ such that $\kappa = \underset{i=0}{\overset{k-1}{\sum }} \kappa _i \alpha ^i$ [23]. Since $\mathrm {Tr}_{p^k/p}(\cdot )$ is an ${\mathbb {F}}_p$-linear function,

$$\begin{aligned} \mathrm {Tr}_{p^k/p}(\kappa \beta )&= \sum _{i=0}^{k-1} \kappa _i \mathrm {Tr}_{p^k/p}(\alpha ^i \beta ) = 0.~(\because \kappa _i \in {\mathbb {F}}_p) \end{aligned}$$

(106)

If $\beta \ne 0$, as by closure property of ${\mathbb {F}}_{p^k}$, $\{\kappa \beta \}_{\kappa \in {\mathbb {F}}_{p^k}} = \{\kappa '\}_{\kappa '\in {\mathbb {F}}_{p^k}}$, from Eq. (106), $\mathrm {Tr}_{p^k/p}(\kappa ')=0$ for every $\kappa '\in {\mathbb {F}}_{p^k}$, which is a contradiction as the trace of all the elements of ${\mathbb {F}}_{p^k}$ cannot be 0. Thus, $\beta =0$. $\square$

B Proof of Theorem 3

Proof

We consider two classical codes $D_1[n,k_1,d_1]$ and $D_2[n,k_2,d_2]$ over ${\mathbb {F}}_{p^k}$ with parity check matrices $H_{d1}$ and $H_{d2}$, respectively. Based on $H_{d1}$ and $H_{d2}$, we construct the matrix $\mathcal {H_R}$ as in Eq. (11), similar to the check matrix of an entanglement-unassisted CSS code in Eq. (8).

After the extension of ${\mathcal {R}}$ to S, the number of stabilizer generators is $\rho = k((n-k_1)+(n-k_2)) = k(2n-k_1-k_2)$. Thus, the dimension of the EA CSS code is $p^{(kn+n_e'-\rho )} = p^{(kn+n_e'-k(2n-k_1-k_2))}6 =p^{(k(k_1+k_2-n)+n_e')}$ .

Similar to the qubit case [29], the distance of the EA stabilizer code depends only on the matrix $\mathcal {H_R}$ as the added subqudits are assumed to be error-free at the receiver end. Similar to the minimum distance of the CSS codes in Theorem 1, we obtain the minimum distance $d'$ of the EA CSS code to be $d' \ge \mathrm {min}(d_1,d_2)$.

$\textit{Computation of }\underline{\varvec{n_e'}}$:

From Eq. (9), the number of entangled subqudits needed for the EA CSS code is $n_e' = \mathrm {rank}\left( {\mathcal {T}}\left( {\mathcal {H}}_{\mathrm {X}}{\mathcal {H}}_{\mathrm {Z}}^{\mathrm {T}}\right) \right)$, where ${\mathcal {H}}_{\mathrm {X}} = [(H_{d1})^{\mathrm {T}}~\alpha (H_{d1})^{\mathrm {T}}\dots \alpha ^{(k-1)}(H_{d1})^{\mathrm {T}}]^{\mathrm {T}}$ and ${\mathcal {H}}_{\mathrm {Z}}= [(H_{d2})^{\mathrm {T}}$ $~\alpha (H_{d2})^{\mathrm {T}}\dots \alpha ^{(k-1)}(H_{d2})^{\mathrm {T}}]^{\mathrm {T}}$. We first compute ${\mathcal {H}}_{\mathrm {X}}{\mathcal {H}}_{\mathrm {Z}}^{\mathrm {T}}$ as follows:

$$\begin{aligned} {\mathcal {H}}_{\mathrm {X}}\!{\mathcal {H}}_{\mathrm {Z}}^{\mathrm {T}}&= \begin{bmatrix} H_{d1}H_{d2}^{\mathrm {T}} &{}\! \alpha H_{d1}H_{d2}^{\mathrm {T}} &{}\!\!\! \cdots &{}\!\!\! \alpha ^{(k-1)}H_{d1}H_{d2}^{\mathrm {T}}\\ \alpha H_{d1}H_{d2}^{\mathrm {T}} &{}\! \alpha ^2 H_{d1}H_{d2}^{\mathrm {T}} &{}\!\!\! \cdots &{}\!\!\! \alpha ^{k}H_{d1}H_{d2}^{\mathrm {T}}\\ \vdots &{}\! \vdots &{}\!\!\! \ddots &{}\!\!\! \vdots \\ \alpha ^{(k-1)}H_{d1}H_{d2}^{\mathrm {T}} &{}\! \alpha ^k H_{d1}H_{d2}^{\mathrm {T}} &{}\!\!\! \cdots &{} \!\!\! \alpha ^{(2k-2)}H_{d1}H_{d2}^{\mathrm {T}} \end{bmatrix}\nonumber \\&=\begin{bmatrix} Q &{}\! \alpha Q &{}\!\! \cdots &{}\!\! \alpha ^{(k-1)}Q\\ \alpha Q &{}\! \alpha ^2 Q &{}\!\! \cdots &{}\!\! \alpha ^{k}Q\\ \vdots &{}\! \vdots &{}\!\! \ddots &{}\!\! \vdots \\ \alpha ^{(k-1)}Q &{}\! \alpha ^k Q &{}\!\! \cdots &{} \!\! \alpha ^{(2k-2)}Q \end{bmatrix}, \end{aligned}$$

(107)

where $Q=H_{d1}H_{d2}^{\mathrm {T}}$.

We note that the matrix $H_{d1}H_{d2}^{\mathrm {T}}$ need not be a full-rank matrix. Let $Q'$ be a submatrix of Q such that the rows of $Q'$ are a set of maximally linearly independent rows of Q. Then, $Q'$ is a full-rank matrix and $\mathrm {rank}(Q)=\mathrm {rank}(Q')$. Let the size of $Q'$ be $r\times c$. We note that the number of columns of $Q'$ and Q are the same.

We note that every element in ${\mathbb {F}}_{p^k}$ can be written as an ${\mathbb {F}}_p$-linear combination of 1, $\alpha$, $\dots$, $\alpha ^{k-1}$. From the definition of $Q'$, the rows of Q can be obtained from the linear combination of the rows of $Q'$. Thus, the rows of Q, $\alpha Q$, $\dots$, $\alpha ^{(k-1)}Q$ can be obtained from the ${\mathbb {F}}_p$-linear combination of the rows of $Q'$, $\alpha Q'$, $\dots$, $\alpha ^{(k-1)}Q'$.

In Eq. (107), considering only those rows of ${\mathcal {H}}_{\mathrm {X}}\!{\mathcal {H}}_{\mathrm {Z}}^{\mathrm {T}}$ that have the rows of $Q'$ in the blocks instead of Q, we obtain the matrix

$$\begin{aligned} P = \begin{bmatrix} Q' &{}\! \alpha Q' &{}\!\! \cdots &{}\!\! \alpha ^{(k-1)}Q'\\ \alpha Q' &{}\! \alpha ^2 Q' &{}\!\! \cdots &{}\!\! \alpha ^{k}Q'\\ \vdots &{}\! \vdots &{}\!\! \ddots &{}\!\! \vdots \\ \alpha ^{(k-1)}Q' &{}\! \alpha ^k Q' &{}\!\! \cdots &{} \!\! \alpha ^{(2k-2)}Q' \end{bmatrix}. \end{aligned}$$

(108)

From Eqs. (107) and (108), we note that the rows of ${\mathcal {H}}_{\mathrm {X}}\!{\mathcal {H}}_{\mathrm {Z}}^{\mathrm {T}}$ can also be obtained from the ${\mathbb {F}}_p$-linear combination of the rows of P as the rows of Q, $\alpha Q$, $\dots$, $\alpha ^{(k-1)}Q$ can be obtained from the ${\mathbb {F}}_p$-linear combination of the rows of $Q'$, $\alpha Q'$, $\dots$, $\alpha ^{(k-1)}Q'$.

Let $q_{ij}$ be the $(i,j)^{\mathrm {th}}$ element of $Q'$, where $i\in \{1,\dots ,r\}$ and $j\in \{1,\dots ,c\}$. We next prove that the matrix ${\mathcal {T}}(P)$ is a full-rank matrix.

If the matrix ${\mathcal {T}}(P)$ is not a full-rank matrix, there exists a non-trivial linear combination of the rows of ${\mathcal {T}}(P)$ that sum up to zero. For $l,m\in \{0,\dots ,k-1\}$, the $(i,j)^{\mathrm {th}}$ elements in the $l^{\mathrm {th}}$ and $m^{\mathrm {th}}$ column subblock of P are $\alpha ^{(l+m)}q_{ij}$. Thus, there exists $g_{il}\in {\mathbb {F}}_{p}$, where $i\in \{1,\dots ,r\}$ and $l\in \{0,\dots ,k-1\}$, such that

$$\begin{aligned} \overset{r}{\underset{i=1}{\sum }}\overset{k-1}{\underset{l=0}{\sum }} g_{il}\mathrm {Tr}_{p^k/p}(\alpha ^{(l+m)}q_{ij}) = 0, \end{aligned}$$

(109)

for all $m\in \{0,\dots ,k-1\}$ and $j\in \{1,\dots ,c\}$. As the field trace is ${\mathbb {F}}_{p}$-linear, from Eq. (109), for all $m\in \{0,1,\dots ,k-1\}$ and $j\in \{1,\dots ,c\}$, we have

$$\begin{aligned} \mathrm {Tr}_{p^k/p}(\alpha ^{m}\overset{r}{\underset{i=1}{\sum }}\overset{k-1}{\underset{l=0}{\sum }} \alpha ^{l}q_{ij}g_{il})&= 0 \Rightarrow \overset{r}{\underset{i=1}{\sum }}\overset{k-1}{\underset{l=0}{\sum }} \alpha ^{l}q_{ij}g_{il} = 0,\nonumber \\&\quad (\text {From Lemma} 1) \Rightarrow \overset{r}{\underset{i=1}{\sum }}q_{ij}\overset{k-1}{\underset{l=0}{\sum }} \alpha ^{l}g_{il} = 0. \end{aligned}$$

(110)

As the rows of $Q'$ are linearly independent, only the trivial linear combinations of $\{q_{ij}\}_{i=1}^{r}$ is zero. Thus, from Eq. (110), we obtain

$$\begin{aligned} \overset{k-1}{\underset{l=0}{\sum }} \alpha ^{l}g_{il} = 0, \forall i\in \{1,\dots ,r\}. \end{aligned}$$

(111)

We note that $f(x) = \overset{k-1}{\underset{l=0}{\sum }} x^{l}g_{il}$ is a polynomial of degree less than k with coefficients from ${\mathbb {F}}_p$. Thus, $f(\alpha ) = \overset{k-1}{\underset{l=0}{\sum }} \alpha ^{l}g_{il}$ cannot be equal to zero unless all $g_{il}$s are zero as $\alpha$ is the primitive element and its minimal polynomial is the primitive polynomial of the field having degree k. Thus, each coefficient of f(x) is zero. Thus, from Eq. (111), we obtain

$$\begin{aligned} g_{il}&= 0, \forall i,l. \end{aligned}$$

(112)

From Eqs. (109) and (112), only the trivial linear combination of the rows of ${\mathcal {T}}(P)$ results in the all-zero vector. Thus, ${\mathcal {T}}(P)$ is a full-rank matrix.

