Generating quantum channels from functions on discrete sets

Abstract

Using the recent ability of quantum computers to initialize quantum states rapidly with high fidelity, we use a function operating on a discrete set to create a simple class of quantum channels. Fixed points and periodic orbits, that are present in the function, generate fixed points and periodic orbits in the associated quantum channel. Phenomenology such as periodic doubling is visible in a 6 qubit dephasing channel constructed from a truncated version of the logistic map. Using disjoint subsets, discrete function-generated channels can be constructed that preserve coherence within subspaces. Error correction procedures can be in this class as syndrome detection uses an initialized quantum register. A possible application for function-generated channels is in hybrid classical/quantum algorithms. We illustrate how these channels can aid in carrying out classical computations involving iteration of non-invertible functions on a quantum computer with the Euclidean algorithm for finding the greatest common divisor of two integers.

Appendices

Appendix A: Number of functions on $\mathbb {Z}_n$ up to permutation

The number of possible functions on a discrete set, equivalent up to permutation, determines the potential range of phenomena exhibited by the class of quantum channel defined in Definition 1. Using work by [10], we find that we can give a expression for the total number of possible functions of $\mathbb {Z}_n$ in the equivalence class.

Let $n\in \mathbb {Z}^+$, and let X denote the set of functions from $\mathbb {Z}_n$ to itself. The permutation group on $\mathbb {Z}_n$, which we denote by $S_n$, acts on X via conjugation:

$$\begin{aligned} \sigma \circ f\circ \sigma ^{-1}, \end{aligned}$$

(A1)

with $\sigma \in S_n$. If $f\in X$, then the orbit of f under this action is the set:

$$\begin{aligned} orb(f)=\{g\in X:\exists \sigma \in S_n\text { s.t }g=\sigma \circ f\circ \sigma ^{-1}\}. \end{aligned}$$

(A2)

If $f,g\in X$, we say f is equivalent to g up to permutation, and write $f\sim g$, if and only if there is a $\sigma \in S_n$ such that $g=\sigma \circ f\circ \sigma ^{-1}$. The relation $\sim$ is an equivalence relation. Moreover, the orbit of any $f\in X$ is precisely the set of all functions equivalent to f up to permutation. We count the number of functions from $\mathbb {Z}_n$ to itself considered up to permutation equivalence, i.e., count the total number of distinct orbits under this group action.

For a fixed $\sigma \in S_n$, we define

$$\begin{aligned} X_\sigma =\{f\in X:\sigma \circ f\circ \sigma ^{-1}=f\} \end{aligned}$$

(A3)

which is the set of all functions from $\mathbb {Z}_n$ to itself which commute with permutation $\sigma$.

Any permutation $\sigma \in S_n$ can be written uniquely as a product of disjoint cycles (of possibly differing lengths). Keeping with the notation of [10], we let $\lambda _i$ denote the number of cycles of length i for $\sigma$ written in cyclic notation in terms of disjoint cycles. Proposition 2.6 from [10] states that the number $N_{\textrm{pairs}}$ of ordered pairs $(\tau ,f)$, with permutation $\tau \in S_n$, function $f \in X$, and such that the permutation commutes with the function: $\tau \circ f=f\circ \tau$, is

$$\begin{aligned} N_{\textrm{pairs}} =&\!\!\!\!\!\ \sum \limits _{\begin{array}{c} 1\lambda _1+\cdots +n\lambda _n=n, \\ \,\,\lambda _i\ge 0 \end{array}}\frac{n!}{\lambda _1!\cdot ...\cdot \lambda _n!1^{\lambda _1}\cdot ...\cdot n^{\lambda _n}}\nonumber \\&\qquad \qquad \times \prod _{k=1}^{n}\left( \sum _{j|k}j\lambda _j\right) ^{\lambda _k}. \end{aligned}$$

(A4)

