Mutually unbiased bases in dimension six: The four most distant bases

I Introduction

This maximum degree of incompatibility between two bases weyl27 ; schwinger60 states that the corresponding nondegenerate observables are complementary. Indeed, the technical formulation of Bohr’s Principle of Complementarity bohr27 that is given in Ref. sew91 relies on the unbiasedness of the pair of bases. Textbook discussions of this matter can be found in Refs. schwinger01 ; englert06 , and Ref. debz10 is a recent review on mutually unbiased bases (MUB), which are sets of bases that are pairwise unbiased.

In addition to playing a central role in quantum kinematics, we note that MUB are important for quantum state tomography ivanovic81 ; wootters89 , for quantifying wave-particle duality in multi-path interferometers ekkc08 , and for various tasks in the area of quantum information, such as quantum key distribution gisin02 or quantum teleportation and dense coding durt04b ; klimov09 ; revzen09 . In the context of quantum state tomography, $d+1$ von Neumann measurements provide ${d-1}$ independent data each in the form of $d$ probabilities with unit sum, so that in total one has the required ${d^{2}-1}$ real numbers that characterize the quantum state. A set of ${d+1}$ MUB is optimal, in a certain sense wootters89 , for these measurements—if there is such a set. Such a set is termed maximal; there cannot be more than ${d+1}$ MUB, since there are at most ${d+1}$ ${(d-1)}$-dimensional orthogonal subspaces in a ${(d^{2}-1)}$-dimensional real vector space wootters89 .

Ivanovic ivanovic81 gave a first construction of maximal sets of MUB if the dimension $d$ is a prime, and Wootters and Fields wootters89 succeeded in constructing maximal sets when $d$ is the power of a prime. These two cases have been rederived in various ways; see Refs. bandyopadhyay02 ; klappenecker04 ; durt04a , for example. For other finite values of $d$, maximal sets of MUB are unknown, but it is always possible to have at least three MUB (see debz10 and references therein).

The smallest non-prime-power dimension is ${d=6}$. Little is known for sure about the six-dimensional case, for which Zauner has conjectured that no more than three MUB exist zauner99 . Numerical studies seem to support Zauner’s conjecture grassl04 ; weigert08 . Computer-aided analytical methods, like Gröbner bases or SemiDefinite Programming, have also been applied to this problem weigert10 , but limitations in computational power have so far prevented any definitive answer.

Recently, Bengtsson et al. bengtsson06 introduced a distance between two bases for a quantification of the notion of “unbiasedness.” The distance vanishes when the two bases are identical and attains its maximal value of unity when they are unbiased. One can then consider the average squared distance (ASD) between several bases and search for its maximal value. Importantly, this ASD is unity if the bases are pairwise unbiased, and only then. A numerical search for the maximum of the ASD between four bases in dimension six can be performed. Actually, a numerical study on essentially the same quantity was recently carried out by Butterley and Hall butterley07 . In terms of the ASD, they found the surprisingly large but strictly-less-than-one maximal value of $0.9983$. This is strong evidence that no more than three MUB exist in dimension six. However, the set of bases behind this maximum value is not reported in Ref. butterley07 .

It is the objective of the present paper to close this gap. In Sec. II we review the notion of Bengtsson et al. for the distance between bases. We perform a numerical search for the maximum ASD between four bases in dimension six and report, in Sec. III, our results which confirm the maximum found by Butterley and Hall. We then provide a two-parameter family of three bases which, together with the canonical basis, reaches the numerically-found maximum, for which we give a closed expression. We study this family in detail in Sec. IV and conclude with a summary and outlook. Some matters of a technical nature are reported in the Appendix.

II A distance between bases

The main goal of this paper is twofold. First we numerically search for the maximum value of the ASD between four bases in dimension six and see that we cannot obtain four MUB. And second, we provide a two-parameter family of three bases which, together with the canonical basis, reaches the numerically-found maximum.