As $\mathrm {rank}(Q') = \mathrm {rank}(Q)$ and $Q'$ is a full-rank matrix, the number of rows r of $Q'$ is $\mathrm {rank}(Q)$. Thus, the number of rows of ${\mathcal {T}}(P)$ is $k~\mathrm {rank}(Q)$. As ${\mathcal {T}}(P)$ is a full-rank matrix, the rank of ${\mathcal {T}}(P)$ is $k~\mathrm {rank}(Q) =k~\mathrm {rank}(H_{d1}H_{d2}^{\mathrm {T}})$.

We note that the rank of the matrix does not change if linearly dependent rows are added to the matrix. Thus, as all the rows of ${\mathcal {H}}_{\mathrm {X}}{\mathcal {H}}_{\mathrm {Z}}^{\mathrm {T}}$ can be obtained from the ${\mathbb {F}}_p$-linear combination of the rows of P, $\mathrm {rank}({\mathcal {T}}({\mathcal {H}}_{\mathrm {X}}{\mathcal {H}}_{\mathrm {Z}}^{\mathrm {T}})) = \mathrm {rank}({\mathcal {T}}(P)) = k~\mathrm {rank}(H_{d1}H_{d2}^{\mathrm {T}})$. Thus, from Eq. (9), we obtain $n_e' = k~\mathrm {rank}(H_{d1}H_{d2}^{\mathrm {T}}).$ $\square$

C Proof of Proposition 1

We will first state two results in Lemmas 5 and 6 necessary to prove Proposition 1.

Lemma 5

The generator matrix of an $[n,n-d_m+1,d_m]$ classical RS code C with parity check matrix $H_b$ is obtained from $H_{n-b+1}$ based on Eq. (12), by replacing $d_m$ with $n-d_m+2$.

Proof

As the dimension of the code C is $n-d_m+1$, the generator matrix has $n-d_m+1$ linearly independent rows. Let A be the matrix obtained from $H_{n-b+1}$ based on Eq. (12), by replacing $d_m$ with $n-d_m+2$. The $(i,j)^{\mathrm {th}}$ element of A is $\alpha ^{(n-b+i)(j-1)}$, where $i \in \{1,\dots ,n-d_m+1\}$ and $j \in \{1,\dots ,n\}$. As the set of rows of A is a subset of the rows of a Vandermonde matrix, A is a full-rank matrix with $n-d_m+1$ linearly independent rows.

We next show that the rowspace of A is equal to the nullspace of $H_b$. As the $(i,j)^{\mathrm {th}}$ element of $H_b$ and A are $\alpha ^{(i+b-1)(j-1)}$ and $\alpha ^{(n-b+i)(j-1)}$, respectively, we obtain

$$\begin{aligned} (H_bA^{\mathrm {T}})_{i,j}&= \underset{g=1}{\overset{n}{\sum }}\alpha ^{(i+b-1)(g-1)}\alpha ^{(n-b+j)(g-1)}=\underset{l=0}{\overset{n-1}{\sum }}\alpha ^{(n+i+j-1)l},\text { where }l=g-1, \end{aligned}$$

(113)

where $i \in \{1,\dots ,d_m-1\}$ and $j \in \{1,\dots ,n-d_m+1\}$. As $n+1\le n+i+j-1 \le 2n-1$, we obtain $(n+i+j-1)~\mathrm {mod}~n \ne 0$. Thus, from Eqs. (17) and (113), we obtain $(H_bA^{\mathrm {T}})_{i,j} =0 ~\forall$ $i \in \{1,\dots ,d_m-1\}$ and $j \in \{1,\dots ,n-d_m+1\}$. Thus, the rowspace of A lies in the null space of $H_b$. As the rank of $H_b$ is $d_m-1$, by the rank-nullity theorem [32], its null space has a dimension of $n-d_m+1$. As the rank of A is $n-d_m+1$ and $H_bA^{\mathrm {T}}=0$, the rowspace of A is equal to the nullspace of $H_b$ and A is the generator matrix of the RS code C. $\square$

Lemma 6

An EA RS code with minimum distance $d'$ constructed from two classical codes $C_1$ and $C_2$ is a non-degenerate code if every element in $C_1\cap C_2^{\perp }$ and $C_2\cap C_1^{\perp }$ has a weight of at least $d'$.

Proof

A stabilizer code is a non-degenerate code if there does not exist two correctable errors $E_a$ and $E_b$ such that $E_aE_b^{\dagger }$ is a stabilizer [1]. A stabilizer code is a non-degenerate code if the weight of each stabilizer is at least the minimum distance $d'$ of the code [33]. For an EA RS code, as the errors $E_a$ and $E_b$ operate only on the first n transmitter end qudits, the operation of $E_aE_b^{\dagger }$ on the receiver end qudits is the identity operation. Thus, for non-degeneracy, we only consider those stabilizers that are obtained by extending the elements of ${\mathcal {R}}$ with identity operators (field isomorphic equivalent is $\varvec{0}$) in Theorem 5.

In Theorem 5, only the elements of ${\mathcal {R}}$ that commute with all elements of ${\mathcal {R}}$ have been extended with identity operators. These elements of ${\mathcal {R}}$ correspond to the indices of the rows of $H_{b1}$ and $H_{b2}$ in sets $N_{\mathrm {OX}}$ and $N_{\mathrm {OZ}}$, respectively. We note that $H_{b1}$ and $H_{b2}$ generate $C_1^{\perp }$ and $C_{2}^{\perp }$ and their nullspaces are $C_1$ and $C_2$, respectively. As $N_{\mathrm {OX}}$ and $N_{\mathrm {OZ}}$ correspond to the rows of $H_{b1}$ and $H_{b2}$ that are orthogonal to all rows of $H_{b2}$ and $H_{b1}$, respectively, we consider the vectors in $C_1\cap C_2^{\perp }$ and $C_2\cap C_1^{\perp }$ to determine whether the code is non-degenerate.

The elements in ${\mathcal {R}}$ corresponding to the vectors in $C_1\cap C_2^{\perp }$ and $C_2\cap C_1^{\perp }$ have the same weight as the vectors. As these elements in ${\mathcal {R}}$ are extended by identity operators, the weight of their corresponding stabilizers is equal to the weight of the corresponding vectors in $C_1\cap C_2^{\perp }$ or $C_2\cap C_1^{\perp }$. From the definition of a non-degenerate code, we obtain an EA RS code to be a non-degenerate code if every element in $C_1\cap C_2^{\perp }$ and $C_2\cap C_1^{\perp }$ has a weight of at least $d'$.

Proposition 1

The EA RS code satisfies the following properties:

• a. The EA RS code is a non-degenerate code.

• b. The minimum distance of the EA RS code is $d_m$.

• c. The EA RS code saturates the quantum Singleton bound.

Proof

As the EA RS code is an $((n,p^{(k(n-2d_m+2+n_e))},d'\ge d_m))_{p^k}$ code, the minimum distance $d'$ of the code is atleast $d_m$, i.e., $d' \ge d_m$. We can write the code $((n,p^{(k(n-2d_m+2+n_e))},d'$ $\ge d_m))_{p^k}$ as an $[[n,n-2d_m+2+n_e,d']]_{p^k}$ code. The minimum distance $d'$ of an EA CSS code⁵ obtained from $C_1$ and $C_2$ is $\mathrm {min}(d_{\mathrm {H}}(C_1\setminus C_2^{\perp }),d_{\mathrm {H}}(C_2\setminus C_1^{\perp }))$, where $d_{\mathrm {H}}$ is the minimum hamming weight of the vectors in the set [28].

We consider the following two cases:

Case A: $\mathbf {2(d_m-1) < n}$

As the minimum distance of both $C_1$ and $C_2$ is $d_m$, the codewords in $C_1$ and $C_2$ have a weight of at least $d_m$. Thus, the vectors in $C_1\setminus C_2^{\perp }$ and $C_2\setminus C_1^{\perp }$ also have a weight of at least $d_m$. As the dual of an $[n,n-d_m+1,d_m]$ RS code is an $[n,d_m-1,n-d_m+2]$ RS code, the minimum distance of both $C_1^{\perp }$ and $C_2^{\perp }$ is $n-d_m+2$ [34]. Thus, the codewords in $C_1^{\perp }$ and $C_2^{\perp }$ have weight of at least $n-d_m+2$. As $2(d_m-1) < n$, we obtain $d_m < n-d_m+2$. Thus, all codewords in $C_1$ and $C_2$ of weight $d_m$ belong to $C_1\setminus C_2^{\perp }$ and $C_2\setminus C_1^{\perp }$, respectively. Thus, the minimum distance $d'$ of the EA RS code is $\mathrm {min}(d_{\mathrm {H}}(C_1\setminus C_2^{\perp }),d_{\mathrm {H}}(C_2\setminus C_1^{\perp })) = d_m$.

As the minimum distance of $C_1^{\perp }$ and $C_2^{\perp }$ is $n-d_m+2 > d_m$, the weight of elements in $C_1\cap C_2^{\perp }$ and $C_2\cap C_1^{\perp }$ have a weight of at least $n-d_m+2$ and greater than $d_m$. From Lemma 6, the EA RS code is a non-degenerate code as all elements in $C_1\cap C_2^{\perp }$ and $C_2\cap C_1^{\perp }$ have weight of at least $d'=d_m$. The quantum Singleton bound for an $[[n,l,d']]_{p^k}$ non-degenerate EA stabilizer code is $2(d'-1) \le n+n_e-l$ [16]. Thus, the $[[n,n-2d_m+2+n_e,d_m]]_{p^k}$ non-degenerate EA RS code saturates the quantum Singleton bound.

Case B: $\mathbf {2(d_m-1)\ge n}$

As the minimum distance to $C_1$ and $C_2$ are both $d_m$, the vectors in $C_1\setminus C_2^{\perp }$ and $C_2\setminus C_1^{\perp }$ have a weight of at least $d_m$. The codewords in the $[n,d_m-1,n-d_m+2]$ RS codes $C_1^{\perp }$ and $C_2^{\perp }$ have a weight of at least $n-d_m+2$. As $C_1$, $C_2$, $C_1^{\perp }$, and $C_2^{\perp }$ are RS codes of length n, their codewords are generated by the rows of the Vandermonde matrix. The codewords of $C_1^{\perp }$ and $C_2^{\perp }$ are generated by the matrices $H_{b1}$ and $H_{b2}$. From Lemma 5, the codewords of $C_1$ and $C_2$ are generated by the matrices obtained from $H_{n-b_1+1}$ and $H_{n-b_2+1}$, by replacing $d_m$ with $n-d_m+2$.

