Quasi-orthogonal subalgebras of matrix algebras

Introduction

We consider pairwise quasi-orthogonal subalgebras ${\cal A}_{1},{\cal A}_{2},\ldots,{\cal A}_{l}$ in $M_{p^{kn}}$ which are isomorphic to $M_{p^{k}}$ for $k\geq 1$, $n\geq 2$ and a prime number $p$ with $p\geq 3$. The aim of this paper is to obtain the maximum $l$. The case $p=2$, $n=2$ and $k=1$ is shown in and the maximum is $4$.

The motivations of this problem are followings. If a total system $M_{n}\otimes M_{m}$ has a statistical operator $\rho$, we can reconstruct the reduced density $\rho_{n}^{(1)}={\rm Tr}_{m}\rho$ in the subsystem $M_{n}$, where ${\rm Tr}_{m}$ is a partial trace onto $M_{n}$. In order to get more information, we change the density $\rho$ by an interaction. For a Hamiltonian $H$, the new state is

after the interaction. The new reduced density is $\rho_{n}^{(2)}={\rm Tr}_{m}W_{1}\rho W_{1}^{*}$. By using other interactions, we have a sequence of reduced states $\rho_{n}^{(1)},\rho_{n}^{(2)},\ldots,\rho_{n}^{(k)}$. We want to determine the minimum $k$ such that this sequence of reduced densities determines $\rho$.

Another reason comes from the relation to the mutually unbiased bases (MUB) problem. Given a orthonormal basis of an $n$-dimensional Hilbert space ${\cal H}$, the linear operators diagonal in this basis form a maximal Abelian subalgebra of $M_{n}\simeq{\cal B}({\cal H})$. Conversely if $|e_{i}\rangle\langle e_{i}|$ are minimal projections in a maximal Abelian subalgebra, then $(|e_{i}\rangle)_{i}$ is a orthonormal basis. Two maximal Abelian subalgebras of $M_{n}$ are quasi-orthogonal if and only if the two corresponding bases $(\xi_{i})$ and $(\zeta_{j})$ are mutually unbiased (see Proposition 2.2), that is,

Mutually unbiased bases are very interesting from many point of view and the maximal number of such bases is not known for arbitrary $n$. We want to study mutually unbiased bases in terms of quasi-orthogonal subalgebras (see ).

In Section 2, we consider quasi-orthogonal subalgebras in $M_{p^{n}}$ which are isomorphic to $M_{p}$. In Section 3, quasi-orthogonal subalgebras in $M_{p^{kn}}$ which are isomorphic to $M_{p^{k}}$ are investigated.

${\cal A}$ is a finite dimensional $C^{*}$-algebra with the usual trace ${\rm Tr}$ and is considered as a Hilbert space under the inner product

Two subalgebras ${\cal A}_{1}$ and ${\cal A}_{2}$ of ${\cal A}$ are called quasi-orthogonal if

The equivalent conditions of this definition are followings :

(a) For any $A_{1}\in{\cal A}_{1}$ and $A_{2}\in{\cal A}_{2}$,

(b) For any $A_{1}\in{\cal A}_{1}$ and $A_{2}\in{\cal A}_{2}$ with ${\rm Tr}(A_{1})={\rm Tr}(A_{2})=0$,

The theory of quasi-orthogonal subalgebras is related to the theory of mutually unbiased bases. This is stated as follows:

From (a), we conclude that $\{\xi_{i}\}$ and $\{\zeta_{j}\}$ are mutually unbiased bases if and only if the algebras generated by $\{P_{i}\}$ and $\{Q_{j}\}$ are quasi-orthogonal. This prove the assertion. $\square$

Now we consider the quasi-orthogonal subalgebras in $M_{p^{2}}=M_{p}\otimes M_{p}$ which are isomorphic to $M_{p}$, where $p$ is a prime number with $p\geq 3$. We will show that we can construct $p^{2}+1$ pairwise quasi-orthogonal subalgebras.

Define the unitary operators $W$ and $S$ in $M_{p}$ by

where $\lambda=e^{2\pi i/p}$. Then these two unitary operators have following properties:

The set $\{S^{i}W^{j}\}_{0\leq i,j\leq p-1}$ is a natural orthogonal basis of $M_{p}$.

From (1), $S^{k_{1}}W^{l_{1}}$ and $S^{k_{2}}W^{l_{2}}$ commute if and only if $k_{1}l_{2}=k_{2}l_{1}$ mod $p$.