Clearly, this distance is symmetrical, $D_{ab}=D_{ba}$ and vanishes when the bases are the same, that is: when the two sets of projectors $\{|a_{i}\rangle\langle a_{i}|\}$ and $\{|b_{j}\rangle\langle b_{j}|\}$ are identical; the maximal distance is unity, $D_{ab}\leq 1$; and this maximum is reached if the bases are unbiased, $|\langle a_{i}|b_{j}\rangle|^{2}=1/d$, and only then.

For a set of $k$ bases, we have the ASD between the $k(k-1)/2$ pairs of bases, given by

Owing to the normalization, we have $\overline{D^{2}}\leq 1$ with $\overline{D^{2}}=1$ only if the $k$ bases are pairwise unbiased.

With this notion of distance at hand, we can numerically search for the maximum ASD between four bases in dimension six and see whether we obtain ${\overline{D^{2}}=1}$, or in other words, if we can find four MUB. This search is the subject matter of the next section.

III Numerical Study

Our numerical approach relies on the mapping between 1-qudit operators and 2-qudit states established in Chapter 3 of debz10 . Plus we use the steepest-ascent algorithm to find the maximum ASD between four bases in dimension six. Details of the numerical method are presented in Appendix A. Our numerical results are reported below.

A similar numerical study was recently performed by Butterley and Hall butterley07 who minimized ${1-\overline{D^{2}}}$ with the so-called Levenberg-Marquadt algorithm. Our approach confirms the extremal value they found, and we also exhibit the structure of the four bases that maximize $\overline{D^{2}}$ for ${d=6}$.

We have used our code not only in dimension ${d=6}$ but also for other $d$ values as a mean of benchmarking. We have run our code 2,500 times for the dimensions two to five, 10,000 times for the dimension six and 300 times for the dimension seven, both for ${k=d+1}$ bases and for four bases. Our results are summarized in Table 1.

Only in two cases, the maximum ASD does not reach the upper bound of ${\overline{D^{2}}=1}$. They are the cases of four bases in dimension two and six.

Since we consider four bases, there are six pairs of bases and their respective distances are not without interest. Indeed, it turns out that one basis is unbiased with the three remaining bases. And these three remaining bases are themselves equidistant. The immediate implication is that the privileged basis can be chosen to be the computational basis while the three remaining bases are Hadamard bases, that is: the unitary matrices composed of the columns that represent the basis kets with reference to the computational basis are complex Hadamard matrices divided by $\sqrt{6}$. We recall here that a complex Hadamard matrix is a $d$-dimensional square matrix satisfying the two conditions of unimodularity and orthogonality hadamard93

Therefore, the unitary matrix $H/\sqrt{d}$ has matrix elements that can be related to a pair of unbiased bases: ${\langle a_{i}|b_{j}\rangle=H_{ij}/\sqrt{d}}$.

Harking back to Table 1, we note that the best set of seven bases in dimension six has an ASD of $0.9849$, short of unity by a mere one-and-a-half percent. For all practical purposes—those of state tomography, say—these seven bases are marginally worse than the imaginary seven MUB that no one has managed to find.

IV The two-parameter family

Following Karlsson karlsson10 , we express the two-parameter family in terms of $2\times 2$ block matrices where each of the nine blocks is itself a complex Hadamard matrix. Such $2\times 2$ block matrices are called $H_{2}$-reducible. The two-parameter family contains three bases, the fourth basis being the canonical basis. We will see that these three Hadamard bases are equidistant, that their determinants are identical, and that they belong to the so-called Fourier transposed family $F_{6}^{T}$. Finally, we will show that together with the canonical basis they reach the numerically-found maximum of the ASD.

We begin by defining a few quantities. We will need the third root of unity $\omega=\exp(i\,2\pi/3)$ as well as the following $2\times 2$ matrices:

where $t=\exp(i\theta_{t})$ and $x=\exp(i\theta_{x})$ are two phases. Let us notice that $T$ and $F_{2}$ are themselves Hadamard matrices.

The Hadamard matrices for the three bases are given by

In the above parameterization, we have introduced the matrices $X_{i}$ and $N_{i}$, $i=1,2,3$, which we will address as dephasing and central matrices, respectively. The derivation of this parameterization is explained in Appendix B.

