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On mutually unbiased unitary bases in prime-dimensional Hilbert spaces

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Akin to the idea of complete sets of mutually unbiased bases for prime-dimensional Hilbert spaces, \(\mathcal {H}_d\), we study its analogue for a d-dimensional subspace of \(M (d,\mathbb {C})\), i.e. mutually unbiased unitary bases (MUUBs) comprising of unitary operators. We note an obvious isomorphism between the vector spaces and beyond that, we define a relevant monoid structure for \(\mathcal {H}_d\) isomorphic to one for the subspace of \(M (d,\mathbb {C})\). This provides us not only with the maximal number of such MUUBs, but also a recipe for its construction.

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  1. This can be understood as a composition of the isomorphism from \(\mathcal {H}_d\) to \(\mathbb {C}^d\) and from the latter to \(M_s\).

  2. We should note that the word ‘corresponding’ is used to highlight the simple fact that one cannot determine the nature of entanglement simply based on states in \( \text {End}(\mathbb {C}^d)\).


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Appendix A The monoid \((\mathcal {H}_d,\bullet ,|0\rangle )\)

We show that \((\mathcal {H}_d,\bullet ,|0\rangle )\) is a monoid. We begin first by showing that \(\bullet \) is closed on \(\mathcal {H}_d\).

\( \vert \varphi \rangle \bullet \vert \phi \rangle =\sum _{i} a_i \vert i \rangle \bullet \sum _{j} b_j \vert j \rangle =\sum _{m} \alpha _m \vert m \rangle \). Note that \(\alpha _m \in \mathbb {C}~, \forall ~m\) , therefore \(\sum _{m} \alpha _m \vert m \rangle \in \mathcal {H}_d\), hence the set is closed under the operation \(\bullet \).

We next show that the operation is associative on \(\mathcal {H}_d\).

$$\begin{aligned} (\vert \varphi \rangle \bullet \vert \phi \rangle ) \bullet \vert \varPhi \rangle =\sum _{m} \alpha _m \vert m \rangle \bullet \sum _{k} c_k \vert k \rangle \end{aligned}$$

where \(\alpha _m=\sum _{i,j} a_i b_j ~and~ i \oplus j =m\). Then,

$$\begin{aligned} \sum _{m} \alpha _m \vert m \rangle \bullet \sum _{k} c_k \vert k \rangle =\sum _{n} \beta _n \vert n \rangle \end{aligned}$$

where \(\beta _n=\sum _{m,k} \alpha _m c_k ~ and~ m \oplus k =n\) or \(\beta _n=\sum _{i,j,k} a_i b_j c_k\) with \(i \oplus j \oplus k =n\).

$$\begin{aligned} \sum _{n} \beta _n \vert n \rangle&=\sum _{i, j, k} a_i b_j c_k \vert i \oplus j \oplus k \rangle \nonumber \\&=\sum _{i,j,k} ( a_i)( b_j c_k) \vert i \oplus j \oplus k \rangle \nonumber \\&=\left( \sum _{i} a_i \vert i \rangle \right) \bullet \left( \sum _{j,k} b_j c_k \vert j \oplus k \rangle \right) \nonumber \\&=\vert \varphi \rangle \bullet (\vert \phi \rangle \bullet \vert \varPhi \rangle ) \end{aligned}$$

where \(\vert \phi \rangle \bullet \vert \varPhi \rangle =\sum _{j,k} b_j c_k \vert j \oplus k \rangle \). Hence, \(\bullet \) is associative.

To complete the requirement of being a monoid, we show the existence of an identity in \(\mathcal {H}_d\) with respect to the operation \(\bullet \). Let \(\vert e \rangle =\sum _{i} a_i \vert i \rangle \) and \(\sum _{i,j} a_i b_j \vert i \oplus j \rangle \) and \( i \oplus j=m\). Then, \(\vert e \rangle \) is an identity if

$$\begin{aligned} a_i = {\left\{ \begin{array}{ll} 1 &{} \text {if }i=0 \\ 0 &{} \text {if }i \ne 0 ~. \end{array}\right. } \end{aligned}$$

This effectively means \(|e\rangle =|0\rangle \). We write it in the above form to ensure that the operation with any state in \(\mathcal {H}_d\) follows from the definition for \(\bullet \). Thus \(\vert e \rangle \bullet \sum _j b_j \vert j \rangle =\sum _{i} a_i \vert i \rangle \bullet \sum _j b_j \vert j \rangle = \vert 0 \rangle \bullet \sum _j b_j \vert j \rangle =\sum _j b_j \vert 0 \oplus j \rangle =\sum _m b_m \vert m \rangle \). On the other hand, \(\sum _j b_j \vert j \rangle \bullet \vert e \rangle =\sum _j b_j \vert j \rangle \bullet \sum _{i} a_i \vert i \rangle =\sum _j b_j \vert j \rangle \bullet \vert 0 \rangle =\sum _j b_j \vert j \oplus 0 \rangle = \sum _m b_m \vert m \rangle \). Hence \(\vert e \rangle =\vert 0 \rangle \) is an identity element of \(\mathcal {H}_d\).