We first show that there exists at least one vector in $C_1\setminus C_2^{\perp }$ and $C_2\setminus C_1^{\perp }$ with weight $d_m$. We prove the result for $C_1\setminus C_2^{\perp }$. The proof is similar for $C_2\setminus C_1^{\perp }$. In Table 1, three exhaustive cases have been provided for the EA RS code based on the relation between the parameters $b_1$, $b_2$, $d_m$, and n. As $2(d_m-1) \ge n$, we do not consider Case 1 from Table 1. As Case 3 from Table 1 results in a zero-rate code⁶, we only consider Cases 2(a) and 2(b). We show that every vector in $C_1\cap C_2^{\perp }$ has weight greater than $d_m$; hence, there exists a vector of weight $d_m$ in $C_1\setminus C_2^{\perp }$. We note that all rows of the generator matrix $A=[\alpha ^{(n-b_1+i)(z-1)}]$ of $C_1$ and all rows of the generator matrix $H_{b2}$ of $C_2^{\perp }$ are rows of the $(n\times n)$-Vandermonde matrix. The rows of the Vandermonde matrix that generate $C_1\cap C_2^{\perp }$ are the rows common to both A and $H_{b2}$. As $A=[\alpha ^{(n-b_1+i)(z-1)}]$ and $H_{b2}=[\alpha ^{(b_2-1+j)(z-1)}]$, the rows are equal if $n-b_1+i = b_2-1+j$, where $i \in \{1,\dots ,n-d_m+1\}$ and $j \in \{1,\dots ,d_m-1\}$. Thus, as $v=(b_1+b_2-2)~\mathrm {mod}~n$, we obtain

$$\begin{aligned}&n-b_1+i=b_2-1+j \Rightarrow i = b_1+b_2-1+j-n = j-(n-v)+1~\mathrm {mod}~n,\nonumber \\&\text {As } 1 \le i \le n-d_m+1,\text { we obtain } 1 \le j-(n-v)+1 \le n-d_m+1,\nonumber \\&\Rightarrow n-v \le j \le 2n-v-d_m. \end{aligned}$$

(114)

For Case 2(a) in Table 1, as $n-v < d_m \Rightarrow n-v \le d_m-1$ and $j \in \{1,\dots ,d_m-1\}$, from Eq. (114), we obtain the range of j for the rows common to both A and $H_{b2}$ to be $j\in \{n-v,\dots ,d_m-1\}$. The set $\{n-v,\dots ,d_m-1\}$ corresponds to $(d_m-1)-(n-v)+1 = d_m-(n-v)$ rows. Thus, the dimension of $C_1\cap C_2^{\perp }$ is $d_m-n+v$. As the rows of the Vandermonde matrix generate $C_1\cap C_2^{\perp }$, it is an $[n,d_m-n+v,2n-d_m-v+1]$ RS code⁷. As $2(d_m-1)+v < 2n$ for Case 2(a) of Table 1, we obtain $2(d_m-1)+v \le 2n-1 \Rightarrow 2n-d_m-v+1 \ge d_m$. When $2n-d_m-v+1 = d_m$, from Theorem 4, $n_e = n-v-1$ and the EA RS code obtained is a zero-rate code as $n+n_e-2(d_m-1)=0$. As this code has only one basis codeword⁸, there is no concept of minimum distance. When $2n-d_m-v+1 > d_m$, there is no vector of weight $d_m$ in $C_1\cap C_2^{\perp }$ as the minimum distance of $C_1\cap C_2^{\perp }$ is $2n-d_m-v+1$. As the minimum distance of $C_1$ is $d_m$, there exists at least one vector in $C_1$ of weight $d_m$. Thus, there exists at least one vector of weight $d_m$ in $C_1\setminus C_2^{\perp }$ and the minimum distance of the EA RS code is $d_m$.

For Case 2(b), as $n-v\ge d_m$ and $j \in \{1,\dots ,d_m-1\}$, from Eq. (114), we obtain the range of j for the rows common to both A and $H_{b2}$ to be $j\in \{1,\dots ,n-v-d_m\}$ (as the row indices are modulo n) that corresponds to $n-v-d_m$ rows. Similar to Case 2(a), we obtain $C_1\cap C_2^{\perp }$ to be an $[n,n-v-d_m,v+d_m+1]$ RS code. As $v+1 >0$, we obtain the minimum distance $v+d_m+1$ of $C_1\cap C_2^{\perp }$ to be greater than $d_m$, i.e., $v+d_m+1> d_m$; hence, $C_1\cap C_2^{\perp }$ does not contain any vector of weight $d_m$. As $C_1$ has minimum distance of $d_m$, there exists a vector of weight $d_m$ in $C_1\setminus C_2^{\perp }$; hence the minimum distance of the EA RS code is $d_m$.

For both Cases 2(a) and 2(b), the minimum distance of $C_1\cap C_2^{\perp }$ is greater than the minimum distance $d'=d_m$ of the EA RS code. From Lemma 6, we obtain the EA RS code to be a non-degenerate code. We note that the $[[n,n-2d_m+2+n_e,d_m]]_{p^k}$ non-degenerate EA RS code saturates the quantum Singleton bound. [16]. $\square$

D Summation of powers of $\varvec{\omega }$ containing $\varvec{\mathrm {Tr}_{p^k/p}}$ operator

Lemma 7

For $\gamma \in {\mathbb {F}}_{p^k}$, $\underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\omega ^{\mathrm {Tr}_{p^k/p}(\gamma \theta )} = \delta _{\gamma ,0}p^k$.

Proof

We consider the two cases of $\gamma \ne 0$ and $\gamma =0$ separately to prove our claim.

Case 1: $\varvec{\gamma \ne 0}.$

Every $\theta \in {\mathbb {F}}_{p^k}$ can be written as $\theta = \underset{i=0}{\overset{k-1}{\sum }}\theta _i\alpha ^i$, where $\theta _0, \theta _1, \dots , \theta _{k-1} \in {\mathbb {F}}_p$. Let $m = \mathrm {Tr}_{p^k/p}(\gamma \theta ) \in {\mathbb {F}}_p$.

Case (a): $\varvec{m=0}$

If $m=0$, then for every $f \in {\mathbb {F}}_p$,

$$\begin{aligned} \mathrm {Tr}_{p^k/p}(\gamma (f\theta )) = \mathrm {Tr}_{p^k/p}(f\gamma \theta ) = f\mathrm {Tr}_{p^k/p}(\gamma \theta ) = fm = 0, \end{aligned}$$

as $\mathrm {Tr}_{p^k/p}(\cdot )$ is an ${\mathbb {F}}_p$-linear function and $f \in {\mathbb {F}}_p$.

We consider the set $M_0 = \{\theta \in {\mathbb {F}}_{p^k} : \mathrm {Tr}_{p^k/p}(\gamma \theta )=0\}$. To obtain the cardinality of the set $M_0$, we note that

$$\begin{aligned}&\mathrm {Tr}_{p^k/p}(\gamma \theta ) = 0 \Rightarrow \mathrm {Tr}_{p^k/p}\bigg (\gamma \underset{i=0}{\overset{k-1}{\sum }}\theta _i\alpha ^i\bigg ) = \underset{i=0}{\overset{k-1}{\sum }}\theta _i\mathrm {Tr}_{p^k/p}(\gamma \alpha ^i) = 0, \end{aligned}$$

(115)

as $\mathrm {Tr}_{p^k/p}(\cdot )$ is an ${\mathbb {F}}_p$-linear function and $\theta _i \in {\mathbb {F}}_p$, where $i\in \{0,\dots , k-1\}$. From Eq. (115), we obtain

$$\begin{aligned}&[\theta _0~\dots ~\theta _{k-1}][\mathrm {Tr}_{p^k/p}(\gamma )~\dots \mathrm {Tr}_{p^k/p}(\gamma \alpha ^{k-1})]^{\mathrm {T}} = 0. \end{aligned}$$

(116)

As every $\theta = \underset{i=0}{\overset{k-1}{\sum }}\theta _i\alpha ^i \in M_0$ satisfies Eq. (116), the cardinality $|M_0|$ is equal to the size of the space orthogonal to the space generated by $[\mathrm {Tr}_{p^k/p}(\gamma )~\dots \mathrm {Tr}_{p^k/p}(\gamma \alpha ^{k-1})]^{\mathrm {T}}$. The dimension of the space generated by $[\mathrm {Tr}_{p^k/p}(\gamma )~\dots \mathrm {Tr}_{p^k/p}(\gamma \alpha ^{k-1})]^{\mathrm {T}}$ is 1. The dimension of the space ${\mathbb {F}}_{p}^{k}$, to which $[\theta _0~\dots ~\theta _{k-1}]^{\mathrm {T}}$ belongs, is k. Thus, the dimension of the orthogonal space is $(k-1)$ and

$$\begin{aligned} |M_0| = p^{(k-1)}, \end{aligned}$$

(117)

as the elements of the vector $[\theta _0~\dots ~\theta _{k-1}]$ belong to ${\mathbb {F}}_{p}$.

Case (b): $\varvec{m\ne 0}$

If $m \ne 0$, then, for $f\in {\mathbb {F}}_p\backslash \{0\}$, we obtain $\mathrm {Tr}_{p^k/p}(\gamma (f\theta )) = f\mathrm {Tr}_{p^k/p}(\gamma \theta )$, and

$$\begin{aligned} \Rightarrow&\overset{p-1}{\underset{f=1}{\sum }}\omega ^{\mathrm {Tr}_{p^k/p}(\gamma (f\theta ))} = \overset{p-1}{\underset{f=1}{\sum }}\omega ^{(f\mathrm {Tr}_{p^k/p}(\gamma \theta ))}=\overset{p-1}{\underset{f=1}{\sum }}\omega ^{fm} = \underset{l \in {\mathbb {F}}_p \backslash \{0\}}{\sum }\omega ^{l}, \end{aligned}$$

(118)

where $l= fm \in {\mathbb {F}}_p\backslash \{0\}$, because, by the closure property of ${\mathbb {F}}_p$, $\{fm\}_{f\in {\mathbb {F}}_p\backslash \{0\}} = \{f\}_{f\in {\mathbb {F}}_p\backslash \{0\}}$ as $m\!\ne \! 0$ and $f,\!m\! \in \!{\mathbb {F}}_p$; hence, the summation of $\omega ^{fm}$ over f is equal to the summation of $\omega ^l$ over l.

As the sum of all the $p^{\mathrm {th}}$ roots of unity is 0, we obtain

$$\begin{aligned}&\underset{l \in {\mathbb {F}}_p}{\sum }\omega ^{l} = 0 \Rightarrow \omega ^0 +\!\! \underset{l \in {\mathbb {F}}_p \backslash \{0\}}{\sum }\!\!\omega ^{l} = 0\Rightarrow \underset{l \in {\mathbb {F}}_p \backslash \{0\}}{\sum }\!\!\omega ^{l} = -\omega ^0 = -1,\nonumber \\ \Rightarrow&\overset{p-1}{\underset{f=1}{\sum }}\omega ^{fm} \!= \!\!\!\underset{l \in {\mathbb {F}}_p \backslash \{0\}}{\sum }\!\!\!\!\!\omega ^{l} \!=\! -1. ~(\text {From Eq.~} 118) \end{aligned}$$

(119)

For every $\theta \in {\mathbb {F}}_{p^k} \setminus M_0$, we can combine $(p-1)$ elements to form the set ${\mathcal {B}}_{\theta } :=\{f\theta \}_{f=1}^{p-1}$. For some $j\in {\mathbb {F}}_{p} \setminus \{0\}$, let us consider $\theta ' = j\theta$. Then, the set ${\mathcal {B}}_{\theta '}$ is

$$\begin{aligned} {\mathcal {B}}_{\theta '} = \{f\theta '\}_{f=1}^{p-1} = \{fj\theta \}_{f=1}^{p-1}. \end{aligned}$$

(120)

As $f,j\in {\mathbb {F}}_p \backslash \{0\}$, by closure property of ${\mathbb {F}}_p$, $\{fj\}_{f=1}^{p-1} = \{f\}_{f=1}^{p-1}$. From Eq. (120), by considering $\{fj\}_{f=1}^{p-1} = \{f\}_{f=1}^{p-1}$, we obtain

$$\begin{aligned} {\mathcal {B}}_{\theta '} = \{fj\theta \}_{f=1}^{p-1} = \{f\theta \}_{f=1}^{p-1} = {\mathcal {B}}_{\theta }. \end{aligned}$$

(121)

From Eq. (121), the set ${\mathcal {B}}_{\theta '}$ considered for any element $\theta ' = f\theta$ in the set ${\mathcal {B}}_{\theta }$ will produce the same set ${\mathcal {B}}_{\theta }$.

