# Interpreting symplectic linear transformations in a two-qubit phase space

William K. Wootters  Department of Physics, Williams College, Williamstown,
Massachusetts 01267, USA

###### Abstract

For the continuous Wigner function and for certain discrete Wigner functions,
permuting the values of the Wigner function in accordance with a symplectic
linear transformation is equivalent to performing a certain unitary
transformation on the state. That is, performing this unitary transformation
is simply a matter of moving Wigner-function values around in phase space.
This result holds in particular for the simplest discrete Wigner function
defined on a $d\times d$ phase space when the Hilbert-space dimension $d$ is
odd. It does not hold for a $d\times d$ phase space if the dimension is even.
Here we show, though, that a generalized version of this correspondence does
apply in the case of a two-qubit phase space. In this case, a symplectic
linear permutation of the points of the phase space, together with a certain
reinterpretation of the Wigner function, is equivalent to a unitary
transformation.

###### pacs:

##  I Introduction

In classical mechanics, the concept of a phase space, with dimensions
representing coordinates and momenta, is utterly central. While it is less
central in quantum mechanics, phase-space representations of quantum states
such as the Wigner function have proved fruitful in a wide variety of
applications Hillery ; Weinbub . A particular point of contact between
classical phase space and quantum phase space is the role played by symplectic
transformations. In the quantum case, a symplectic linear transformation of
the phase space corresponds to a unitary transformation in the Hilbert space
Arvind ; Ferraro . For example, to perform a unitary squeezing operation on
the Wigner function representing a quantum state, one simply squeezes the
Wigner function itself: its values are moved around in phase space in
accordance with the appropriate linear transformation. In the case of
classical mechanics, a canonical transformation is typically not linear but
its Jacobian at every point is a symplectic matrix.

To a certain extent, the strong connection between symplectic linear
transformations in phase space and a particular class of unitary
transformations in Hilbert space carries over to the discrete Wigner function,
which is used to represent states of a system with a finite-dimensional
Hilbert space. The discrete phase space we consider in this paper is a two-
dimensional vector space over a finite field Gibbons ; Klimov ; Vourdas ;
Gross . Such a phase space, when the field has $d$ elements, is appropriate
for representing a system with a $d$-dimensional Hilbert space. If $d$ is odd,
it is possible to define on this phase space, essentially uniquely, a discrete
Wigner function that exhibits the same kind of correspondence between
symplectic transformations and unitary transformations that one sees in the
continuous Wigner function Klimov ; Vourdas ; Gross ; Bengtsson ; Zhu3 . That
is, if $L$ is a symplectic linear transformation of the phase space—in our
setting this simply means that $L$ can be represented by a unit-determinant
$2\times 2$ matrix with entries in the finite field—then there exists a
unitary transformation $U_{L}$ corresponding to $L$ such that

$W_{\alpha}(U_{L}\rho U_{L}^{\dagger})=W_{L^{-1}\alpha}(\rho).$ (1)

Here $W_{\alpha}(\rho)$ is the Wigner function representing the density matrix
$\rho$, evaluated at the point $\alpha$ in the discrete phase space. Thus, in
order to find the value of the post-transformation Wigner function at the
point $\alpha$, one simply asks what point in phase space was sent to $\alpha$
by $L$. The value of the pre-transformation Wigner function at that point,
$L^{-1}\alpha$, has now been moved to the point $\alpha$.

The connection (1) between symplectic linear transformations and unitary
transformations does not hold for a $d\times d$ phase space when the dimension
$d$ is a power of 2. This is unfortunate, since systems of qubits are of great
interest in quantum information science.

We can see how Eq. (1) fails in the case of a single qubit. In that case, we
adopt the following definition of the discrete Wigner function Feynman ;
Wootters . First, the phase-space points are labeled by
$\alpha=(\alpha_{q},\alpha_{p})$, where $\alpha_{q}$ and $\alpha_{p}$ take
values in the two-element field ${\mathbb{F}}_{2}=\\{0,1\\}$. (We actually
treat $\alpha$ as a column vector—a matrix acting on it acts from the left—but
we sometimes write it as a row vector to make the typesetting less awkward.)
We picture $\alpha_{q}$ as the horizontal coordinate and $\alpha_{p}$ as the
vertical coordinate, and we picture the origin, $\alpha=(0,0)$, in the lower
left-hand corner of the $2\times 2$ grid. For the qubit density matrix $\rho$,
the corresponding Wigner function is defined as

$W_{\alpha}(\rho)=\tfrac{1}{2}\hbox{Tr}(A_{\alpha}\rho),$ (2)

where the phase point operators $A_{\alpha}$ are given by

$A_{(0,0)}=\tfrac{1}{2}[I+X+Y+Z]$ (3)

and

$\begin{split}&A_{(1,0)}=XA_{(0,0)}X,\\\ &A_{(1,1)}=YA_{(0,0)}Y,\\\
&A_{(0,1)}=ZA_{(0,0)}Z.\end{split}$ (4)

Here $X$, $Y$, and $Z$ are the Pauli matrices. Note that the $A$’s constitute
an orthogonal basis for the real vector space of Hermitian $2\times 2$
matrices: $\hbox{Tr}(A_{\alpha}A_{\beta})=2\delta_{\alpha\beta}$. With this
definition, the property of the Wigner function expressed in Eq. (1) is
equivalent to the following property of the phase point operators:

$U_{L}A_{\alpha}U^{\dagger}_{L}=A_{L\alpha}.$ (5)

There are exactly six symplectic matrices $L$ for this simple phase space.
Three of the six are of the form $R^{s}$, with $s=0,1,2$, where $R$ is the
“rotation” matrix

$R=\left(\begin{array}[]{cc}1&1\\\ 1&0\end{array}\right).$ (6)

This transformation cyclically permutes the three non-zero points of the phase
space in a counter-clockwise sense:
$(1,0)\rightarrow(1,1)\rightarrow(0,1)\rightarrow(1,0)$. For $R$, and
therefore for its powers, Eq. (5) does hold: there exists a unitary
transformation $U_{R}$, unique up to a phase factor, such that

$U_{R}A_{\alpha}U^{\dagger}_{R}=A_{R\alpha}.$ (7)

Specifically,

$U_{R}=\tfrac{1}{2}e^{i\pi/4}[I-i(X+Y+Z)]=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&-i\\\
1&i\end{array}\right).$ (8)

This transformation satisfies Eq. (7) because, when applied by conjugation to
the Pauli matrices, it effects the permutation

$X\rightarrow Y\rightarrow Z\rightarrow X.$ (9)

In fact, Eq. (7) requires this permutation, as one can see by writing the
Pauli matrices as linear combinations of the $A$’s. The other three symplectic
matrices are of the form $R^{s}F$, where $F$ is the axis-interchanging
transformation

$F=\left(\begin{array}[]{cc}0&1\\\ 1&0\end{array}\right).$ (10)

In order to satisfy Eq. (5) with $L$ equal to $F$, we would need a unitary
that effects the permutation

$X\rightarrow Z\rightarrow X$ (11)

while leaving $Y$ entirely unchanged. This transformation would be a
reflection of the Bloch sphere through a certain plane, which is not unitary.
Thus there is no correspondence between symplectic matrices and unitary
operators in the case of a single qubit, and a similar problem arises for any
number of qubits Zhu ; Raus1 ; Feldmann .

One way to deal with the problematic aspects of a system of qubits is to
modify the phase space. This approach has been taken in Ref. Raussendorf2 ,
which employs an expanded phase space with an overcomplete set of phase point
operators, and in Ref. Kocia , in which the points of phase space are labeled
by elements of a Grassmann algebra. An earlier example of an expanded phase
space for a system of qubits is the $2d\times 2d$ phase space proposed by
Leonhardt Leonhardt1 ; Leonhardt2 , for which the associated Wigner function
has been shown to exhibit symplectic covariance Hashimoto . In the present
paper we leave the $d\times d$ phase space unchanged but allow the possibility
of reinterpreting its points and lines after a transformation. We focus on a
specific case, namely, a two-qubit system, and work out what I expect is the
simplest generalization of the property (1) that one can hope to apply to the
phase space ${\mathbb{F}}_{d}^{2}$ when $d$ is a power of two.

The idea can be illustrated in the single-qubit case by considering the above
axis-interchanging transformation $F$. Let us associate with this
transformation the unitary operator

$U_{F}=\frac{1}{\sqrt{2}}(Z-X)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{cc}1&-1\\\
-1&-1\end{array}\right).$ (12)

Starting with a density matrix $\rho$, if we apply $U_{F}$ to $\rho$, the
effect of $U_{F}$ is equivalent to permuting the values of $W(\rho)$ in phase
space according to $F$, and reinterpreting these values in accordance with a
different definition of the Wigner function. Specifically, one can show that

$\widetilde{W}_{\alpha}(U_{F}\rho U_{F}^{\dagger})=W_{F^{-1}\alpha}(\rho),$
(13)

where the definition of $\widetilde{W}$ is closely analogous to that of $W$:

$\widetilde{W}_{\alpha}(\rho)=\tfrac{1}{2}\hbox{Tr}(\tilde{A}_{\alpha}\rho),$
(14)

with

$\tilde{A}_{(0,0)}=\tfrac{1}{2}[I-X-Y-Z],$ (15)

and

$\begin{split}&\tilde{A}_{(1,0)}=X\tilde{A}_{(0,0)}X,\\\
&\tilde{A}_{(1,1)}=Y\tilde{A}_{(0,0)}Y,\\\
&\tilde{A}_{(0,1)}=Z\tilde{A}_{(0,0)}Z.\end{split}$ (16)

That is, one obtains $\tilde{A}_{\alpha}$ for nonzero $\alpha$ just as before
from the phase point operator associated with the origin, but the latter
operator is now defined differently. Thus it is still the case that we can
express the effect of the unitary transformation $U_{F}$ by letting $L$ tell
us how to permute the values of the Wigner function, but we must reinterpret
those values with respect to a different operator basis, that is, a different
set of phase point operators.

For a single qubit, there is no need for more than the two operator bases we
have seen here. For the case of two qubits, which is the main focus of this
paper, we show below that we can do something analogous using exactly twelve
operator bases, that is, twelve distinct sets of phase point operators. Each
such set can be generated by specifying the phase point operator associated
with the origin and then using standard displacement operators—generalizations
of the $X$, $Y$, and $Z$ used above—to get the other $A$’s.

