All Mutually Unbiased Product Bases in Dimension Six

Introduction

Complete sets of MU bases also exist for quantum systems with dimension $d=p^{n}$, where $n$ is a positive integer. However, for “composite” dimensions such as $d\equiv d_{1}d_{2}\in\{6,10,12,\ldots\}$ complete sets of MU bases seem to be absent. In spite of considerable numerical searches , computer-algebraic efforts , and numerical calculations with rigorous error bounds, only three MU bases have been found in dimension six, four less than the maximally allowed number . Thus, the six-dimensional state space of a qubit-qutrit system appears to differ structurally from the state space of a pair of qubits ($d=4)$ or a pair of qutrits $(d=9)$.

One of the few known results in dimension $d=6$ is the impossibility to extend, by more that one further MU basis, the pair of MU bases consisting of the standard basis and its dual, the Fourier basis . Thus, triples of MU bases are the largest sets to be found in this way. Another, more recent result states that the Fourier family of Hadamard matrices together with the identity cannot be extended to a MU quadruple. These initial pairs, after non-local equivalence transformations, consist of product states only, a fact which has received little attention.

Upon reflection, it seems worthwhile to systematically study MU bases in composite dimensions which contain only product states. In the present paper we carry out a comprehensive study of MU product bases in dimensions six, complementing studies devoted to the entanglement structure of complete sets of MU bases .

Readers mainly interested in the results relevant to dimension six are advised to immediately proceed to Sec. 6 after having familiarized themselves with the concept of mutually unbiased product bases presented in Sec. 2.

MU product bases

which are the conditions for bases to be MU in a space of dimension $pq$.

One can construct MU product bases of the type given in Eq. $(\ref{eq:product bases})$ using Heisenberg-Weyl (HW) operators. In dimension $p$, with $p$ prime, the HW cyclic shift (modulo $p$) and phase operators $X_{p}$ and $Z_{p}$, respectively, are defined as

a local unitary transformation $\hat{u}\otimes\hat{U}$ effecting

which leaves invariant the value of all scalar products;

the multiplication of all states within a basis by possibly different phase factors such that

these transformations exploit the fact that the overall phase of a quantum state has no physical significance and automatically drops out from the conditions defining MU bases. It is worth noting that a single phase factor $e^{i\phi}$ can dephase both states of a product: let $\phi\equiv\phi^{\prime}+\phi^{\prime\prime}$ to find $e^{i\phi}|n_{\rho},N_{\rho}\rangle=(e^{i\phi^{\prime}}|n_{\rho}\rangle)\otimes(e^{i\phi^{\prime\prime}}|N_{\rho}\rangle)$;

permutations of the product states within each basis; as an example, consider the permutation of states $|n_{\rho},N_{\rho}\rangle$ and $|n^{\prime}_{\rho},N^{\prime}_{\rho}\rangle$ in the $\rho^{\text{th}}$ basis

which amounts to relabelling the elements within each basis;

swaps all scalar products resulting from the first factors without changing their numerical values;

pairwise exchanges of two bases, which amounts to relabelling the bases.

2 MU bases in dimensions two and three

where $\{|\pm\rangle\}\equiv\{|j_{x}\rangle\}$ is the $x$-eigenbasis, and the operator $\hat{r}_{\lambda},\lambda\in[0,\pi)$, represents a rotation by an angle $\lambda$ about the $z$-axis. Since any such rotation leaves the standard basis $\{|j_{z}\rangle\}$ unchanged, the second MU basis can be transformed into $\{|j_{x}\rangle\}$. The matrix representation of the resulting pair of MU bases reads

In dimension three, one of two given MU bases can always be mapped to the standard basis $\{|J_{z}\rangle,J=0,1,2\}$, so that the second basis consists of states of the form

exploiting the fact that the overall phase of a quantum state has no physical meaning. One can construct three states of this form which are pairwise orthogonal: writing

the condition $\langle A|A^{\perp}\rangle=0$ implies $\gamma+\delta=-1$. A geometric argument in the complex plane implies either $\gamma=\omega$ and $\delta=\omega^{2}$, or $\gamma=\omega^{2}$ and $\delta=\omega$, where $\omega=e^{2\pi i/3}$ is a third root of unity. We denote the resulting basis by

which are also MU with respect to each other. The matrices $H_{x},H_{y}$ and $H_{w}$ are complex $(3\times 3)$ Hadamard matrices, i.e. they are unitary and the moduli of all their entries are equal to $1/\sqrt{3}$.

