All Mutually Unbiased Bases in Dimensions Two to Five

Introduction

Position and momentum of a classical non-relativistic particle are intimately linked since the momentum variable generates spatial translations. Mathematically, this important relation is embodied in the structure of the Galilei group. It turns out to be even more fundamental for a quantum mechanical particle, where it takes the form of the commutation relation of its position and momentum operators.

where $b,b^{\prime}=0,1,\ldots,d$. Such complete sets of MU bases are ideally suited to reconstruct quantum states while sets of up to $(d+1)$ MU bases have applications in quantum cryptography and in the solution of the Mean King’s problem , for example.

The methods to construct complete sets of MU bases typically deal with all prime or prime-power dimensions simultaneously. They either make use of the Heisenberg-Weyl group , exploit identities from number theory , or they are couched in the language of finite fields . All these methods are constructive and effectively lead to the same bases. The existence of other, inequivalent sets of complete MU bases remains unclear.

In this paper, we choose a different method to study MU bases in dimensions two to five. It allows us to directly conclude that the corresponding complete sets of $(d+1)$ MU bases are unique (up to some irrelevant equivalence, cf. below). What is more, we are also able to exhaustively list all inequivalent classes of $d$ or less MU bases. This approach is attractive because it uses elementary methods only, except in dimension five. In order to present the details of this method we need to briefly discuss the relation between MU bases and complex Hadamard matrices.

The approach we take is inspired by recent works aimed at dimension six . Starting with a pair of MU bases $\{I,H\}$ we will find all vectors $v$ which are MU to $I$ and $H$. Only these vectors represent candidates to form additional bases and by analysing their inner products we will be able to obtain all MU bases in low dimensions.

In dimension six it is not trivial to determine all candidate vectors. It becomes necessary to use Gröbner bases or to discretise the underlying space . For $d\leq 5$, however, we find that elementary properties of the complex plane are sufficient to solve (most of) the relevant equations in closed form.

The task to find all MU bases is complicated by the fact that, actually, many sets of apparently different MU bases are identical to each other. For the desired classification, it is sufficient to enumerate all dephased sets of $(r+1)$ MU bases. This standard form is given by

Dimensions d=2𝑑2d=2 and d=3𝑑3d=3

In this section, we construct all sets of MU bases in dimensions two and three using only simple properties of the complex plane. The direct approach to construct all MU bases for $d=4$ in Sec. 3 will be based on similar arguments.

The matrices consisting of the eigenvectors of the Heisenberg-Weyl operators form a set of three MU bases in dimension two which are unique up to the equivalences specified in Appendix A. We present a simple proof of this well-known fact.

Let us begin by noting that there is only one dephased complex Hadamard matrix in $d=2$ (up to equivalences), the discrete $(2\times 2)$ Fourier matrix

These equations hold simultaneously only if $e^{i\alpha}=\pm i$. Thus, there are only two vectors which are MU to both $I$ and $F_{2}$, given by $v_{\pm}=\left(1,\pm i\right)^{T}/\sqrt{2}$. Since this is a pair of orthogonal vectors, they form a Hadamard matrix $H_{2}=(v_{+}|v_{-})$ and, therefore, the three sets

represent all (equivalence classes of) one, two, or three MU bases in dimension two.

2 Dimension d=3𝑑3d=3

In dimension three there is also only one dephased complex Hadamard matrix up to equivalence. It is given by the $(3\times 3)$ discrete Fourier matrix

defining $\omega=e^{2\pi i/3}.$ Again, we search for dephased vectors $v=(1,e^{i\alpha},e^{i\beta})^{T}/\sqrt{3}$, $0\leq\alpha,\beta\leq 2\pi$ , which are MU with respect to the matrix $F_{3}$. This leads to the following three conditions

Removing an overall factor of $e^{i\alpha/2},$ they can be rewritten

where $2\zeta=e^{i(\beta-\alpha/2)}$. By considering a plot in the complex plane, Fig. 1, we see that these three equations hold simultaneously only if two of the cosine terms are equal. This implies that the only possible values of the parameter $\alpha$ are $0,\pi/3,$ or $2\pi/3$, leading to the requirement $\pm 1/2=\cos\beta$. Consequently, the Eqs. (7) have exactly six solutions which give rise to vectors