The next section is devoted to proving the three properties earlier mentioned. Before we turn to the proofs, we wish to point out that an additional relation between the two phases $x$ and $t$ exists,

It reduces the two-parameter family to a single-parameter family. Of course, as a subfamily, it conserves all the fundamental properties of the two-parameter family. Furthermore it still reaches the maximum ASD.

IV.2 Properties

A significant property of the three proposed Hadamard matrices is their equidistance. The relevant terms that appear in the distance between the two bases $M_{a}$ and $M_{b}$ (i.e., $|\langle a_{i}|b_{j}\rangle|$) are the elements of the product matrix $M_{a}^{\dagger}M_{b}^{\ }$ (i.e., $\langle a_{i}|b_{j}\rangle$) in absolute value. Therefore, if the three product matrices $M_{1}^{\dagger}M_{2}^{\ }$, $M_{2}^{\dagger}M_{3}^{\ }$ and $M_{3}^{\dagger}M_{1}^{\ }$ have equal coefficients in absolute value, then the three bases $M_{1}$, $M_{2}$ and $M_{3}$ are equidistant. This is exactly what happens here. Indeed, we have the following cyclic structure:

where, on the one hand, the $2\times 2$ submatrices $a_{1}$, $b_{2}$ and $c_{3}$ have the same coefficients in absolute value and, on the other hand, $a_{2}$, $a_{3}$, $b_{1}$, $b_{3}$, $c_{1}$ and $c_{2}$ have the same coefficients in absolute value. More precisely, these matrices have the following forms (where the symbol $\check{\ }$ stands for swapping the two diagonal elements). First,

The various coefficients in Eqs. (9) and (10) can be expressed in terms of the two angles $\theta_{x}$ and $\theta_{t}$,

When Eq. (7) is fulfilled, we have $\epsilon=\omega^{*}\delta^{*}$ and a few simplifications arise. We obtain

IV.2.2 Determinant

Accordingly, the three Hadamard bases share the same determinant

However, although the determinants are equal, there seems to be no simple relation between the three matrices $M_{1}$, $M_{2}$, and $M_{3}$. In particular, they do not have the same spectrum and are, therefore, not related by unitary operators.

IV.2.3 Fourier transposed family

The Fourier transposed family, first studied by Haagerup haagerup96 , is parameterized by Karlsson in the form karlsson10

where the $2\times 2$ Hadamard matrices $T_{1}$ and $T_{2}$ are given by

The equivalence relation in Eq. (15) means equality up to left and right dephasing and left and right permutations. In other words, the central matrix is the fundamental object that specifies the equivalence class. In the form of Eq. (6), it is clear that the three matrices $N_{1}$, $N_{2}$, and $N_{3}$ belong to the Fourier transposed family. As a result, the two-parameter family itself belongs to the Fourier transposed family.

Let us note here that only the right equivalence is natural for more than two bases as it states that bases are defined up to permutations and global phases of their basis states. In particular, the distance between bases is invariant under right equivalence but not under left equivalence.

IV.3 Average distance

Let us now compute the global maximum of the ASD between the three bases. Since the three bases are equidistant, we only have to compute the distance between, say, $M_{1}$ and $M_{2}$. A direct calculation leads to the following expression

which agrees with the numerically-found maximum ASD within the machine precision.

Furthermore, the distance ${D_{12}}$ vanishes for

As can be verified from the parameterization (6) or from the matrix products (8), the bases are indeed identical up to global phases and permutations for these values of the two phases $\theta_{x}$ and $\theta_{t}$.

We can also consider the single-parameter family that we obtain when eliminating $\theta_{t}$ by using Eq. (7). Since Eq. (19) is equivalent to Eq. (7), this single-parameter family reaches the maximum of the ASD — and also the minimum since Eq. (19) is obeyed by $(\theta_{x},\theta_{t})=(\pi/2,2\pi/3)$. This is illustrated in Fig. 2, a contour plot of $\overline{D^{2}}$ for the two-parameter family of Hadamard bases, with the location of the $(\theta_{x},\theta_{t})$ values of the single-parameter family indicated. The location of one of the eight maxima is marked, and the locations of the other seven follow from the symmetry properties of the contours.