Two triples of MU bases now result from adding either $\{|J_{y}\rangle\}$ or $\{|J_{w}\rangle\}$ to the pair $\{|J_{z}\rangle;|J_{x}\rangle\}$. These triples are equivalent to each other as follows from taking the complex conjugate (defined in the $z$-basis) of the triple $\{|J_{z}\rangle;|J_{x}\rangle;|J_{y}\rangle\}$: the complex conjugation only affects the ordering of states within $\{|J_{x}\rangle\}$ while $\{|J_{y}\rangle\}$ turns into $\{|J_{w}\rangle\}$. Thus we conclude that the triples are indeed equivalent which we express formally by writing

Constructing product bases in dimensions four and six

with $|\psi_{2}\rangle=|a^{\perp}\rangle$ being the unique state orthogonal to $|a\rangle$. Now we need to consider two separate cases: we can have either $|\psi_{3}\rangle=|a\rangle$ (or, equivalently, $|\psi_{3}\rangle=|a^{\perp}\rangle$) or $|\psi_{3}\rangle=|b\rangle$ such that $0<|\langle a|b\rangle|<1$, meaning that the state $|b\rangle$ is neither a multiple of the state $|a\rangle$ nor orthogonal to it; we call such a vector $|b\rangle$ skew to $|a\rangle$.

By a simple argument using the restrictions imposed by the orthogonality conditions, one finds that three different bases result:

The basis $\mathcal{B}_{0}$ is a direct product basis while the bases $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are not. After performing suitable LETs, we can thus summarise the complete list of product bases in dimension four as follows.

The symmetry becomes particularly obvious if we represent the bases of Lemma 1 by quantum circuits. The idea is to visualise the operation needed to map the states of the standard product basis $\mathcal{I}_{0}$ into the desired product basis by means of a quantum gate. This is always possible since any two orthonormal bases are connected by a unitary operation. Obviously, the trivial gate, described by the identity $\hat{I},$ maps the four vectors of the standard product basis to itself. Fig. (1) shows that (non-local) controlled-$\hat{u}$ and controlled-$\hat{v}$ gates are required to output the bases $\mathcal{I}_{1}$ and $\mathcal{I}_{2}$, respectively. As expected, the two circuits are identical upon swapping the qubits.

2 All product bases in d=6𝑑6d=6

Without any restrictions on the five unitary operators $\hat{u},\ldots,\hat{V}$ some product bases would occur more than once in this list. For example, if $\hat{U}\equiv\hat{I}$, the basis $\mathcal{I}_{1}$ turns into $\mathcal{I}_{0}$; similarly, the bases associated with $\hat{U}$ and $\hat{U}^{*}$ are identical. We could remove such multiple occurrences by appropriately restricting the unitary operators but it is rather cumbersome to do so and not particularly informative.

Adding MU product states to sets of orthogonal product vectors

In this section we derive a theorem which will play a crucial role in the construction of all pairs and triples of MU product bases in dimension four and six. This theorem is inspired by a constraint on two direct product bases to be MU, obtained in :

Two [direct] product bases $\{|j_{a},J_{a}\rangle\}$ and $\{|k_{b},K_{b}\rangle\}$ in dimension $d=pq$ are MU if and only if $|j_{a}\rangle$ is MU to $|k_{b}\rangle$ in dimension $p$ and $|J_{a}\rangle$ is MU to $|K_{b}\rangle$ in dimension $q$.

This result covers the Lemma given at the beginning of this section. To see this, group the basis $\{|j_{a},J_{a}\rangle\}$ into $q$ sets of $p$ orthonormal vectors $\{|j_{a},1_{a}\rangle\}$, $\{|j_{a},2_{a}\rangle\}\ldots$ $\{|j_{a},q_{a}\rangle\}$; then, by Lemma 3, any product state $|\phi,\Phi\rangle$ is mutually unbiased to each set of vectors if and only if the state $|\phi\rangle$ is MU to all states $|j_{a}\rangle$, and the state $|\Phi\rangle$ is MU to all states $|J_{a}\rangle$. By replacing the state $|\phi,\Phi\rangle$ with a vector from the basis $\{|k_{a},K_{a}\rangle\}$ and repeating the argument for all states in this basis, one arrives at the Lemma for direct product bases.

The following generalisation uses the fact that we know all direct and indirect product bases in dimensions four and six.