Let K be the set of $\theta$s for which ${\mathcal {B}}_{\theta }$s are distinct. If $\theta$ satisfies $\mathrm {Tr}_{p^k/p}(\gamma \theta )\ne 0$, then it belongs to one of these K sets. For each $\theta \in K$, from Eq. (119), we have

$$\begin{aligned}&\underset{f \in {\mathbb {F}}_p\setminus \{0\}}{\sum }\!\!\!\!\!\! \omega ^{\mathrm {Tr}_{p^k/p}(\gamma f\theta )} =\! \underset{f \in {\mathbb {F}}_p\setminus \{0\}}{\sum }\!\!\!\!\!\! \omega ^{(f\mathrm {Tr}_{p^k/p}(\gamma \theta ))}= \underset{f \in {\mathbb {F}}_p\setminus \{0\}}{\sum }\!\!\!\!\!\! \omega ^{fm} = -1.~(\because m = \mathrm {Tr}_{p^k/p}(\gamma \theta )) \end{aligned}$$

(122)

The size of each ${\mathcal {B}}_{\theta }$ is $p-1$ as ${\mathcal {B}}_{\theta } = \{f\theta \}_{f=1}^{p-1}$. Using Eq. (117), the number of distinct sets ${\mathcal {B}}_{\theta }$, where $\theta \in {\mathbb {F}}_{p^k} \setminus M_0$, is equal to

$$\begin{aligned} |K|&= \frac{|{\mathbb {F}}_{p^k}| - |M_0|}{p-1}= \frac{p^{k}-p^{k-1}}{p-1} = p^{k-1}. \end{aligned}$$

(123)

From Case (a) and Case (b) in this proof, we obtain

$$\begin{aligned} \underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\!\!\omega ^{\mathrm {Tr}_{p^k/p}(\gamma \theta )}\!&=\!\! \underbrace{\underset{\theta \in M_0}{\sum }\!\omega ^{\mathrm {Tr}_{p^k/p}(\gamma \theta )}}_{\theta \in M_0} \!+\!\! \underbrace{\underset{\theta \in {\mathbb {F}}_{p^k} \setminus M_0}{\sum }\!\!\!\omega ^{\mathrm {Tr}_{p^k/p}(\gamma \theta )}}_{\theta \notin M_0},\nonumber \\&= \underbrace{\underset{\theta \in M_0}{\sum }\!\omega ^{\mathrm {Tr}_{p^k/p}(\gamma \theta )}}_{\theta \in M_0} + \underbrace{\underset{\theta \in K}{\sum }~\underset{f \in {\mathbb {F}}_p\setminus \{0\}}{\sum }\! \omega ^{\mathrm {Tr}_{p^k/p}(\gamma f\theta )}}_{\theta \notin M_0}, \end{aligned}$$

(124)

where the first part of the summation corresponds to the elements of $M_0$ and the second part of the summation corresponds to the $p^{k-1}$ sets $B_{\theta }$ for $\theta \in K$.

Substituting Eqs. (122) and (123) in Eq. (124), and as $\mathrm {Tr}_{p^k/p}(\gamma \theta ) = 0$ for $\theta \in M_0$, we obtain

$$\begin{aligned}&\underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\!\omega ^{\mathrm {Tr}_{p^k/p}(\gamma \theta )}= \underset{\theta \in M_0}{\sum }\!\omega ^{0} + \underset{\theta \in K}{\sum }~ (-1)= |M_0| + (-1)|K|= p^{k-1}-p^{k-1} = 0. \end{aligned}$$

(125)

Case 2: $\varvec{\gamma = 0.}$

$$\begin{aligned} \underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\omega ^{\mathrm {Tr}_{p^k/p}(\gamma \theta )}&= \underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\omega ^{\mathrm {Tr}_{p^k/p}(0)} = |{\mathbb {F}}_{p^k}|=p^k. \end{aligned}$$

(126)

From Eqs. (125) and (126),$\underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\omega ^{\mathrm {Tr}_{p^k/p}(\gamma \theta )} \!=\! \delta _{\gamma ,0}p^k\!\!.$ $\square$

Proofs of relations (R1)–(R10)

In this appendix, we provide a few relations with proofs that have been used in the main body of this paper. For $\beta \in {\mathbb {F}}_{p^k}$, the following relations hold true:

(R1):

$\varvec{\mathrm {Q}(\mathrm {FFFT})\mathrm {X}_i^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))}^{\varvec{-1}}\varvec{=}\underset{\varvec{j=0}}{\overset{\varvec{n-1}}{\varvec{\otimes }}}\varvec{\mathrm {X}}^{\varvec{(p^k)}}\varvec{(\beta \alpha }^{\varvec{j(i-1)}}\varvec{),}~\varvec{\forall }~\varvec{i\in }\{\varvec{1,}\dots ,\varvec{n}\}$ .

Proof

For $i\in \{1,\dots ,n\}$ and $\beta \in {\mathbb {F}}_{p^k}$, we have

$$\begin{aligned} \mathrm {Q}(\mathrm {FFFT})\mathrm {X}_i^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))^{-1}&= \!\!\!\left( \!\!\underset{\varvec{\theta _1}\in {\mathbb {F}}_{p^k}^{n}}{\sum } \vert {\mathrm {FFFT}}({\varvec{\theta _1}})\rangle \!\langle {\varvec{\theta _1}}\vert \right) \!\!\mathrm {X}_i^{(p^k)}\!(\beta )\!\!\left( \!\underset{\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}{\sum } \vert {\varvec{\theta _2}}\rangle \!\langle {\mathrm {FFFT}}({\varvec{\theta _2}})\vert \right) , \nonumber \\&=\underset{\varvec{\theta _1},\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}{\sum } \!\!\!\!\vert \mathrm {FFFT}(\varvec{\theta _1})\rangle \!\langle \varvec{\theta _1}|\varvec{\theta _2} +\beta {\mathbf {e}}_i^{\mathrm {T}}\rangle \langle {\mathrm {FFFT}(\varvec{\theta _2})}\vert , \nonumber \\&=\underset{\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\!\vert {\mathrm {FFFT}}({\varvec{\theta _2}}+ \beta {\mathbf {e}}_i^{\mathrm {T}})\rangle \langle {\mathrm {FFFT}}({\varvec{\theta _2}})\vert , \nonumber \\&=\underset{\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\! \vert {\mathrm {FFFT}}(\varvec{\theta _2})+{\mathrm {FFFT}}(\beta {\mathbf {e}}_i^{\mathrm {T}})\rangle \langle \mathrm {FFFT}(\varvec{\theta _2})\vert , \end{aligned}$$

(127)

as $\mathrm {FFFT}(\cdot )$ is a linear function .

Let $\varvec{\theta _2'} = \mathrm {FFFT}(\varvec{\theta _2})$, then $\varvec{\theta _2} = \mathrm {FFFT}^{-1}(\varvec{\theta _2'})$. As $\mathrm {FFFT}$ is an invertible function, for every $\varvec{\theta _2} \in {\mathbb {F}}_{p^k}^n$ we obtain a unique $\varvec{\theta _2'} = \mathrm {FFFT}(\varvec{\theta _2}) \in {\mathbb {F}}_{p^k}$. Thus, the set $\{\varvec{\theta _2'} = \mathrm {FFFT}(\varvec{\theta _2})\}_{\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}$ contains $p^{kn}$ unique elements in ${\mathbb {F}}_{p^k}^{n}$ as the set $\{\varvec{\theta _2}\}_{\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}$ contains $p^{kn}$ elements. As the field ${\mathbb {F}}_{p^k}^n$ has only $p^{kn}$ elements, the set $\{\varvec{\theta _2'} = \mathrm {FFFT}(\varvec{\theta _2})\}_{\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}$ contains all the elements of ${\mathbb {F}}_{p^k}^{n}$.

By substituting $\varvec{\theta _2'}$ in Eq. (127), the summation over $\varvec{\theta _2}$ taking all the values of ${\mathbb {F}}_{p^k}^n$ transforms into a summation over $\varvec{\theta _2'}$ taking all the values of ${\mathbb {F}}_{p^k}^n$. We will use this idea throughout the paper to transform summations when we change summation variables.

From Eq. (2), $\mathrm {FFFT}(\beta {\mathbf {e}}_i^{\mathrm {T}}) = [\beta ~\beta \alpha ^{(i-1)}~\dots ~\beta \alpha ^{(n-1)(i-1)}]$. Substituting $\varvec{\theta _2'} = \mathrm {FFFT}(\varvec{\theta _2})$ and $\mathrm {FFFT}(\beta {\mathbf {e}}_i^{\mathrm {T}}) = [\beta ~\beta \alpha ^{(i-1)}~\dots ~\beta \alpha ^{(n-1)(i-1)}]$ in Eq. (127), we obtain

$$\begin{aligned} \!\!\!\!\mathrm {Q}(\mathrm {FFFT})\mathrm {X}_i^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))^{-1}\!&=\!\underset{\varvec{\theta _2'}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\!\vert \varvec{\theta _2'}\!+\![\beta ~\beta \alpha ^{(i-1)}\dots \beta \alpha ^{(n-1)(i-1)}]\rangle \langle \varvec{\theta _2'}\vert ,\nonumber \\&=\!\left( \underset{j=0}{\overset{n-1}{\otimes }}\mathrm {X}^{(p^k)}\!(\beta \alpha ^{j(i-1)})\!\!\right) \!\!\underset{\varvec{\theta _2'}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\! \vert \varvec{\theta _2'}\rangle \!\langle \varvec{\theta _2'}\vert . \end{aligned}$$

(128)

By using the completeness relation of quantum mechanics, $\underset{\varvec{\theta _2'}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\! \vert \varvec{\theta _2'}\rangle \!\langle \varvec{\theta _2'}\vert = \mathrm {I}_{p^k}^{\otimes n}$, and we obtain

$$\begin{aligned} \mathrm {Q}(\mathrm {FFFT})\mathrm {X}_i^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))^{-1}\!=\!\underset{j=0}{\overset{n-1}{\otimes }}\mathrm {X}^{(p^k)}(\beta \alpha ^{j(i-1)}).\qquad \qquad \qquad \qquad \qquad \qquad \qquad \end{aligned}$$

$\square$

(R2) :

$\varvec{\mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_i^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))}^{\varvec{-1}}\!\varvec{=}\!\!\underset{\varvec{j=0}}{\overset{\varvec{n-1}}{\varvec{\otimes }}}\varvec{\mathrm {Z}}^{\varvec{(p^k)}}\varvec{((p\!\!-\!1)\beta \alpha }^{\varvec{j(n-i+1)}}\varvec{),}~\varvec{\forall }\varvec{i\in }\!\{\varvec{1,}\dots ,\varvec{n}\}$.