Examining their inner products shows that there is only one way to arrange them (after normalization) into two orthonormal bases, namely $H_{3}^{(1)}=(v_{1}|v_{2}|v_{3})$ and $H_{3}^{(2)}=(v_{4}|v_{5}|v_{6})$. It is useful to note that one can write

where $D={\mbox{diag}(1,\omega,\omega)}$ is a diagonal unitary matrix with entries identical to the components of the vector $v_{1}$, i.e. the first column of $H_{3}^{(1)}$. The triples obtained from adding either $H_{3}^{(1)}$ or $H_{3}^{(2)}$ to the pair $\left\{I,F_{3}\right\}$ are equivalent,

as follows from first applying the unitary $D^{2}$ globally from the left, rephasing the first basis with $D^{-2}\equiv D$, and finally rearranging the last two bases. We therefore conclude that the sets constitute a complete classification of all sets of MU bases in dimension $d=3$.

Dimension d=4𝑑4d=4

In dimension $d=4$, a one-parameter family of complex Hadamard matrices exists,

After dephasing, any vector MU to the standard basis takes the form $v=(1,e^{i\alpha^{\prime}},e^{i\beta^{\prime}},$ $e^{i\gamma^{\prime}})^{T}/2$ where $0\leq\alpha^{\prime},\beta^{\prime},\gamma^{\prime}<2\pi$. For convenience, we will use an enphased variant of $v$. Multiplying through by the phase factor $e^{-i\alpha^{\prime}/2}$ and defining $\alpha=\alpha^{\prime}/2\in\left[0,\pi\right]$, $\beta=\beta^{\prime}-\alpha^{\prime}/2\in[0,2\pi]$, and similarly for $\gamma$, we consider the parametrization $v=(e^{-i\alpha},e^{i\alpha},e^{i\beta},$ $e^{i\gamma})^{T}/2$ instead. The conditions for $v(\alpha,\beta,\gamma)$ to be MU to the columns of $F_{4}(x)$ lead to four equations,

where complex numbers $\zeta_{\pm}=(e^{i\beta}\pm e^{i\gamma})/2$ have been introduced. We will now construct all solutions of these equations as a function of the value of $x$. We treat the cases (i) $\alpha=0$, (ii) $\alpha=\pi/2$, and (iii) $\alpha\neq 0,\alpha\neq\pi/2$ separately since the Eqs. (32) and (33) take different forms for these values.

(i): α=0𝛼0\alpha=0

Eqs. (33) simplify to the pair $\left|\pm e^{-ix}\zeta_{-}\right|=1\,,$ which only hold simultaneously if $|\zeta_{-}|=1$ or $e^{i\gamma}=-e^{i\beta}$, implying that $\zeta_{+}=0$ so that Eqs. (32) are satisfied automatically. Thus, solutions exist for any value of $x$ whenever $\beta=\gamma+\pi\rm{\,\,mod\,}{2\pi}$, and the resulting vectors can be written as $v(\beta)=\left(1,1,e^{i\beta},-e^{i\beta}\right)^{T}/2$, with $\beta\in[0,2\pi]$. It will be convenient to divide this family of states into two sets,

introducing $y=\beta$ and $y^{\prime}=\pi+\beta$.

(ii): α=π/2𝛼𝜋2\alpha=\pi/2

Eqs. (32) and (33) now reverse their roles: the conditions $|\pm\zeta_{+}|=1$ require $e^{i\gamma}=e^{i\beta}$, with (32) being satisfied since $|\zeta_{-}|=1$ follows immediately. Hence, there is another one-parameter family of mutually unbiased vectors for all values of $x$ if $\beta=\gamma$. This family can be written as $v(\varphi)=\left(1,-1,e^{i\varphi},e^{i\varphi}\right)^{T}/2$, $\varphi\in[0,2\pi]$, after dephasing and absorbing a factor of $i$ in the definition of the phase, $\varphi=\pi/2+\beta$. Again, we express these solutions as a set of pairs,

where $z=\varphi$ and $z^{\prime}=\pi+\varphi$.

