On the Impossibility to Extend Triples of Mutually Unbiased Product Bases in Dimension Six

Introduction

Much is known about MU bases of quantum systems if the number $d$ of discrete levels equals a prime or prime power, i.e. $d=p^{n}$ (cf. ). In these cases, complete sets of $(p^{n}+1)$ MU bases can be constructed while their (non-) existence for “composite” dimensions such as $d=6,10,12\ldots$ remains unknown.

Considerable efforts have been devoted to dimension six, the smallest dimension where a complete set is elusive. Triples of MU bases have been constructed in a number of ways but only piecemeal progress has been made towards a proof of the conjecture that only three MU bases exist for $d=6$ . For example, increasingly strong numerical evidence supports this view, and the use of computer-algebraic methods allows one to rule out certain MU bases from being part of a complete set.

Only a few analytic results are known for sets of MU bases in composite dimensions such as $d=6$ or $d=10$. Let us briefly summarize them: (i) complete sets of MU bases are equivalent to orthogonal decompositions of Lie algebras ; (ii) in specific composite dimensions, more than $(q^{a}+1)$ MU bases can be constructed using Latin squares – here $q^{a}$ is the smallest factor in the prime decomposition of $d$; (iii) given a “nice unitary error basis” in dimension six, none of its partitions gives rise to more than three MU bases ; (iv) a complete set of MU bases contains a fixed amount of entanglement which implies that, in dimension six, any such set will contain at most three MU product bases ; (v) an exhaustive list of inequivalent pairs and triples of MU product bases has been established in dimension six ; (vi) no complete set of MU bases in dimension six contains both the standard basis and the Fourier basis .

This paper is set out as follows. We start Sec. 2 by recalling the list of all MU product triples in dimension six, followed immediately by a proof of the main theorem. In Sec. 3 we marginally improve the theorem by introducing MU product constellations and discuss the limitations of this approach. We summarise and discuss our results in Sec. 4.

MU product triples in d=6𝑑6d=6

The starting point of our derivation is the fact that, in dimension six, no more than two triples of MU product bases exist which are inequivalent under specific unitary or anti-unitary transformations, defined by the requirements to respect the product structure of the states and to leave invariant the modulus of their inner products, as explained in . The transformations include a unitary map acting on all bases simultaneously, the multiplication of any state by an arbitrary phase factor, the permutation of states within a basis, and the complex conjugation of all bases; in addition one can re-order the bases arbitrarily.

The following lemma lists all triples of MU product bases, up to equivalence defined by the transformations just described.

For later reference we now write out the matrix representations of the MU bases ${\cal B}_{y}$ and ${\cal B}_{w}$ in the computational basis,

The columns of each matrix are orthogonal, and since the modulus of each entry equals $1/\sqrt{3}$, the matrices $H_{y}$ and $H_{w}$ are complex $(3\times 3)$ Hadamard matrices. Thus, the complete set of MU bases in dimension $d=3$ can be written as the set of $(3\times 3)$ matrices $\{I,F_{3},H_{y},H_{w}\}$, where $I$ is the identity and $F_{3}$ is the Fourier matrix,

2 Excluding MU product triples from a complete set of MU bases

We now present the main result of this paper, which is an analytic proof of the following theorem.

No triple of MU product bases in dimension six can be extended by a single MU vector.

In other words, any triple of MU product bases acts like a cul-de-sac when attempting to construct a complete set of MU bases in dimension six. The first proof of this result in depends on an exact computer-algebraic search: Lemma 1 guarantees that any triple of MU product bases contains the pair $\{|j_{z},J_{z}\rangle\}$ and $\{|j_{x},J_{x}\rangle\}$ (or is equivalent to such a pair) and, according to , only 48 vectors exist which are MU to this pair . However, there is no subset of these 48 vectors which, when combined with $\{|j_{z},J_{z}\rangle\}$ and $\{|j_{x},J_{x}\rangle\}$, would give rise to more than a triple of MU bases. Consequently, no state can be MU to either $\mathcal{T}_{0}$ or $\mathcal{T}_{1}$ which implies Theorem 1.

We now proceed to prove Theorem 1 analytically. Due to Lemma 1, it is sufficient to show that no vector is MU to either of the triples $\mathcal{T}_{0}$ or $\mathcal{T}_{1}$; any other MU product triple can be transformed to one of these two triples using the equivalence transformations described above.

not all of which are independent. Similarly, the state $|\psi\rangle$ is mutually unbiased to the product triple $\mathcal{T}_{1}$ if and only if

It will take us three steps to show that each of these two sets of equations is contradictory. In other words, there is no state $|\psi\rangle$ satisfying either the constraints (4) or (5).