V Summary and outlook

We performed a numerical search for the maximum ASD between four bases in dimension six. We found that it is strictly smaller than unity and so confirmed the recent study by Butterley and Hall butterley07 . We regard this result as strong evidence that no four MUB exist in dimension six.

Next, we went beyond this numerical result by providing the four bases behind the numerically-found maximum. More specifically, we found a two-parameter family of three bases, which together with the canonical basis, reaches the maximum of the ASD. We characterized this two-parameter family in full. We proved its inclusion in the Fourier transposed family and shown that the three base are equidistant. Furthermore, we analytically computed the maximum ASD between these three Hadamard bases and the canonical basis to show that it reproduces the numerical result.

Two directions might be relevant for an extension of the present study. First, it would be interesting to see if the optimality of our solution can be extended to a larger family of bases, for example, to the whole Fourier transposed family. Second and complementarily, there might exist an argument to restrict the search for the maximum ASD between the canonical basis and three Hadamard bases to the Fourier transposed family, instead of the entire Hadamard family which, so far, has not been fully parameterized. In this context, however, it should be noted that—as follows from the findings of Jaming et al. Jaming+4:09 —there are no four MUB if one restricts the search to members of the Fourier family.

Finally, if there is no complete set of seven MUB in dimension six, the optimal measurement for state tomography, in terms of statistical errors, remains to be found.

Appendix A Numerical method

As discussed in Sec. 3.1 of Ref. debz10 , for any ket $|\varphi\rangle$ or bra $\langle\phi|$ in a $d$-dimensional Hilbert space $\mathcal{H}$ or $\mathcal{H}^{\dagger}$, respectively, there is a conjugate bra or ket

This mapping is not unique, but two different realizations differ at most by a unitary transformation. As a rule, $\langle\phi^{*}|$ and $\langle\phi|=|\phi\rangle^{\dagger}$ are different bras.

Once a particular choice of mapping has been made, there is a one-to-one correspondence between one-qudit operators and two-qudit kets,

In particular, for an orthonormal basis of kets in $\mathcal{H}$, a=\bigl{\{}|a_{1}\rangle,|a_{2}\rangle\dots,|a_{d}\rangle\bigr{\}}, we have the conjugate basis a^{*}=\bigl{\{}|a^{*}_{1}\rangle,|a^{*}_{2}\rangle,\dots,|a^{*}_{d}\rangle\bigr{\}}, and jointly they are used in defining the two-qubit state

which has the $d$-fold eigenvalue $1/d$ and the $(d^{2}-d)$-fold eigenvalue zero.

We normalize the Hilbert-Schmidt inner product of two-qudit operators in accordance with

so that $(\rho_{a},\rho_{a})=1$ and $(\rho_{a},\rho_{b})=1/d$ for a pair of unbiased bases. For the two-qudit states associated with two single-qudit bases, we then have

with the distance $D_{ab}$ of Eq. (2), where the identity \langle a^{*}_{j}a^{\ }_{j}|b^{*}_{k}b^{\ }_{k}\rangle=\bigl{|}\langle a_{j}|b_{k}\rangle\bigr{|}^{2} is used. It follows that $D_{ab}$ can be expressed in terms of the Hilbert-Schmidt norm of ${\rho_{a}-\rho_{b}}$,

with $|\!|A|\!|=\sqrt{(A,A)}$. This tells us something important: If $a\neq b$, then $\rho_{a}\neq\rho_{b}$, so that the mapping $a\leftrightarrow\rho_{a}$ is one-to-one.

In passing, we note the following challenge. Clearly, not all two-qudit states with $\rho=d\rho^{2}$ correspond to a single-qudit basis in the sense of Eq. (27). But which additional criteria identify the set of two-qudit states that do?

We are interested in finding the maximum value of the ASD between $k$ bases in dimension $d$. The numerical search begins with a randomly chosen initial set of bases, and then modifies the bases in each iteration round such that $\overline{D^{2}}$ is systematically increased.