(iii) α≠0,α≠π/2formulae-sequence𝛼0𝛼𝜋2\alpha\neq 0,\alpha\neq\pi/2

Similarly, Eqs. (33) for $\sin\alpha\neq 0$ imply that $e^{-ix}\zeta_{-}=\pm i\cos\alpha$. Using $\gamma=\pi-\beta\rm{\,\,mod\,}{2\pi}$ in the definition of $\zeta_{-}$, we find $\zeta_{-}=\cos\beta$, so that

The right-hand-side of this equation only vanishes if $x=\pi/2$: both $\beta=\pi/2$ and $\beta=3\pi/2$ would, according to (37), require $\alpha=\pi/2$ which we currently exclude. Therefore, solutions to Eqs. (32,33) with $\alpha\neq 0$ or $\alpha\neq\pi/2$ only exist for $x=\pi/2$ if a second relation between $\alpha$ and $\beta$ holds,

The form of the additional MU vectors is determined by Eqs. (37) and (39) which have four solutions. First, for $\beta=\alpha$ we obtain MU vectors of the form $(e^{-i\alpha},e^{i\alpha},e^{i\alpha},$ $-e^{ia})^{T}/2$ or $\left(1,e^{2i\alpha},e^{2i\alpha},-1\right)^{T}/2$ after dephasing. Splitting this family into two subsets as before, we find

Similarly, the choice $\beta=\pi+\alpha\rm{\,\,mod\,}{2\pi}$ leads to two sets of dephased MU vectors,

Next, when proceeding in an entirely analogous manner for the remaining two choices $\beta=\pi-\alpha\rm{\,\,mod\,}{2\pi}$ and $\beta=2\pi-\alpha\rm{\,\,mod\,}{2\pi}$, we obtain the following four families of dephased vectors MU to $F_{4}(\pi/2)$,

with $0\leq r,r^{\prime},s,s^{\prime}<\pi$.

2 Forming MU bases

Knowing all vectors that are MU to both the identity and $F_{4}$, we now determine those combinations which form other bases.

In other words, there is a three parameter-family of triplets of MU bases $\left\{I,F_{4}(x),H_{4}(y,z)\right\}$ in dimension $d=4$. This family was not known before.

If $x=\pi/2$, additional MU vectors $j_{1},\ldots,j_{4}$, and $k_{1},\ldots,k_{4}$, have been identified, cf. Eqs. (40-42). Calculating the scalar products within each group, one sees that two further orthonormal two-parameter bases emerge,

if the conditions $r=r^{\prime}$, $s=s^{\prime}$, and $t=t^{\prime}$, $u=u^{\prime}$, respectively, are satisfied. No other combinations of the MU vectors can form inequivalent bases so that the matrices in Eqs. (43-53) represent all possible choices of a MU basis. Permuting appropriate rows and columns of the matrices $J_{4}$ and $K_{4}$ transforms them into $H_{4}$; thus, the triples $\left\{I,F_{4}(\pi/2),J_{4}(r,s)\right\}$ and $\left\{I,F_{4}(\pi/2),K_{4}(t,u)\right\}$ are equivalent to $\left\{I,F_{4}(\pi/2),H_{4}(y,z)\right\}$.

Let us begin by noting that sets of four MU bases cannot exist away from $x=\pi/2$. No two matrices $H_{4}(y,z)$ and $H_{4}(y^{\prime},z^{\prime})$ are MU since

however, these equations have no solution for any values of $y$ and $y^{\prime}$. A similar argument shows that there are no values of $z$ and $z^{\prime}$ such that the matrices $H_{4}(y,z)$ and $H_{4}(y,z^{\prime})$ are MU.

We now show that for $x=\pi/2$ the bases $H_{4}(y,z),J_{4}(r,s)$ and $K_{4}(t,u)$ give rise to four and five MU bases if the free parameters are chosen appropriately. An argument similar to the one just presented shows that no two bases within either the family $J_{4}(r,s)$ or $K_{4}(t,u)$ are MU. Thus, any quadruple of MU bases must contain bases from different families.