Proof

For $i\in \{1,\dots ,n\}$, $\beta \in {\mathbb {F}}_{p^k}$ ,and $\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^n$, we note that

$$\begin{aligned} \mathrm {Z}_i^{(p^k)}\!(\beta )\vert \varvec{\theta _2}\rangle = \omega ^{\mathrm {Tr}_{p^k/p}(\varvec{\theta _2}\beta {\mathbf {e}}_i^{\mathrm {T}})}\vert \varvec{\theta _2}\rangle , \end{aligned}$$

(129)

where $\omega ^{\mathrm {Tr}_{p^k/p}(\varvec{\theta _2}\beta {\mathbf {e}}_i^{\mathrm {T}})}$ is obtained as the operator $\mathrm {Z}_i^{(p^k)}\!(\beta )$ performs $\mathrm {Z}^{(p^k)}\!(\beta )$ on the $i^{\mathrm {th}}$ qudit.

For $i\in \{1,\dots ,n\}$, we consider

$$\begin{aligned} \mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_i^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))^{-1}&= \!\left( \!\underset{\varvec{\theta _1}\in {\mathbb {F}}_{p^k}^{n}}{\sum } \vert \mathrm {FFFT}(\varvec{\theta _1})\rangle \langle \varvec{\theta _1}\vert \!\!\right) \!\!\mathrm {Z}_i^{(p^k)}\!(\beta ) \!\!\left( \!\underset{\varvec{\theta _2} \in {\mathbb {F}}_{p^k}^{n}}{\sum } \vert \varvec{\theta _2}\,\rangle \,\,\!\!\langle \mathrm {FFFT}(\varvec{\theta _2})\vert \!\!\right) \!\!, \nonumber \\&=\!\underset{\varvec{\theta _1},\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}{\sum } \vert \mathrm {FFFT} (\varvec{\theta _1})\rangle \langle \varvec{\theta _1}|\mathrm {Z}_i^{(p^k)} (\beta )|\varvec{\theta _2}\rangle \langle \mathrm {FFFT}(\varvec{\theta _2})\vert . \end{aligned}$$

(130)

Substituting Eq. (129) in Eq. (130), we obtain

$$\begin{aligned} \mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_i^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))^{-1}&=\underset{\varvec{\theta _1},\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\!\!\!\omega ^{\mathrm {Tr}_{p^k/p}(\varvec{\theta _2}\beta {\mathbf {e}}_i^{\mathrm {T}})}\vert \mathrm {FFFT}(\varvec{\theta _1})\rangle \!\langle \varvec{\theta _1}|\varvec{\theta _2}\rangle \!\langle \mathrm {FFFT}(\varvec{\theta _2})\vert , \nonumber \\&=\underset{\varvec{\theta _2}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\omega ^{\mathrm {Tr}_{p^k/p}(\varvec{\theta _2}\beta {\mathbf {e}}_i^{\mathrm {T}})}\vert \mathrm {FFFT}(\varvec{\theta _2})\rangle \langle \mathrm {FFFT}(\varvec{\theta _2})\vert . \end{aligned}$$

(131)

Let $\varvec{\theta _2'} = \mathrm {FFFT}(\varvec{\theta _2})$. Let $\varvec{\theta _2'}=[t_0~t_1~\dots ~t_{n-1}]$. From Eq. (41), we have

$$\begin{aligned}&\varvec{\theta _2} \!=\! \mathrm {FFFT}^{-1}(\varvec{\theta _2'})\!=\! \left[ \underset{j=0}{\overset{n-1}{\sum }}(p-1)t_j\alpha ^{j(n-m)}\right] _{m=0}^{n-1}. \end{aligned}$$

(132)

Let $l = (m+1)$, then l ranges from 1 to n as m ranges from 0 to $n-1$. Substituting $l=(m+1)$ in Eq. (132), we obtain

$$\begin{aligned} \varvec{\theta _2}&= \mathrm {FFFT}^{-1}(\varvec{\theta _2'})= \left[ \underset{j=0}{\overset{n-1}{\sum }}(p-1)t_j\alpha ^{j(n-l+1)}\right] _{l=1}^{n},\nonumber \\ \Rightarrow \mathrm {Tr}_{p^k/p}(\varvec{\theta _2}\beta {\mathbf {e}}_i^{\mathrm {T}})&= \mathrm {Tr}_{p^k/p}(\mathrm {FFFT}^{-1}(\varvec{\theta _2'})\beta {\mathbf {e}}_i^{\mathrm {T}})\nonumber \\&=\mathrm {Tr}_{p^k/p}\left( \left[ \underset{j=0}{\overset{n-1}{\sum }}(p-1)t_j\alpha ^{j(n-l+1)}\right] _{l=1}^{n}\!\!\!\beta {\mathbf {e}}_i^{\mathrm {T}}\right) ,\nonumber \\&=\mathrm {Tr}_{p^k/p}\left( \underset{j=0}{\overset{n-1}{\sum }}(p-1)\beta t_j\alpha ^{j(n-i+1)}\right) . \end{aligned}$$

(133)

Substituting $\varvec{\theta _2'} = \mathrm {FFFT}(\varvec{\theta _2})$ and Eq. (133) in Eq. (131), we obtain

$$\begin{aligned}&\mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_i^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))^{-1}\nonumber \\&\quad =\underset{\varvec{\theta _2'}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\!\omega ^{\mathrm {Tr}_{p^k/p}(\mathrm {FFFT}^{-1}(\varvec{\theta _2'})\beta {\mathbf {e}}_i^{\mathrm {T}})}\vert \varvec{\theta _2'}\rangle \langle {\varvec{\theta _2'}}\vert ,\nonumber \\&\quad =\!\!\!\!\!\!\underset{t_0,\dots ,t_{n-1}\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\!\!\!\!\!\omega ^{\!\mathrm {Tr}_{\!p^k\!/\!p}\!\left( \underset{j=0}{\overset{n-1}{\sum }}\!(p-\!1)\beta t_j\alpha ^{j(n-i+1)}\!\!\right) }\!\!\vert t_0\! \dots \! t_{n-1}\rangle \langle {t_0\! \dots \! t_{n-1}}\vert , (\because \varvec{\theta _2'}\!=\![t_0~t_1~\dots ~t_{n-1}])\nonumber \\&\quad =\!\!\!\!\!\!\underset{t_0,\dots ,t_{n-\!1}\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\left( \underset{j=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)}\!(-\beta \alpha ^{j(n-i+1)}\!)\!\!\right) \!\!\vert t_0\! \dots \! t_{n-1}\rangle \langle {t_0\! \dots \! t_{n-1}}\vert ,(\because (-1)~\mathrm {mod}~p \!=\! (p\!-\!1))\nonumber \\&\quad =\!\!\left( \underset{j=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)} \!(-\beta \alpha ^{j(n-i+1)}\!)\!\!\right) \!\!\!\! \underset{t_0,\dots ,t_{n-\!1}\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\!\!\!\vert t_0\! \dots \! t_{n-1}\rangle \langle t_0\! \dots \! t_{n-1}|,\nonumber \\&\quad =\left( \underset{j=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)}\!(-\beta \alpha ^{j(n-i+1)}\!)\right) \underset{\varvec{\theta _2'}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\! \vert \varvec{\theta _2'}\rangle \langle \varvec{\theta _2'}\vert . \end{aligned}$$

(134)

By using the completeness relation, $\underset{\varvec{\theta _2'}\in {\mathbb {F}}_{p^k}^{n}}{\sum }\!\!\!\! \vert \varvec{\theta _2'}\rangle \langle \varvec{\theta _2'}\vert = \mathrm {I}_{p^k}^{\otimes n}$. Thus, from Eq. (134),

$$\begin{aligned} \mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_i^{(p^k)}\!(\beta )(\mathrm {Q}(\mathrm {FFFT})\!)^{\!-1}\!\!\!=\!\!\underset{j=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)}\!(\!-\!\beta \alpha ^{j(n-i+1)}). \qquad \qquad \qquad \qquad \qquad \qquad \end{aligned}$$

$\square$

(R3) :

$\varvec{\mathrm {Q(FFFT)}}\left( \underset{\varvec{j=0}}{\overset{\varvec{n-1}}{\varvec{\otimes }}}\varvec{\mathrm {X}}^{\varvec{(p^k)}}\varvec{(\beta \alpha }^{\varvec{ij}})\right) \varvec{(\mathrm {Q(FFFT))}}^{\varvec{-1}} =\varvec{\mathrm {X}}_{\varvec{(n-i+1)}}^{\varvec{(p^k)}}\varvec{((p-1)\beta ),}~\varvec{\forall }~\varvec{i\in }\{\varvec{1,}\dots ,\varvec{n}\}$.

Proof

For $i\in \{1,\dots , n\}$ and $\beta \in {\mathbb {F}}_{p^k}$, we have

$$\begin{aligned} \!\!\!\underset{j=0}{\overset{n-1}{\otimes }}\!\mathrm {X}^{(p^k)}\!(\beta \alpha ^{ij})\!\!=\! \mathrm {X}^{(p^k)}\!(\beta \alpha ^{(0)i})\! \otimes \!\cdots \! \otimes \! \mathrm {X}^{(p^k)}\!(\beta \alpha ^{(n-1)i}).\!\!\! \end{aligned}$$

(135)