Then we will show (Step 2) that the states $|D\rangle$ and $|D^{\perp}\rangle$ are given by two states either of the basis ${\cal B}_{y}$ or of ${\cal B}_{w}$, displayed in Eqs. (2). Calling these states $|H\rangle$ and $|H^{\perp}\rangle$, a total of twelve candidates remains, namely

However, any state $|\psi\rangle$ of the form (7) will turn out to be incompatible with some MU conditions not used so far (Step 3).

Step 1: Fix the values of $j$ and $a$ in Eqs. (4). Summing over $J$ leads to six equations

which are sufficient to determine the components of the Bloch vector $\bf n$ of $\hat{\rho}_{A}=({\hat{I}}_{A}+{\bf n}\cdot\hat{\sigma})/2$. Since the spin components are given by $\hat{\sigma}_{a}=|0_{a}\rangle\langle 0_{a}|-|1_{a}\rangle\langle 1_{a}|$, one finds that

which means that the smaller subsystem must reside in the maximally mixed state,

Summing Eqs. (5) over $J$, with $j$ and $a$ fixed, results in the same six equations $\langle j_{a}|\hat{\rho}_{A}|j_{a}\rangle=1/2$, hence Eq. (10) holds in this case as well.

This result agrees with a known result: if a complete set of seven MU bases in dimension six contains three MU product bases then all states of the remaining four MU bases are maximally entangled .

Step 2: Now consider the reduced density matrix for the larger subsystem (with label $B$),

which has eigenvalues $(1/2,1/2,0)$, in agreement with those of $\hat{\rho}_{A}$ in (10), except for a padded zero. The requirement that the state $|\psi\rangle$ be MU to the states \bigl{\{}|j_{z},J_{z}\rangle\bigr{\}} and \bigl{\{}|j_{x},J_{x}\rangle\bigr{\}}, which appear in both triples, imposes restrictions on the states $|D\rangle$ and $|D^{\perp}\rangle$. Summing the conditions in (4) and $(\ref{eq: c6 conditions on psi 2})$ over all values of $j$ while keeping $J$ fixed, one obtains six further constraints now on the density matrix $\hat{\rho}_{B}$,

where $J=0,1,2$ and $a=x,z$, similar in spirit to Eqs. (8). However, these expectation values are not sufficient to reconstruct the reduced density matrix $\hat{\rho}_{B}$. Nevertheless, one can draw the important conclusion that

and calculate its expectation value in the state $|J_{a}\rangle$.

with two real parameters $\vartheta\in[0,\pi]$, and $\phi\in[0,2\pi)$. Projecting the candidate $|D\rangle$ given in (17) onto the states $|0_{z},J_{z}\rangle,J=0,1,2$, produces three constraints on the free parameters:

Using \bigl{|}\langle J_{z}|H\rangle\bigr{|}^{2}=\bigl{|}\langle J_{z}|{H}^{\perp}\rangle\bigr{|}^{2}=1/3, this equation leads to the conditions

where the relation $\langle H|J_{z}\rangle\langle J_{z}|{H}^{\perp}\rangle\equiv(1/3)\,e^{i\mu_{J}}$ defines the angles $\mu_{J}\in[0,2\pi),J=0,1,2$. However, the states $|H\rangle$ and $|{H}^{\perp}\rangle$ are orthogonal, which implies that

with some constant $\mu\in[0,2\pi)$. Therefore, Eqs. (19) require either $\sin\vartheta\equiv 0$ or

Since the zeros of the cosine function occur at intervals of length $\pi$ (not $2\pi/3$), we conclude that $\vartheta/2\in\left\{0,\pi/2\right\}$ are the only values allowed in (17). An entirely analogous argument leads to the same conclusion if we consider the state $|D^{\perp}\rangle$ defined in (17) instead of $|D\rangle$.