An infinitesimal variation of a ket in basis $a$ is given by

where $\epsilon_{a}$ is an infinitesimal hermitian operator acting on the basis $a$. We have one such hermitian $\epsilon$ operator for each basis. The resulting response of $\overline{D^{2}}$ is

is the $a$th component of the gradient. If bases $a$ and $b$ are unbiased, there is no contribution to $G_{a}$ from basis $b$ and, therefore, there is no gradient for a set of MUB. But the converse is not true: We can have a vanishing gradient although the bases are not pairwise unbiased.

When the gradient has nonzero components, we choose $\epsilon_{a}=\kappa G_{a}$ with a common $\kappa>0$ that specifies the step size. This guarantees $\delta\overline{D^{2}}>0$ if $\kappa$ is not too large, and maximization along the line specified by the direction of the gradient can be done by optimizing the value of $\kappa$. The line optimization is a necessary ingredient if conjugate gradients are used for accelerating the convergence; see Ref. shewchuck94 , for instance.

The finite unitary change of basis $a$, $|a_{j}\rangle\to V_{a}|a_{j}\rangle$, is then accomplished by

or yet other ones, whichever of them is convenient to use. All three $V_{a}$s equal $\mathbf{1}+i\epsilon_{a}$ to first order in $\epsilon_{a}$ and differ in the higher-order terms. Note that a high-precision evaluation of the infinite product in the third version of $V_{a}$ requires very few terms. This makes the third version a viable alternative if the computation of the exponential in the first version or of the inverse operator in the second version is time consuming or imprecise.

The iteration is terminated, when all components of the gradient vanish (in the numerical sense specified by the machine precision). We repeat this steepest-ascent search many times to ensure that we find the global maximum. As Fig. 1 shows for $(d,k)=(6,4)$, the iteration gets stuck in local maxima for about three attempts in ten and, see Table 1, only four in ten trials are successful for $(d,k)=(6,7)$.

Appendix B Derivation of the two-parameter family

The $d\times d$ matrix $U_{ab}$ composed of the transition amplitudes $\langle a_{j}|b_{k}\rangle$ of two orthonormal bases is unitary,

The columns and the rows of $U_{ab}$ are representations of the kets $|b_{k}\rangle$ and the bras $\langle a_{j}|$, respectively. The unitary matrices associated with a set of bases have a composition law for consecutive basis changes: $U_{ab}=U_{ac}U_{cb}$, $U_{aa}=\openone$. In particular, $\sqrt{d}\,U_{ab}$ is a complex Hadamard matrix if the bases $a$ and $b$ are unbiased; see the paragraph containing Eq. (III).

Now, from the numerical search we know that one of the bases that maximize the ASD between four bases in dimension six is unbiased with the other three bases. We identify this privileged basis as the canonical basis and refer to it as the zeroth basis, and characterize the set of four bases by the three $6\times 6$ transition matrices

so that the columns of $M_{i}$ are composed of the probability amplitudes of the kets of the $i$th basis with respect to the privileged basis.

When multiplied by $\sqrt{6}$, the matrices $M_{1}$, $M_{2}$, and $M_{3}$ are $6\times 6$ Hadamard matrices, for which we use Karlsson’s parameterization karlsson10 .

His parameterization applies to $H_{2}$-reducible Hadamard matrices that can be written in the form $H=X_{L}P_{L}NP_{R}X_{R}$, where the left and right $X$ matrices only contain phases on the diagonal, the $P$ matrices are permutation matrices, and the central matrix has the form

where $F_{2}$ is the unnormalized two-dimensional Fourier matrix of Eqs. (5) and the $2\times 2$ $T_{i}$ matrices are those of Eq. (16),

with a unitary and hermitian $2\times 2$ matrix $\Lambda$. It turns out that our Hadamard matrices are indeed $H_{2}$-reducible since they can be written as ${M_{i}=X_{L_{i}}P_{L_{i}}N_{i}P_{R_{i}}X_{R_{i}}}$ with the central matrices given by

As in Eqs. (5), we choose to express the matrix $T$ with factors of $\omega=\exp(i2\pi/3)$,

to exhibit the crucial dependence on the phase factor $t$. The left permutation matrices are all equal, $P_{L_{1}}=P_{L_{2}}=P_{L_{3}}=P_{L}$.