We prove this statement by considering the cases $d=4$ and $d=6$ separately:

$\bullet\,d=4$: All product bases in dimension four are given by the bases $\mathcal{I}_{0},\mathcal{I}_{1}$ and $\mathcal{I}_{2}$, collected in Lemma 1. Each of these bases can be divided into groups of states of the form $\{|\psi_{j},\Psi\rangle,j=1,2\}$, or $\{|\psi,\Psi_{j}\rangle,j=1,2\}$. Thus, Theorem 1 follows immediately from Lemma 3.

$\bullet\,d=6$: It is sufficient to consider the four families of bases given in Lemma 2. Each of the bases $\mathcal{I}_{0}$, $\mathcal{I}_{1}$ and $\mathcal{I}_{3}$ can be split into sets of the form required to apply Lemma 3; thus, Theorem 1 holds for these bases. To complete the proof, we need to consider the basis $\mathcal{I}_{2}$ which has no such decomposition. To begin, suppose that the basis $\mathcal{I}_{2}$ is MU to the state $|\phi,\Phi\rangle$. According to Lemma 3 this state is MU to the pair $\{|j_{z},0_{z}\rangle\}$ if both $|\langle\phi|0_{z}\rangle|^{2}=|\langle\phi|1_{z}\rangle|^{2}=1/2$ and $|\langle\Phi|0_{z}\rangle|^{2}=1/3$ hold. The state $|\phi,\Phi\rangle$ also needs to satisfy

Using $\sum_{J}|\langle\Phi|J_{z}\rangle|^{2}=1$, i.e. the completeness relation of the basis $\{|J_{z}\rangle\}$, and $|\langle\Phi|0_{z}\rangle|^{2}=1/3$, we find that $|\langle\Phi|1_{z}\rangle|^{2}+|\langle\Phi|2_{z}\rangle|^{2}=2/3$. Substituting this identity into (30) leaves us with $|\langle\phi|\hat{u}0_{z}\rangle|^{2}=1/2$, so that $|\langle\Phi|1_{z}\rangle|^{2}=|\langle\Phi|2_{z}\rangle|^{2}=1/3$ as well. A similar argument applied to the pair $\{|\hat{u}1_{z},\hat{V1_{z}}\rangle,|\hat{u}1_{z},\hat{V}2_{z}\rangle\}$ shows that indeed $|\langle\phi|\hat{u}1_{z}\rangle|^{2}=1/2$ and $|\langle\Phi|\hat{V}1_{z}\rangle|^{2}=|\langle\Phi|\hat{V}2_{z}\rangle|^{2}=1/3$, which confirms that the state $|\phi,\Phi\rangle$ is of the desired form. The converse direction of the statement is straightforward.

We conjecture Theorem 1 to hold for all product dimensions $d\equiv pq$, i.e. $d=4,6,9,10,\ldots$ However, a proof similar to the one for $d=4,6,$ would rely on the structure of all product bases in composite dimensions $d>6$ – which is not known to us.

MU product bases in dimension four

where $j,k=0,1$, and the unitary operator $\hat{r}_{\nu}$ rotates the basis $\{|j_{x}\rangle\}\equiv\{|k_{x}\rangle\}\equiv\{|\pm\rangle\}$ into the $xy$-plane according to $\hat{r}_{\nu}|\pm\rangle=(|0_{z}\rangle\pm e^{i\nu}|1_{z}\rangle)/\sqrt{2}$ for $\nu\in(0,\pi)$; the operator $\hat{s}_{\mu}$ generates rotations about the $x$-axis, i.e. $\hat{s}_{\mu}|k_{z}\rangle=(|0_{x}\rangle+(-1)^{k_{z}}e^{i\mu}|1_{x}\rangle)/\sqrt{2}$ for $\mu\in[0,\pi)$.

The pair $\mathcal{P}_{0}^{(4)}$ is the Heisenberg-Weyl pair consisting of two direct product bases. The pair of MU bases $\mathcal{P}_{1}^{(4)}$ is a two-parameter family and may contain direct and indirect product bases. Notice that the operator $\hat{s}_{\mu}$ can act as the identity since the first basis of $\mathcal{P}_{1}^{(4)}$ may be the standard basis $\{|j_{z},k_{z}\rangle\}$.

The pair $\mathcal{P}_{1}^{(4)}$ turns out to be equivalent under non-local transformations to the Fourier basis as follows from mapping the first basis to the standard basis $\{|j_{z},k_{z}\rangle\}$. Thus, we have obtained all known pairs of MU bases in dimension four (cf. Sec. 3 of ) in spite of limiting ourselves initially to MU product bases only.