Discrete Wigner functions have been applied with considerable success to
problems in quantum computation. Typically—for example in Refs. Raus1 ;
Feldmann ; Veitch1 ; Howard ; Veitch2 ; Delfosse ; Liu —the phase space one
uses for such applications, when the system’s dimension is $d=p^{n}$ with $p$
a prime number, is the $2n$-dimensional vector space ${\mathbb{Z}}_{p}^{2n}$,
as proposed in Refs. Wootters ; Gross ; Galetti . This choice contrasts with
the phase space we use here, that is, the two-dimensional vector space
${\mathbb{F}}_{d}^{2}$ over the $d$-element field. For many purposes the
difference between the two phase spaces is inconsequential Gross ; however, it
does make a difference in the interpretation of Eq. (1). As noted above, in
this paper we take the matrix $L$ in Eq. (1) to be a $2\times 2$ matrix with
entries in ${\mathbb{F}}_{d}$, whereas the higher-dimensional phase space
naturally suggests making $L$ a $2n\times 2n$ matrix with entries in
${\mathbb{Z}}_{p}$. As a consequence, our focus will be on what has been
called the “restricted Clifford group” rather than the full Clifford group
Bengtsson ; Zhu1 ; Zhu ; Appleby . (We say more about this distinction in
Section II below.) Because of this difference, it is not immediately clear how
the results we obtain in this paper might be relevant to the works cited above
on quantum computation. On the other hand, other authors have used the
${\mathbb{F}}_{d}^{2}$ phase space to analyze quantum computation Galvao ;
Paz1 ; Paz2 ; vanDam . Moreover, there are other studies in which Eq. (1),
with $L$ interpreted as we interpret it here, has played an important role
Braasch1 ; Braasch2 , and the fact that that equation does not apply to
systems of qubits has limited the scope of those studies. So there are reasons
to look for a generalization of the equation in the ${\mathbb{F}}_{d}^{2}$
setting.

It is a pleasure to offer this paper as a contribution to a special issue
celebrating the birthday of Alexander Holevo. In fact one can identify a
genuine historical chain connecting the work presented in this paper to some
of Prof. Holevo’s work. Much of his early work concerns the extraction of
classical information from quantum systems Holevo1 ; Holevo2 ; Holevo3 ;
Holevo4 ; Holevo5 , and these early results certainly motivated other
researchers to focus their attention on this problem. Thinking about
extracting classical information from quantum systems is precisely what led a
number of authors to study mutually unbiased bases, that is, orthonormal bases
for a Hilbert space such that each vector in any one of the bases is an equal-
magnitude superposition of the vectors in any of the other bases. Such bases
are useful when one wants to extract classical information from a large
collection of identically prepared quantum systems Ivanovic ; Fields ; Petz .
Finally, the discrete phase space we use in this paper is intimately connected
to mutually unbiased bases: as we describe in the following section, each
striation of the phase space, that is, each complete set of parallel lines,
defines an orthogonal basis for the Hilbert space, and the bases associated
with different striations are mutually unbiased Gibbons ; Bandyopadhyay ;
Pittenger .

Our construction in this paper follows the basic strategy of Ref. Gibbons ,
but we introduce, in Section III, a new element that makes the labeling of the
possible phase point operators more systematic. First, though, we describe in
Section II the discrete phase space and the displacement operators for a
system of two qubits, and we show how the action of each linear symplectic
transformation $L$ on the points of this phase space can be mimicked by the
action of a corresponding unitary operator $U_{L}$ on the displacement
operators. Section III shows how to construct, non-uniquely, a discrete Wigner
function satisfying two requirements: (i) the Wigner function must be
covariant under displacements, and (ii) the sums of the Wigner function over
the lines in phase space having a particular slope must equal the
probabilities of the outcomes of the orthogonal measurement associated with
that slope. It is at this point that we introduce our systematic labeling
scheme, and we single out one particular definition of the Wigner function to
call “standard.” To keep track of the effect of the unitaries $U_{L}$ on the
phase point operators, we introduce in Section IV the notion of the “index” of
a phase point operator. Shifting to a different definition of the Wigner
function, that is, shifting to a different interpretation of the Wigner-
function values, is a matter of changing the index of the phase point operator
associated with the origin of phase space. We describe such reinterpretations
in Section V and give a couple of examples. Finally, we summarize and discuss
the results in Section VI.

Though in this paper we focus on the case of just two qubits, there is no
reason to doubt that many of our results can be extended to an arbitrary
number of qubits. Our construction relies heavily on the existence of a
symplectic linear transformation that cycles through all the possible slopes
of phase-space lines—we think of it as a rotation—and it is known that such a
transformation exists for the $d\times d$ finite-field phase space whenever
$d$ is a power of two Chau ; Sussman . So it is reasonable to hope such an
extension can be found. In the meantime, even our results for a pair of qubits
give us a novel way of representing operations on this simple but conceptually
interesting quantum system.

##  II A two-qubit phase space

Again, the phase space we consider here is a two-dimensional vector space over
a finite field. For a system of two qubits, the relevant field is the four-
element field, ${\mathbb{F}}_{4}$. Let the elements of this field be called
$\\{0,1,\omega,\bar{\omega}\\}$. Addition and multiplication in this field are
given by the following tables.

$+$ | 0 | 1 | $\omega$ | $\bar{\omega}$
---|---|---|---|---
0  | 0 | 1 | $\omega$ | $\bar{\omega}$
1  | 1 | 0 | $\bar{\omega}$ | ${\omega}$
$\omega$ | $\omega$ | $\bar{\omega}$ | 0 | 1
$\bar{\omega}$ | $\bar{\omega}$ | ${\omega}$ | 1 | 0
$\times$ | 0 | 1 | $\omega$ | $\bar{\omega}$
---|---|---|---|---
0  | 0 | 0 | 0 | 0
1  | 0 | 1 | ${\omega}$ | $\bar{\omega}$
$\omega$ | 0 | ${\omega}$ | $\bar{\omega}$ | 1
$\bar{\omega}$ | 0 | $\bar{\omega}$ | 1 | $\omega$

As in the Introduction, a point $\alpha$ in the phase space has horizontal and
vertical coordinates $\alpha_{q}$ and $\alpha_{p}$; these coordinates now take
values in ${\mathbb{F}}_{4}$. So the two-qubit phase space can be pictured as
a $4\times 4$ grid.

The field arithmetic allows us to identify lines and parallel lines in the
phase space: a line consists of points $\alpha$ satisfying an equation of the
form

$a\alpha_{q}+b\alpha_{p}=c,$ (17)

where $a$, $b$, and $c$ are in ${\mathbb{F}}_{4}$ with $a$ and $b$ not both
zero. Two lines are parallel if they can be represented by equations of this
form that differ only in the value of $c$. A complete set of parallel lines is
called a striation—in this $4\times 4$ phase space there are exactly five
striations. It will also be useful to refer to the rays in phase space, which
are the lines running through the origin (so that $c$ equals zero).

In order to distinguish our two qubits, we choose, for each of the two phase-
space axes, a basis for the field, that is, a pair of field elements
$(e_{1},e_{2})$ such that any element $x$ of the field can be written as

$x=x_{1}e_{1}+x_{2}e_{2},$ (18)

$x_{1}$ and $x_{2}$ being elements of ${\mathbb{F}}_{2}=\\{0,1\\}$. The
coefficients $x_{1}$ and $x_{2}$ will be associated with the first and second
qubits, respectively, in a sense to be explained in the following paragraph.
In order to guarantee the connection with mutually unbiased bases that we rely
on in our definition of the Wigner function, we need to choose the bases we
assign to the two axes to be the duals of each other, at least up to a
multiplicative constant Gibbons ; Bengtsson ; Appleby . For the rest of this
paper we adopt the specific basis $(e_{1},e_{2})=(\bar{\omega},{\omega})$ for
each axis. This basis happens to be self-dual, thus giving us a certain
symmetry between the two axes.

A displacement of the phase space by the vector $\beta$ is a mapping that
takes each point $\alpha$ of phase space to the point $\alpha+\beta$. With
each such displacement we associate a unitary transformation as follows. Let
$\beta=(\beta_{q},\beta_{p})$ be the displacement vector. Here $\beta_{q}$ and
$\beta_{p}$ are elements of ${\mathbb{F}}_{4}$, and they can be expanded in
the basis $(\bar{\omega},{\omega})$:

$\beta_{q}=\beta_{q1}\bar{\omega}+\beta_{q2}{\omega},\hskip
28.45274pt\beta_{p}=\beta_{p1}\bar{\omega}+\beta_{p2}{\omega}.$ (19)

We define the unitary displacement operator $D_{\beta}$ to be

$D_{\beta}=e^{i\phi}X^{\beta_{q1}}Z^{\beta_{p1}}\otimes
X^{\beta_{q2}}Z^{\beta_{p2}}$ (20)

with a phase $\phi$ that may depend on $\beta$. To fix this phase, we choose
each displacement operator to be of the form
$D_{\beta}=\sigma_{j}\otimes\sigma_{k}$, where the $\sigma$’s are the matrices
$\\{I,X,Y,Z\\}$. Thus, when Eq. (20) calls on us to use the product $XZ$ for,
say, the first qubit, we write $Y$ in place of $XZ$. This choice makes each
displacement operator Hermitian in addition to being unitary. Note that the
displacement operators satisfy the composition rule

$D_{\alpha}D_{\beta}\propto D_{\alpha+\beta},$ (21)

where the proportionality sign indicates that there may be a nontrivial phase
factor relating $D_{\alpha}D_{\beta}$ to $D_{\alpha+\beta}$. In fact the phase
factor will always be a power of $i$ Calderbank . The following table shows
the discrete phase space, with the axes labeled by the values of $\beta_{q}$
and $\beta_{p}$ and with each displacement operator $D_{\beta}$ indicated at
the location of the point $\beta$.

$\beta_{p}$ |  |  | $\bar{\omega}$
---
$\omega$
1
0
| $Z\otimes I$ | $Y\otimes X$ | $Z\otimes X$ | $Y\otimes I$
---|---|---|---
$I\otimes Z$ | $X\otimes Y$ | $I\otimes Y$ | $X\otimes Z$
$Z\otimes Z$ | $Y\otimes Y$ | $Z\otimes Y$ | $Y\otimes Z$
$I\otimes I$ | $X\otimes X$ | $I\otimes X$ | $X\otimes I$
|  | 0 | 1 | $\omega$ | $\bar{\omega}$
---|---|---|---
|  $\beta_{q}$

Throughout this paper, we consistently label the phase-space axes as they are
labeled in this table, with the field values $0,1,\omega,\bar{\omega}$ always
appearing in that order.

Now consider any ray. Referring to the above table, we see that each ray is
associated with a set of four displacement operators, one of them being the
identity. One can check that these four operators commute with each other. For
example, the ray with slope $\omega$ consists of the points
$(0,0),(1,\omega),(\omega,\bar{\omega})$, and $(\bar{\omega},1)$, and the
associated displacement operators are $I\otimes I,X\otimes Y,Z\otimes X$, and
$Y\otimes Z$. Since the operators commute, there exists an orthonormal basis
of simultaneous eigenvectors of these operators. In fact, for each of the five
rays, the basis is unique up to phase factors. We associate this basis with
the given ray as well as with the striation that includes that ray. One can
show that the five bases generated in this way are mutually unbiased—this
result in fact generalizes to all prime-power values of the Hilbert-space
dimension Gibbons ; Bandyopadhyay ; Pittenger . Moreover, for any dimension
$d$, the number of mutually unbiased bases cannot be greater than $d+1$
Ivanovic . So the five bases associated with the five striations constitute a
complete set of mutually unbiased bases for the four-dimensional Hilbert
space. Again, these mutually unbiased bases play a central role in defining a
Wigner function for two qubits, as we discuss in the following section.