The form of Proposition 2.6 stated by [10] is defined in terms of functions on $\{1,\ldots ,n\}$; however, the version of Proposition 2.6 stated above is equivalent as $\{1,\ldots ,n\}$ and $\mathbb {Z}_n$ are finite sets of the same size. Using this proposition, we find the following:

Theorem 1

Let $n\in \mathbb {Z}^+$. The number of functions from $\mathbb {Z}_n$ to itself, considered up to permutation equivalence, is

$$\begin{aligned} \sum \limits _{\begin{array}{c} 1\lambda _1+\cdots +n\lambda _n=n,\\ \,\,\lambda _i\ge 0 \end{array}}\frac{1}{\lambda _1!\cdot ...\cdot \lambda _n!1^{\lambda _1}\cdot ...\cdot n^{\lambda _n}}\prod _{k=1}^{n}\left( \sum _{j|k}j\lambda _j\right) ^{\lambda _k} \end{aligned}$$

(A5)

Table 7 Properties of function-generated channels for three-state systems

Proof

We denote X to be the set of all functions from $\mathbb {Z}_n$ to itself. The number of functions $f \in X$, equivalent up to permutation, is equal to the number r of distinct orbits under the conjugation action of the permutation group $S_n$ on X. This is a setting where we can apply Burnside’s lemma for the number of orbits, giving

$$\begin{aligned} r=\frac{1}{n!}\sum _{\sigma \in S_{\mathbb {Z}_n}}|X_{\sigma }|. \end{aligned}$$

(A6)

Here, we have used the fact that $n! = |S_n|$ and $X_{\sigma }$ as defined in Eq. (A3). The sum $\sum _{\sigma \in S_n}|X_{\sigma }|$ is precisely the number of all ordered pairs $(\tau ,f)$ such that $\tau \circ f=f\circ \tau$, where permutation $\tau \in S_n$ and f is a function from $\mathbb {Z}_n$ to itself. Applying Proposition 2.6 from [10] (via Eq. A4) with Eq. (A6), we obtain the desired result of Eq. (A5). $\square$

Appendix B: Properties of function-generated channels for three-state systems

In Table 7, we include the eigenvalues and left and right eigenvectors of the channels for three-state systems with Kraus operators listed in Table 5 and illustrated in Fig. 8. By r-eigenvector, we mean right eigenvector and by l-eigenvector we mean left eigenvector. We list left eigenvectors if the dual channel differs from the channel; ${\mathcal {E}}$ differs from ${{\mathcal {E}}}^\ddagger$. For eigenvalues of modulus 1, we choose orthogonal right eigenvectors, and conserved quantities or left eigenvectors that satisfy a biorthogonality condition via the Frobenius inner product. These examples help illustrate different cases in the proof of Appendix C.

Left eigenvectors with eigenvalues of 1 are conserved quantities. The identity is always a conserved quantity. For channel $F_{2a}$ $\left| {0}\right\rangle \!\left\langle {0}\right| + \left| {1}\right\rangle \!\left\langle {1}\right|$ is also a conserved quantity, which implies that the trace of the subspace spanned by $\left| {0}\right\rangle , \left| {1}\right\rangle$ is preserved via the channel. The $D_{2a}$ channel has conserved quantity $\left| {0}\right\rangle \!\left\langle {1}\right| + \left| {1}\right\rangle \!\left\langle {0}\right|$ in addition to $\left| {0}\right\rangle \!\left\langle {0}\right| + \left| {1}\right\rangle \!\left\langle {1}\right|$. The two conserved quantities within the subspace spanned by $\{ \left| {0}\right\rangle , \left| {1}\right\rangle \}$ are present because the channel gives a unitary transformation within this subspace.

Appendix C: Eigenvalues of a discrete function-generated channel

Theorem 2

Consider ${{\mathcal {L}}}_{{\mathcal {E}}}$ the matrix representation of a channel ${{\mathcal {E}}}_{S_D,f}$ generated from a function f and collection of disjoint sets $S_D$ defined as in Definition 1. All eigenvalues of ${{\mathcal {L}}}_{{\mathcal {E}}}$ are either 0 or have modulus unity.