2 All triples of MU product bases

Now we are in a position to derive all triples of MU product bases in dimension $d=4$: we need to determine which of the pairs of MU product bases given in Proposition 1 can be extended by a third MU product basis.

It is easy to see that the MU pair $\mathcal{P}_{0}^{(4)}\equiv\{|j_{z},k_{z}\rangle;\,|j_{x},k_{x}\rangle\}$ can be extended by adjoining a third direct product basis, namely $|j_{y},k_{y}\rangle$, resulting in the standard Heisenberg-Weyl triple. This is the only possibility, as follows immediately from Theorem 1: a product state $|\phi,\Phi\rangle$ is MU to both $\{|j_{z},k_{z}\rangle\}$ and {$|j_{x},k_{x}\rangle\}$ only if $|\phi\rangle$ is MU both to $\{|j_{z}\rangle\}$ and {$|j_{x}\rangle\}$, and if $|\Phi\rangle$ is MU both to $\{|k_{z}\rangle\}$ and {$|k_{x}\rangle\}$.

Using Theorem 1 again, the non-existence of even a single product state MU to the triple $\mathcal{T}_{0}^{(4)}$ follows immediately—all states MU to the triple must be entangled.

MU product bases in dimension six

We will now construct all pairs of MU product bases in dimension six following the method used in dimension four (cf. Sec. 3.1). To obtain a MU pair we take each basis listed in Lemma 2 and go through all possibilities of adding one of the product bases $\mathcal{B}_{0}$ to $\mathcal{B}_{3}$ (cf. Eqs. (47,46,50,51) of Appendix A).

When constructing pairs of MU product bases, it is not necessary to include the basis $\mathcal{I}_{2}$ in Lemma 2. We will show now that the operator $\hat{V}$ must either act as the identity on the pair of states $\{|1_{z}\rangle,|2_{z}\rangle\}$ or swap them, i.e. only $\alpha=0$ or $\beta=0$ are allowed in the expression $\hat{V}|1_{z}\rangle=\alpha|1_{z}\rangle+\beta|2_{z}\rangle$. However, in both cases the simplified product basis $\mathbf{\mathcal{I}}_{2}$ turns into a special case of $\mathbf{\mathcal{I}}_{3},$ given in (28).

Now using the explicit expressions of the states $|A\rangle$ and $|A^{\perp}\rangle$ given in Eq. (16) and the identity $|\alpha|^{2}+|\beta|^{2}=1$, the first equality leads to

which implies that either $\alpha\equiv 0$ or $\beta\equiv 0$. Thus, for the construction of pairs it is sufficient to use the restricted basis

instead of $\mathbf{\mathcal{I}}_{2}$ given in Lemma 2. All bases of this form, however, are contained in $\mathcal{I}_{3}$ if one chooses $\hat{v}=\hat{w}\equiv\hat{u}$ in (28). This simplification also holds for the basis $\mathcal{B}_{2}$ when occurring in a pair of product bases.

The actual derivation of all MU product bases in dimension six is lengthy but straightforward. The calculations have been relegated to Appendix B except for the pairing of the basis $\mathcal{I}_{1}$ with $\mathcal{B}_{1}$, which gives rise to the pair $\mathcal{P}_{3}$. The proof that no other (non-trivial) pair of MU product bases results from $\{\mathcal{I}_{1};\mathcal{B}_{1}\}$ has been obtained by A. Sudbery, and it is given in Appendix C. We now summarise the results derived in these two appendices.

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.

As before, the ranges of the parameters are assumed to be such that no MU product pair occurs more than once in the list. The pairs $\mathcal{P}_{0}$ and $\mathcal{P}_{2}$ have no parameter dependence, the pair $\mathcal{P}_{1}$ depends on two parameters, while $\mathcal{P}_{3}$ is a four-parameter family.

Theorem 2 represents the first main result of this paper. It states that there are continuously many possibilities to select pairs of MU bases which, however, can be listed exhaustively. In the remainder of this paper we will proceed by analytically constructing all triples of MU bases which exist in $d=6$. This will lead to our most important result, namely Theorem 4 in Sec. 7 which states the impossibility to extend any MU product triple by even a single MU vector. Thus, complete sets of MU bases in $d=6$ will contain at most pairs of MU product bases.

An alternative method to exploit Theorem 2 has been pursued in . Upon using suitable non-local unitary transformations and known results obtained by computer-algebraic methods, the strongest possible statement about MU product bases is then derived: if a complete set of seven MU bases exists, it will contain at most one product basis – which may be chosen to be the standard basis.