For now, we turn our attention to symplectic linear transformations,
represented by unit-determinant $2\times 2$ matrices with entries in
${\mathbb{F}}_{4}$. For each such matrix $L$, there exists a unitary
transformation $U$ such that

$UD_{\alpha}U^{\dagger}\propto D_{L\alpha},$ (22)

where the possible phase factor indicated by the proportionality sign is now
$\pm 1$ Calderbank . In fact, for each $L$ there are exactly 16 such unitary
transformations $U$ differing from each other by more than just a phase
factor. If $U$ and $U^{\prime}$ are two such unitaries for a given $L$, then
there exists a displacement operator $D_{\beta}$ such that $U^{\prime}\propto
D_{\beta}U$ Calderbank ; Zhu . The number 16 thus comes from the fact that
there are 16 displacement operators.

When the Hilbert-space dimension $d$ is odd, one can write the analog of Eq.
(22) with an equals sign rather than a proportionality sign, and the unitary
corresponding to $L$ is then unique up to a phase factor Appleby .
Unfortunately, within our two-qubit framework the equals sign is not possible.
To see this, it is enough to consider again the axis-interchanging
transformation $\left(\begin{array}[]{cc}0&1\\\ 1&0\end{array}\right)$. If
there were an equals sign in Eq. (22), the unitary $U$ associated with this
matrix would have to satisfy

$\begin{split}&U(X\otimes I)U^{\dagger}=Z\otimes I,\\\ &U(Z\otimes
I)U^{\dagger}=X\otimes I,\\\ &U(Y\otimes I)U^{\dagger}=Y\otimes I.\end{split}$
(23)

But by multiplying the first two of these equations together, we get
$U(Y\otimes I)U^{\dagger}=-Y\otimes I$, which contradicts the third; moreover,
this argument would be unaffected by any additional phase factors we might
attach to the displacement operators. So for us there is not a unique choice
of unitary for each $L$. (One can achieve equality in Eq. (22) for a system of
qubits by letting $L$ be a $2\times 2$ matrix with entries in
${\mathbb{Z}}_{2d}$, which for the two-qubit case would be ${\mathbb{Z}}_{8}$
ApplebyJMP . But this choice would entail a rather different framework from
the one in which we are working.)

For definiteness, though, we now choose one particular unitary matrix
satisfying Eq. (22) for each symplectic $L$, and we distinguish this unitary
with the symbol $U_{L}$. We make our choices in such a way that the $U_{L}$’s
also satisfy the equation

$U_{L_{1}L_{2}}\propto U_{L_{1}}U_{L_{2}}.$ (24)

We begin by identifying a unitary matrix for each of the four symplectic
matrices $H_{x}$ of the form $H_{x}=\left(\begin{array}[]{cc}1&0\\\
x&1\end{array}\right)$ with $x\in{\mathbb{F}}_{4}$. We can think of these
symplectic transformations as vertical shears, as they shift the points within
each vertical line while leaving the line itself unchanged.

$\displaystyle\begin{split}H_{0}=I=\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right)\hskip
28.45274pt&U_{H_{0}}=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&1&0&0\\\
0&0&1&0\\\ 0&0&0&1\end{array}\right),\\\
\fixTABwidth{T}H_{1}=\left(\begin{array}[]{cc}1&0\\\
1&1\end{array}\right)\hskip
28.45274pt&\fixTABwidth{T}U_{H_{1}}=\parenMatrixstack{0&0&0&-i\\\ 0&0&1&0\\\
0&1&0&0\\\ i&0&0&0},\\\
\fixTABwidth{T}H_{\omega}=\left(\begin{array}[]{cc}1&0\\\
\omega&1\end{array}\right)\hskip
28.45274pt&\fixTABwidth{T}U_{H_{\omega}}=\parenMatrixstack{0&-i&0&0\\\
i&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0},\\\
\fixTABwidth{T}H_{\bar{\omega}}=\left(\begin{array}[]{cc}1&0\\\
\bar{\omega}&1\end{array}\right)\hskip
28.45274pt&\fixTABwidth{T}U_{H_{\bar{\omega}}}=\parenMatrixstack{0&0&-i&0\\\
0&0&0&1\\\ i&0&0&0\\\ 0&1&0&0}.\end{split}$ (25)

We also choose a unitary operator to associate with the following symplectic
matrix $R$, which we think of as a rotation, since it cycles through all five
striations.

$R=\left(\begin{array}[]{cc}\bar{\omega}&1\\\ 1&0\end{array}\right)\hskip
28.45274pt\fixTABwidth{T}\setstacktabbedgap{1.5mm}U_{R}=\frac{1}{2}\parenMatrixstack{i&1i-1\\\
i-1i1\\\ i1-i1\\\ i-1-i-1}.$ (26)

One can verify that the following sixty matrices are distinct and therefore
represent all sixty of the $2\times 2$ symplectic matrices with entries in
${\mathbb{F}}_{4}$.

$\begin{split}&H_{0}R^{s},\;\;s=0,\ldots,4,\\\
&H_{\bar{\omega}}R^{s},\;\;s=0,\ldots,4,\\\
&R^{r}H_{1}R^{s},\;\;r,s=0,\ldots,4,\\\
&R^{r}H_{\omega}R^{s},\;\;r,s=0,\ldots,4.\\\ \end{split}$ (27)

($H_{\bar{\omega}}$ has the special property that
$RH_{\bar{\omega}}R=H_{\bar{\omega}}$. So it would be redundant to multiply
$H_{\bar{\omega}}$ by $R$ on both sides.) For each such matrix $L$, we define
its associated unitary $U_{L}$ to be the operator obtained by replacing each
$H$ with the appropriate $U_{H}$ and replacing each $R$ with $U_{R}$. Thus,
for example, for the matrix

$G=R^{2}H_{1}R=\left(\begin{array}[]{cc}\bar{\omega}&0\\\
0&\omega\end{array}\right),$ (28)

the associated unitary is

$\fixTABwidth{T}U_{G}=U_{R}^{2}U_{H_{1}}U_{R}=\parenMatrixstack{-1&000\\\
0010\\\ 000-1\\\ 0100}.$ (29)

That the $U_{L}$’s defined in this way satisfy Eq. (24) is shown in the
Appendix.

If we are given a symplectic matrix $L$, then in order to apply the above
definition of $U_{L}$, it would evidently be helpful to know how to express
$L$ in one of the four forms seen in Eq. (27). One way to do this is as
follows. First, to find the value of $x$ in $L=R^{r}H_{x}R^{s}$, we can make
use of the fact that $RH_{\bar{\omega}}R=H_{\bar{\omega}}$, from which one can
show that

$\hbox{Tr}(L^{T}H_{\bar{\omega}}LH_{\bar{\omega}})=x^{2}.$ (30)

(The $H_{\bar{\omega}}$’s in this equation essentially remove all the $R$’s
that appear in $L$, leaving only $H_{x}$.) Then, if $x$ turns out to be $0$ or
$\bar{\omega}$, we can find the value of $s$ in $L=H_{x}R^{s}$ by computing
the vector $\mu=L\left(\begin{array}[]{c}1\\\ 0\end{array}\right)$. The ratio
$\mu_{p}/\mu_{q}$, which takes five possible values, including “infinity” when
$\mu_{q}$ is zero, turns out to be perfectly correlated with the value of $s$:

$x+\mu_{p}/\mu_{q}$ |  $\omega$ |  $\bar{\omega}$ |  $0$ |  $1$ | $\,\infty\,$
---|---|---|---|---|---
$s$ |  1 |  3 |  0 |  2 | 4

($x=0$ or $\bar{\omega}$).

(Here $x+\infty$ is understood to be $\infty$.) If $x$ comes out to be $1$ or
$\omega$, we can isolate the integer $s$ in $L=R^{r}H_{x}R^{s}$ by computing
the vector

$\nu=\left[L^{T}H_{\bar{\omega}}L\left(\begin{array}[]{c}1\\\
0\end{array}\right)\right]-\left(\begin{array}[]{c}1\\\ x\end{array}\right),$
(31)

whose “slope” $\nu_{p}/\nu_{q}$ is likewise perfectly correlated with the
value of $s$:

$\nu_{p}/\nu_{q}$ |  $\omega$ |  $\bar{\omega}$ |  $0$ |  $1$ | $\,\infty\,$
---|---|---|---|---|---
$s$ |  0 |  1 |  2 |  3 | 4

($x=1$ or ${\omega}$).

And we can similarly get the value of $r$ from the slope of

$\tau=\left[LH_{\bar{\omega}}L^{T}\left(\begin{array}[]{c}1\\\
0\end{array}\right)\right]-\left(\begin{array}[]{c}1\\\ x\end{array}\right),$
(32)

in accordance with the following table:

$\tau_{p}/\tau_{q}$ |  $\omega$ |  $\bar{\omega}$ |  $0$ |  $1$ | $\,\infty\,$
---|---|---|---|---|---
$r$ |  1 |  2 |  3 |  4 | 0

($x=1$ or ${\omega}$).

Once we know the values of $x$, $r$, and $s$—here $r$ is taken to be zero if
$x=0$ or $x=\bar{\omega}$—we identify the unitary $U_{L}$ as

$U_{L}=U_{R}^{r}U_{H_{x}}U_{R}^{s}.$ (33)

The set of unitary operators of the form $D_{\beta}U_{L}$, modulo phase
factors, constitutes the restricted Clifford group mentioned in the
Introduction Bengtsson ; Zhu1 ; Zhu ; Appleby . It is a subgroup of the full
Clifford group, differing from this latter group in that Eq. (22) respects the
finite-field arithmetic—in our case the arithmetic of ${\mathbb{F}}_{4}$. For
example, the controlled-not operation permutes the displacement operators and
is thus a Clifford operation, but the permutation is not linear in
${\mathbb{F}}_{4}^{2}$. So the controlled-not operation is not an element of
our restricted Clifford group.