In the following, we show that the operation is also distributive on \(\mathcal {H}_d\) with respect to vector sum. Consider

$$\begin{aligned} \vert \varphi \rangle \bullet (\vert \phi \rangle + \vert \varPhi \rangle )&=\sum _{i} a_i \vert i \rangle \bullet \left( \sum _{j} b_j \vert j \rangle + \sum _{j} c_j \vert j \rangle \right) =\sum _{s} \omega _s \vert s \rangle \end{aligned}$$

where \(\omega _s=\sum _{i,j} a_i (b_j + c_j)\) and \(i \oplus j = s\).

$$\begin{aligned} \sum _{s} \omega _s \vert s \rangle&=\sum _{i,j} a_i (b_j + c_j) \vert i \oplus j \rangle \nonumber \\&=\sum _{i,j} a_i b_j \vert i \oplus j \rangle + \sum _{i,j} a_i c_j \vert i \oplus j \rangle \nonumber \\&=\left( \sum _{i} a_i \vert i \rangle \bullet \sum _{j} b_j \vert b_j \rangle \right) +\left( \sum _{i} a_i \vert i \rangle \bullet \sum _{j} c_j \vert j \rangle \right) \nonumber \\&=(\vert \varphi \rangle \bullet \vert \phi \rangle ) +( \vert \varphi \rangle \bullet \vert \varPhi \rangle )~. \end{aligned}$$

Thus, \(\bullet \) is distributive.

Appendix B

As noted from the text, the subspace spanned by \(\{|\mathcal {U}_0\rangle ,\ldots ,|\mathcal {U}_{d-1}\rangle \}\) is isomorphic to \(\mathbb {C}^d\) and a basis for the latter can be written as \(\{|i_U\rangle \}_{i=0}^{d-1}\). Hence, \(\{|i_U\rangle \langle j_U|~|~i,j=0,\ldots ,d-1\}\) is an orthonormal basis for the \(d^2\) space of \(\text {End}(\mathbb {C}^d)\). In order to ascertain if any of these states, \(\rho \in \text {End}(\mathbb {C}^d)\) is (actually) a MES, we need to reconsider the subsystems, of the state. In this case, the subsystems are two d-dimensional states and we refer to the subsystems later as systems 1 and 2. Let us thus consider the entangled state from the Choi isomorphism from the unitary \(U_i\) given as \(|\mathcal {U}_i\rangle \),

$$\begin{aligned} |\mathcal {U}_i\rangle =\left( \sum _{m}\sum _n \langle n|U_i|m\rangle |m\rangle |n\rangle \right) /\sqrt{d}~. \end{aligned}$$

The set \(\{|\mathcal {U}_i\rangle \langle \mathcal {U}_j|~|~i,j=0,\ldots ,d-1\}\) is an orthonormal basis for a \(d^2\) subspace of \(\text {End}(\mathbb {C}^d\otimes \mathbb {C}^d)\). Let us call this subspace \(M_{d^2}\).

We show that a function \(F:\text {End}\left( \mathbb {C}^d\right) \rightarrow M_{d^2}\) with the rule

$$\begin{aligned} F\left( \sum _{i,j}a_{i,j}|i_U\rangle \langle j_U|\right) =\sum _{i,j}a_{i,j}|\mathcal {U}_i\rangle \langle \mathcal {U}_j| \end{aligned}$$

is an isomorphism. Note that F is injective,

$$\begin{aligned}&F\left( \sum _{i,j}a_{i,j}|i_U\rangle \langle j_U|\right) =F\left( \sum _{i,j}b_{i,j}|i_U\rangle \langle j_U|\right) \nonumber \\&\quad \Rightarrow \sum _{i,j}a_{i,j}|\mathcal {U}_i\rangle \langle \mathcal {U}_j|=\sum _{i,j}b_{i,j}|\mathcal {U}_i\rangle \langle \mathcal {U}_j|\nonumber \\&\quad \therefore \sum _{i,j}a_{i,j}|\mathcal {U}_i\rangle \langle \mathcal {U}_j|-\sum _{i,j}b_{i,j}|\mathcal {U}_i\rangle \langle \mathcal {U}_j|=0\nonumber \\&\quad \therefore \forall i,j,~a_{i,j}=b_{i,j}\Rightarrow \sum _{i,j}a_{i,j}|i_U\rangle \langle j_U|=\sum _{i,j}b_{i,j}|i_U\rangle \langle j_U|~. \end{aligned}$$