2 All triples of MU product bases

It is straightforward to enlarge the existing pairs of MU product bases in Theorem 2 to triples: simply add the MU product bases listed in Lemma 2, one after the other, to each of the pairs $\mathcal{P}_{0}$ to $\mathcal{P}_{3}$ and check whether a valid MU product triple results.

$\bullet\,\{\mathcal{P}_{0};\mathcal{B}_{0}\}$: If we choose the third basis to be of the form $\mathcal{B}_{0}$, there are only two choices, $\{|j_{y},J_{y}\rangle\}$ or $\{|j_{y},J_{w}\rangle\}$. Using the local complex conjugation $\hat{I}\otimes\hat{K}$, the resulting triples are found to be equivalent,

consequently, all triples of this type are equivalent to the Heisenberg-Weyl triple

$\bullet\,\{\mathcal{P}_{0};\mathcal{B}_{1}\}$: If we extend $\mathcal{P}_{0}$ by an indirect product basis of the form $\mathcal{B}_{1}$, there are only two choices, $\{|0_{y},J_{y}\rangle,|1_{y},J_{w}\rangle\}$ or $\{|0_{y},J_{w}\rangle,|1_{y},J_{y}\rangle\}$. Again, a local complex conjugation $\hat{k}\otimes\hat{I}$ maps one of the triples into the other,

Now turning to the pair $\mathcal{P}_{1}$, we again attempt to obtain a triple by adding either $\mathcal{B}_{0}$ or $\mathcal{B}_{1}$.

$\bullet\,\{\mathcal{P}_{1};\mathcal{B}_{0}\}$ or $\{\mathcal{P}_{1};\mathcal{B}_{1}\}$: First, extend the pair $\mathcal{P}_{1}$ by a direct product basis, resulting in either $\{|j_{z},J_{z}\rangle;|0_{x},J_{x}\rangle,|1_{x},J_{y}\rangle;|j_{y},J_{w}\rangle\}$ or $\{|j_{z},J_{z}\rangle;|0_{x},J_{x}\rangle,|1_{x},J_{w}\rangle;|j_{y},J_{y}\rangle\}$. It is not difficult to apply suitable LETs to transform them into the triple $\mathcal{T}_{1}$. Now extend the pair $\mathcal{P}_{1}$ by an indirect product basis $\mathcal{B}_{1}$. This leads to a contradiction since we would need the states $\{|\hat{R}_{\xi,\eta}J_{x}\rangle\}$ in $\mathcal{P}_{1}$ to coincide with $\{|J_{x}\rangle\}$, which is not allowed.

This completes the construction of all MU product triples in dimension six, leading to the second main result of this paper.

According to Theorem 1, neither of these triples can be extended by a single MU product state. Thus, any complete set of seven MU bases in dimension six will contain at most three product bases, and if it does, the triple must be equivalent to one of those in Theorem 3. In the following section we will obtain an even stronger result.

Excluding triples of MU product bases from complete sets

In this section we derive the third main result of this paper.

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

In other words, no complete set of seven MU bases in $d=6$ contains a triple of MU product bases. This result relies on a computer-algebraic proof in , which finds a total of 48 vectors MU to the pair of eigenbases of the Heisenberg-Weyl operators $X_{6}$ and $Z_{6}$, giving rise to sixteen different orthonormal bases. However, none of these bases allows one to extend the given pair beyond a triple of MU bases.

The present construction of MU product triples effectively produces twelve (and only twelve) product vectors that are MU to the pair $\mathcal{P}_{0}=\{|j_{z},J_{z}\rangle;\,|j_{x},J_{x}\rangle\}$, namely $\{|j_{y},J_{y}\rangle\}$ and $\{|j_{y},J_{w}\rangle$}, and they give rise to the only two inequivalent triples of MU bases, $\mathcal{T}_{0}$ and $\mathcal{T}_{1}$. Since $\mathcal{P}_{0}$ is equivalent to the eigenbases of $X_{6}$ and $Z_{6}$, clearly these twelve product vectors must figure among the 48 vectors given in .