Proof

We work in the orthonormal basis $\left| {i}\right\rangle \!\left\langle {k}\right|$ (with $i,k \in {{\mathbb {Z}}}_N$) for operators with respect to the Hilbert–Schmidt or Frobenius inner product. Here, N is the dimension of the Hilbert space. We consider how the channel affects diagonal basis operators first.

1.1 1. Diagonal basis vectors

For every $i \in {{\mathbb {Z}}}_N$, i is either a fixed point of the function ($f(i)=i$) and is a member of a cycle ($f^k(i)=i$ for $k>1$) or there exists a $k>1$ such that $f^k(i)$ is either a fixed point or a member of a cycle. We consider each of these cases separately. In each case, we find a right eigenstate of ${{\mathcal {L}}}_{{\mathcal {E}}}$.

If $f(i) = i$ is a fixed point of the function, then $\left| {i}\right\rangle \!\left\langle {i}\right|$ is a fixed point of the channel with eigenvalue 1. This follows as the index i belongs to one of the sets $S_j$, and application of $K_j$ gives ${{\mathcal {E}}}(\left| {i}\right\rangle \!\left\langle {i}\right| ) = K_j \left| {i}\right\rangle \!\left\langle {i}\right| K_j^\dagger = \left| {i}\right\rangle \!\left\langle {i}\right|$.

If i is a member of a k-cycle of f, with $k>1$, then there are k associated eigenvectors and eigenvalues of the channel. We find the set $S_j$ that contains i. Application of the channel ${{\mathcal {E}}}(\left| {i}\right\rangle \!\left\langle {i}\right| )= K_j \left| {i}\right\rangle \!\left\langle {i}\right| K_j^\dagger = \left| {f(i)}\right\rangle \!\left\langle {f(i)}\right|$. By applying the channel a second time, we obtain $\left| {f^2(i)}\right\rangle \!\left\langle {f^2(i)}\right|$. We apply the channel iteratively until we obtain the original state $\left| {i}\right\rangle \left\langle {i}\right|$. A total of k orthogonal (via the Hilbert–Schmidt inner product) right eigenstates of ${{\mathcal {L}}}_{{\mathcal {E}}}$ can be constructed via discrete Fourier transform of the intermediate states in the cycle and using powers of $\omega = e^{2\pi i/k}$ (see Eq. 38). Eigenvalues are complex roots of unity which are integer powers of $\omega$. The members of the cycle (j such that $f^k(j)=j$ with $k>1$) are different from the fixed points of the function (i such that $f(i)=i$), so eigenstates generated from a cycle are orthogonal to eigenstates generated by fixed points. Elements in a cycle or that are fixed points are disjoint from those that are not in a cycle or are a fixed point. Diagonal states that are formed via elements that are neither in a cycle or are a fixed point would generate diagonal states that are perpendicular to those generated from fixed points or elements of cycles. For each k cycle, the quantum Fourier transform gives k linearly independent right eigenvectors that are sums of diagonal operators.

The remaining integers i are those where i is not a fixed point of the function f and i is not a member of a cycle.

If f(i) is a fixed point of f, (satisfying $f^2(i) = f(i)$) but i is not a fixed point ($f(i) \ne i$), then $\left| {i}\right\rangle \!\left\langle {i}\right| - \left| {f(i)}\right\rangle \!\left\langle {f(i)}\right|$ is a right eigenvector with eigenvalue 0. Note that it is not orthogonal (in the sense of the Hilbert–Schmidt inner product) to the eigenstate associated with the fixed point $\left| {f(i)}\right\rangle \!\left\langle {f(i)}\right|$. However, this eigenstate is linearly independent of the right eigenstates of ${{\mathcal {L}}}_{{\mathcal {E}}}$ associated with fixed points and with cycles of the function.