Thus, we have shown that there are only two cases in which the requirements of (4) or (5) are satisfied: we must have either

In both cases, the phase factors may be absorbed into the definition of the state $|{H}^{\perp}\rangle$, which leaves us with two possible candidates being MU to the three product bases in $\mathcal{T}_{0}$ or $\mathcal{T}_{1}$, namely

and the state obtained from swapping $|H\rangle$ with $|{H}^{\perp}\rangle$. Consequently, the requirement of the state $|D^{\perp\!\!\!\perp}\rangle$ to be a member of ${\cal B}_{y}$ or ${\cal B}_{w}$ implies that the states $|D\rangle$ and $|D^{\perp}\rangle$ must coincide with the two other members of the same basis. Overall, we have indeed reduced the possible states mutually unbiased to $\mathcal{T}_{0}$ or $\mathcal{T}_{1}$ to twelve entangled states listed in Eq. (7).

Step 3: Finally, we show that states $|\psi\rangle$ of the form (25) are not MU to the states $|1_{x},J_{x}\rangle,J=0,1,2,$ which are present in both product triples, $\mathcal{T}_{0}$ and $\mathcal{T}_{1}$. The mechanics to produce this contradiction is similar to the one given at the end of Step 2.

To begin, let us consider the state $|\psi\rangle$ in (25): the conditions

Upon writing $\langle H|J_{x}\rangle\langle J_{x}|{H}^{\perp}\rangle\equiv(1/3)\,e^{i\nu_{J}}$, one obtains

where we have used the fact that the orthogonality of the states $|H\rangle$ and $|{H}^{\perp}\rangle$ restricts the values of the phases $\nu_{J}$ in analogy to Eqs. (21). However, the three equations in (28) cannot hold simultaneously, and the state $|\psi\rangle$ in (25) is found not to be MU to the given three product bases. The same contradiction occurs for the other eleven states listed in (7) which completes the proof that there is not a single state mutually unbiased to the triple ${\cal T}_{0}$ or ${\cal T}_{1}$.

An unextendible MU product constellation

A MU constellation is a set of states that contains both orthogonal and mutually unbiased states . MU constellations result, for example, upon removing states from a complete set of MU bases. A constellation which contains only product states will be called a MU product constellation.

The product constellation $\{5,5,4\}^{\otimes}_{6}$ cannot be part of a complete set of seven MU bases.

This result is an immediate consequence of the following lemma, the proof of which will be the main part of this section.

The product constellation $\{5,5,4\}^{\otimes}_{6}$ extends to a triple of MU bases only by adding product states.

If the product constellation $\{5,5,4\}^{\otimes}_{6}$ was part of a complete set of seven MU bases, Lemma 2 would imply that the complete set must contain a triple of MU product bases, contradicting Theorem 1.

To prove Lemma 2, we need the complete list of pairs of MU product bases obtained in :

with $j=0,1$ and $J=0,1,2$. The unitary operator $\hat{R}_{\xi,\eta}$ is defined as $\hat{R}_{\xi,\eta}=|0_{z}\rangle\langle 0_{z}|+e^{i\xi}|1_{z}\rangle\langle 1_{z}|+e^{i\eta}|2_{z}\rangle\langle 2_{z}|\,,$ for $\eta,\xi\in[0,2\pi)$, and $\hat{S}_{\zeta,\chi}$ is defined analogously with respect to the $x$-basis; the unitary operators $\hat{r}_{\sigma}$ and $\hat{r}_{\tau}$ act on the basis $\{|j_{x}\rangle\}\equiv\{|\pm\rangle\}$ according to $\hat{r}_{\sigma}|j_{x}\rangle=(|0_{z}\rangle\pm e^{i\sigma}|1_{z}\rangle)/\sqrt{2}$ for $\sigma\in(0,\pi)$, etc.

The four product states in $\mathcal{S}$ must be MU to one of the pairs listed in Lemma 3. However, we can exclude the pairs $\mathcal{P}_{2}$ and $\mathcal{P}_{3}$ since no product state can be MU to either pair, as follows from a result also derived in :

The states $|1_{y},B^{\perp}\rangle$ and $|1_{y},B^{{\perp\!\!\!\perp}}\rangle$ are orthogonal to the quadruple (30), as are their linear combinations,

with $|\alpha|^{2}+|\beta|^{2}=1$. Hence, adding any two orthogonal states from this family to the set $\mathcal{S}_{1}$ in (30) produces a MU product basis.