From Eq. (135), the operator $\underset{j=0}{\overset{n-1}{\otimes }}\mathrm {X}^{(p^k)}(\beta \alpha ^{ij})$ performs the operation $\mathrm {X}^{(p^k)}(\beta \alpha ^{ij})$ on the $(j+1)^{\mathrm {th}}$ qudit, for all $j\in \{0,\dots ,n-1\}$. For $s\in \{1,\dots , n\}$ and $\gamma \in {\mathbb {F}}_{p^k}$, the operator $\mathrm {X}^{(p^k)}_s(\gamma )$ performs $\mathrm {X}^{(p^k)}(\gamma )$ on the $s^{\mathrm {th}}$ qudit. Thus, from Eq. (135), $\underset{j=0}{\overset{n-1}{\otimes }}\mathrm {X}^{(p^k)}(\beta \alpha ^{ij}) = \overset{n-1}{\underset{j=0}{\prod }} \mathrm {X}^{(p^k)}_{(j+1)}(\beta \alpha ^{ij})$ and we obtain

$$\begin{aligned}&\mathrm {Q(FFFT)}\left( \underset{j=0}{\overset{n-1}{\otimes }}\mathrm {X}^{(p^k)}(\beta \alpha ^{ij})\right) (\mathrm {Q(FFFT))}^{-1}\nonumber \\&\quad =\mathrm {Q(FFFT)}\left( \overset{n-1}{\underset{j=0}{\prod }} \mathrm {X}^{(p^k)}_{(j+1)}(\beta \alpha ^{ij})\right) (\mathrm {Q(FFFT))}^{-1},\nonumber \\&\quad =\mathrm {Q}(\mathrm {FFFT})\mathrm {X}_1^{(p^k)}(\beta )\cdots \mathrm {X}_n^{(p^k)}\!(\beta \alpha ^{i(n-\!1)}\!)(\mathrm {Q}(\mathrm {FFFT}))^{\!-1}\!\!,\nonumber \\&\quad =\mathrm {Q}(\mathrm {FFFT})\mathrm {X}_1^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))^{\!-1}\mathrm {Q}(\mathrm {FFFT})\mathrm {X}_2^{(p^k)}(\beta \alpha ^{i})\nonumber \\&\qquad (\mathrm {Q}(\mathrm {FFFT}))^{\!-1}\!\!\cdots \mathrm {Q}(\mathrm {FFFT})\mathrm {X}_n^{(p^k)}\!(\beta \alpha ^{i(n-\!1)}\!)(\mathrm {Q}(\mathrm {FFFT}))^{\!-1}\!\!,\nonumber \\&\quad = \underset{j=0}{\overset{n-1}{\prod }}\mathrm {Q(FFFT)}\mathrm {X}_{(j+1)}^{(p^k)}(\beta \alpha ^{ij})(\mathrm {Q(FFFT)})^{-1},\nonumber \\&\quad = \underset{j=0}{\overset{n-1}{\prod }}\underset{m=0}{\overset{n-1}{\otimes }}\mathrm {X}^{(p^k)}(\beta \alpha ^{ij}\alpha ^{m((j+1)-1)})= \underset{m=0}{\overset{n-1}{\otimes }}\underset{j=0}{\overset{n-1}{\prod }}\mathrm {X}^{(p^k)}(\beta \alpha ^{j(m+i)}),~~~(\text {From}~(R1))\nonumber \\&= \underset{m=0}{\overset{n-1}{\otimes }}\mathrm {X}^{(p^k)}\!\left( \underset{j=0}{\overset{n-1}{\sum }}\beta \alpha ^{j(m+i)}\!\!\right) ,~(\because \! \mathrm {X}^{(p^k)}\!(\beta )\mathrm {X}^{(p^k)}\!(\gamma ) \!=\! \mathrm {X}^{(p^k)}(\beta \!+\!\gamma ))\nonumber \\&= \underset{g=1}{\overset{n}{\otimes }}\mathrm {X}^{(p^k)}\!\!\left( \!\beta \underset{j=0}{\overset{n-1}{\sum }}\alpha ^{j(g+i-1)}\!\right) \!,\!\text { where }g \!= \!m\!+\!1. \end{aligned}$$

(136)

As $\alpha ^n = \alpha ^{p^k-1} = 1$, we note that

$$\begin{aligned}&\underset{j=0}{\overset{n-1}{\sum }}\alpha ^{j(g+i-1)}\nonumber \\&\quad = {\left\{ \begin{array}{ll} \underset{j=0}{\overset{n-1}{\sum }}\alpha ^{0} &{} (g\!+\!i\!-\!1)~\mathrm {mod}~n=0\\ \underset{j=0}{\overset{n-1}{\sum }}\alpha ^{j(g+i-1)} &{} (g\!+\!i\!-\!1)~\mathrm {mod}~n\ne 0 \end{array}\right. }= {\left\{ \begin{array}{ll} n &{} (g\!+\!i\!-\!1)~\mathrm {mod}~n=0\\ 0 &{} (g\!+\!i\!-\!1)~\mathrm {mod}~n\ne 0\end{array}\right. }, \end{aligned}$$

(137)

as $\alpha ^{n}=1$ and the roots of unity sum to 0. Thus, from Eq. (137), we have

$$\begin{aligned}&\underset{j=0}{\overset{n-1}{\sum }}\alpha ^{j(g+i-1)} = n\delta _{(g+i-1),n} = n\delta _{g,(n-i+1)}. \end{aligned}$$

(138)

Substituting Eq. (138) in Eq. (136), we obtain

$$\begin{aligned}&\mathrm {Q(FFFT)}\!\!\left( \underset{j{=}0}{\overset{n{-}1}{\otimes }}\mathrm {X}^{(p^k)}(\beta \alpha ^{ij})\!\right) \!\!(\mathrm {Q(FFFT)\!)}^{-1}\!\!=\!\! \underset{g{=}1}{\overset{n}{\otimes }}\mathrm {X}^{(p^k)}\!\left( \beta n\delta _{g,n{-}i{+}1}\right) \!=\! \mathrm {X}_{(n{-}i{+}1)}^{(p^k)}\!\left( (p\!-\!1)\beta \right) , \end{aligned}$$

because $n=p^k-1=p-1 ~(\text {in }{\mathbb {F}}_{p^k})$, and in the tensor product $\underset{g=1}{\overset{n}{\otimes }}\mathrm {X}^{(p^k)}\left( \beta n\delta _{g,n-i+1}\right)$, only the $(n-i+1)^{\mathrm {st}}$ element is non-identity ($\because g= (n-i+1)$), while the rest become identity ($\because g\ne (n-i+1) \Rightarrow \mathrm {X}^{(p^k)}(0)=\mathrm {I}_{p^k}$). We note that $i \in \{1,\dots ,n\} \Rightarrow (n-i+1)\in \{1,\dots ,n\}$. $\square$

(R4) :

$\varvec{\mathrm {Q(FFFT)}}\left( \underset{\varvec{j{=}0}}{\overset{\varvec{n{-}1}}{\varvec{\otimes }}}\varvec{\mathrm {Z}}^{\varvec{(p^k)}}\varvec{(\beta \alpha }^{\varvec{ij}})\right) \varvec{(\mathrm {Q(FFFT))}}^{\varvec{-1}} {=}\varvec{\mathrm {Z}}_{\varvec{(i+1)}}^{\varvec{(p^k)}}\varvec{(\beta ),}~\varvec{\forall }~\varvec{i\in }\{\varvec{0,}{\dots },\varvec{n-1}\}$.

Proof

For $i\in \{0,\dots , n-1\}$ and $\beta \in {\mathbb {F}}_{p^k}$, we have

$$\begin{aligned} \!\!\!\underset{j=0}{\overset{n-1}{\otimes }}\!\mathrm {Z}^{(p^k)}\!(\beta \alpha ^{ij})\!=\! \mathrm {Z}^{(p^k)}\!(\beta \alpha ^{(0)i}) \!\!\otimes \!\cdots \! \otimes \! \mathrm {Z}^{(p^k)}\!(\beta \alpha ^{(n-1)i}).\!\!\! \end{aligned}$$

(139)

From Eq. (139), the operator $\underset{j=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)}(\beta \alpha ^{ij})$ performs the operation $\mathrm {Z}^{(p^k)}(\beta \alpha ^{ij})$ on the $(j+1)^{\mathrm {th}}$ qudit, for all $j\in \{0,\dots ,n-1\}$. For $s\in \{1,\dots , n\}$ and $\gamma \in {\mathbb {F}}_{p^k}$, the operator $\mathrm {Z}^{(p^k)}_s(\gamma )$ performs $\mathrm {Z}^{(p^k)}(\gamma )$ on the $s^{\mathrm {th}}$ qudit. Thus, from Eq. (135), $\underset{j=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)}(\beta \alpha ^{ij}) = \overset{n-1}{\underset{j=0}{\prod }} \mathrm {Z}^{(p^k)}_{(j+1)}(\beta \alpha ^{ij})$ and we obtain

$$\begin{aligned}&\mathrm {Q(FFFT)}\left( \underset{j=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)}(\beta \alpha ^{ij})\right) (\mathrm {Q(FFFT))}^{-1}\nonumber \\&\quad =\mathrm {Q(FFFT)}\left( \overset{n-1}{\underset{j=0}{\prod }} \mathrm {Z}^{(p^k)}_{(j+1)}(\beta \alpha ^{ij})\right) (\mathrm {Q(FFFT))}^{-1},\nonumber \\&\quad =\mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_1^{(p^k)}(\beta )\cdots \mathrm {Z}_n^{(p^k)}\!(\beta \alpha ^{i(n-\!1)}\!)(\mathrm {Q}(\mathrm {FFFT}))^{\!-1}\!\!,\nonumber \\&\quad =\mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_1^{(p^k)}(\beta )(\mathrm {Q}(\mathrm {FFFT}))^{\!-1}\mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_2^{(p^k)}(\beta \alpha ^{i})\nonumber \\&\quad (\mathrm {Q}(\mathrm {FFFT}))^{\!-1}\!\!\cdots \mathrm {Q}(\mathrm {FFFT})\mathrm {Z}_n^{(p^k)}\!(\beta \alpha ^{i(n-\!1)}\!)(\mathrm {Q}(\mathrm {FFFT}))^{\!-1}\!\!,\nonumber \\&\quad = \underset{j=0}{\overset{n-1}{\prod }}\mathrm {Q(FFFT)}\mathrm {Z}_{(j+1)}^{(p^k)}(\beta \alpha ^{ij})(\mathrm {Q(FFFT)})^{-1},\nonumber \\&= \underset{j=0}{\overset{n-1}{\prod }}\underset{m=0}{\overset{n-1}{\otimes }}\!\!\mathrm {Z}^{(p^k)}((p\!-\!1)\beta \alpha ^{ij}\alpha ^{m(n-(j+1)+1)}),(\text {From}~(R2))\nonumber \\&= \underset{m=0}{\overset{n-1}{\otimes }}\underset{j=0}{\overset{n-1}{\prod }}\mathrm {Z}^{(p^k)}((p-1)\beta \alpha ^{j(i-m)}),\nonumber \\&= \underset{m=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)}\left( \underset{j=0}{\overset{n-1}{\sum }}(p-1)\beta \alpha ^{j(i-m)}\right) ~(\because \mathrm {Z}^{(p^k)}(\beta )\mathrm {Z}^{(p^k)}(\gamma ) = \mathrm {Z}^{(p^k)}(\beta +\gamma ))\nonumber \\&= \!\!\underset{g=1}{\overset{n}{\otimes }}\!\mathrm {Z}^{(p^k)}\!\!\!\left( \!\!(p\!-\!1)\beta \underset{j=0}{\overset{n-1}{\sum }}\alpha ^{j(i-g+1)}\!\!\right) \!\!,\!\text {where }g \!=\!m\!+\!1. \end{aligned}$$

(140)

Similar to Eq. (137), we obtain

$$\begin{aligned}&\underset{j=0}{\overset{n-1}{\sum }}\alpha ^{j(i-g+1)}= {\left\{ \begin{array}{ll} n &{} (i-g+1)~\mathrm {mod}~n=0\\ 0 &{} (i-g+1)~\mathrm {mod}~n\ne 0\end{array}\right. }, \end{aligned}$$

(141)

Thus, from Eq. (141), we have

$$\begin{aligned}&\underset{j=0}{\overset{n-1}{\sum }}\alpha ^{j(i-g+1)} = n\delta _{(i-g+1),n}= n\delta _{g,(i+1)}, \end{aligned}$$

(142)

as $g\in \{1,\dots ,n\}$ and $(n+g) \equiv g$ in the range 1 to n.