To show this, we must first deal with a difference in our definition of the HW operators. The HW pair used in does not have the same form as $\mathcal{P}_{0}$ since the $x$-basis in is the eigenbasis of the operator $X_{6}$, whereas we have used the eigenbasis of the operator $X_{2}\otimes X_{3}$ (cf. Eq. (5)). Nevertheless, both pairs of bases turn out to be equivalent using a non-local unitary transformation. By writing the operators as matrices, we find that $X_{2}\otimes X_{3}=P_{25}X_{6}P_{25}$, where $P_{25}$ is a permutation matrix permuting rows two and five. This non-local transformation brings the eigenbasis of $X_{6}$ into product form, i.e. $\{|j_{x},J_{x}\rangle\}$, by multiplying it with $P_{25}$ from the left.

The same transformation must also be applied to the list of $48$ vectors so that they are MU to the pair $\mathcal{P}_{0}$. After multiplying each of these vectors by the matrix $P_{25}$ from the left, one easily identifies the twelve product vectors, numbered by $1,2,5,6,9,10,13,14,17,18,21$ and $22$ in the Appendix of the updated version of . For example, the vector labelled (1) transforms as follows:

where $\alpha=e^{2\pi i/12}$ and $\omega=e^{2\pi i/3}$. This vector is the product state $|1_{y},1_{y}\rangle$.

The twelve vectors give rise to four of the sixteen orthonormal bases which are MU to the original pair. These product bases are covered by the product bases we construct when extending the Heisenberg-Weyl pair $\mathcal{P}_{0}$ to a triple; however, only two of the four triples are locally inequivalent as follows from exploiting suitable local equivalence transformations.

Summary and discussion

Theorem 3 allows us to partly replicate results obtained by means of a computer-algebraic method. Out of the 48 vectors mutually unbiased to the Heisenberg-Weyl pair $\mathcal{P}_{0}$, found in , we successfully recover twelve, and they are shown to be equivalent to product vectors.

A similar situation has been described in where a different class of MU bases is studied. Given a “nice unitary error basis”, consisting of $d^{2}$ suitable matrices, one can search for MU bases within these sets. In the case of dimension six, it is shown that any partition of a nice error basis gives rise to no more than three MU bases. This limitation and the non-existence of more that three MU product bases are independent results: MU product bases and MU bases arising from nice unitary error bases are structurally different. For example, our construction reproduces the continuous family $\mathcal{P}_{1}^{(4)}$ of MU product pairs in $d=4$, and it is known that some of the pairs in this family are inequivalent to MU bases stemming from nice unitary error bases .

Let us conclude by formulating a conjecture which emerges naturally from our results: we expect Theorem 1 to hold for all composite dimensions $d=pq\geq 4$, not only for $d=4$ and $d=6$. Our pedestrian proof in these dimensions relies on enumerating all orthonormal product bases. However, the set of product bases in composite dimensions is likely to possess a certain structure which, once spelled out, should allow for a more elegant proof applicable to arbitrary composite dimensions.

The proof of Theorem 5, presented in Appendix C, has been found by A. Sudbery; we gratefully acknowledge his permission to reproduce it here. We thank S. Brierley, M. Grassl, and A. Sudbery for comments and suggestions. This work has been supported by EPSRC.

References

Appendix A Appendix

Case 1: If all three bases coincide, we have

also introducing an arbitrary second triple of orthogonal states. If the two triples coincide, we find the important special case of a direct product basis

These three cases complete the construction of all product bases in dimension six. Using local equivalence transformations in analogy to the procedure used in Sec. 3.1, one can write the four sets of product bases as displayed in Lemma 2.

Appendix B Appendix

In this Appendix we derive all pairs of MU product bases in dimension six by pairwise combining the orthonormal product bases $\mathcal{B}_{0}$ to $\mathcal{B}_{3}$, defined in Eqs. (47,46,50,51). In principle, we need to look at only 10 of the 16 pairs $\{\mathcal{B}_{i};\mathcal{B}_{j}\},\,i,j=0\ldots 3$, since the order of the bases does not matter: the pairs $\{\mathcal{B}_{i};\mathcal{B}_{j}\}$ and $\{\mathcal{B}_{j};\mathcal{B}_{i}\}$ are equivalent for $i\neq j$. Using local equivalence transformations, each pair can be brought to the form $\{\mathcal{I}_{i};\mathcal{B}_{j}\},\,i\leq j$, where the bases $\mathcal{I}_{0}$ to $\mathcal{I}_{3}$ are those listed in Lemma 2. As shown in the main text, it is not actually necessary to consider the bases $\mathcal{I}_{2}$ and $\mathbf{\mathcal{B}}_{2}$ at all, reducing the number of cases to six. Parameter ranges are assumed so that no pair occurs more than once.