We consider the commutativity condition iv in the context of a vector space over $Z_{p}$, where $Z_{p}$ is a finite field with $p$ elements. Let $Z_{p}^{4}=\{(k_{1},l_{1},k_{2},l_{2})\,|\,k_{1},l_{1},k_{2},l_{2}\in Z_{p}\}$ be a vector space over $Z_{p}$ and define a natural homomorphism $\pi$ (up to scalar multiple) from $Z_{p}^{4}$ to $M_{p}\otimes M_{p}$ by

where $u=(k_{1},l_{1},k_{2},l_{2})$ and $u^{\prime}=(k_{1}^{\prime},l_{1}^{\prime},k_{2}^{\prime},l_{2}^{\prime})$. From (1),

Hence $\pi(u)$ and $\pi(u^{\prime})$ commute if and only if their symplectic product equals zero.

If $\pi(u)$ and $\pi(u^{\prime})$ are not commutative for $u=(k_{1},l_{1},k_{2},l_{2}),\,u^{\prime}=(k_{1}^{\prime},l_{1}^{\prime},k_{2}^{\prime},l_{2}^{\prime})\in Z_{p}^{4}$, then the algebra ${\cal A}$ generated by $\pi(u)$ and $\pi(u^{\prime})$ is isomorphic to $M_{p}$.

Proof. From the assumption, $u\circ u^{\prime}\neq 0$. Define a map $\rho$ from $\{S,W^{u\circ u^{\prime}}\}$ to ${\cal A}$ by

From (2) and $SW^{u\circ u^{\prime}}=\lambda^{-u\circ u^{\prime}}W^{u\circ u^{\prime}}S$, the commutativity condition of $\pi(u)$, $\pi(u^{\prime})$ and that of $S$, $W^{u\circ u^{\prime}}$ are same. Therefore $\rho$ can be extended to an isomorphism from $M_{p}$ generated by $S$ and $W^{u\circ u^{\prime}}$ to ${\cal A}$. $\square$

From this lemma, we need to find such $u$ and $u^{\prime}$. Let $D$ be a non-zero integer in $Z^{p}$ with the requirement that $D\neq k^{2}$ mod $p$ for all $k$ in $Z_{p}$, i.e., $D$ is not a quadratic residue of $p$. For any $a_{0},a_{1}\in Z_{p}$, we define a subspace of $Z_{p}^{4}$ by

where scalar multiplication and addition are defined by a natural way. Moreover put

The only vector common to any pair of above subspaces is $(0,0,0,0)$. In particular, the subspaces expect $(0,0,0,0)$ partition $Z_{p}^{4}\backslash\{(0,0,0,0)\}$.

Proof. Since there are $p^{2}+1$ subspaces and each subspace has $p^{2}$ elements, it is enough to prove that the intersection of any two subspaces is $\{(0,0,0,0)\}$. It is easy to see $C_{a_{0},a_{1}}\cap C_{\infty}=\{(0,0,0,0)\}$. Therefore we prove that $C_{a_{0},a_{1}}\cap C_{a_{0}^{\prime},a_{1}^{\prime}}=\{(0,0,0,0)\}$ if $a_{0}\neq a_{0}^{\prime}$ or $a_{1}\neq a_{1}^{\prime}$.

Assume $b_{0}(1,a_{1},0,a_{0})+b_{1}(0,a_{0},-1,a_{1}D)=b_{0}^{\prime}(1,a_{1}^{\prime},0,a_{0}^{\prime})+b_{1}^{\prime}(0,a_{0}^{\prime},-1,a_{1}^{\prime}D)$, then from the first and third components we have $b_{0}=b_{0}^{\prime}$ and $b_{1}=b_{1}^{\prime}$. Similarly, from the second and fourth components we are led to the equations

These equations can be rewritten as a matrix equation

If $b_{0}=b_{1}=0$, then the common element is $(0,0,0,0)$. Therefore we assume $b_{0}\neq 0$ or $b_{1}\neq 0$. Then the above matrix is invertible, indeed

Here we use that $b_{1}^{2}D\neq b_{0}^{2}$ mod $p$ from the assumption of $D$. This implies $a_{0}=a_{0}^{\prime}$ and $a_{1}=a_{1}^{\prime}$ which is a contradiction. $\square$

For any $a\in Z_{p}$, define subspaces by

The only vector common to any pair of above subspaces is $(0,0,0,0)$. Moreover we have

Proof. The proof of the first assertion is same as Lemma 2.4. To show the second assertion, it is enough to prove $D_{a},D_{\infty}\subset\bigcup_{a_{1}\in Z_{p}}C_{0,a_{1}}\cup C_{\infty}$. Indeed, since each subspaces has $p^{2}$ elements and the intersection is trivial, the numbers of elements in both sets are equal.