Substituting Eq. (142) in Eq. (140), we obtain

$$\begin{aligned}&\mathrm {Q(FFFT)}\!\!\left( \underset{j=0}{\overset{n-1}{\otimes }}\mathrm {Z}^{(p^k)}(\beta \alpha ^{ij})\!\!\right) \!\!(\mathrm {Q(FFFT))}^{-1}\!\!=\!\! \underset{g=1}{\overset{n}{\otimes }}\mathrm {Z}^{(p^k)}\left( (p-1)\beta n\delta _{g,i+1}\right) \!=\! \mathrm {Z}_{(i+1)}^{(p^k)}\left( \beta \right) , \end{aligned}$$

because $(p-1)n=(p-1)(p^k-1)=(p-1)(p-1) = 1 ~(\text {in }{\mathbb {F}}_{p^k})$, and in the tensor product $\underset{g=1}{\overset{n}{\otimes }}\mathrm {Z}^{(p^k)}\left( (p-1)\beta n\delta _{g,i+1}\right)$, only the $(i+1)^{\mathrm {st}}$ element is non-identity ($\because g= (i+1)$), while the rest become identity ($\because g\ne (i+1) \Rightarrow \mathrm {Z}^{(p^k)}(0)=\mathrm {I}_{p^k}$). We note that as $i \in \{0,\dots ,n-1\}$, we obtain $(i+1)\in \{1,\dots ,n\}$. $\square$

(R5) :

$\varvec{\mathrm {X}}^{\varvec{(p^k)}}\varvec{(\beta )}\varvec{\vert \epsilon \rangle = \vert \epsilon \rangle ,}\!\mathbf{ where }\varvec{\vert \epsilon \rangle =\mathrm {DFT}}^{\varvec{-1}}\varvec{\vert 0\rangle }$

Proof

From Eq. (77),

$$\begin{aligned} \vert \epsilon \rangle&=\mathrm {DFT}^{-1}\vert 0\rangle = \frac{1}{\sqrt{p^k}}\underset{\theta _1\in {\mathbb {F}}_{p^k}}{\sum }\vert \theta _1\rangle . \end{aligned}$$

(143)

For $\beta \in {\mathbb {F}}_{p^k}$, we analyse the action of $\mathrm {X}^{(p^k)}(\beta )$ on $\vert \epsilon \rangle$ from Eq. (143) as follows:

$$\begin{aligned} \!\!\!\!\mathrm {X}^{(p^k)}(\beta )\vert \epsilon \rangle \!\!=\! \mathrm {X}^{(p^k)}(\beta )\frac{1}{\sqrt{p^k}}\underset{\theta _1\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\vert \theta _1\rangle \!=\! \!\frac{1}{\sqrt{p^k}}\underset{\theta _1\in {\mathbb {F}}_{p^k}}{\sum }\!\!\mathrm {X}^{(p^k)}(\beta )\vert \theta _1\rangle \!=\!\! \frac{1}{\sqrt{p^k}}\!\underset{\theta _1\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\vert \theta _1+\beta \rangle . \end{aligned}$$

(144)

Performing the change of variable from $\theta _1$ to $\theta _1' = \theta _1+\beta$ in Eq. (144) and using Eq. (143), we obtain

$$\begin{aligned} \mathrm {X}^{(p^k)}(\beta )\vert \epsilon \rangle&= \frac{1}{\sqrt{p^k}}\underset{\theta _1'\in {\mathbb {F}}_{p^k}}{\sum }\vert \theta _1'\rangle = \vert \epsilon \rangle . \end{aligned}$$

$\square$

(R6) :

$\varvec{\mathrm {Z}}^{\varvec{(p^k)}}\varvec{(\beta )}\varvec{\vert 0\rangle = \vert 0\rangle .}$

Proof

For $\beta \!\in \! {\mathbb {F}}_{p^k}$, $\mathrm {Z}^{(p^k)}(\beta )\!\vert 0\rangle \!\!=\!\omega ^{\mathrm {Tr}_{p^k/p}(\beta .0)}\vert 0\rangle \!=\!\! \vert 0\rangle$. $\square$

(R7) :

$\varvec{(\mathrm {X}^{(p^k)}((p-1)\beta ) \otimes \mathrm {X}^{(p^k)}(\beta ))\vert \eta \rangle = \vert \eta \rangle }$.

Proof

For $\beta \in {\mathbb {F}}_{p^k}$, from Eq. (78), we have

$$\begin{aligned}&(\mathrm {X}^{(p^k)}((p-1)\beta ) \otimes \mathrm {X}^{(p^k)}(\beta ))\vert \eta \rangle \\&\quad =(\mathrm {X}^{(p^k)}((p\!-\!1)\beta ) \otimes \mathrm {X}^{(p^k)}(\beta ))\!\!\left( \frac{1}{\sqrt{p^k}}\underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\!\vert \theta \rangle \vert (p-1)\theta \rangle \!\right) ,\nonumber \\&\quad =\left( \frac{1}{\sqrt{p^k}}\underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\vert \theta +(p-1)\beta \rangle \vert (p-1)\theta +\beta \rangle \right) ,\nonumber \\&\quad =\left( \frac{1}{\sqrt{p^k}}\underset{\theta '\in {\mathbb {F}}_{p^k}}{\sum }\!\!\vert \theta '\rangle \vert (p-1)\theta '\rangle \right) \!,\text {where }\theta '\!=\!\theta +(p-1)\beta ,\nonumber \\&\quad = \vert \eta \rangle . \end{aligned}$$

(R8) :

$\varvec{(\mathrm {Z}^{(p^k)}(\beta )\otimes \mathrm {Z}^{(p^k)}(\beta ))\vert \eta \rangle = \vert \eta \rangle }$

Proof

For $\beta \in {\mathbb {F}}_{p^k}$, from Eq. (78), we have

$$\begin{aligned}&(\mathrm {Z}^{(p^k)}(\beta ) \otimes \mathrm {Z}^{(p^k)}(\beta ))\vert \eta \rangle =(\mathrm {Z}^{(p^k)}(\beta ) \otimes \mathrm {Z}^{(p^k)}(\beta ))\left( \frac{1}{\sqrt{p^k}}\underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\vert \theta \rangle \vert (p-1)\theta \rangle \right) ,\nonumber \\&\quad =\frac{1}{\sqrt{p^k}}\underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\omega ^{\mathrm {Tr}_{p^k/p}(\beta \theta )}\omega ^{\mathrm {Tr}_{p^k/p}((p-1)\beta \theta )}\vert \theta \rangle \vert (p-1)\theta \rangle \\&\quad =\frac{1}{\sqrt{p^k}}\underset{\theta \in {\mathbb {F}}_{p^k}}{\sum }\vert \theta \rangle \vert (p-1)\theta \rangle = \vert \eta \rangle . \end{aligned}$$

$\square$

(R9) :

$\varvec{\mathrm {DFT}_{\varvec{p^k}}~\mathrm {X}}^{\varvec{(p^k)}}\varvec{(\beta )}\varvec{\mathrm {DFT}}_{\varvec{p^k}}^{\varvec{-1}} \varvec{=\mathrm {Z}}^{\varvec{(p^k)}}\varvec{(\beta )}$

Proof

For $\beta \in {\mathbb {F}}_{p^k}$, from Eq. (75), we have

$$\begin{aligned}&\mathrm {DFT}_{p^k}~\mathrm {X}^{(p^k)}(\beta )\mathrm {DFT}_{p^k}^{-1}\\&\quad = \!\left( \!\frac{1}{\sqrt{p^k}}\underset{\theta _1,\theta _2\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\!\!\omega ^{\mathrm {Tr}_{p^k/p}(\theta _1\theta _2)}\vert \theta _1\rangle \!\langle \theta _2\vert \!\right) \!\!\mathrm {X}^{(p^k)}(\beta )\!\!\left( \!\!\frac{1}{\sqrt{p^k}}\underset{\theta _3,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\!\omega ^{\mathrm {Tr}_{p^k/p}(-\theta _3\theta _4)}\vert \theta _3\rangle \!\langle \theta _4\vert \right) ,\\&\quad = \!\frac{1}{p^k}\underset{\theta _1,\theta _2,\theta _3,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\!\!\!\!\!\! \omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _1\theta _2-\theta _3\theta _4)\!\!}\, \vert \theta _1\rangle \!\langle \theta _2|\theta _3+\beta \rangle \!\langle \theta _4\vert ,\\&\quad = \!\frac{1}{p^k}\underset{\theta _1,\theta _3,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\!\! \omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _1(\theta _3+\beta )-\theta _3\theta _4)\!\!}\,\,\, \vert \theta _1\rangle \!\langle \theta _4\vert ,\\&\quad = \!\frac{1}{p^k}\underset{\theta _1,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\! \omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _1\beta )\!\!}\,\,\, \vert \theta _1\rangle \!\langle \theta _4\vert \underset{\theta _3\in {\mathbb {F}}_{p^k}}{\sum }\omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _3(\theta _1-\theta _4))\!\!},\\&\quad = \!\!\frac{1}{p^k}\!\underset{\theta _1,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _1\beta )}\,\,\, \vert \theta _1\rangle \!\langle \theta _4\vert p^k\delta _{\theta _1-\theta _4,0},\!(\text {From Lemma }7)\\&\quad =\!\!\! \underset{\theta _1\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _1\beta )}\vert \theta _1\rangle \!\langle \theta _1\vert \!=\! \mathrm {Z}^{(p^k)}(\beta )\!\!\underset{\theta _1\in {\mathbb {F}}_{p^k}}{\sum } \vert \theta _1\rangle \!\langle \theta _1\vert \!=\! \mathrm {Z}^{(p^k)}(\beta ), \end{aligned}$$

because $\underset{\theta _1\in {\mathbb {F}}_{p^k}}{\sum } \vert \theta _1\rangle \langle \theta _1\vert = \mathrm {I}_{p^k}$ by completeness relation. $\square$

(R10) :

$\varvec{\mathrm {DFT}_{p^k}~\mathrm {Z}}^{\varvec{(p^k)}}\varvec{(\beta )}\varvec{\mathrm {DFT}}_{p^k}^{\varvec{-1}} \varvec{=\mathrm {X}}^{\varvec{(p^k)}}\varvec{(-\beta )}$

