Generalizations of Pauli channels

Introduction

As a generalization of the Pauli channel on a qubit, we define a linear mapping $\alpha:M_{n}\to M_{n}$ such that

Let $\sigma_{0}=I$ and $\sigma_{1},\sigma_{2},\sigma_{3}$ be Pauli matrices, i.e.,

and let ${\cal E}:M_{2}\to M_{2}$ be defined as

Density matrices are sent to density matrices if and only if

It is not difficult to compute the representing block matrix $X:=\sum_{i,j}{\cal E}(E_{ij})\otimes E_{ij}$, we have

According to Choi’s theorem the positivity of this matrix is equivalent to the complete positivity of ${\cal E}$. $X$ is unitarily equivalent to the matrix

This matrix is obviously positive if and only if

This is necessary and sufficient condition of complete positivity. $\square$

It is not obvious that condition (3) is symmetric in the three variables $\lambda_{1},\lambda_{2},\lambda_{3}$. Condition (3) actually determines the tetrahedron which is the convex hull of the points $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$ and $(-1,-1,1)$.

Now we show the idea leading to the generalization. The mapping ${\cal E}$ in (2) has the form

From the expansion of ${\cal E}(\sigma_{j})$ we can get equations and the solution is the following:

If $\mu_{i}\geq 0$ for every $i$, then ${\cal E}$ is a completely positive mapping. Therefore,

or together this is (3). (Actually, this argument gives that (3) is a sufficient condition for the complete positivity.)

Pauli channels form an important and popular subject in quantum information theory . The mappings (1) were studied in the paper in the case when the subalgebras are maximal Abelian and pairwise complementary. Our method is different and we allow non-commutative subalgebras as well.

The mapping (1) restricted to ${\cal A}_{i}$ has the form

on density matrices $D$. If $0\leq\lambda_{i}\leq 1$, then we can say that $D$ does not change with probability $\lambda_{i}$ and with probability $1-\lambda_{i}$ it is sent to the tracial state. Such mappings are usually called as depolarizing channels .

A simple example including non-commutative subalgebras is the following.

Consider $M_{4}=M_{2}\otimes M_{2}$ and the complementary F-subalgebras ${\cal A}_{1},\dots,{\cal A}_{4}$ generated by the following triplets of unitaries:

We take also the M-subalgebra ${\cal A}_{5}$ generated by $\sigma_{1}\otimes\sigma_{1},\sigma_{2}\otimes\sigma_{2},\sigma_{3}\otimes\sigma_{3}$. The conditional expectations $E_{j}:M_{4}\to{\cal A}_{j}$ are convex combinations of automorphisms

where $U_{j1}=I$ and $U_{ji}$’s are orthogonal unitaries from ${\cal A}_{j}^{\prime}$. Since ${\cal A}_{5}$ is an M-subalgebra, ${\cal A}_{5}^{\prime}={\cal A}_{5}$. The subalgebras ${\cal A}_{1}^{\prime},\dots,{\cal A}_{4}^{\prime}$ are F-subalgebras generated by the following unitaries:

(The above triplets generating ${\cal A}_{j}$ and ${\cal A}_{j}^{\prime}$ ($1\leq j\leq 4$) are Pauli triplets, see for details.) Moreover,

The linear mapping (1) has the concrete form

where the conditional expectations $E_{j}$ is expressed by the commutant, see (4). (The condition for complete positivity of $\alpha$ is in Theorem 4.) $\square$

Our main result is the necessary and sufficient condition for the complete positivity of mappings like (1) which can be called generalized Pauli channel.

Generalized Pauli channels

Let ${\cal A}$ be a (unital *-) subalgebra of $M_{n}$. Our aim is to describe the conditional expectation onto ${\cal A}$ by means of an orthogonal system in the commutant.

Up to unitary equivalence, a subalgebra ${\cal A}$ of $M_{n}$ can be written as

The commutant ${\cal A}^{\prime}$ in $M_{n}$ is

Let $N=\sum_{i=1}^{k}n_{i}^{2}$ and let $P_{i}$ be a minimal central projection of ${\cal A}$, that is, $P_{i}=I_{n_{i}}\otimes I_{m_{i}}$.

Let $\{U_{i}\}_{i=1}^{N}$ be an orthonormal basis of ${\cal A}$. Then the completely positive map $F$ from $M_{n}$ onto ${\cal A}^{\prime}$ given by

In particular, if all $n_{i}/m_{i}$ are equal, then $\frac{n}{{\rm dim}{{\cal A}}}F$ is the trace-preserving conditional expectation from $M_{n}$ onto ${\cal A}^{\prime}$.