##  III A two-qubit Wigner function

As we have noted in the preceding section, each striation of the phase space
is associated with an orthogonal basis of the Hilbert space, and the five
bases obtained in this way are mutually unbiased. The Wigner functions of Ref.
Gibbons  are defined specifically so that the sums of $W_{\alpha}$ over the
lines of any striation yield the probabilities of the outcomes of the
measurement defined by the associated Hilbert-space basis. An analogous
property holds for the original, continuous Wigner function Wootters ;
Bertrand . Another key idea of Ref. Gibbons  is that the Wigner function
should be covariant under displacements. That is,

$W_{\alpha}(D_{\beta}\rho D_{\beta}^{\dagger})=W_{\alpha-\beta}(\rho).$ (34)

(The dagger in this equation is unnecessary, since our $D_{\beta}$’s are
Hermitian. We include it so that the equation would still hold if one were to
adopt a different phase convention for the $D_{\beta}$’s.) These requirements
still allow many distinct definitions of the Wigner function. The following
algorithm will generate every Wigner-function definition compatible with these
requirements Gibbons .

  1. 1.

For each ray $n$, choose arbitrarily a vector from the orthogonal basis
associated with the ray. (In Fig. 2 below, we label the rays and striations
with the values $n=0,\ldots,4$, but we do not need the details of that
labeling scheme here.) Let the vectors that have been chosen be called
$|v^{(n)}\rangle$.

  2. 2.

Construct the operator

$A_{0}=\sum_{n}|v^{(n)}\rangle\langle v^{(n)}|-I,$ (35)

where the sum is over all the rays. Take this operator to be the phase point
operator associated with the origin of phase space.

  3. 3.

Define the other phase point operators by the following equation:

$A_{\alpha}=D_{\alpha}A_{0}D^{\dagger}_{\alpha}.$ (36)

  4. 4.

The Wigner function $W_{\alpha}(\rho)$ is then defined by

$W_{\alpha}(\rho)=\tfrac{1}{4}\hbox{Tr}(A_{\alpha}\rho).$ (37)

Now, for each of the five rays, there are four possible vectors to choose in
step 1, and after these choices are made the definition is fully determined.
So there are $4^{5}=1024$ possible definitions of the two-qubit Wigner
function.

It will be helpful to have a way of labeling these possible definitions.
Alternatively, we can say we are labeling the 1024 possible choices of the
“origin operator” $A_{0}$. And it will be helpful to make this labeling very
systematic.

To this end, we return to the unitary operator $U_{R}$ of Eq. (26). We use
$U_{R}$ to construct a labeling scheme for the vectors of the five special
mutually unbiased bases. We start with the standard basis, which we refer to
as the zeroth basis. Let us call the vectors in this basis
$(|b^{(0)}_{0}\rangle,|b^{(0)}_{1}\rangle,|b^{(0)}_{\omega}\rangle,|b^{(0)}_{\bar{\omega}}\rangle)$,
where $|b^{(0)}_{0}\rangle$ is the column vector with components $(1,0,0,0)$
and $|b^{(0)}_{k}\rangle=D_{(k,0)}|b^{(0)}_{0}\rangle$. We then obtain the
other bases, and their labels, by repeatedly applying $U_{R}$. For
$n=0,\ldots,4$, which labels the bases, and for $k=0,1,\omega,\bar{\omega}$,
which labels the vectors within each basis, we define

$|b^{(n)}_{k}\rangle=U_{R}^{n}|b^{(0)}_{k}\rangle=U_{R}^{n}D_{(k,0)}|b^{(0)}_{0}\rangle.$
(38)

It is worth writing down explicitly the vectors of the five bases, along with
their labeling. We do this in Fig. 1.

Figure 1: Labeling scheme for the vectors of the five mutually unbiased bases.
The integer $n$ at the left labels the bases, and the field element $k_{n}$ at
the top labels the vectors of each basis. (The vectors shown here sometimes
differ from $|b_{k}^{(n)}\rangle$ by an overall phase factor.)

Again, the origin operator $A_{0}$ is obtained by choosing one vector from
each of these five bases, and it can now be expressed as

$A_{0}=\sum_{n=0}^{4}|b^{(n)}_{k_{n}}\rangle\langle b^{(n)}_{k_{n}}|-I.$ (39)

Let us choose a particular definition of the origin operator $A_{0}$, and
hence of the Wigner function, and distinguish this definition with the
adjective “standard”: it is the one for which $A_{0}$ has
$k_{0}=k_{1}=k_{2}=k_{3}=k_{4}=0$. Notationally, we distinguish the standard
Wigner function, and its associated phase point operators, with a boldfaced
${\bf W}$ and ${\bf A}$.

We would also like to label the lines of phase space. We do so in a way that
parallels our labeling of the basis vectors. Fig. 2, which shows all the
lines, was constructed as follows: at the top are the vertical lines, which we
label $\lambda^{(0)}_{k}$ with $k\in{\mathbb{F}}_{4}$. Each successive entry
of the figure is obtained from the entry above it by applying our “rotation”
matrix $R$.

Figure 2: Striations obtained by repeatedly applying the linear transformation
$R$, starting with the vertical striation. The integer $n$ at the left labels
the striations, and the field element $k_{n}$ at the top labels the lines
within each striation.

Thus we define $\lambda^{(n)}_{k}$ to be $R^{n}\lambda^{(0)}_{k}$. (Applying
$R$ to a line means applying it to each point in the line.)

##  IV Behavior of the “index” of a phase point operator or set of lines

Regardless of which choice we have made for $A_{0}$—whether “standard” or
not—any two-qubit phase point operator $A$ can be expressed as in Eq. (39):

$A=\sum_{n=0}^{4}|b^{(n)}_{k_{n}}\rangle\langle b^{(n)}_{k_{n}}|-I.$ (40)

We define the index of the operator $A$, denoted ${{\mathcal{I}}}(A)$, to be
the five-component column vector whose components are the $k_{n}$’s in the
above equation.

The concept of an index can also be applied to a collection of lines. Let
$\Lambda$ be any ordered set of lines of the form
$\Lambda=(\lambda^{(0)}_{k_{0}},\lambda^{(1)}_{k_{1}},\lambda^{(2)}_{k_{2}},\lambda^{(3)}_{k_{3}},\lambda^{(4)}_{k_{4}})$;
that is, $\Lambda$ picks out one line from each striation. We say that the
index of $\Lambda$ is the five-component column vector whose entries are the
$k_{n}$’s. Note, in particular, that any point $\alpha$ of phase space defines
such a set $\Lambda$: it picks out the line in each striation on which
$\alpha$ lies. So we can also speak of the index of a point. For example, by
consulting Fig. 2, one can see that if $\alpha$ is the point $(1,\omega)$, its
index is

${\mathcal{I}}(\alpha)=\left(\begin{array}[]{c}1\\\ \omega\\\ 0\\\ \omega\\\
1\end{array}\right).$ (41)

We now ask how the index, of a collection of lines or of a phase point
operator, is affected by (i) a displacement and (ii) a symplectic linear
transformation.

###  IV.1 The effect of a displacement

We begin with the case of a displacement, and for now we focus on a collection
$\Lambda$ of lines. Let $\Delta_{\beta}$ represent a displacement by $\beta$.
($\Delta_{\beta}$ acts on phase-space points, whereas $D_{\beta}$ acts on
Hilbert-space vectors.) How does the index of $\Lambda$ change under a
displacement? A displacement will move each line to a line in the same
striation; so the $k$-value of the line might change but the $n$-value will
not. We have

$\begin{split}\Delta_{\beta}\lambda^{(n)}_{k}&=\Delta_{\beta}R^{n}\lambda^{(0)}_{k}\\\
&=R^{n}(R^{-n}\Delta_{\beta}R^{n})\lambda^{(0)}_{k}\\\
&=R^{n}\Delta_{R^{-n}\beta}\lambda^{(0)}_{k}.\end{split}$ (42)

Now, the $k$-value of a vertical line is not changed by a vertical
displacement; so we need only the horizontal component of $R^{-n}\beta$:

$\Delta_{\beta}\lambda^{(n)}_{k}=R^{n}\lambda^{(0)}_{k+(R^{-n}\beta)_{q}}=\lambda^{(n)}_{k+(R^{-n}\beta)_{q}}.$
(43)

For each value of $n$, the expression $(R^{-n}\beta)_{q}$ is linear in
$\beta$, so it must be of the form

$(R^{-n}\beta)_{q}=\beta_{q}Q_{n}+\beta_{p}P_{n},$ (44)

where $Q$ and $P$ are five-component column vectors with entries in
${\mathbb{F}}_{4}$. In fact one finds that

$Q=\left(\begin{array}[]{c}1\\\ 0\\\ 1\\\ \bar{\omega}\\\
\bar{\omega}\end{array}\right),\hskip 28.45274ptP=\left(\begin{array}[]{c}0\\\
1\\\ \bar{\omega}\\\ \bar{\omega}\\\ 1\end{array}\right).$ (45)

We thus have

${\mathcal{I}}(\Delta_{\beta}\Lambda)={\mathcal{I}}(\Lambda)+\beta_{q}Q+\beta_{p}P.$
(46)

We can make essentially the same argument to find the effect of a unitary
displacement operator $D_{\beta}$ on the index of a phase point operator, and
we obtain essentially the same answer. Specifically, we can write

$\begin{split}D_{\beta}|b^{(n)}_{k}\rangle&=D_{\beta}U_{R}^{n}|b^{(0)}_{k}\rangle\\\
&=U_{R}^{n}(U^{-n}_{R}D_{\beta}U^{n}_{R})|b^{(0)}_{k}\rangle\\\ &\propto
U_{R}^{n}D_{R^{-n}\beta}|b^{(0)}_{k}\rangle.\\\ \end{split}$ (47)

But $|b^{(0)}_{k}\rangle$ is an eigenstate of any unitary displacement
operator associated with a vertical displacement. So we need to consider only
the horizontal component of $R^{-n}\beta$:

$\begin{split}D_{\beta}|b^{(n)}_{k}\rangle\propto
U_{R}^{n}|b^{(0)}_{k+(R^{-n}\beta)_{q}}\rangle&=|b^{(n)}_{k+(R^{-n}\beta)_{q}}\rangle\\\
&=|b^{(n)}_{k+\beta_{q}Q_{n}+\beta_{p}P_{n}}\rangle.\end{split}$ (48)

Since each of our phase point operators is built out of the vectors
$|b^{(n)}_{k}\rangle$ in accordance with Eq. (40), Eq. (48) leads to the
equation

${\mathcal{I}}(D_{\beta}A_{\alpha}D^{\dagger}_{\beta})={\mathcal{I}}(A_{\alpha})+\beta_{q}Q+\beta_{p}P.$
(49)

We see, then, that the change in the index of either a set of lines or a phase
point operator upon displacement by $\beta$ is fairly straightforward: we
simply add $\beta_{q}Q+\beta_{p}P$.