Proof

For $\beta \in {\mathbb {F}}_{p^k}$, from Eq. (75), we have

$$\begin{aligned}&\mathrm {DFT}_{p^k}~\mathrm {Z}{(p^k)}(\beta )\mathrm {DFT}_{p^k}^{-1}\\&\quad = \!\left( \!\frac{1}{\sqrt{p^k}}\underset{\theta _1,\theta _2\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\!\!\omega ^{\mathrm {Tr}_{p^k/p}(\theta _1\theta _2)}\vert \theta _1\rangle \!\langle \theta _2\vert \!\!\right) \!\!\mathrm {Z}^{(p^k)}(\beta )\!\!\left( \!\!\frac{1}{\sqrt{p^k}}\underset{\theta _3,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\omega ^{\mathrm {Tr}_{p^k/p}(-\theta _3\theta _4)}\,\,\, \!\!\vert \theta _3\rangle \! \langle \theta _4\vert \right) ,\\&\quad = \frac{1}{p^k}\!\!\!\underset{\theta _1,\theta _2,\theta _3,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _1\theta _2-\theta _3\theta _4+\theta _3\beta )}\vert \theta _1\rangle \!\langle \theta _2|\theta _3\rangle \! \langle \theta _4\vert \\&\quad = \frac{1}{p^k}\!\underset{\theta _1,\theta _3,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _1\theta _3-\theta _3\theta _4+\theta _3\beta )}\vert \theta _1\rangle \!\langle \theta _4\vert ,\\&\quad = \!\!\frac{1}{p^k}\!\!\underset{\theta _1,\theta _4\in {\mathbb {F}}_{p^k}}{\sum } \vert \theta _1\rangle \!\langle \theta _4\vert \underset{\theta _3\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\omega ^{\!\mathrm {Tr}_{p^k\!/\!p}(\theta _3(\theta _1-\theta _4+\beta ))\!\!}= \!\!\frac{1}{p^k}\,\,\,\!\underset{\theta _1,\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\!\!\!\!\!\!\!\! \vert \theta _1\rangle \!\langle \theta _4\vert p^k\delta _{\theta _1-\theta _4+\beta ,0},\!(\text {From Lemma }7)\\&\quad =\!\! \underset{\theta _4\in {\mathbb {F}}_{p^k}}{\sum } \vert \theta _4-\beta \rangle \!\langle \theta _4\vert \!=\! \mathrm {X}^{(p^k)}(-\beta )\!\!\!\underset{\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\vert \theta _4\rangle \!\langle \theta _4\vert \!=\! \mathrm {X}^{(p^k)}(-\beta ), \end{aligned}$$

because $\underset{\theta _4\in {\mathbb {F}}_{p^k}}{\sum }\vert \theta _4\rangle \!\langle \theta _4\vert = \mathrm {I}_{p^k}$ by completeness relation. $\square$

F Computing the elements of set $P_{\mathrm {O}}$

In this appendix, we show that the set $P_{\mathrm {O}}$ is based on the ranges provided in Table 2.

We recall that the position index sets are

$$\begin{aligned} P_{\mathrm {X}}&=\{n-b_1-d_m+3,\dots ,n-b_1+1\}, \end{aligned}$$

(145)

$$\begin{aligned} P_{\mathrm {Z}}&= \{b_2+1,\dots ,b_2+d_m-1\}, \end{aligned}$$

(146)

$$\begin{aligned} P_{\mathrm {O}}&= P_{\mathrm {X}} \cap P_{\mathrm {Z}}. \end{aligned}$$

(147)

From Eqs. (145) and (146), when elements s and t belong to $P_{\mathrm {X}}$ and $P_{\mathrm {Z}}$, respectively, s and t satisfy the following inequalities:

$$\begin{aligned}&n-b_1-d_m+3 \le s \le n-b_1+1 \Rightarrow n-(b_1-2+(d_m-1))\\&\quad \le s \le n-(b_1-2+1),\\&b_2+1 \le t \le b_2+d_m-1 \Rightarrow b_2+1\le t \le b_2+(d_m-1). \end{aligned}$$

Thus, the general form of $s \in P_{\mathrm {X}}$ and $t \in P_{\mathrm {Z}}$ is

$$\begin{aligned} s&= n-(b_1-2+i),\text { where }i\in \{1,\dots ,(d_m-1)\}, \end{aligned}$$

(148)

$$\begin{aligned} t&=b_2+j,~\text { where }j\in \{1,\dots ,(d_m-1)\}. \end{aligned}$$

(149)

The value of s and t in Eqs. (148) and (149) are considered modulo n and belong to the range 1 to n.

We next consider an element r that belongs to $P_{\mathrm {O}}$ from Eq. (147). As $P_{\mathrm {O}} = P_{\mathrm {X}} \cap P_{\mathrm {Z}}$, the element r belongs to both $P_{\mathrm {X}}$ and $P_{\mathrm {Z}}$. Thus, from Eqs. (148) and (149), for some $i,j \in \{1,\dots , (d_m-1)\}$, we obtain

$$\begin{aligned}&r = n-(b_1-2+i)\text { and }r = b_2+j, \end{aligned}$$

(150)

$$\begin{aligned} \Rightarrow&(b_1+b_2-2)+i+j=n. \end{aligned}$$

(151)

As the general form of the elements of $P_\mathrm {X}$ and $P_\mathrm {Z}$ in Eqs. (148) and (149) is considered modulo n, $r = (n-(b_1-2+i))$ and $r = (b_2+j)$ are also considered modulo n. From Eq. (18), as $v=(b_1+b_2-2)~\mathrm {mod}~n$, Eq. (151) is same as the condition $(v+i+j)~\mathrm {mod}~n=0$ obtained in Sect. 4 to find the value of $n_e$. Thus, i and j in Eq. (151) belong to the ranges in Table 2 that was obtained using the condition $(v+i+j)~\mathrm {mod}~n=0$. From Table 2, let the range of i be from $i_l$ to $i_m$, then, from Eq. (150), we obtain $P_{\mathrm {O}} = \{n-(b_1-2+i)\}_{i=i_l}^{i_m}$. We note that the elements in the set $P_{\mathrm {O}}$ are modulo n and range from 1 to n. The set $P_O$ for various cases in Table 1 is provided in Supplementary material.

G Proof of Lemma 4

Proof

Let $L_{i,j}$ be the $(i,j)^{\mathrm {th}}$ element of the lower triangular matrix L. For $i\in \{1,\dots ,n\}$, let $L_i$ be the matrix obtained by replacing $i^{\mathrm {th}}$ row of an identity matrix with $i^{\mathrm {th}}$ row of L.

$$\begin{aligned} L_i = \begin{bmatrix} 1 &{} \dots &{} 0 &{} 0 &{} 0 &{} \dots &{} 0\\ \vdots &{} \ddots &{} \vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} \dots &{} 1 &{} 0 &{} 0 &{} \dots &{} 0 \\ L_{i,1} &{} \dots &{} L_{i,(i-1)} &{} L_{i,i} &{} 0 &{} \dots &{}0\\ 0 &{} \dots &{} 0 &{} 0 &{} 1 &{} \dots &{} 0\\ \vdots &{} \ddots &{}\vdots &{} \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0 &{} \dots &{} 0 &{} 0 &{} 0 &{} \dots &{} 1 \end{bmatrix}. \end{aligned}$$

(152)

We consider the matrix $F=L_s$ and $G=L_t$, where $s,t\in \{1,\dots ,n\}$ and $s\ne t$. Let $F_{i,j}$ and $G_{i,j}$ be the $(i,j)^{\mathrm {th}}$ elements of F and G, respectively. Based on the definition of $L_s$ and $L_t$, and from Eq. (152), we have

$$\begin{aligned} F_{i,m}&\!=\! {\left\{ \begin{array}{ll} L_{i,m} &{} i\!=\!s\\ 1 &{} i \!\ne \! s \text { and }i \!=\!m\\ 0 &{} \text {else} \end{array}\right. }\!\Rightarrow \! F_{i,m}\!=\!{\left\{ \begin{array}{ll} L_{i,m} &{} i\!=\!s\\ \delta _{i,m} &{} i \!\ne \! s \end{array}\right. }, \end{aligned}$$

(153)

$$\begin{aligned} G_{m,j}&\!=\! {\left\{ \begin{array}{ll} L_{m,j} &{} m\!=\!t\\ 1 &{} m \!\ne \! t \text { and }m \!=\!j\\ 0 &{} \text {else} \end{array}\right. }\!\Rightarrow \! G_{m,j} \!=\! {\left\{ \begin{array}{ll} L_{m,j} &{} m\!=\!t\\ \delta _{m,j} &{} m \!\ne \! t \end{array}\right. }. \end{aligned}$$

(154)

From Eq. (153), the value of the $(i,j)^{\mathrm {th}}$ element of FG is

$$\begin{aligned} (FG)_{i,j}&= \underset{m=1}{\overset{n}{\sum }}F_{i,m}G_{m,j}=\underset{m=1}{\overset{s}{\sum }}F_{i,m}G_{m,j}. \end{aligned}$$

(155)

We consider three cases, where we consider the $s^{\mathrm {th}}$ row, $t^{\mathrm {th}}$ row, and the rest of the rows, respectively.

Case 1: $\varvec{i=s}$

From Eq. (155), we obtain

$$\begin{aligned} (FG)_{s,j}&=\underset{m=1}{\overset{s}{\sum }}F_{s,m}G_{m,j}=\underset{m=1}{\overset{s}{\sum }}L_{s,m}G_{m,j},~(\text {From Eq.~}153)\\&=\underset{m=1}{\overset{s}{\sum }}L_{s,m}\delta _{m,j}= L_{s,j},~(\text {From Eq.~}154\text { as }m \ne t). \end{aligned}$$

Thus, the $s^{\mathrm {th}}$ row of FG is equal to the $s^{\mathrm {th}}$ row of L.

Case 2: $\varvec{i=t}$

From Eq. (155), we obtain

$$\begin{aligned} (FG)_{t,j}&=\underset{m=1}{\overset{s}{\sum }}F_{t,m}G_{m,j}=\underset{m=1}{\overset{s}{\sum }}\delta _{t,m}G_{m,j},~(\text {From Eq.~}153\text { with }i \ne s)\\&=G_{t,j}= L_{t,j}.~(\text {From Eq.~}154\text { as }m=t) \end{aligned}$$

Thus, the $t^{\mathrm {th}}$ row of FG is equal to the $t^{\mathrm {th}}$ row of L.

Case 3: $\varvec{i\ne s}$ and $\varvec{i \ne t}$

From Eq. (155), we obtain

$$\begin{aligned} (FG)_{i,j}&=\underset{m=1}{\overset{s}{\sum }}F_{i,m}G_{m,j}=\underset{m=1}{\overset{s}{\sum }}\delta _{i,m}G_{m,j},~(\text {From Eq.~}153 \text {with }i\ne s)\\&=G_{i,j}=\delta _{i,j}. ~(\text {From Eq.~}154 \text {with }m\ne t) \end{aligned}$$

Thus, the $i^{\mathrm {th}}$ row of FG, where $i\ne s$ and $i \ne t$, is equal to the $i^{\mathrm {th}}$ row of the identity matrix $\mathrm {I}_{n}$.

Thus, the matrix $L_sL_t = FG$ is the matrix obtained by replacing the $s^{\mathrm {th}}$ and $t^{\mathrm {th}}$ row of the identity matrix with the $s^{\mathrm {th}}$ and $t^{\mathrm {th}}$ row of L. Thus, $L_1L_2\dots L_n$ is the matrix obtained by replacing the $1^{\mathrm {st}}$, $2^{\mathrm {nd}}$, $\dots$, $n^{\mathrm {th}}$ row of the identity matrix by the $1^{\mathrm {st}}$, $2^{\mathrm {nd}}$, $\dots$, $n^{\mathrm {th}}$ row of L, respectively, which is equal to L. $\square$