If $f^2(i)$ is a fixed point of f ($f^3(i) = f^2(i)$) but f(i) is not a fixed point ($f^2(i) \ne f(i)$), then consider the operator $b = \left| {i}\right\rangle \!\left\langle {i}\right| - \left| {f^2(i)}\right\rangle \!\left\langle {f^2(i)}\right|$. We operate on b with the channel giving

$$\begin{aligned} {{\mathcal {E}}}(b)&= \left| {f(i)}\right\rangle \!\left\langle {f(i)}\right| - \left| {f^2(i)}\right\rangle \!\left\langle {f^2(i)}\right| \\ {{\mathcal {E}}}^2(b)&= 0. \end{aligned}$$

The operator ${{\mathcal {E}}}(b)$ is a right eigenvector of ${{\mathcal {L}}}_{{\mathcal {E}}}$ with eigenvalue 0. This means b is an generalized right eigenvector associated with right eigenvector ${{\mathcal {E}}}(b)$ that has a zero eigenvalue. This generalized eigenstate is linearly independent of the previously identified right eigenstates associated with fixed points and cycles of the function f.

If f(i) is a member of a k-cycle of f with $k>1$, and i is not a member of the cycle, then $\left| {i}\right\rangle \!\left\langle {i}\right| - \left| {f^{k-1}(i)}\right\rangle \!\left\langle {f^{k-1}(i) }\right|$ is a right eigenvector with eigenvalue 0. This eigenstate is linearly independent of the right eigenstates we discussed above that are associated with fixed points and cycles of the function.

If $f^2(i)$ is a member of a k-cycle of f with $k>1$, and f(i) is not in the cycle, then consider the state $b = \left| {i}\right\rangle \!\left\langle {i}\right| - \left| {f(i)}\right\rangle \!\left\langle {f(i)}\right|$. We operate on b with the channel giving ${{\mathcal {E}}}(b) = \left| {f(i)}\right\rangle \!\left\langle {f(i)}\right| - \left| {f^2(i)}\right\rangle \!\left\langle {f^2(i)}\right|$. This is the form previously considered above and is a right eigenvector which has an eigenvalue of 0. This implies that b is a generalized eigenvector associated with right eigenvector ${{\mathcal {E}}}(b)$ that has a zero eigenvalue.

If $f^k(i)$ is the smallest positive k with $k>1$ such that $f^k(i)$ is either a member of a cycle or is a fixed point, then $b = \left| {i}\right\rangle \!\left\langle {i}\right| - \left| {f(i)}\right\rangle \!\left\langle {f(i)}\right|$ is a generalized eigenvector. This follows iteratively using the construction used above as ${{\mathcal {E}}}(b)$ is either a right eigenvector with eigenvalue 0 or a generalized eigenvector with eigenvalue 0.

Altogether we can find a unique eigenvector comprised of states on the diagonal for each i value. If we are careful not to over count members of cycles, we find that the set contains N linearly independent eigenvectors and generalized eigenvectors comprised of sums of diagonal operators. The eigenvectors are independent of the disjoint subsets $\{ S_j\}$ used to generate the channel. The types of eigenvectors are illustrated in the graph shown in Fig. 12 of a function which makes it clearer why there are N eigenvalues. The construction of each type discussed above shows that they are linearly independent. All the generalized eigenvectors are associated with eigenvalues of 0. Only members of cycles or fixed points generate eigenvectors with eigenvalues that are roots of unity.