The inner products $\left|h_{1}^{\dagger}(y)j_{1}(r)\right|,\left|h_{1}^{\dagger}(y)j_{2}(r)\right|,\left|h_{2}^{\dagger}(y)j_{1}(r)\right|$ and $\left|h_{2}^{\dagger}(y)j_{2}(r)\right|$ have modulus $1/2$ if there are values for $y$ and $r$ such that the equations

hold simultaneously. Upon introducing a factor of $e^{-ir/2}$, Eqs. (56) are equivalent to the constraints

Consequently, one must have $r=\pi/2$ , and thus $e^{-iy}=\pm i$ or $y=\pi/2$ since $0\leq r,y<\pi$. An entirely analogous argument restricts the values of $s$ and $z$: checking the inner products $\left|h_{1}(r)^{\dagger}j_{3}(z)\right|,\left|h_{1}(r)^{\dagger}j_{4}(z)\right|$ etc. tells us that the matrices $H_{4}(y,z)$ and $J_{4}(r,s)$ are mutually unbiased only if $y=z=r=s=\pi/2$. We also find that the pairs $\{J_{4}(r,s),$ $K_{4}(t,u)\}$ and $\left\{K_{4}(t,u),H_{4}(x,y)\right\}$ are MU only when all six parameters take the value $\pi/2.$

There is one three-parameter family of triples consisting of the one-parameter Fourier family $F_{4}(x)$ combined with two-parameter set $H_{4}(y,z)$; neither $J_{4}(r,s)$ nor $K_{4}(t,u)$ give rise to other triples since each of these sets of Hadamard matrices is equivalent to $\{I,F(\pi/2),H_{4}(y,z)\}$. The three-dimensional set (59) of MU bases in dimension $d=4$ may be visualized as a cuboid defined by $0\leq x,y<\pi$ and $0\leq z<\pi/2$. The reduction in the parameter range of $z$ is due to the equivalence $\{I,F_{4}(x),H_{4}(y,z)\}\,\sim\{I,F_{4}(\pi-x),H_{4}(\pi-y,\pi-z)\}\,$ which follows from an overall complex conjugation. Each of the points in the cuboid corresponds to one triple while both the quadruple and the quintuple are located at the point, $x=y=z=\pi/2$.

Only one set of four MU bases exists, $\{I,F_{4}(\pi/2),H_{4}(\pi/2,\pi/2),J_{4}(\pi/2,\pi/2)\}$, since the other two candidates obtained by combining $K_{4}(\pi/2,\pi/2)$ with either $J_{4}(\pi/2,\pi/2)$ or $H_{4}(\pi/2,\pi/2)$ are permutations of this quadruple. Finally, there is a unique way to a construct five MU bases which is easily seen to be equivalent to the standard construction of a complete set of MU bases in dimension four.

Dimension d=5𝑑5d=5

As in dimensions two and three, there is a unique choice of a $(5\times 5)$ dephased complex Hadamard matrix ,

equal to the discrete $(5\times 5)$ Fourier matrix, with $\omega=\exp(2\pi i/5)$ denoting a fifth root of unity. The uniqueness of $F_{5}$ is obtained by an analytical method that is far from elementary .

We have not found an elementary method to obtain a list of all vectors which are MU to the Fourier matrix $F_{5}$. Instead, we will rely on earlier work where those vectors have been constructed analytically by means of a computer program.

defining $\alpha_{0}\equiv 0$. According to , the solutions of these equations give rise to 20 vectors which can be arranged in four MU bases,

Each of the four matrices in (83) is related to the Fourier matrix in a remarkably simple manner. In analogy to the unitary diagonal matrix used in Eq. (29), define a diagonal unitary matrix

with entries given by the first column of $H_{5}^{(1)}$ and you find that

Using this observation, we can express the unique complete set of six MU bases for dimension $d=5$ as follows

Select one of the four matrices given in (83) and adjoin it to the pair $\{I,F_{5}\}$. You obtain four triples of MU bases with two immediate equivalences, namely,

on the other. The equivalence (88) follows from multiplying the set $\{I,F_{5},DF_{5}\}$ with $D^{4}$ from the left, rephasing the first basis with $D$ from the right, using $D^{5}=I$ and swapping the last two matrices. A similar argument establishes the equivalence (89), using $D^{3}$ instead of $D^{4}$.