First consider a element $(0,0,b_{0},b_{1})$ in $D_{\infty}$. If $b_{0}=0$, then $(0,0,0,b_{1})\in C_{\infty}$. If $b_{0}\neq 0$, then

Hence $D_{\infty}\subset\bigcup_{a_{1}\in Z_{p}}C_{0,a_{1}}\cup C_{\infty}$. Next consider the element $b_{0}(1,1,-a,aD)+b_{1}(1,2,-a,2aD)$ in $D_{a}$. If $b_{0}+b_{1}=0$, then

Therefore we obtain $D_{a}\subset\bigcup_{a_{1}\in Z_{p}}C_{0,a_{1}}\cup C_{\infty}$. $\square$

Since $(1,1,-a,aD)\circ(1,2,-a,2aD)=1-a^{2}D\neq 0$ by the assumption of $D$ and $(0,0,1,0)\circ(0,0,0,1)=1$, we obtain

by Lemma 2.3. Consequently we have the next theorem.

There are $p^{2}+1$ pairwise quasi-orthogonal subalgebras in $M_{p^{2}}$ which are isomorphic to $M_{p}$.

Consider a $9$-level quantum system $M_{3}\otimes M_{3}$. We list $10$ pairwise quasi-orthogonal subalgebras in $M_{9}$ which are isomorphic to $M_{3}$:

$M_{p^{n}}$ contains $N_{n}$ pairwise quasi-orthogonal subalgebras which are isomorphic to $M_{p}$.

We define a homomorphism $\hat{\pi}$ (up to scalar multiple) from ${Z}_{p}^{4}$ to ${\cal A}\otimes{\cal C}$ by

From a similar method of the proof of Lemma 2.4, the only vector common to any pair of above subspaces is $(0,0,0,0)$. Moreover since $W$ and $S$ satisfy (1) and ${\cal C}$ is commutative, we have

Therefore the algebra ${\rm span}\{\hat{\pi}(\hat{C}_{a_{0},a_{1}})\}$ generated by $\hat{\pi}(1,0,a_{0},a_{1})$ and $\hat{\pi}(0,1,a_{1}D,a_{0})$ is isomorphic to the algebra generated by $S$ and $W$, that is, $M_{p}$. Furthermore the algebra ${\rm span}\{\hat{\pi}(\hat{C}_{\infty})\}$ generated by $\hat{\pi}(0,0,1,0)$ and $\hat{\pi}(0,0,0,1)$ is $I\otimes{\cal C}$. Consequently ${\cal A}\otimes{\cal C}$ can be decomposed by ${\cal A}\otimes I={\rm span}\{\hat{\pi}(\hat{C}_{0,0})\}$, $I\otimes{\cal C}$ and $p^{2}-1$ pairwise quasi-orthogonal subalgebras which are isomorphic to $M_{p}$. We denote the $p^{2}-1$ pairwise quasi-orthogonal subalgebras by $\{{\cal B}_{{\cal A},{\cal C}}^{k}\}_{k=1}^{p^{2}-1}$.

pairwise quasi-orthogonal subalgebras in $M_{p^{n}}$ which are isomorphic to $M_{p}$. $\square$

In this section, we consider quasi-orthogonal subalgebras in $M_{p^{kn}}$ which are isomorphic to $M_{p^{k}}$. First we consider the case $n=2$.

$GF(p^{k})$ denotes a finite field with $p^{k}$ elements. Up to isomorphisms, $GF(p^{k})$ is unique and is defined using a polynomial

that is irreducible over the field ${Z}_{p}$. Then we can write

for $a=(a^{(1)}(t),a^{(2)}(t),a^{(3)}(t),a^{(4)}(t))$ and $b=(b^{(1)}(t),b^{(2)}(t),b^{(3)}(t),b^{(4)}(t))$ in $GF(p^{k})^{4}$. Similarly, symplectic product on $Z_{p}^{4k}$ is denoted by

for $u=(u_{i}^{(j)})_{1\leq i\leq k,1\leq j\leq 4}$ and $v=(v_{i}^{(j)})_{1\leq i\leq k,1\leq j\leq 4}$ in $Z_{p}^{4k}=(Z_{p}^{k})^{4}$. From , there exist a linear functional $\varphi$ on $GF(p^{k})$ and a linear isomorphism $\pi_{1}$ from $GF(p^{k})^{4}$ to $Z_{p}^{4k}$ such that

where we consider $GF(p^{k})^{4}$ and $Z_{p}^{4k}$ as vector spaces over $Z_{p}$. Define a natural homomorphism (up to scalar multiple) $\pi_{2}$ from $Z_{p}^{4k}$ to $M_{p^{2k}}=\bigotimes_{i=1}^{k}M_{p}\otimes\bigotimes_{i=1}^{k}M_{p}$ by