###  IV.2 The effect of a symplectic linear transformation

We now turn our attention to the case of a symplectic linear transformation
$L$. We begin by considering the effect of $L$ on the index of a set of lines
$\Lambda$. Unlike a displacement, a linear transformation can take a striation
to a different striation. But at least it will not split up a striation: if
two lines are parallel, their transformed images will also be parallel. For a
given linear transformation $L$, the $n$th striation (where $n=0,\ldots,4$)
will be sent by $L$ to some striation whose label we call $m$. (The value of
$m$ depends on $n$, but we leave this dependence implicit.) The transformation
$L$ will send rays to rays, so

$L\lambda^{(n)}_{0}=\lambda^{(m)}_{0}.$ (50)

To figure out the effect of $L$ on the labels of the other lines, we use the
fact that

$\Delta_{\beta}\lambda^{(n)}_{0}=\lambda^{(n)}_{\beta_{q}Q_{n}+\beta_{p}P_{n}}.$
(51)

Let us now apply $L$ to the right-hand side of this equation.

$\begin{split}L\lambda^{(n)}_{\beta_{q}Q_{n}+\beta_{p}P_{n}}&=L\Delta_{\beta}\lambda^{(n)}_{0}\\\
&=(L\Delta_{\beta}L^{-1})L\lambda^{(n)}_{0}\\\
&=\Delta_{L\beta}L\lambda^{(n)}_{0}\\\ &=\Delta_{L\beta}\lambda^{(m)}_{0}\\\
&=\lambda^{(m)}_{(L\beta)_{q}Q_{m}+(L\beta)_{p}P_{m}}.\end{split}$ (52)

For each value of $n$, any value of $k$ can be written as
$\beta_{q}Q_{n}+\beta_{p}P_{n}$ for at least one value of $\beta$. So the
above equation is sufficient to tell us what $L$ does to each of the lines.
Moreover, to multiply $k$ by any factor, we can multiply $\beta$ by the same
factor, and the equation tells us that the subscript of $\lambda^{(m)}$ is
also multiplied by that same factor. So for a set of lines $\Lambda$, the
$m$th component of the index of $L\Lambda$ must be simply proportional to the
$n$th component of the index of $\Lambda$ (where the proportionality constant
is an element of ${\mathbb{F}}_{4}$). According to Eq. (52), the relation must
be expressible as

${\mathcal{I}}(L\Lambda)_{m}=\left(\frac{(L\beta)_{q}Q_{m}+(L\beta)_{p}P_{m}}{\beta_{q}Q_{n}+\beta_{p}P_{n}}\right){\mathcal{I}}(\Lambda)_{n}$
(53)

for any value of $\beta$ that makes the denominator nonzero. (The value of the
fraction is the same for any such value of $\beta$.) We can write this
relation as

${\mathcal{I}}(L\Lambda)=S_{L}{\mathcal{I}}(\Lambda),$ (54)

where $S_{L}$ is a $5\times 5$ matrix with entries in ${\mathbb{F}}_{4}$. The
only nonzero entries of $S_{L}$ are the five entries $(S_{L})_{mn}$, and they
are given by

$(S_{L})_{mn}=\frac{(L\beta)_{q}Q_{m}+(L\beta)_{p}P_{m}}{\beta_{q}Q_{n}+\beta_{p}P_{n}},$
(55)

where, again, any choice of $\beta$ is allowed as long as the fraction is well
defined. (We label the rows and columns of $S_{L}$ with integers
$0,\ldots,4$.) A special case of Eq. (54) is the analogous equation for the
index of a phase-space point:

${\mathcal{I}}(L\alpha)=S_{L}{\mathcal{I}}(\alpha).$ (56)

Let us call $S_{L}$ the “index operator” for the linear transformation $L$.

As an example, consider the linear transformation we have called $H_{1}$:

$H_{1}=\left(\begin{array}[]{cc}1&0\\\ 1&1\end{array}\right).$ (57)

By looking at the rays of Fig. 2, one can see how $H_{1}$ maps striations to
other striations. The mapping is given by the following table.

$n$ | 0 | 1 | 2 | 3 | 4
---|---|---|---|---|---
$m$ | 0 | 3 | 4 | 1 | 2

Again, the only nonzero elements of $S_{H_{1}}$ are $(S_{H_{1}})_{mn}$, and we
can use Eq. (55), with $L$ set equal to $H_{1}$, to evaluate those elements.
In this way one finds that

$S_{H_{1}}=\left(\begin{array}[]{ccccc}1&0&0&0&0\\\ 0&0&0&\omega&0\\\
0&0&0&0&\bar{\omega}\\\ 0&\bar{\omega}&0&0&0\\\
0&0&{\omega}&0&0\end{array}\right).$ (58)

We now turn our attention to phase point operators. How does the index of a
phase point operator change when we apply the unitary transformation $U_{L}$
corresponding to a symplectic matrix $L$? We begin with the origin operator
${\bf A}_{0}$ of our standard Wigner function. For any $L$, one can simply
work out what $U_{L}$ does to each of the basis vectors $|b^{(n)}_{0}\rangle$
of which ${\bf A}_{0}$ is composed. Consider, for example, $L=H_{1}$, for
which our chosen unitary transformation is (see Eq. (25))

$\fixTABwidth{T}U_{H_{1}}=\parenMatrixstack{0&00-i\\\ 0010\\\ 0100\\\ i000}.$
(59)

Applying this matrix to each of the vectors $|b^{(n)}_{0}\rangle$ in the left-
most column of Fig. 1, we see that

${\mathcal{I}}(U_{H_{1}}{\bf
A}_{0}U^{\dagger}_{H_{1}})=\left(\begin{array}[]{c}1\\\ 0\\\ 1\\\ 0\\\
\omega\end{array}\right).$ (60)

Let us label this last column vector $f_{H_{1}}$. In a similar way, we can
obtain an analogous vector $f_{L}$ for each symplectic matrix $L$. That is, we
will have

${\mathcal{I}}(U_{L}{\bf A}_{0}U^{\dagger}_{L})=f_{L}.$ (61)

We call $f_{L}$ the “shift vector” for $L$. Unlike the index operator $S_{L}$,
the shift vector $f_{L}$ depends on our particular choice of the unitary
$U_{L}$. If instead of $U_{L}$ we had associated with $L$ the unitary operator
$D_{\beta}U_{L}$, then instead of $f_{L}$ on the right-hand side of Eq. (61)
we would have had the vector $f_{L}+\beta_{q}Q+\beta_{p}P$.

Note that we are seeing here a difference between the action of $U_{L}$ on the
$|b^{(n)}_{0}\rangle$’s and the action of $L$ on the rays $\lambda^{(n)}_{0}$.
Whereas $L$ merely permutes the rays among themselves, the unitary operator
$U_{L}$—which effects the same permutation of the rows of the table—can also
shift $|b^{(m)}_{0}\rangle$ to a different place in the row (that is, it can
change the value of $k$). This shift is precisely what we can avoid when the
dimension of the Hilbert space is an odd prime power. But it cannot be avoided
when the dimension is a power of 2. It is the shift by $f_{L}$ that will force
us to interpret the Wigner-function values with respect to a different set of
phase point operators.

From the effect of $U_{L}$ on the index of ${\bf A}_{0}$, we can work out its
effect on the index of any allowed phase point operator, that is, any operator
of the form (40). We begin with the following relation, which is analogous to
the one in Eq. (52).

$\begin{split}U_{L}|b^{(n)}_{\beta_{q}Q_{n}+\beta_{p}P_{n}}\rangle\langle
b^{(n)}_{\beta_{q}Q_{n}+\beta_{p}P_{n}}|U^{\dagger}_{L}&=U_{L}D_{\beta}|b^{(n)}_{0}\rangle\langle
b^{(n)}_{0}|D^{\dagger}_{\beta}U^{\dagger}_{L}\\\
&=D_{L\beta}U_{L}|b^{(n)}_{0}\rangle\langle
b^{(n)}_{0}|U^{\dagger}_{L}D^{\dagger}_{L\beta}.\end{split}$ (62)

But now, in place of the last step of Eq. (52), we need to take into account
the shift expressed in Eq. (61):

$U_{L}|b^{(n)}_{\beta_{q}Q_{n}+\beta_{p}P_{n}}\rangle\langle
b^{(n)}_{\beta_{q}Q_{n}+\beta_{p}P_{n}}|U^{\dagger}_{L}=D_{L\beta}|b^{(m)}_{(f_{L})_{m}}\rangle\langle
b^{(m)}_{(f_{L})_{m}}|D^{\dagger}_{L\beta}.$ (63)

As before, the $m$ in this equation labels the striation that is the image
under $L$ of the striation labeled by $n$. And $(f_{L})_{m}$ is the $m$th
component of the vector $f_{L}$. Meanwhile, Eq. (48) tells us that

$D_{L\beta}|b^{(m)}_{(f_{L})_{m}}\rangle=|b^{(m)}_{(f_{L})_{m}+(L\beta)_{q}Q_{m}+(L\beta)_{p}P_{m}}\rangle.$
(64)

Now consider a phase point operator $A$ whose index is given by
${\mathcal{I}}(A)_{n}=\beta_{q}Q_{n}+\beta_{p}P_{n}$. It follows from Eqs.
(63) and (64) that

${\mathcal{I}}(U_{L}AU^{\dagger}_{L})_{m}=(L\beta)_{q}Q_{m}+(L\beta)_{p}P_{m}+(f_{L})_{m}.$
(65)

Recall, though, that the matrix $S_{L}$ is defined so as to take the index
with components $\beta_{q}Q_{n}+\beta_{p}P_{n}$ to the index with components
$(L\beta)_{q}Q_{m}+(L\beta)_{p}P_{m}$. So we have

${\mathcal{I}}(U_{L}AU^{\dagger}_{L})=S_{L}{\mathcal{I}}(A)+f_{L}.$ (66)

Note the difference between this formula and the analogous formula for a set
of lines, Eq. (54): in Eq. (54) there is no need for the shift vector $f_{L}$.