The surjectiveness of F is obvious (any element in \(M_{d^2}\) can be written as \(\sum _{i,j}k_{i,j}|\mathcal {U}_i\rangle \langle \mathcal {U}_j|=F\left( \sum _{i,j}k_{i,j}|i_U\rangle \langle j_U|\right) \). To show the linearity of F;

$$\begin{aligned}&F\left( \alpha \sum _{i,j}a_{i,j}|i_U\rangle \langle j_U|+\sum _{i,j}b_{i,j}|i_U\rangle \langle j_U|\right) \nonumber \\&\quad =F\left( \sum _{i,j}(\alpha a_{i,j}+b_{i,j})|i_U\rangle \langle j_U|\right) =\sum _{i,j}(\alpha a_{i,j}+b_{i,j})|\mathcal {U}_i\rangle \langle \mathcal {U}_j|\nonumber \\&\quad =\sum _{i,j}\alpha a_{i,j}|\mathcal {U}_i\rangle \langle \mathcal {U}_j|+b_{i,j}|\mathcal {U}_i\rangle \langle \mathcal {U}_j|\nonumber \\&\quad =\alpha F\left( \sum _{i,j}a_{i,j}|i_U\rangle \langle j_U|\right) +F\left( \sum _{i,j}b_{i,j}|i_U\rangle \langle j_U|\right) ~. \end{aligned}$$

Let \(\rho \in \text {End}(\mathbb {C}^d)\) and \(\rho _M=F(\rho )\in \text {End}(\mathbb {C}^d\otimes \mathbb {C}^d)\). It is easy to see that to ascertain whether \(\rho \) corresponds to some MES, we consider if \(\text {Tr}_1(\rho _M)=\text {Tr}_2(\rho _M)=\mathbb {I}_d/d\) where the subscripts 1 and 2 refer to the subsystems the partial trace is taken over. Let us refer to the space \(H_{M}\) as the smallest subspace of Hermitian operators \(M_{d^2}\) which contains all MES in \(M_{d^2}\). Following [9], we can say that all MES in \(M_{d^2}\) is a convex set of the subspace

$$\begin{aligned} H_M=\{J\in \text {End}(\mathbb {C}^d\otimes \mathbb {C}^d)~|~\text {Tr}_1(J)=\text {Tr}_2(J)= \text {Tr}(J)\mathbb {I}_d/d\}~. \end{aligned}$$

Note that every element in \(M_{d^2}\) can be written as a \(\sum r_{ik}\lambda _i\otimes \lambda _k\) where \(\{\lambda _i\}_{i=0}^{d^2-1}\) is an orthonormal Hermitian operator basis for \(\text {End}(\mathbb {C}^d)\) with \(\lambda _0=\mathbb {I}/\sqrt{d}\) and \(r_{ik}\in \mathbb {R}\). The subspace \(H_M\), however, requires \(r_{k0}=r_{0k}=0\) which implies that every element in \(H_M\) is also an element in \(M_{d^2}\), but not vice versa. Thus \(H_M\) is a proper subset of \(M_{d^2}\); and \(\text {dim}~H_M<d^2\).

Finally, given the isomorphism F, we can identify the smallest subspace of all operators in \(\text {End}(\mathbb {C}^d)\) containing operators corresponding to MES, call it \(H_{mes}\), with dimensionality equal to \(\text {dim}~H_M\).

Appendix C

Let d be some prime number, \(a_1,a_2=0,1,\ldots ,d-1\) and p be integers. Let \(l=a_1p\mod d=a_2p\mod d\). There exists integers \(n_1,n_2\) such that,

$$\begin{aligned} a_1 p-l&=n_1 d\nonumber \\ a_2 p-l&=n_2 d~. \end{aligned}$$

With \(p\ne 0\), we have

$$\begin{aligned} (a_1-a_2)=\left( \dfrac{n_1-n_2}{p}\right) d~. \end{aligned}$$

We shall show that Eq. (39) is only true when \(a_1=a_2\). Let us first consider the case \(p > |n_1-n_2| \), implying that \((n_1-n_2)/p\) is a fraction. As d is prime greater than 2, therefore \(a_1- a_2\) is a fraction. However, as \(a_1, a_2 \in [0, d-1]\), therefore \(a_1-a_2\) cannot be a fraction. Thus, p cannot be greater than \(|n_1-n_2|\).