If $a\circ b\neq 0$ for $a,b\in GF(p^{k})^{4}$, then the algebra ${\cal A}$ generated by

is isomorphic to $M_{p^{k}}$, where $p(t)a$ is defined by

Proof. If $\{p(t)a\,|\,p(t)\in GF(p^{k})\}\cap\{q(t)b\,|\,q(t)\in GF(p^{k})\}\neq\{(0,0,0,0)\}$, there exists $r(t)\in GF(p^{k})$ such that $a=r(t)b$. This shows $a\circ b=r(t)(b\circ b)=0$. Hence we can assume that $\{p(t)a\}\cap\{q(t)b\}=\{(0,0,0,0)\}$.

For $q(t)\in GF(p^{k})$, let $\psi_{q(t)}$ be a functional on $GF(p^{k})$ defined by

where we consider $GF(p^{k})$ as a vector space over $Z_{p}$. By (3), we have

Since $a\circ b\neq 0$, $\psi_{q^{(1)}(t)}=\psi_{q^{(2)}(t)}$ implies $q^{(1)}(t)=q^{(2)}(t)$. Therefore we obtain $\{\psi_{q(t)}\,|\,q(t)\in GF(p^{k})\}=GF(p^{k})^{*}$, where $GF(p^{k})^{*}$ is a dual space of $GF(p^{k})$.

Let $\{p^{(i)}(t)\}_{1\leq i\leq k}$ be a basis of the vector space $GF(p^{k})$. Then there exists a basis $\{q^{(j)}\}_{1\leq j\leq k}\subset GF(p^{k})$ such that

For $1\leq i,j\leq k$, let $S_{i}=I^{\otimes i-1}\otimes S\otimes I^{\otimes k-i}$ and $W_{j}=I^{\otimes j-1}\otimes W\otimes I^{\otimes k-j}$ in $M_{p^{k}}=\bigotimes^{k}M_{p}$. We define a map $\rho$ from $\{S_{i},W_{j}\,|\,1\leq i,j\leq k\}$ to ${\cal A}$ by

Then the commutativity condition of $\{S_{i},W_{j}\}$, that is,

and the commutativity condition of $\{\pi_{2}\pi_{1}(p^{(i)}(t)a),\pi_{2}\pi_{1}(q^{(j)}(t)b)\,|\,1\leq i,j\leq k\}$, that is,

are same. Hence $\rho$ can be extended to an isomorphism from $M_{p^{k}}$ generated by $\{S_{i},W_{j}\}_{1\leq i,j\leq k}$ to ${\cal A}$. $\square$

From this lemma, we need to find such $a$ and $b$. Let $D$ be a non-zero element in $GF(p^{k})$ with the requirement that $D\neq p(t)^{2}$ for all $p(t)$ in $GF(p^{k})$. For any $a(t),b(t)\in GF(p^{k})$, define a subspace of $GF(p^{k})^{4}$ by

The only vector common to any pair of above subspaces is $(0,0,0,0)$. In particular, the subspaces expect $(0,0,0,0)$ partition $GF(p^{k})^{4}\backslash\{(0,0,0,0)\}$.

Proof. This proof is same as Lemma 2.4. $\square$

Since $(1,b(t),0,a(t))\circ(0,a(t),-1,b(t)D)=2a(t)$, if $a(t)\neq 0$ then

For any $a(t)\in GF(p^{k})$, define subspaces by

The only vector common to any pair of above subspaces is $(0,0,0,0)$. Moreover we have

Proof. This proof is same as Lemma 2.5. $\square$

Since $(1,1,-a(t),a(t)D)\circ(1,2,-a(t),2a(t)D)=1-a(t)^{2}D\neq 0$ by the assumption of $D$ and $(0,0,1,0)\circ(0,0,0,1)=1$, we have

by Lemma 3.1. Consequently we have the next theorem.

There are $p^{2k}+1$ pairwise quasi-orthogonal subalgebras in $M_{p^{2k}}$ which are isomorphic to $M_{p^{k}}$.