Thus, it remains to check whether the triples $\mathcal{T}^{(1)}\equiv\{I,F_{5},H_{5}^{(1)}\}$ and $\mathcal{T}^{(2)}\equiv\{I,F_{5},$ $H_{5}^{(2)}\}$ are equivalent to each other. It turns out that these two triples are, in fact, inequivalent. More explicitly, this means that no unitary matrix $U$ and no monomial matrices $M_{0},M_{1}$ and $M_{2}$ can be found which would map $\mathcal{T}^{(1)}$ into $\mathcal{T}^{(2)}$ according to

A proof of this statement is given in Appendix B.

There are six possibilities to form quadruples by selecting two of the four matrices in Eq. (83) and adding them to the pair $\{I,F_{5}\}$. Recalling that $H_{5}^{(k)}=D^{k}F_{5}$, we identify the following equivalences which relate three quadruples each,

Interestingly, these two classes of MU bases are equivalent to each other leaving us with a single equivalence class of quadruples in dimension five, with representative $\{I,F_{5},H_{5}^{(1)},H_{5}^{(2)}\}$, say. To show this equivalence, we multiply the first quadruple with the adjoint of $F_{5}$ from the left

using the identity $F^{\dagger}=FP$, with some permutation matrix $P$, and swapping the first two bases. The action of $F_{5}^{\dagger}$ on the other two elements is surprisingly simple: the Hadamard matrix $H_{5}^{(1)}$ is mapped to itself,

up to a monomial matrix $M$, while $H_{5}^{(2)}$ is sent to $H_{5}^{(3)}$,

again up to some monomial matrix $M^{\prime}$. Both relations simply follow from working out the product on the left and factoring the result, e.g.

where the $k^{th}$ entry of the diagonal matrix $D^{(2)}$ is given by the sum of the $k^{th}$ column of $H_{5}^{(2)}$, denoted by $s(k)=\sum_{i}H_{ik}^{(2)}$, and $P$ permutes the columns. Using these identities in Eq. (93) we find that

Let us show now that this representative, which has been obtained by leaving out $H_{5}^{(4)}$, is equivalent to the set $\{I,F_{5},H_{5}^{(1)},H_{5}^{(2)},H_{5}^{(4)}\}$, for example. Indeed, the equivalence

follows immediately from multiplying the second set by $D$ from the left and using $D^{5}=I$,

Reordering the set of five matrices on the right reveals the desired equivalence with the quintuple $\{I,F_{5},H_{5}^{(1)},H_{5}^{(2)},H_{5}^{(3)}\}$. Effectively, the four matrices different from $I$ undergo a cyclic shift under multiplication with $D$, and the remaining equivalences follow from shifts induced by $D^{2}$ and $D^{3}$, respectively.

Summary and Discussion

We have constructed all inequivalent sets of mutually unbiased bases in dimension two to five. Our approach is based on the fact that all complex Hadamard matrices are known in these dimensions. For dimensions up to $d=4$, elementary arguments suffice to classify the existing sets of MU bases while dimension five requires some analytic results which have been found earlier using algebraic computer software.

The first four columns of Table 1 summarize the results obtained in this paper. All pairs of MU bases in dimensions two to five are listed in the first row, effectively reflecting the known classification of inequivalent Hadamard matrices; a continuous (one-parameter) set of inequivalent MU pairs only exists in dimension four.

The main results concern triples of MU bases in dimension four where we find a three-parameter family and in dimension five where we obtain two inequivalent triples.

Finally, we have shown that there is only one class of both MU quadruples and MU quintuples in dimensions four and five. In all dimensions considered, there is a unique $d$-tuple which can be extended to a complete set of $(d+1)$ MU bases using a construction presented in .