Fig. 12

We show graphs of a function f and how its structure is related to right eigenvectors that are comprised of diagonal operators. The right eigenvectors are those of ${{\mathcal {L}}}_{{\mathcal {E}}}$ the matrix representation of a quantum channel that is generated by a function and a set of disjoint sets. Each arrow begins at an element x and points to another element f(x). Elements shown as red circles generate a right eigenvector with eigenvalue 0. Elements shown as blue stars generate right eigenvectors with eigenvalue that have $|\lambda |=1$. Elements shown with green square generate generalized eigenvectors with eigenvalue 0. The right eigenvectors comprised of diagonal elements are insensitive to the disjoint sets $\{S_j\}$ (Color figure online)

1.2 2. Off-diagonal basis vectors

We now consider how the channel affects operators comprised of off-diagonal terms. We consider how the channel operates on $\left| {i}\right\rangle \!\left\langle {k}\right|$ where $i\ne k$. We consider two cases, i, k, are both contained in one disjoint set $S_j$ and where there is no disjoint set in $S_D$ containing both i, k.

Suppose there is no disjoint set $S_j \in S_D$ that contains both i, k. We find that $K_j \left| {i}\right\rangle \!\left\langle {k}\right| K_j^\dagger = 0$ for all j and this implies that ${{\mathcal {E}}}(\left| {i}\right\rangle \left\langle {k}\right| ) = 0$. The operator $\left| {i}\right\rangle \!\left\langle {k}\right|$ is a right eigenvector of ${{\mathcal {L}}}_{{\mathcal {E}}}$ with eigenvalue 0.

Fig. 13

We show graphs of a function f and how its structure is related to right eigenvectors that are comprised of off-diagonal operators. The dashed lines show different disjoint sets. On the left $\left| {k}\right\rangle \!\left\langle {l}\right|$ (with $k \ne l$) and its transpose have eigenvalue 0 and $\left| {i}\right\rangle \!\left\langle {j}\right|$ ($i \ne j$) and its transpose are generalized eigenvectors with eigenvalue 0. On the right $\left| {a}\right\rangle \!\left\langle {b}\right|$ (with $a\ne b$) and its transpose are generalized eigenvectors with eigenvalue 0 and $\left| {c}\right\rangle \!\left\langle {d}\right|$ ($c\ne d$) and its transpose are right eigenvectors with eigenvalue 0. The operators $\left| {e}\right\rangle \!\left\langle {f}\right|$, $\left| {e}\right\rangle \!\left\langle {g}\right|$, $\left| {e}\right\rangle \!\left\langle {h}\right|$, and $\left| {e}\right\rangle \!\left\langle {h}\right|$ (with e, f, g, h all distinct) and are in a cycle and so generate 4 right eigenvectors with eigenvalues that are complex roots of unity

Consider the second case for i, k where $i \ne k$ and both are contained in a disjoint set $S_j$.

$$\begin{aligned} {{\mathcal {E}}}( \left| {i}\right\rangle \!\left\langle {k}\right| ) \!&=\! K_j \left| {i}\right\rangle \!\left\langle {k}\right| K_j^\dagger \\&=\! (\left| {f(i)}\right\rangle \!\left\langle {i}\right| \!+\! \left| {f(k)}\right\rangle \!\left\langle {k}\right| ) \left| {i}\right\rangle \!\left\langle {k}\right| (\left| {i}\right\rangle \!\left\langle {f(i)}\right| \! +\! \left| {k}\right\rangle \!\left\langle {f(k)}\right| )\\&= \left| {f(i)}\right\rangle \! \left\langle {f(k)}\right| . \end{aligned}$$