Proof: If all $n_{i}/m_{i}$ are equal, then their ratio is equal to ${{\rm dim}{\cal A}\over{n}}$. Therefore it is sufficient to prove the first assertion.

Let $\{e_{ij}^{(l)}\}_{i,j=1}^{n_{l}}$ and $\{f_{ij}^{(l)}\}_{i,j=1}^{m_{l}}$ be matrix units of $M_{n_{l}}$ and $M_{m_{l}}$, respectively. Then $U_{i}$ is written by

for $1\leq i\leq N$, $1\leq l\leq k$ and $1\leq s,t\leq n_{l}$ is unitary. Indeed, $W$ can be considered as the matrix which takes the orthonormal basis $\{{1\over\sqrt{m_{l}}}e_{st}^{(l)}\}$ of ${\cal A}$ into the orthonormal basis $\{U_{i}\}$. Hence we have

Let $T$ be a partial isometry with $T^{*}T=e_{s_{1}s_{1}}^{(l_{1})}\otimes f_{t_{1}t_{1}}^{(l_{1})}$ and $TT^{*}=e_{s_{2}s_{2}}^{(l_{2})}\otimes f_{t_{2}t_{2}}^{(l_{2})}$. Then we obtain

by (5) so that $F$ maps the off-diagonal part to , that is, if $l_{1}\neq l_{2}$ then $F(T)=0$.

Now let $T=e_{s_{2}s_{1}}^{(l)}\otimes f_{t_{2}t_{1}}^{(l)}$. Then we obtain

which shows the first assertion. $\square$

The commutant of M-subalgebras and F-subalgebras are again M-subalgebras and F-subalgebras, and in both types it is possible to choose an orthogonal basis consisting of unitaries, only. Thus by an application of the previous proposition, for such a subalgebra ${\cal A}$, the trace-preserving conditional expectation is the convex combination of automorphisms:

where $\{U_{i}^{\prime}\}$ is an orthogonal basis of ${\cal A}^{\prime}$ consisting of unitaries. Bases consisting of unitaries are important also in quantum state teleportation .

Let $\{U_{i}:1\leq i\leq m\}$ be an orthonormal system in $M_{n}$. Then the linear mapping

is completely positive if and only if $\mu_{i}\geq 0$ for every $1\leq i\leq m$.

Proof: If $\mu_{i}\geq 0$ for every $1\leq i\leq m$, it is clear that $\alpha$ is completely positive. To prove the converse, we first show that

is a projection for every $1\leq k\leq m$. To show that they are pairwise orthogonal, we compute the trace of $P_{k}P_{l}$:

is positive, therefore $\mu_{k}\geq 0$. $\square$

Proof: Since both sides are bilinear in the variables $X$ and $Y$, it is enough to check the case $X=E_{ab}$ and $Y=E_{cd}$. Simple computation gives that left-hand-side is $\delta_{ab}\delta_{cd}$. A physicist might make a different proof of the lemma:

We also need the next lemma; the proof can be found in .

Let $V_{1},V_{2},\dots,V_{n^{2}}$ be matrices in $M_{n}$. Then the following conditions are equivalent:

The next result includes important particular cases which are formulated afterwards.

Let ${\cal A}_{1},\dots,{\cal A}_{r}$ be pairwise complementary subalgebras of $M_{n}$ such that their commutants ${\cal A}_{1}^{\prime},\dots,{\cal A}_{r}^{\prime}$ are pairwise complementary as well. Then the trace-preserving conditional expectations $E_{j}:M_{n}\to{\cal A}_{j}$ can be expressed by the orthonormal bases $U_{j1},U_{j2},\dots,U_{jn(j)}\in{\cal A}_{j}^{\prime}$, where $U_{j1}={1\over\sqrt{n}}I$, via the formula

and the generalized Pauli channel (1) is completely positive if and only if

To prove this theorem we prepare the following proposition.

Let ${\cal A}_{1}$ and ${\cal A}_{2}$ be complementary subalgebras of $M_{n}$. Then ${\cal A}_{1}^{\prime}$ and $A_{2}^{\prime}$ are complementary if and only if ${\cal A}_{1}{\cal A}_{2}$ linearly spans $M_{n}$. Moreover, in this case the trace-preserving conditional expectation $E_{1}:M_{n}\to{\cal A}_{1}^{\prime}$ can be expressed as

where $\{U_{i}\}$ is an orthonormal basis of ${\cal A}_{1}$.