The last column of Table 1 contrasts these results with dimension six where the classification of all complex Hadamard matrices is not known to be complete. The first entry shows that there is a three-parameter family of pairs of MU bases (it has been conjectured that the parameter space has, in fact, four dimensions ). Furthermore, it is possible to construct a two-parameter family of triples using an idea taken from . There is strong numerical and analytical evidence to suggest that triples are the largest sets of MU bases in dimension six but this problem remains open.

The notion of equivalence used in this paper (see Appendix A) is mathematical in nature; it captures all possible operations that leave invariant the conditions (1) for two bases to be mutually unbiased. Motivated by experiments, there is a finer equivalence of complete sets of MU bases based on the entanglement structure of the states contained in each basis . For dimensions that are a power of two, a complete set of MU bases can be realized using Pauli operators acting on each two-dimensional subsystem. Two sets of MU bases are then called equivalent when they can be factored into the same number of subsystems. For $d=2,4$ this notion of equivalence also leads to a unique set of $(d+1)$ MU bases. However, for $d=8,16,\ldots$ complete sets of MU bases can have different entanglement structures even though they are equivalent up to an overall unitary transformation .

The traditional approach to find complete sets of MU bases in prime-power dimensions via the Heisenberg-Weyl group or by using finite fields is constructive and, therefore, does not exclude the existence of other inequivalent complete sets. The present approach is, in contrast, exhaustive: we are able to affirm that the known complete sets for $2\leq d\leq 5$ are unique (up to equivalence). Their uniqueness has been shown earlier for $d\leq 4$ while contains a proof for $2\leq d\leq 5$ in a Lie algebraic setting. We find it appealing that it is possible to prove the uniqueness of complete sets of MU bases in low dimensions by elementary methods.

Acknowledgements

The authors would like to thank M. Kibler for pointing out the complex conjuation equivalence relation given in Eq. (105). SB and SW gratefully acknowledge financial support through QNET, a network funded by the EPSRC, and IB has been supported by the Swedish Research Council.

References

Appendix A Equivalent sets of MU bases

Many sets of MU bases are identical to each other. To simplify the enumeration of all sets of MU bases we introduce equivalence classes and a standard form of sets of MU bases.

if they can be transformed into each other by a succession of the following four transformations:

an overall unitary transformation $U$ applied from the left,

which leaves invariant the value of all scalar products;

$(r+1)$ diagonal unitary transformations $D_{\rho}$ from the right which attach phase factors to each column of the $(r+1)$ matrices,

these transformations exploit the fact that the overall phase of a quantum state drops out of from the conditions of MU bases;

$(r+1)$ permutations of the elements within each basis,

which amount to relabeling the elements within each basis by means of unitary permutation matrices $P_{n}$ satisfying $PP^{T}=I$;

which leaves the values of all scalar products invariant.

We show that the two classes of triples of MU bases given by $\mathcal{T}^{(1)}\equiv\{I,F_{5},H_{5}^{(1)}\}$ and $\mathcal{T}^{(2)}\equiv\{I,F_{5},H_{5}^{(2)}\}$ are inequivalent. In a first step, we explain that it is sufficient to search for equivalence transformations generated by matrices of a special form. In a second step we show that a contradiction arises if one assumes that the triples $\mathcal{T}^{(1)}$ and $\mathcal{T}^{(2)}$ are equivalent.

with a unitary $U$ and $(r+1)$ monomial matrices $M_{i}$ being a product of diagonal unitaries with permutation matrices; to keep the notation simple we do not reorder the ($r+1$) bases within $\mathcal{M}^{\prime}$. For the set $\mathcal{M}^{\prime}$ to be in standard form, one of the bases in $\mathcal{M}$, say $B_{\rho}$, must be mapped to the identity. As a consequence, the overall unitary transformation $U$ must have a particular form, namely

where $N$ is some monomial matrix and $B_{\rho}$ is one of the matrices contained in the set $\mathcal{M}$.