The matrices $\left| {i}\right\rangle \!\left\langle {k}\right| , \left| {k}\right\rangle \!\left\langle {i}\right|$ are generalized right eigenvectors associated with eigenvalue 0 unless both f(i), f(j) are in the same disjoint set $f(i), f(k) \in S_m$ for some m. Here, $S_m$ could be the same disjoint set as $S_j$. Note that $f(i) \ne f(k)$ is due to a requirement on the nature of the disjoint sets in the definition of the channel (Definition 1). Thus, ${{\mathcal {E}}}(\left| {i}\right\rangle \!\left\langle {k}\right| )$ must give an off-diagonal matrix. Suppose $f(i)=i$ and $f(k) = k$ and $i,j \in S_j$. This case $\left| {i}\right\rangle \!\left\langle {k}\right|$ and $\left| {k}\right\rangle \!\left\langle {i}\right|$ are both right eigenvectors with eigenvalue of 1. If both $f(i), f(k) \in S_m$ (and i, k are not fixed points and $i,j \in S_j$), we can apply the channel iteratively. The next iteration gives ${{\mathcal {E}}}^2(\left| {i}\right\rangle \!\left\langle {k}\right| ) = \left| {f^2(i)}\right\rangle \!\left\langle {f^2(k)}\right| .$ Again, if both $f^2(i), f^2(k)$ are contained in different disjoint sets, then $\left| {i}\right\rangle \!\left\langle {k}\right|$ is a generalized eigenvector associated with eigenvalue 0.

If $f^m(i)=i$, $f^m(k)=k$ for a positive integer $m>1$ and for all positive integers n, there exists a disjoint set $S_j$ such that both $f^n(i), f^n(k) \in S_j$, and then, $\left| {i}\right\rangle \!\left\langle {k}\right|$ generates a cycle. From $\left| {i}\right\rangle \!\left\langle {k}\right|$ and iterates of it, we can construct a set of right eigenvectors (comprised only of sums of off-diagonal operators) with eigenvalues of modulus 1 using a Fourier transform like that given in Eq. (38). The resulting right eigenvectors are linearly independent and would be linearly independent of off-diagonal operators that have 0 eigenvalues or that are generalized eigenvectors associated with a 0 eigenvalue. The only other way iteration of $\left| {i}\right\rangle \!\left\langle {k}\right|$ can terminate (other than terminating via a cycle) is when an iteration of i and k by the function gives a pair of elements that are not contained in the same disjoint set. In this case, we have a series of generalized right eigenvectors associated with eigenvalue 0. Examples of the construction of eigenvectors for off-diagonal operators are shown in Fig. 13.

Each off-diagonal term is either a right eigenvector with eigenvalue 0, a generalized right eigenvector associated with eigenvalue 0, a fixed point, or generates a cycle giving eigenvectors with eigenvalues that are complex roots of unity. In total we can generate $N(N-1)$ linearly independent eigenvectors from each pair off-diagonal elements. The union of the N eigenvectors comprised of diagonal operators and the $N(N-1)$ eigenvectors comprised of off-diagonal operators is a set of $N^2$ linearly independent right eigenvectors.

We have shown iteratively that we can find a full set ($N^2$) of eigenvalues of ${{\mathcal {L}}}_{{\mathcal {E}}}$ where eigenvalues either are roots of unity, arising from the fixed points and cycles of f, or they have eigenvalue 0. Zero eigenvalues are associated with both right eigenvectors and generalized right eigenvectors. In both cases, they arise from orbits of f that, after iteration, have end states that are fixed points or cycles. Right eigenvectors or generalized right eigenvectors are sometimes are not orthogonal (in the sense of the Hilbert–Schmidt inner product) to each other. However, we have identified a set of $N^2$ right eigenvectors and generalized right eigenvectors that are linearly independent. The Jordan form of ${{\mathcal {L}}}_{{\mathcal {E}}}$ consists of Jordan blocks of dimension 1 that contain eigenvalues of modulus 1, and Jordan blocks that can have a larger dimension that have eigenvalues of 0. As the non-trivial Jordan blocks are associated with chains of f that have end states that are fixed points or cycles, the maximum dimension of any Jordan block is the maximum link number $k_c$ (as defined in Eq. 7). As every Jordan block of ${{\mathcal {L}}}_{{\mathcal {E}}}$ generates an invariant subspace of the ${{\mathcal {L}}}_{{\mathcal {E}}}$, and nonzero eigenvalues have modulus of 1, the asymptotic subspace of the channel is spanned by the set of right eigenvectors with nonzero eigenvalues. $\square$