$\bullet\,\{\mathcal{I}_{0};\mathcal{B}_{0}\}$: First we extend $\mathcal{I}_{0}$ to a pair of MU bases by combining it with

$\bullet\,\{\mathcal{I}_{0};\mathcal{B}_{1}\}$ and $\{\mathcal{I}_{0};\mathcal{B}_{3}\}$: These cases will be covered by the pairs $\{\mathcal{I}_{1};\mathcal{B}_{1}\}$ and $\{\mathcal{I}_{1};\mathcal{B}_{3}\}$, respectively, since we can treat the basis $\mathcal{I}_{0}$ as a subset of $\mathcal{I}_{1}$.

$\bullet\,\{\mathcal{I}_{1};\mathcal{B}_{1}\}$: In a first step, we act with a local unitary on the second basis

to rotate the $a$-basis of states that are MU to $\{|j_{z}\rangle\}$ into the basis $\{|j_{x}\rangle\}$ while the $A$-basis turns into $\{|J_{x}\rangle\}$, as before. This maps $\mathcal{B}_{1}$ to

$\bullet\,\{\mathcal{I}_{1};\mathcal{B}_{3}\}$: The second basis reads explicitly

which involve two rotations of the basis $\{|j_{x}\rangle\}$ about the $z$-axis, $\hat{r}_{\sigma}$ and $\hat{r}_{\tau}$. The operator $\hat{U}$ in $\mathcal{I}_{1}$ must be chosen such that $\{|\hat{U}J_{z}\rangle\}$ is MU to the $x$-basis. All such $U(3)$-rotations are given by the two-parameter family

diagonal in the $x$-basis, and defined in analogy to $\hat{R}_{\xi,\eta}$ in Eq. (15). Altogether, we obtain a four-parameter family of MU product pairs,

$\bullet\,\{\mathcal{I}_{3};\mathcal{B}_{3}\}$: No pair results when we combine the product basis $\mathcal{B}_{3}$ with $\mathcal{I}_{3}$. The standard transformations to simplify $\mathbf{\mathcal{B}}_{3}$ lead to

Appendix C Appendix

Here we report a proof by A. Sudbery that the conditions of Eq. (56) in Appendix B are only satisfied if the bases in (at least) one pair coincide or all four bases are mutually unbiased. If $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are orthonormal bases, we write $\mathcal{B}_{1}\,\mu\,\mathcal{B}_{2}$ to mean “$\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ are mutually unbiased”.

Then either $\mathcal{B}_{0}$ and $\mathcal{B}_{1}$ are equivalent bases or $\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ are equivalent bases or all four bases are mutually unbiased.

where $D$ and $D^{\prime}$ are diagonal and $\displaystyle F\equiv F_{3}$ is the Fourier matrix defined in Eq. (16).

The condition $\mathcal{B}_{2}\,\mu\,\mathcal{B}_{0}$ implies the unitary $V$ is a Hadamard matrix, and since $F^{\dagger}=FP$, the basis $\mathcal{B}_{2}$ is equivalent to a basis represented by $V=DF$. Similarly, $\mathcal{B}_{3}$ is equivalent to a basis represented by $W=D^{\prime}F$ where $D^{\prime}$ is diagonal. Now

where $K$, $L$, $K^{\prime}$ and $L^{\prime}$ are diagonal and $F^{(i)}$ is either $F$ or $F^{\dagger}$ ($i=1,2$). Hence

We will now examine the relationship between $U$ and the diagonal matrices $D,K,L$ in the two cases $U=DFKFL$ and $U=DFKF^{\dagger}L$, respectively. We can assume the leading entries of $D$ and $L$ to be $d_{11}=l_{11}=1$ by absorbing two phase factors in the diagonal matrix $K$.

Suppose $U=DFKFL$ where $D,K,L$ are diagonal unitary matrices with $D=\text{diag}(1,\alpha,\beta)$. Then either $U=PE$ where $P$ is a permutation matrix and $E$ is diagonal, or the matrix elements of $U$ are all non-zero and satisfy

Let $K=\text{diag}(\gamma,\delta,\epsilon)$ and $L=\text{diag}(1,\zeta,\eta)$. Then

Suppose one of $a,b,c$ were zero, say $a=0$. Then, since $\gamma,\delta,\epsilon$ all have modulus $1$, they must form an equilateral triangle in the complex plane, so either $\delta=\omega\gamma$ and $\epsilon=\omega^{2}\gamma$, when $b=0$ and $c=\gamma$, or $\delta=\omega^{2}\gamma$ and $\epsilon=\omega\gamma$, when $b=\gamma$ and $c=0$. In both cases $U$ is of the form $PE$.