Now, If \(p \le |n_1-n_2|,\) then this implies that \(|a_1-a_2| \ge d\). However, note that as \(|a_1-a_2|< d-1\), hence p is neither less than nor equal to \(|n_1-n_2|\). Thus Eq. (39) is true if \(a_1-a_2=n_1-n_2=0\). In turn, if \(a_1\ne a_2\), therefore we have \(a_1p\mod d\ne a_2p\mod d\). Further to that, \(l=ap\mod d\) is defined for all values of \(a=0,\ldots ,d-1\), thus \(l=0,\ldots ,d-1\).

Appendix D

In showing that \((\mathbb {X}_d^b\mathbb {Z}_d^a)^i(\mathbb {X}_d^b\mathbb {Z}_d^a)^j=(\mathbb {X}_d^bZ_d^a)^{i\oplus j}\), we first need to prove the following identity,

$$\begin{aligned} (\mathbb {X}_d^b\mathbb {Z}_d^a)^n=\omega ^{ab(n^2-n)/2} \mathbb {X}_d^{bn}\mathbb {Z}_d^{an}~. \end{aligned}$$

The proof is by induction. It is obvious that for \(n=1\)

$$\begin{aligned} (\mathbb {X}_d^b\mathbb {Z}_d^a)^1=\omega ^{ab(1^2-1)/2}\mathbb {X}_d^{b} \mathbb {Z}_d^{a}=\mathbb {X}_d^b\mathbb {Z}_d^a~. \end{aligned}$$

Now assume this is true for \(n=k\)

$$\begin{aligned} (\mathbb {X}_d^b\mathbb {Z}_d^a)^k=\omega ^{ab(k^2-k)/2}\mathbb {X}_d^{bk} \mathbb {Z}_d^{ak}~, \end{aligned}$$

we need to determine for \(n=k+1\),

$$\begin{aligned} (\mathbb {X}_d^b\mathbb {Z}_d^a)^{k+1}=\omega ^{ab(k^2-k)/2}\mathbb {X}_d^{bk} \mathbb {Z}_d^{ak}\mathbb {X}^b\mathbb {Z}_d^a~. \end{aligned}$$

As we swap the positions of \(\mathbb {X}_d^b\) and \(\mathbb {Z}_d^ak\), we will make a total number of abk swaps and thus introduce the term \(\omega ^{abk}\). Hence

$$\begin{aligned} (\mathbb {X}_d^b\mathbb {Z}_d^a)^{k+1}&=\omega ^{ab(k^2-k)/2}\omega ^{abk} \mathbb {X}_d^{b(k+1)}\mathbb {Z}_d^{a(k+1)}\nonumber \\&=\omega ^{ab(k^2-k+2k)/2}\mathbb {X}_d^{b(k+1)}\mathbb {Z}_d^{a(k+1)}\nonumber \\&=\omega ^{ab[(k+1)^2-(k+1)]/2}\mathbb {X}_d^{b(k+1)}\mathbb {Z}_d^{a(k+1)}~. \end{aligned}$$

Now, if \(n=d>2\) (the case for \(d=2\) is trivial), we have

$$\begin{aligned} (\mathbb {X}_d^b\mathbb {Z}_d^a)^d=\omega ^{ab(d^2-d)/2}\mathbb {X}_d^{bd} \mathbb {Z}_d^{ad}~. \end{aligned}$$

As \(ab(d^2-d)/2=abd(d-1)/2\), and \(d-1\) is an even number (because d is prime), hence \(abd(d-1)/2\) is some positive multiple, M, of d or \(abd(d-1)/2=Md\)). Note that \(\omega ^{Md}=(\omega ^d)^{M}=1\). Thus

$$\begin{aligned} (\mathbb {X}_d^b\mathbb {Z}_d^a)^d=\mathbb {X}_d^{bd}\mathbb {Z}_d^{ad}= (\mathbb {X}_d^{d})^b(\mathbb {Z}_d^{d})^a=\mathbb {I} \end{aligned}$$

as \(\mathbb {X}_d^b=\mathbb {Z}_d^a=\mathbb {I}.\) Noting that, \(i+j=Nd+(i\oplus j)\), for some positive integer N, \((\mathbb {X}_d^b\mathbb {Z}_d^a)^i(\mathbb {X}_d^b\mathbb {Z}_d^a)^j=(\mathbb {X}_d^bZ_d^a)^{i\oplus j}\) follows straightforwardly.

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Nasir, R.N.M., Shaari, J.S. & Mancini, S. On mutually unbiased unitary bases in prime-dimensional Hilbert spaces. Quantum Inf Process 18, 178 (2019).

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