For the special case of the standard Wigner function, we can express Eq. (66)
in a particularly simple way. To see how, we need to show that the vectors
$|b_{k_{n}}^{(n)}\rangle$ used to construct ${\bf A}_{\alpha}$ have the same
values of $k_{n}$ as the lines that pass through the point $\alpha$; that is,
${\mathcal{I}}({\bf A}_{\alpha})={\mathcal{I}}(\alpha)$. We can see this by
first observing that the equation holds when $\alpha$ is the origin: the rays,
which are the lines passing through the origin, all have $k_{n}=0$, and by
definition, ${\bf A}_{0}$ is constructed from the vectors
$|b_{k_{n}}^{(n)}\rangle$ with $k_{n}=0$. We now apply a displacement $\alpha$
and use Eqs. (46) and (49):

$\begin{split}{\mathcal{I}}({\bf A}_{\alpha})&={\mathcal{I}}({\bf
A}_{0})+\alpha_{q}Q+\alpha_{p}P\\\
&={\mathcal{I}}(\hbox{origin})+\alpha_{q}Q+\alpha_{p}P={\mathcal{I}}(\alpha).\end{split}$
(67)

So we can write

$S_{L}{\mathcal{I}}({\bf
A}_{\alpha})=S_{L}{\mathcal{I}}(\alpha)={\mathcal{I}}(L\alpha)={\mathcal{I}}({\bf
A}_{L\alpha}).$ (68)

Thus Eq. (66) becomes

${\mathcal{I}}(U_{L}{\bf A}_{\alpha}U^{\dagger}_{L})={\mathcal{I}}({\bf
A}_{L\alpha})+f_{L}.$ (69)

Returning now to the case of a general phase point operator, it will be
helpful to use the formula (66) to work out the effect of a composition of two
$U_{L}$’s. Suppose we apply to a phase point operator $A$ the transformation
$U_{L_{1}}$ followed by $U_{L_{2}}$. The effect on the index is

$\begin{split}{\mathcal{I}}(U_{L_{2}}U_{L_{1}}AU^{\dagger}_{L_{1}}U^{\dagger}_{L_{2}})&=S_{L_{2}}{\mathcal{I}}(U_{L_{1}}AU^{\dagger}_{L_{1}})+f_{L_{2}}\\\
&=S_{L_{2}}\left[S_{L_{1}}{\mathcal{I}}(A)+f_{L_{1}}\right]+f_{L_{2}}\\\
&=S_{L_{2}}S_{L_{1}}{\mathcal{I}}(A)+S_{L_{2}}f_{L_{1}}+f_{L_{2}}.\\\
\end{split}$ (70)

We use this formula immediately in the next subsection.

###  IV.3 Finding the shift vectors $f_{L}$

The shift vectors are an important part of our story, so it will be good to
have a simple way of finding them. First, by applying the unitaries
$U_{H_{x}}$ of Eq. (25) to the $|b^{(n)}_{0}\rangle$’s, we can find the four
shift vectors $f_{H_{x}}$ for $x\in{\mathbb{F}}_{4}$:

$H_{0}=\left(\begin{array}[]{cc}1&0\\\ 0&1\end{array}\right)\hskip
28.45274ptf_{H_{0}}=\left(\begin{array}[]{c}0\\\ 0\\\ 0\\\ 0\\\
0\end{array}\right),$ (71) $H_{1}=\left(\begin{array}[]{cc}1&0\\\
1&1\end{array}\right)\hskip 28.45274ptf_{H_{1}}=\left(\begin{array}[]{c}1\\\
0\\\ 1\\\ 0\\\ \omega\end{array}\right),$ (72)
$H_{\omega}=\left(\begin{array}[]{cc}1&0\\\ \omega&1\end{array}\right)\hskip
28.45274ptf_{H_{\omega}}=\left(\begin{array}[]{c}\omega\\\ 1\\\
\bar{\omega}\\\ {\omega}\\\ \bar{\omega}\end{array}\right),$ (73)
$H_{\bar{\omega}}=\left(\begin{array}[]{cc}1&0\\\
\bar{\omega}&1\end{array}\right)\hskip
28.45274ptf_{H_{\bar{\omega}}}=\left(\begin{array}[]{c}\bar{\omega}\\\
\bar{\omega}\\\ \bar{\omega}\\\ \bar{\omega}\\\
\bar{\omega}\end{array}\right).$ (74)

Recall that all the other symplectic matrices can be obtained by multiplying
these four on the right and left by powers of the “rotation” $R$ (see Eq.
(27))—we saw in Section II how to find $x$, $r$, and $s$ in the expression
$L=R^{r}H_{x}R^{s}$. We will use Eq. (70) to see how multiplication by $R$
affects the shift vectors. First note that the index operator and shift vector
for the matrix $R$ are

$S_{R}=\left(\begin{array}[]{ccccc}0&0&0&0&1\\\ 1&0&0&0&0\\\ 0&1&0&0&0\\\
0&0&1&0&0\\\ 0&0&0&1&0\end{array}\right)\hskip 14.22636pt\hbox{and}\hskip
14.22636ptf_{R}=\left(\begin{array}[]{c}0\\\ 0\\\ 0\\\ 0\\\
0\end{array}\right).$ (75)

Now suppose we multiply one of the $H$’s on the right by $R$. Then according
to Eq. (70), with $A$ set equal to ${\bf A}_{0}$, we have

$f_{HR}=S_{H}f_{R}+f_{H}=f_{H},$ (76)

since $f_{R}$ is the zero vector. So multiplying on the right by $R$ does not
change the shift vector. Now suppose we multiply one of the $H$’s on the left
by $R$. We have

$f_{RH}=S_{R}f_{H}+f_{R}=S_{R}f_{H}.$ (77)

That is, we cyclically shift $f_{H}$ one step downward to get $f_{RH}$.
Continuing in this way, we can find the shift vector for each of our 60
symplectic matrices as expressed in Eq. (27). These 60 matrices can be sorted
into 12 sets of 5 each, such that the matrices within each set all have the
same shift vector (since multiplying on the right by $R$ does not change $f$).
Thus for our 60 $L$’s, there are only 12 distinct shift vectors $f_{L}$.

In fact we can obtain a formula that yields $f_{L}$ directly for any $L$,
bypassing the explicit decomposition of $L$ into the form $R^{r}H_{x}R^{s}$.
We begin by noting that the following 12 vectors constitute the full set of
nonzero vectors $\mu$ in ${\mathbb{F}}_{4}^{2}$ whose slopes $\mu_{p}/\mu_{q}$
are not infinite.

$\begin{split}&H_{x}H_{\bar{\omega}}^{T}H_{x}^{T}\left(\begin{array}[]{c}1\\\
0\end{array}\right),\;\;x=0\;\hbox{or}\;\bar{\omega},\\\
&R^{t}H_{x}H_{\bar{\omega}}^{T}H_{x}^{T}R^{t}\left(\begin{array}[]{c}1\\\
0\end{array}\right),\;\;x=1\;\hbox{or}\;{\omega},\end{split}$ (78)

where $t=0,\ldots,4$. Now, for any symplectic matrix $L$ and for
$n=0,\ldots,4$, consider the vector $\mu^{(n)}_{L}$ defined by

$\mu^{(n)}_{L}=R^{-n}LH_{\bar{\omega}}^{T}L^{T}R^{-n}\left(\begin{array}[]{c}1\\\
0\end{array}\right).$ (79)

The slope of this vector—which is not infinite since $\mu^{(n)}_{L}$ reduces
to one of the vectors in Eq. (78)—turns out to be equal to the value of
$(f_{L})_{n}$:

$(f_{L})_{n}=\frac{\big{(}\mu_{L}^{(n)}\big{)}_{p}}{\big{(}\mu^{(n)}_{L}\big{)}_{q}}.$
(80)

To see this, one can first check that the relation agrees with Eqs. (71)–(74)
when $L=H_{x}$ with $x\in{\mathbb{F}}_{4}$. From there it suffices to make
sure the formula behaves correctly when we multiply $L$ on the left or right
by $R$. But this is indeed the case:

$\begin{split}&\mu^{(n)}_{LR}=\mu^{(n)}_{L}\hskip 2.84526pt\hbox{since
$RH_{\bar{\omega}}^{T}R=H_{\bar{\omega}}^{T}$ },\\\
&\mu^{(n)}_{RL}=R^{-n+1}LH_{\bar{\omega}}^{T}L^{T}R^{-n+1}\left(\begin{array}[]{c}1\\\
0\end{array}\right)=\mu^{(n-1)}_{L},\end{split}$ (81)

where the superscript on $\mu$ is to be interpreted mod 5. Thus Eqs. (79) and
(80) constitute a valid formula for $f_{L}$.

##  V Reinterpreting Wigner-function values

We now ask how the effect of $U_{L}$ is to be described in phase space. Let
the initial state of our two qubits be $\rho$, and let us express this state
as a Wigner function, adopting for now the “standard” definition ${\bf W}$:

${\bf W}_{\alpha}(\rho)=\tfrac{1}{4}\hbox{Tr}({\bf A}_{\alpha}\rho).$ (82)

We perform on $\rho$ one of our unitary transformations $U_{L}$. What does the
action of $U_{L}$ look like in phase space?

Let $\rho^{\prime}$ be the transformed density matrix:
$\rho^{\prime}=U_{L}\rho U^{\dagger}_{L}$. Then
$\rho=U^{\dagger}_{L}\rho^{\prime}U_{L}$, and we can write

${\bf W}_{\alpha}(\rho)=\tfrac{1}{4}\hbox{Tr}({\bf
A}_{\alpha}U^{\dagger}_{L}\rho^{\prime}U_{L})=\tfrac{1}{4}\hbox{Tr}\left[(U_{L}{\bf
A}_{\alpha}U^{\dagger}_{L})\rho^{\prime}\right].$ (83)

Let us define the “$f$-Wigner function” $W^{f}_{\alpha}$ such that it has as
its origin operator not ${\bf A}_{0}$, whose index is the zero vector, but
rather the operator whose index is $f$:

$A^{f}_{0}=\sum_{n}|b^{(n)}_{f_{n}}\rangle\langle b^{(n)}_{f_{n}}|-I.$ (84)

The other phase point operators are defined in the usual way:

$A^{f}_{\alpha}=D_{\alpha}A^{f}_{0}D^{\dagger}_{\alpha},$ (85)

and the $f$-Wigner function itself is given by

$W^{f}_{\alpha}=\tfrac{1}{4}\hbox{Tr}(A^{f}_{\alpha}\rho).$ (86)

Let us now examine the index of the operator in parentheses on the right-hand
side of Eq. (83).

$\begin{split}{\mathcal{I}}(U_{L}{\bf
A}_{\alpha}U^{\dagger}_{L})&=S_{L}{\mathcal{I}}({\bf A}_{\alpha})+f_{L}\\\
&=S_{L}(\alpha_{q}Q+\alpha_{p}P)+f_{L}\\\
&=(L\alpha)_{q}Q+(L\alpha)_{p}P+f_{L}\\\
&={\mathcal{I}}(A^{f_{L}}_{L\alpha}).\end{split}$ (87)

Thus the operator in parentheses on the right-hand side of Eq. (83) is
$A^{f_{L}}_{L\alpha}$, and the trace in that equation gives us the Wigner
function $W^{f_{L}}_{L\alpha}(\rho^{\prime})$. So we have

${\bf W}_{\alpha}(\rho)=W^{f_{L}}_{L\alpha}(U_{L}\rho U^{\dagger}_{L}).$ (88)

We can write this equation in a form analogous to that of Eq. (1) or Eq. (13):

$W^{f_{L}}_{\alpha}(U_{L}\rho U^{\dagger}_{L})={\bf W}_{L^{-1}\alpha}(\rho).$
(89)

That is, performing the unitary $U_{L}$ is equivalent to moving the values of
the Wigner function in phase space according to $L$, and reinterpreting those
values as values of the new Wigner function $W^{f_{L}}$.