Any orthonormal state extending the set $\mathcal{S}_{2}$ in (31) can be written as

which is entangled unless $|A^{\perp\!\!\!\perp}\rangle=|B^{\perp\!\!\!\perp}\rangle$ or one of the constants $\alpha$ and $\beta$ is zero. We now show that the state $|\psi_{2}\rangle$ cannot be entangled if it is to satisfy the MU conditions

Write $|A^{\perp\!\!\!\perp}\rangle=(\omega_{0}|0_{z}\rangle+\omega_{1}|1_{z}\rangle+\omega_{2}|2_{z}\rangle)/\sqrt{3}$ and $|B^{\perp\!\!\!\perp}\rangle=(\omega_{0}^{\prime}|0_{z}\rangle+\omega_{1}^{\prime}|1_{z}\rangle+\omega_{2}^{\prime}|2_{z}\rangle)/\sqrt{3}$, where $\omega_{0}=\omega_{0}^{\prime}=1$ and each of the four coefficients $\omega_{1},\ldots,\omega_{2}^{\prime}$ is a third root of unity such that the states $|A^{\perp\!\!\!\perp}\rangle$ and $|B^{\perp\!\!\!\perp}\rangle$ coincide with any two different states of the bases $\mathcal{B}_{y}$ and $\mathcal{B}_{w}$. Then, the MU conditions in (34) turn into

However, these constraints on the phase factors cannot be satisfied by any allowed choice of the pair of states $|A^{\perp\!\!\!\perp}\rangle$ and $|B^{\perp\!\!\!\perp}\rangle$ with $|A^{\perp\!\!\!\perp}\rangle\neq|B^{\perp\!\!\!\perp}\rangle$. Thus, either $\alpha$ or $\beta$ must equal zero, and we conclude that $|\psi_{2}\rangle$ is a product state. This completes the proof of Lemma 2.

Concluding Remarks

The main result of this paper is an analytical proof that no vector is MU to any triple of MU product bases, i.e. Theorem 1. Our approach exploits the structure of MU product bases in a novel fashion, and it is entirely independent of any computer-aided results. Thus, we consider it to be a worthy addition to the few existing analytic results on MU bases in dimension six.

Results stronger than Theorem 1 are known which exclude a wider class of MU bases from complete sets; however, the numerical searches for MU bases with rigorous error bounds and the proof that the Heisenberg-Weyl pair of bases $\{|j_{z},J_{z}\rangle\}$ and $\{|j_{x},J_{x}\rangle\}$ is MU to at most one further basis rely on a computer in one way or another. Interestingly, the latter result immediately provides an alternative (computer-aided) proof of Theorem 1: firstly, the pair $\{|j_{z},J_{z}\rangle\}$ and $\{|j_{x},J_{x}\rangle\}$ is present in both product triples $\mathcal{T}_{0}$ and $\mathcal{T}_{1}$ and, secondly, any product triple is known to be equivalent to one of these two triples.

A recent analytic result employs combinatorial and Fourier analytic arguments to prove that no complete set of MU bases in dimension six will contain both the standard and Fourier basis. As a consequence, no complete set of MU bases will contain triples of MU product bases. Whilst this is also a consequence of Theorem 1, the result presented here is different to the result in since we have shown the impossibility to extend a product triple by a single MU vector. The result in does not seem to forbid such an extension.

In order to strengthen Theorem 1 we also considered a MU product constellation that is slightly smaller than MU product triples. The resulting Theorem 2 states that the product constellation $\{5,5,4\}^{\otimes}_{6}$ cannot be part of a complete set of seven MU bases. Its derivation relies on an enumeration of all pairs of MU product bases in dimension six which was given in .

To make any stronger statements regarding product constellations seems to be surprisingly difficult. For example, we are not able to show whether a complete set of seven MU bases may (or may not) contain the MU constellation $\{5,4,4\}^{\otimes}_{6}$, consisting of one MU product basis and two sets of four orthogonal MU product states. The main difficulty is that the proof of Theorem 2 relies on Lemma 4 which does not apply to the case of the constellation $\{5,4,4\}^{\otimes}_{6}$.

It is worth recalling that a hypothetical complete set of MU bases in dimension six will contain at most one product basis . While this is a stronger statement than the one obtained here, the proof depends on a numerical search with rigorous error bounds . We hope that the analytic proof given here will be extended to cover this stronger result and, ultimately, will help put to rest the existence question of complete sets of MU bases in composite dimensions – or at least in dimension six, for a start.

We thank Maurice Kibler for carefully reading a draft of this paper. This work has been supported by EPSRC.

References