$M_{p^{kn}}$ contains $N_{(k,n)}$ pairwise quasi-orthogonal subalgebras which are isomorphic to $M_{p^{k}}$.

Define a homomorphism $\hat{\pi}_{2}$ (up to scalar multiple) from $Z_{p}^{4k}$ to ${\cal A}\otimes{\cal C}$ by

for $(u_{i}^{(j)})_{1\leq i\leq k,1\leq j\leq 4}\in Z_{p}^{4k}$. We denote other symplectic products by

for $a=(a^{(1)}(t),a^{(2)}(t),a^{(3)}(t),a^{(4)}(t))$ and $b=(b^{(1)}(t),b^{(2)}(t),b^{(3)}(t),b^{(4)}(t))$ in $GF(p^{k})^{4}$, and

for $u=(u_{i}^{(j)})_{1\leq i\leq k,1\leq j\leq 4}$ and $v=(v_{i}^{(j)})_{1\leq i\leq k,1\leq j\leq 4}$ in $Z_{p}^{4k}$. Since ${\cal C}$ is commutative, we have

Moreover from , there exists a homomorphism $\hat{\pi}_{1}$ from $GF(p^{k})^{4}$ to $Z_{p}^{4k}$ such that

Here for elements $a=(a^{(1)}(t),a^{(2)}(t),0,0)$ and $b=(0,0,b^{(3)}(t),b^{(4)}(t))$ in $GF(p^{k})^{4}$, we can assume

for some $(u_{i}^{(1)}),(u_{j}^{(2)}),(v_{i}^{(3)}),(v_{j}^{(4)})\in Z_{p}^{k}$. Then we can prove that the algebra generated by $\hat{\pi}_{2}\hat{\pi}_{1}(a)$ and $\hat{\pi}_{2}\hat{\pi}_{1}(b)$ is isomorphic to $M_{p^{k}}$, if $a\hat{\circ}b\neq 0$, by using a similar proof of Lemma 3.1.

From the similar method of the proof of Lemma 2.4, the only vector common to any pair of above subspaces is $(0,0,0,0)$. Since $(1,0,a(t),b(t))\circ(0,1,b(t)D,a(t))=1$, we obtain

Consequently ${\cal A}\otimes{\cal C}$ can be decomposed by ${\cal A}\otimes I={\rm span}\{\hat{\pi}_{2}\hat{\pi}_{1}(\hat{C}_{0,0})\}$, $I\otimes{\cal C}={\rm span}\{\hat{\pi}_{2}\hat{\pi}_{1}(\hat{C}_{\infty})\}$ and $p^{2k}-1$ pairwise quasi-orthogonal subalgebras which are isomorphic to $M_{p^{k}}$. We denote the $p^{2k}-1$ pairwise quasi-orthogonal subalgebras by $\{{\cal B}_{{\cal A},{\cal C}}^{k}\}_{k=1}^{p^{2k}-1}$.

for $i_{1}\neq i_{2}$ or $j_{1}\neq j_{2}$, ${\cal B}_{{\cal A}_{i_{1}},{\cal C}_{j_{1}}}^{k_{1}}$ and ${\cal B}_{{\cal A}_{i_{2}},{\cal C}_{j_{2}}}^{k_{2}}$ are quasi-orthogonal if $i_{1}\neq i_{2}$ or $j_{1}\neq j_{2}$ or $k_{1}\neq k_{2}$. Furthermore, ${\cal B}_{{\cal A}_{i_{1}},{\cal C}_{j_{1}}}^{k_{1}}$ is quasi-orthogonal with ${\cal A}_{i_{2}}\otimes I_{M_{p^{2k}}}$ and $I_{M_{p^{k(n-2)}}}\otimes{\cal C}_{j_{2}}$ for any $i_{2}$ and $j_{2}$. Therefore we can decompose $M_{p^{kn}}$ by $M_{p^{k(n-2)}}\otimes I_{M_{p^{2k}}}={\rm span}\{{\cal A}_{i}\otimes I\}$, $I_{M_{p^{k(n-2)}}}\otimes M_{p^{2k}}={\rm span}\{I\otimes{\cal C}_{j}\}$ and $\{{\cal B}_{{\cal A}_{i},{\cal C}_{j}}^{k}\}_{i,j,k}$. In consequence, we get

pairwise quasi-orthogonal subalgebras in $M_{p^{kn}}$ which are isomorphic to $M_{p^{k}}$. $\square$

References