As an example, let us consider the state $\rho=|\psi\rangle\langle\psi|$,
where $|\psi\rangle$ is the product vector

$|\psi\rangle=\mid\uparrow\rangle\otimes\mid\rightarrow\rangle=\left(\begin{array}[]{c}1\\\
0\end{array}\right)\otimes\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}1\\\
1\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}1\\\ 1\\\ 0\\\
0\end{array}\right).$ (90)

Here I am using arrows to represent standard Bloch-sphere vectors. In our
standard Wigner-function representation, this state appears as

${\bf W}(\rho)=\frac{1}{4}$ $\,\,1\,\,$ $\,\,0\,\,$ $\,\,1\,\,$ $\,\,0\,\,$ 0
0 0 0 0 0 $0$ $0$ $\,\,1\,\,$ $\,\,0\,\,$ $1$ $0$ $\leftarrow\rightarrow$
$\rightarrow\leftarrow$ $\leftarrow\leftarrow$ $\rightarrow\rightarrow$

$\uparrow\uparrow$ | $\downarrow\downarrow$ | $\uparrow\downarrow$ | $\downarrow\uparrow$
---|---|---|---

The arrows give the interpretation of the vertical and horizontal lines, in
accordance with the values $k_{0}=0$ and $k_{1}=0$ for the standard Wigner
function. Note that the sums over the vertical and horizontal lines give the
correct probabilities of the states associated with those lines. (The same
holds for the other lines of phase space, which are not labeled in the
diagram.)

We now perform on $\rho$ the unitary transformation $U_{G}$ corresponding to
the symplectic matrix

$G=R^{2}H_{1}R=\left(\begin{array}[]{cc}\bar{\omega}&0\\\
0&\omega\end{array}\right).$ (91)

(See Eqs. (28) and (29).) The resulting state is
$\rho^{\prime}=|\psi^{\prime}\rangle\langle\psi^{\prime}|$, where
$|\psi^{\prime}\rangle$ is the maximally entangled state

$|\psi^{\prime}\rangle=\frac{1}{\sqrt{2}}(\mid\downarrow\rangle\otimes\mid\downarrow\rangle-\mid\uparrow\rangle\otimes\mid\uparrow\rangle)=\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}-1\\\
0\\\ 0\\\ 1\end{array}\right).$ (92)

We would like to express this state as a Wigner function, using the definition
$W^{f_{G}}_{\alpha}$. For the matrix $G$, we can find $f_{G}$ either by
starting with $f_{H_{1}}$, which we found earlier, and shifting it downward by
two steps (since $G$ has two factors of $R$ on the left of $H_{1}$) or by
using the formula in Eqs. (79) and (80). The result is

$f_{G}=\left(\begin{array}[]{c}0\\\ \omega\\\ 1\\\ 0\\\ 1\end{array}\right).$
(93)

This $f_{G}$ tells us how to define the origin operator $A^{f_{G}}_{0}$, and
the other phase point operators are obtained via displacements. One finds that
the $f_{G}$-Wigner function of $\rho^{\prime}$ is

${W}^{f_{G}}(\rho^{\prime})=\frac{1}{4}$ $\,\,0\,\,$ $\,\,0\,\,$ $\,\,0\,\,$
$\,\,0\,\,$ 0 0 0 0 1 1 $0$ $0$ $\,\,1\,\,$ $\,\,1\,\,$ $0$ $0$
$\leftarrow\leftarrow$ $\rightarrow\rightarrow$ $\leftarrow\rightarrow$
$\rightarrow\leftarrow$

$\uparrow\uparrow$ | $\downarrow\downarrow$ | $\uparrow\downarrow$ | $\downarrow\uparrow$
---|---|---|---

Note that the horizontal lines now have a different interpretation than
before; this is because the $n=1$ component of $f_{G}$ is not zero. In effect,
the labels of the horizontal lines have been displaced by $\omega$, since
$(f_{G})_{1}=\omega$. One can check that the sum of the Wigner function over
each vertical and horizontal line yields the correct probability of the
outcome associated with that line when the system is in the state
$\rho^{\prime}$.

One can also check that the locations of the nonzero entries in the above
table are indeed the images of the locations of the nonzero entries in the
earlier table under the operation of the symplectic matrix $G$. In this way,
the example illustrates Eq. (89). (Again, we never change the order of the
field elements $0,1,\omega,\bar{\omega}$ labeling the two axes. It is only the
physical interpretation of those values that changes.)

So far in this section, we have expressed our initial state as a standard
Wigner function ${\bf W}_{\alpha}$. But suppose we are starting by expressing
our initial state $\rho$ in the $f$-Wigner framework for some index $f$. If we
now perform $U_{L}$ on $\rho$ to get $\rho^{\prime}=U_{L}\rho
U^{\dagger}_{L}$, what does this transformation look like in phase space?

We now have, in place of Eq. (83), the equation

$W^{f}_{\alpha}(\rho)=\tfrac{1}{4}\hbox{Tr}\left[(U_{L}A^{f}_{\alpha}U^{\dagger}_{L})\rho^{\prime}\right].$
(94)

As before, we wish to re-express the operator in parentheses as an operator
belonging to a different definition of the Wigner function. We can generalize
Eq. (87) as follows.

$\begin{split}{\mathcal{I}}(U_{L}A_{\alpha}^{f}U^{\dagger}_{L})&=S_{L}{\mathcal{I}}(A_{\alpha}^{f})+f_{L}\\\
&=S_{L}(f+\alpha_{q}Q+\alpha_{p}P)+f_{L}\\\
&=S_{L}f+(L\alpha)_{q}Q+(L\alpha)_{p}P+f_{L}\\\
&={\mathcal{I}}(A^{S_{L}f+f_{L}}_{L\alpha}).\end{split}$ (95)

So the operator in parentheses on the right-hand side of Eq. (94) is
$A^{S_{L}f+f_{L}}_{L\alpha}$, and that equation can thus be rewritten as

$W^{f}_{\alpha}(\rho)=W^{S_{L}f+f_{L}}_{L\alpha}(U_{L}\rho U^{\dagger}_{L}).$
(96)

The form of the equation analogous to Eqs. (1), (13), and (89) is

$W^{S_{L}f+f_{L}}_{\alpha}(U_{L}\rho
U^{\dagger}_{L})=W^{f}_{L^{-1}\alpha}(\rho).$ (97)

So if we are already in the $f$ representation and perform the unitary
$U_{L}$, we let $L$ act on the points of phase space and we change our
representation from $f$ to $S_{L}f+f_{L}$.

To see how this works, let us go back to the above example. Again, we started
with the state $|\psi\rangle=\mid\uparrow\rangle\otimes\mid\rightarrow\rangle$
and applied the unitary operator $U_{G}$ of Eq. (29). The result was an
entangled state. We now take this entangled state as our starting state and
rename it $\rho$. We copy its $W^{f_{G}}$ representation here:

${W}^{f_{G}}(\rho)=\frac{1}{4}$ $\,\,0\,\,$ $\,\,0\,\,$ $\,\,0\,\,$
$\,\,0\,\,$ 0 0 0 0 1 1 $0$ $0$ $\,\,1\,\,$ $\,\,1\,\,$ $0$ $0$
$\leftarrow\leftarrow$ $\rightarrow\rightarrow$ $\leftarrow\rightarrow$
$\rightarrow\leftarrow$

$\uparrow\uparrow$ | $\downarrow\downarrow$ | $\uparrow\downarrow$ | $\downarrow\uparrow$
---|---|---|---

Suppose we now apply $U_{G}$ a second time. The resulting state is

$U_{G}\left[\frac{1}{\sqrt{2}}(\mid\downarrow\rangle\otimes\mid\downarrow\rangle-\mid\uparrow\rangle\otimes\mid\uparrow\rangle)\right]=\;\mid\leftarrow\rangle\otimes\mid\uparrow\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}[]{c}1\\\
0\\\ -1\\\ 0\end{array}\right).$ (98)

Let $\rho^{\prime}$ be the density matrix representing this state. According
to Eq. (97), we should now use the Wigner representation associated with the
index $S_{L}f+f_{L}$. In our present example, this index is
$S_{G}f_{G}+f_{G}$, which works out to be

$S_{G}f_{G}+f_{G}=\left(\begin{array}[]{c}0\\\ 1\\\ 0\\\ \omega\\\
1\end{array}\right).$ (99)

(Alternatively, we can observe that $G^{2}=R^{4}H_{1}R^{3}$ and find
$f_{G^{2}}$ starting with Eq. (72).) With this definition of the Wigner
function, one finds that

${W}^{S_{G}f_{G}+f_{G}}(\rho^{\prime})=\frac{1}{4}$ $\,\,0\,\,$ $\,\,0\,\,$
$\,\,0\,\,$ $\,\,0\,\,$ 1 0 0 1 0 0 $0$ $0$ $\,\,1\,\,$ $\,\,0\,\,$ $0$ $1$
$\rightarrow\leftarrow$ $\leftarrow\rightarrow$ $\rightarrow\rightarrow$
$\leftarrow\leftarrow$

$\uparrow\uparrow$ | $\downarrow\downarrow$ | $\uparrow\downarrow$ | $\downarrow\uparrow$
---|---|---|---

Note that the arrow labels of the horizontal lines have again changed, because
the $n=1$ component of the index has changed. This is part of the
reinterpretation of the Wigner-function values. The labels are consistent,
though, with the state. For example, the probability of finding the first
qubit in the state $\mid\leftarrow\rangle$, as computed by summing the Wigner-
function over the rows labeled $\leftarrow\leftarrow$ and
$\leftarrow\rightarrow$, is one, as it should be. This example illustrates Eq.
(97): the locations of the nonzero entries in the table for $\rho^{\prime}$
are the images, under the symplectic transformation $G$, of the locations of
the nonzero entries in the table for $\rho$.

Suppose now that, beginning with a state expressed in the standard Wigner
function, we execute a series of unitary operations
$U_{L_{1}},U_{L_{2}},\ldots,U_{L_{N}}$. We will typically run through a series
of $f$-Wigner functions, where the various $f$-values arise from repeated
applications of the formula $f\rightarrow S_{L}f+f_{L}$. Or, if we have kept
track of the $L_{j}$’s, then at the $M$th stage, we use the Wigner function
defined by $f_{L_{M}\cdots L_{1}}$. For any such string of unitaries, the only
Wigner-function definitions we will ever need are the 12 $f$-Wigner functions
associated with our 12 possible shift vectors $f_{L}$.

##  VI Discussion

We began this paper by recalling a remarkable fact about discrete Wigner
functions in odd prime-power dimensions: for every symplectic linear
transformation $L$ on the discrete phase space, there exists a unitary
operator $U_{L}$ corresponding to $L$ such that

$W_{\alpha}(U_{L}\rho U_{L}^{\dagger})=W_{L^{-1}\alpha}(\rho).$ (100)

That is, the effect of this unitary is simply to permute the values of the
Wigner function in accordance with the linear transformation $L$. Our work in
this paper was motivated by the fact that there is no such equation for a
$d\times d$ phase space when the dimension $d$ is a power of two.