Proof: Assume ${\cal A}_{1}^{\prime}$ and ${\cal A}_{2}^{\prime}$ are complementary. Let $\{U_{i}^{\prime}\}$ and $\{V_{j}^{\prime}\}$ be orthonormal bases of ${\cal A}_{1}^{\prime}$ and ${\cal A}_{2}^{\prime}$, respectively, which consist of scalar multiple of their matrix units. Then the trace-preserving conditional expectations onto ${\cal A}_{1}$ and ${\cal A}_{2}$ are given by the linear combinations of $U_{i}^{\prime*}(\,\cdot\,)U_{i}^{\prime}$ and $V_{j}^{\prime*}(\,\cdot\,)V_{j}^{\prime}$, respectively, thanks to Proposition 1.

Conversely assume ${\cal A}_{1}{\cal A}_{2}$ linearly spans the whole space $M_{n}$. Since ${\cal A}_{1}$ is a subalgebra of $M_{n}$, ${\cal A}_{1}$ can be written as

Let $Q$ be a minimal central projection in ${\cal A}_{2}$ and let $\{U_{i}^{(s)}\}$ and $\{V_{j}\}$ are orthonormal bases of ${\cal A}_{1}$ and ${\cal A}_{2}$, respectively, with the assumption $U_{i}^{(s)}\in M_{n_{s}}\otimes I_{m_{s}}$. Since ${\cal A}_{1}$ and ${\cal A}_{2}$ are complementary and ${\rm span}\{{\cal A}_{1}{\cal A}_{2}\}=M_{n}$, $\{\sqrt{n}U_{i}^{(s)}V_{j}\}$ is an orthonormal basis of $M_{n}$. Therefore by Lemma 2 and Proposition 1, we have

for some $c>0$. These equations imply, for $1\leq s\leq k$,

where $P_{s}$ is a central projection $I_{n_{s}}\otimes I_{m_{s}}$. Now we take the trace to the above equation. Then we have

Hence ${n_{s}/m_{s}}$ is equal to ${1/c}={{\rm dim}{\cal A}_{1}\over n}$ for all $1\leq s\leq k$ and so

is the trace-preserving conditional expectation onto ${\cal A}_{1}^{\prime}$ by Proposition 1. Similarly,

is the trace-preserving conditional expectation onto ${\cal A}_{2}^{\prime}$. Since

Proof of Theorem 2. The first assertion is already proven in the above proposition. Due to the Lemma 2, we have

and the coefficient of ${A\over n}=\left({1\over\sqrt{n}}I\right)A\left({1\over\sqrt{n}}I\right)$ is

Theorem 1 tells us that completely positivity holds if and only if both are positive. $\square$

Assume that $M_{n}$ contains pairwise complementary M-subalgebras ${\cal A}_{1},\dots,{\cal A}_{r}$. Then the generalized Pauli channel is completely positive if and only if

Bipartite channels

In this section we consider subalgebras of $M_{n}\otimes M_{n}$. A subalgebra isomorphic to $M_{n}$ will be called F-subalgebra. An M-subalgebra is a maximal Abelian subalgebra. Both kinds of subalgebras are subspaces of dimension $n^{2}$.

Assume that ${\cal A}_{1}$ and ${\cal A}_{2}$ are F- or M-subalgebras of $M_{n}\otimes M_{n}$. If they are complementary, then the commutants ${\cal A}_{1}^{\prime}$ and ${\cal A}_{2}^{\prime}$ are complementary as well.

Proof: Since both kinds of subalgebras are subspaces of dimension $n^{2}$, the dimension of ${\cal A}_{1}{\cal A}_{2}$ is $n^{4}$ so that ${\cal A}_{1}{\cal A}_{2}=M_{n}\otimes M_{n}$. Therefore the commutants ${\cal A}_{1}^{\prime}$ and ${\cal A}_{2}^{\prime}$ are complementary by Proposition 2. $\square$

Assume that $M_{n}\otimes M_{n}$ is decomposed to pairwise complementary F- and M-subalgebras ${\cal A}_{i}$ ($1\leq i\leq n^{2}+1$). The trace-preserving conditional expectation of $M_{n}\otimes M_{n}$ onto ${\cal A}_{i}$ is denoted by $E_{i}$. The linear trace-preserving mapping acting as

Proof: Theorem 3 allows to use Theorem 2 and the result follows. $\square$

The theorem can be applied in Example 2. Note that decompositions of $M_{2}\otimes M_{2}$ into F- and M-subalgebras are discussed in , while decomposition of $M_{n}\otimes M_{n}$ into F-subalgebras is constructed in if $n=p^{k}$ with a prime number $p>2$.

Acknowledgement. The authors thank to Professor Tsuyoshi Ando for communication and to the project of JSPS and Hungarian Academy of Sciences for support.

References