Third, we notice that only the left dephasing and permutation matrices are relevant for the distance. Indeed the right dephasing matrices only add global phases to the basis vectors while the right permutation only permute the basis vectors. In other words, two bases $B$ and $BP_{R}X_{R}$ are equivalent in terms of distance. Therefore we can choose to conserve only the relevant structure for our bases, that is, $M_{i}=X_{L_{i}}P_{L_{i}}N_{i}$.

The fourth step is to use the fact that only relative dephasing and permutations of the rows are relevant to the distance. Therefore we define new bases as

To simplify the notations, we again denote the two new diagonal matrices in $P_{L}^{\dagger}X_{2}^{\dagger}X_{1}P_{L}$ and $P_{L}^{\dagger}X_{2}^{\dagger}X_{3}P_{L}$ by $X_{1}$ and $X_{3}$, respectively. We further observe that

Next we add a suitable global phase to $X_{1}$ and $X_{3}$. We multiply $X_{1}$ by $\exp(-i\textrm{Arg}(A_{1}A_{1}/2))$ and $X_{3}$ by $\exp(-i\textrm{Arg}(B_{1}B_{1}/2))$ such that $A_{1}$ and $B_{1}$ take the simple form

for some phase $\phi$. We end up with the remarkable form

and it only remains to find the structure behind the two $2\times 2$ dephasing matrices $A_{2}$ and $B_{3}$.

To do so, we now consider the products $U_{ij}=M_{i}^{\dagger}M_{j}^{\ }$. We obtain

and $F_{3}$ is the standard (unnormalized) 3-dimensional Fourier matrix

The seventh step is to look once more at the numerics. With respect to the product $M_{1}^{\dagger}M_{2}^{\ }$, we see that

Thus we are lead to define the matrix equation

This only represents a system of three equations since $E_{1}=E_{1}$. In the same manner, we have for $M_{2}^{\dagger}M_{3}^{\ }$

and $E_{2}=E_{2}$ so that, here too, only three equations are relevant. Finally, for $M_{3}^{\dagger}M_{1}^{\ }$, we obtain

and, owing to $(\omega^{*}-1)E_{3}=t(1-\omega)E_{3}$, again only three equations are relevant. We should mention here that there are other interesting identities within the products $M_{i}^{\dagger}M_{j}^{\ }$, such as $b_{2}=[a_{1}+a_{1}^{\dagger}+Z(a_{1}-a_{1}^{\dagger})Z]/2$, but they are much more complicated to handle and will not be necessary to achieve our parameterization.

The eighth step is to solve the above nine equations. We obtain

From the numerics, we know that $r=r^{\prime}$ and thus $A_{1}=A_{3}$, which we already found by looking at the dephasing matrix $X_{1}$. Note also that the expression of the complex number $r$ is not required. Furthermore we find

From the numerics, we know that $s=s^{\prime}(=r)$ and thus $B_{1}=\omega B_{2}$, which we already obtained by looking at the dephasing matrix $X_{3}$. The next three equations are much more interesting. Indeed we have

From the numerics, we know that $u=0$ and the last equation reduces to

Since $Y_{2}=\omega A_{1}A_{2}$ and $Y_{3}=B^{*}_{3}A_{1}$, the above equation directly translates into

This last relation can be inserted in $E_{3}$ and $E_{3}$, which become identical and can be written as

A last hint from the numerics is needed. We actually notice that

As $Y_{3}=t^{*2}Y_{2}$, we arrive at $t^{*2}Y_{1}Y_{2}^{2}=-\openone$ so that $\omega t^{*}A_{1}^{2}A_{2}=\pm iU$, where $U^{2}=\openone$, that is, $U=\openone$ or $U=Z$ since it has to be diagonal. With the help of the numerics, we conclude that

The final parametrization of the dephasing matrices is therefore given by

Let us finally come back to Eq. (67). We can now substitute $Y_{1}=A_{1}^{2}$ and $Y_{2}=(i\omega^{*}tZA_{1}^{*2})(\omega A_{1})=itZA_{1}^{*}$ in Eq. (67) and, upon defining $x=\exp(i\theta_{x})$ and $t=\exp(i\theta_{t})$, we arrive at

References