We have found that there is a similar equation for the case $d=4$, but it
requires reinterpreting the values of the Wigner function with respect to a
variety of operator bases. We can write the equation as

$W^{g}_{\alpha}(U_{L}\rho U_{L}^{\dagger})=W^{f}_{L^{-1}\alpha}(\rho),$ (101)

where $g$ and $f$ are five-component vectors with entries in
${\mathbb{F}}_{4}$. The index $g$ depends on both $f$ and $L$ in accordance
with Eq. (97): $g=S_{L}f+f_{L}$. We have noted that the full set of indices
one will need consists of twelve shift vectors, given by Eqs. (71)-(74)
together with the shift rule associated with multiplication on the left by
$R$.

The number twelve as it appears here ties into one of the results of Ref.
Gibbons , as we now explain.

Ref. Gibbons  classifies the allowed definitions of the discrete Wigner
function for a given dimension—or one could say it classifies the allowed
choices of the set of phase point operators $\\{A_{\alpha}\\}$—into
“equivalence classes” and “similarity classes.” Two sets of operators
$\\{A_{\alpha}\\}$ and $\\{A_{\alpha}^{\prime}\\}$ are equivalent if they are
related to each other by displacement:
$A_{\alpha}^{\prime}=D_{\beta}A_{\alpha}D_{\beta}^{\dagger}$. They are similar
if they are related to each other by a symplectic linear transformation and an
associated unitary: $A_{L\alpha}^{\prime}=UA_{\alpha}U^{\dagger}$. Comparing
this last relation with Eq. (95), we see that all the sets of phase point
operators $\\{A^{f}_{\alpha}\\}$ that we will ever need in the
reinterpretation scheme of the preceding section belong to the same similarity
class.

One can also find in Ref. Gibbons  a formula, for the case $d=4$, that tells
us to which similarity class any legitimate set of phase point operators
belongs. This formula is a function whose input is essentially the index of
$A_{0}$ (though the conventions in that paper are different) and whose output
is an element of ${\mathbb{F}}_{4}$ that labels the similarity class. (There
are exactly four similarity classes when $d=4$.) The similarity-class value in
that paper was called $D$ (not to be confused with displacement operators),
and it happens that all of the $f$-Wigner functions emerging from the
construction described in the preceding section belong to the class
$D=\bar{\omega}$. It was noted in Ref. Gibbons  that this particular
similarity class consists of exactly twelve equivalence classes. These twelve
correspond to our twelve canonical shift vectors $f_{L}$. Each of the
associated $f$-Wigner functions is in a different equivalence class.

It is worth rewriting the formula for the similarity-class value $D$ in terms
of the index convention we have introduced in the present paper. And in order
to label the similarity classes in a way that privileges the particular class
we have been using here, let us define $E$ to be $D+\bar{\omega}$. Then the
similarity class considered in this paper is the one with $E=0$. Writing down
the formula for $E$ is a matter of starting with the formula for $D$ in Ref.
Gibbons  and re-expressing it in terms of the $k_{n}$’s of Fig. 1. One finds
that for any shift vector $f$—not necessarily one that arises in our
reinterpretation scheme—the similarity class of the $f$-Wigner function has
the $E$-value given by

$E(f)=r^{T}{f}+{f}^{T}M{f},$ (102)

where

$r={\omega}\left(\begin{array}[]{c}1\\\ 1\\\ 1\\\ 1\\\
1\end{array}\right)\hskip 17.07164pt\hbox{and}\hskip
17.07164ptM=\left(\begin{array}[]{ccccc}0&1&{\omega}&0&0\\\
0&0&1&{\omega}&0\\\ 0&0&0&1&{\omega}\\\ {\omega}&0&0&0&1\\\
1&{\omega}&0&0&0\end{array}\right).$ (103)

Clearly our standard Wigner function, whose $f$ vector consists entirely of
zeroes, has $E=0$. It must also be the case that for any symplectic matrix
$L$, the transformation $f\rightarrow S_{L}f+f_{L}$ preserves the value of
$E$. (Again we are comparing Eq. (95) to the definition of “similar.”) In
particular, for any $L$, we have $E(f_{L})=0$.

Finally, we note that all the $f$-Wigner functions in the similarity class
$E=0$ have a special kind of symmetry. Consider first our standard Wigner
function ${\bf W}$. From Eq. (89) with $L$ set equal to $R$, we find that our
standard Wigner function is covariant with respect to the unitary “rotation”
operator $U_{R}$:

${\bf W}_{\alpha}(U_{R}\rho U^{\dagger}_{R})={\bf W}_{R^{-1}\alpha}(\rho),$
(104)

where we have used the fact that $f_{R}$ is the zero vector. To see how this
relation generalizes to the $f_{L}$-Wigner function, we begin again with Eq.
(89):

$W^{f_{L}}_{\alpha}(U_{L}\rho U^{\dagger}_{L})={\bf W}_{L^{-1}\alpha}(\rho).$
(105)

From Eqs. (104) and (105), it follows that

$W_{\alpha}^{f_{L}}(V_{R_{L}}\rho
V^{\dagger}_{R_{L}})=W^{f_{L}}_{R_{L}^{-1}\alpha}(\rho),$ (106)

where

$R_{L}=LRL^{-1}\hskip 8.53581pt\hbox{and}\hskip
8.53581ptV_{R_{L}}=U_{L}U_{R}U^{\dagger}_{L}.$ (107)

The matrix $R_{L}$ and the unitary $V_{R_{L}}$ have the following two
important properties: (i) $V_{R_{L}}$ is a unitary operator that we can
associate with $R_{L}$, in the sense that

$V_{R_{L}}D_{\beta}V^{\dagger}_{R_{L}}\propto D_{R_{L}\beta},$ (108)

and (ii) the matrix $R_{L}$ has as much right to be regarded as a rotation
matrix as $R$ does, in that it has period five and cycles through all five
striations. Thus, the $f_{L}$-Wigner function has a kind of rotational
symmetry, just as ${\bf W}$ does, but with a different notion of rotation.
Moreover, one can extend this argument to any other $f$-Wigner function with
$E(f)=0$, using the fact that each such $f$ can be obtained from one of the
$f_{L}$’s by a translation.

Indeed, we could have built analogues of Figures 1 and 2 based on the special
rotation matrix and special “rotation unitary” associated with any of these
$f$-Wigner functions, in which case that particular $f$-Wigner function would
have emerged as the natural “standard” Wigner function. So it would be wrong
to suppose that our work in this paper favors a single definition of the
Wigner function for a pair of qubits. What it does, rather, is to show how one
can select a relatively small set of such definitions—exactly twelve—that
collectively exhibit a kind of covariance under the restricted Clifford group.

## Acknowledgements

I thank Billy Braasch for a stimulating conversation that sparked the research
presented in this paper.

## Appendix

Here we show that the unitary matrices $U_{L}$ defined in Section II satisfy
Eq. (24). That is, we show that

$U_{L_{1}}U_{L_{2}}\propto U_{L_{1}L_{2}}$ (109)

for all pairs of symplectic matrices $L_{1}$ and $L_{2}$.

We begin with a few special cases that are easily checked:

$\begin{split}&U_{H_{x}}U_{H_{y}}=U_{H_{x}H_{y}},\;\hbox{for}\;x,y\in{\mathbb{F}}_{4},\\\
&U_{R}U_{H_{\bar{\omega}}}U_{R}=U_{RH_{\bar{\omega}}R}=U_{H_{\bar{\omega}}},\\\
&U_{R}^{5}=U_{R^{5}}=U_{H_{0}}=I.\end{split}$ (110)

We now claim that it is sufficient to show that

$U_{H_{x}}U_{R}^{s}U_{H_{y}}\propto
U_{H_{x}R^{s}H_{y}},\;\hbox{for}\;x,y\in{\mathbb{F}}_{4},\;s=0,\ldots,4.$
(111)

It will then immediately follow that Eq. (109) is true for $L_{1}$ and $L_{2}$
of the forms

$L_{1}=H_{x}R^{s_{1}}\hskip 14.22636pt\hbox{and}\hskip
14.22636ptL_{2}=R^{r_{2}}H_{y}.$ (112)

Extending this result to all $L_{1}$ and $L_{2}$ is a matter of multiplying
the above $L_{1}$ on the left by $R^{r_{1}}$, multiplying $L_{2}$ on the right
by $R^{s_{2}}$, and using the fact that $U_{R^{r}H_{x}R^{s}}$ is defined to be
$U_{R}^{r}U_{H_{x}}U_{R}^{s}$.

So we turn our attention to demonstrating Eq. (111).

If either $x$ or $y$ is zero, the equation reduces to the definition of
$U_{R^{r}H_{x}R^{s}}$ just quoted.

Now suppose that $x=\bar{\omega}$. Then on the left-hand side of Eq. (111), we
can write

$U_{H_{\bar{\omega}}}U_{R}^{s}=U_{R}^{-s}\big{(}U_{R}^{s}U_{H_{\bar{\omega}}}U_{R}^{s}\big{)}=U_{R}^{-s}U_{H_{\bar{\omega}}},$
(113)

so that $U_{H_{\bar{\omega}}}$ directly multiplies $U_{H_{y}}$ and we can
combine the two unitaries as in Eq. (110). Meanwhile, on the right-hand side
of Eq. (111), we can similarly remove the $R$’s lying between
$H_{\bar{\omega}}$ and $H_{y}$ and combine those two matrices into a single
$H$ matrix. And then Eq. (111) again follows from the definition of
$U_{R^{r}H_{x}R^{s}}$. An analogous argument applies when $x$ has an arbitrary
value and $y$ is equal to $\bar{\omega}$.

This leaves us with the cases where $(x,y)$ is equal to $(1,1)$, $(1,\omega)$,
$(\omega,1)$, and $(\omega,\omega)$. In fact, though, it is enough to deal
with the case $(x,y)=(1,1)$, for which Eq. (111) becomes

$U_{H_{1}}U_{R}^{s}U_{H_{1}}\propto U_{H_{1}R^{s}H_{1}}.$ (114)

By multiplying both sides of this equation on the right by
$U_{H_{\bar{\omega}}}$ and using essentially the same argument as in the
preceding paragraph, we can turn this equation into

$U_{H_{1}}U_{R}^{s}U_{H_{\omega}}\propto U_{H_{1}R^{s}H_{\omega}}.$ (115)

And we can make the same transformation on the left-hand sides of the
expressions. All that remains, then, is to check that Eq. (114) holds for
$s=0,\ldots,4$. Upon checking directly, one finds that the equation indeed
holds.

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