In view of Eq. (107) we are lead to determine the action of $F_{5}^{\dagger}$ and $(H_{5}^{(1)})^{\dagger}$ on the triple $\mathcal{T}^{(1)}$ as well as the action of $F_{5}^{\dagger}$ and $(H_{5}^{(2)})^{\dagger}$ on the triple $\mathcal{T}^{(2)}$. It turns out that both triples are invariant under these global transformations as we have the equivalences

The first equivalence in (108) follows from using $F_{5}^{\dagger}=F_{5}P$ and Eq. (94) while the second one also requires the identity

with some monomial matrix $M$. The equivalences (109) are derived in a similar way.

Consequently, we can always remove the effect of the matrices $B_{\rho}^{\dagger}$ in the global transformations (107) which leaves us with

where $N,M_{1}$ and $M_{2}$ are monomial matrices, and up to rearranging terms. The non-zero entries of the monomial matrix $N$ must, in fact, be fifth roots of unity but we will not need this factAssume that $N$ has a nonzero element different from a fifth root, say $e^{i\alpha}$. This makes it impossible to transform $\mathcal{T}^{(1)}$ into standard form using right multiplication by monomial matrices unless the other nonzero elements of $N$ also equal $e^{i\alpha}$. It follows that $N$ must be a permutation matrix $P$ apart from a phase factor, $N=e^{i\alpha}P$. Thus the matrices $M_{\rho}$ must have a common factor of $e^{-i\alpha}$ which, however, is irrelevant for the definition of MU bases.

Using the restricted transformations shown in Eqs. (111), the triples $\{I,F_{5},H_{5}^{(1)}\}$ and $\{I,F_{5},H_{5}^{(2)}\}$ are equivalent to each other only if either

hold for some monomial matrices $M_{1}$ and $M_{2}$. The choice $M_{0}=N^{-1}=N^{\dagger}$ in Eqs. (111) ensures that the identity will be mapped to the identity.

Eqs. (112) will now be shown to imply the identity

Use $H_{5}^{(j)}=D^{j}F_{5},j=1,2$, to express the second equation in (112) as

We now show that Eq. (114) only holds if the matrix $\Delta$ is proportional to the identity. Write the monomial matrix $M$ in (114) in the form

where $P$ is a permutation matrix and $\Delta^{\prime\prime}$ is a diagonal matrix with entries having modulus one only. Denoting the inverse of $\Delta^{\prime\prime}$ by $\Delta^{\prime}$, Eq. (114) takes the form

Let us write $\Delta=\mbox{diag}(\alpha,\beta,\ldots,\epsilon)$ with phase factors $\alpha,\beta$, etc, and similarly for $\Delta^{\prime}$, and consider the simplest case $P\equiv I$. Then the matrix relation (118) reads explicitly

The conditions resulting from the first row immediately imply that the elements on the diagonal of $\Delta^{\prime}$ are all equal to $\alpha^{*}$, or $\Delta^{\prime}=\alpha^{*}I$. The conditions of the first column imply that the matrix $\Delta$ is also a multiple of the identity, namely $\Delta=\alpha I$. This contradicts the fact that the matrix $\Delta$ is different from a multiple of the identity.

Let us now drop the restriction the $P=I$. The effect of $P$ acting on $F_{5}$ from the right is to permute its columns. The first row of $F_{5}$ will not change under this operation. Under the action of $P$, the first column will either stay where is is or it will be mapped to one of the four others. In the first case, we can immediately apply the argument given above to derive a contradiction. In the second case, it it straightforward to see that a similar argument still applies involving the first row of the matrices and that column which is the image of the first column. Thus, all possible choices of the monomial matrix $M$ in (114) require $\Delta$ to be a multiple of the identity—which it is not.

Finally, we consider the action of an overall complex conjugation (105) on either of the triples. We find that the set of three MU bases, $\mathcal{T}^{(1)},$ remains invariant under complex conjugation

Similarly, complex conjugation maps $\mathcal{T}^{(2)}$ to itself, $\mathcal{T}^{(2)\,*}\sim\mathcal{T}^{(2)}$. In summary, we have shown that the equivalence relations (101) to (105) cannot transform the triple $\mathcal{T}^{(1)}$ into $\mathcal{T}^{(2)}$ or vice versa, i.e. these triples are inequivalent.