If none of $a,b,c$ are zero, then all the matrix elements of $U$ are non-zero and equations (68), (69) and (70) follow immediately from (71).

Suppose $U=DFKF^{\dagger}L$ where $D,K,L$ are as in Lemma 4. Then either $U=PE$ where $P$ is a permutation matrix and $E$ is diagonal, or the matrix elements of $U$ are all non-zero and satisfy

while $\alpha$ is given by (69) and $\beta$ by

We now return to eq. (67) and consider the four possibilities for $(F^{(1)},F^{(2)})$.

Case 1: $U=DFKFL=D^{\prime}FK^{\prime}FL^{\prime}$.

Let $D=\text{diag}(1,\alpha,\beta)$, $D^{\prime}=\text{diag}(1,\alpha^{\prime},\beta^{\prime})$. Then, by Lemma 4, either $U$ is of the form $PE$ (when the bases $\mathcal{B}_{0}$ and $\mathcal{B}_{1}$ are equivalent), or

Hence $\alpha^{\prime}=\alpha$ or $\omega\alpha$ or $\omega^{2}\alpha$, so

In each case the columns of $W$ are a permutation of those of $V$. Thus either the bases $\mathcal{B}_{0}$ and $\mathcal{B}_{1}$ are equivalent or $\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ are equivalent.

Case 2: $U=DFKFL=D^{\prime}FK^{\prime}F^{\dagger}L^{\prime}$.

Suppose $U$ is not of the form $PE$. Then both Lemmas 4 and 5 apply, and $U$ has non-zero matrix elements satisfying (68) and (73). As in case 1, let $D=\text{diag}(1,\alpha,\beta)$ and $D^{\prime}=\text{diag}(1,\alpha^{\prime},\beta^{\prime})$. Now $\alpha$ and $\beta$ are given by Lemma 4, but $\alpha^{\prime}$ and $\beta^{\prime}$ are given by Lemma 5. Once again we have $\alpha^{3}=\alpha^{\prime 3}$, but now $\beta^{\prime}/\beta$ is not determined solely by $\alpha^{\prime}/\alpha$:

Hence $\alpha^{\prime}/\alpha$ and $\beta^{\prime}/\beta$ are both cube roots of $1$. Write $\alpha^{\prime}=\phi\alpha$, $\beta^{\prime}=\chi\beta$. If $\chi=\phi^{2}$ then, as shown in Case 1, the columns of $V$ and $W$ are the same, up to permutation, and the bases $\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ are equivalent. If $\chi\neq\phi^{2}$ then two of $1,\chi,\phi$ are equal and the third is different. The same is true of the sets $\{1,\omega\chi,\omega^{2}\phi\}$ and $\{1,\omega^{2}\chi,\omega\phi\}$. Hence the sums $a=1+\chi+\phi$, $b=1+\omega\chi+\omega^{2}\phi$ and $c=1+\omega^{2}\chi+\omega\phi$ all have the same modulus. For $\chi\neq\phi^{2}$, the product

is a Hadamard matrix and hence the bases $\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ are mutually unbiased. Thus in this case, $\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ are either equivalent or mutually unbiased.

Case 3: $U=DFKF^{\dagger}L=D^{\prime}FK^{\prime}FL^{\prime}$.

This is the same as Case 2 with $V$ and $W$ interchanged.

Case 4: $U=DFKF^{\dagger}L=D^{\prime}FK^{\prime}F^{\dagger}L^{\prime}$.

This is similar to Case 1, using Lemma 5 instead of Lemma 4. The conclusion is the same.

We have now shown that in every case, either $\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ are equivalent or $\mathcal{B}_{0}$ and $\mathcal{B}_{1}$ are equivalent or $\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ are mutually unbiased. But the assumptions of the theorem are symmetric between the pairs $\{\mathcal{B}_{0},\mathcal{B}_{1}\}$ and $\{\mathcal{B}_{2},\mathcal{B}_{3}\}$, so we can also prove that if $\mathcal{B}_{2}$ is not equivalent to $\mathcal{B}_{3}$ and $\mathcal{B}_{0}$ is not equivalent to $\mathcal{B}_{1}$, then $\mathcal{B}_{0}$ and $\mathcal{B}_{1}$ are mutually unbiased and therefore all four bases are mutually unbiased.