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ELSEVIER Electronic Notes in Theoretical Computer Science 270 (2) (2011) 37-58

www.elsevier.com/locate/entcs

Classical Representations of Qubit Channels

Tanner Crowder1 Keye Martin2

Naval Research Laboratory Center for High Assurance Computer Systems Washington DC 20375

Abstract

A set of qubit channels has a classical representation when it is isomorphic to the convex closure of a group of classical channels. We prove that there are five such groups, each being either a subgroup of the alternating group on four letters, or a subgroup of the symmetric group on three letters.

Keywords: Group theory, quantum channels, convexity.

1 Introduction

In [7], the Klein four group Z^ is represented as a collection of classical channels with four inputs and four outputs, and it is shown that its convex closure can be embedded into the set of qubit channels. The qubit channels which result from this embedding can be studied, both qualitatively and quantitatively, as though they were classical channels. This has a surprising consequence: by working with classical channels, we can develop a method for completely interrupting quantum communication that only requires us to act in a deterministic manner, despite the fact that the end result of this process is purely random. Stepping back for a minute, a natural question emerges: are there any other classes of qubit channels that have classical representations?

In this paper, we completely answer this question by showing that there are five such classes and no others. Each class corresponds to either a subgroup of A4, the alternating group on four letters, or to a subgroup of S3, the symmetric group on three letters. Let us briefly explain why we find this to be a very surprising result. If we take the stance that classical and quantum channels are very different, then we would expect them to have only trivial structure in common. But the

1 Email: crowder@chacs.nrl.navy.mil

2 Email: kmartin@itd.nrl.navy.mil

1571-0661/$ - see front matter, Published by Elsevier B.V. doi:10.1016/j.entcs.2011.01.022

channels represented by the convex closure of S3 and A4 have structure that is far from trivial. If, on the other hand, we try to dismiss this commonality on the grounds that the definition of a classical representation forces all qubit channels to be unital and that unital channels 'seem' classical, then why aren't there more than five groups capable of providing a classical representation, especially when the set of unital qubit channels contains arbitrarily large groups?

2 Classical representations

Let (m, n) denote the set of stochastic matrices with m rows and n columns, the classical channels with m inputs and n outputs. The set of qubit channels Q consists of all affine transformations of the unit ball in R3 which arise as the Bloch representations of the convex linear, completely positive, trace preserving maps on 2 x 2 density matrices. The convex closure of a set X is denoted as {X).

All subgroups of matrices are assumed to have the identity matrix I as the group identity and to be nontrivial.

Definition 2.1 Let G be a subgroup of (m, n). An embedding of {G) into Q is a function p : {G) ^ Q such that for all x,y £ {G),

• p(I) = I,

• p(xy) = p(x)p(y),

• p(px + (1 — p)y) = pp(x) + (1 — p)p(y) whenever p £ [0,1], and

• p(x) = p(y) ^ x = y.

That is, an embedding is an injective, convex-linear homomorphism. The set of qubit channels p({G)) is then said to have a classical representation.

Proposition 2.2 Let G be a subgroup of (m,n). If there is an embedding p : {G) ^ Q, then n = m, G is finite and the image of p is isomorphic to a convex subset of {SO(3)).

Proof. Since every element of G is an invertible matrix, we have m = n. The only classical channels in which the inverses are also classical channels are the permutation matrices, so G is finite. To prove that the image of G under p is contained in SO (3), let us give a new and elementary proof of the fact that any qubit channel with an inverse that is also a qubit channel must be a rotation of the Bloch sphere.

Let f (x) = Mx + b be a qubit channel with an inverse. Because f is injective, the equation Mx = 0 has a unique solution (x = 0), which means M is invertible. Then the inverse of f

f-l(x) = M-1x — M-lb

is also affine. Thus, f and f-1 are both continuous, and so f is a homeomorphism of the unit ball B3 into itself. By algebraic topology, f must take the boundary of the unit ball into itself:

(Vx £ B3) \x\ = 1 ^\f (x)| = 1.

We now claim that (Mx,b) = 0 for all \x\ = 1. Let \x\ = 1. Then \ — x\ = 1, and since f takes boundaries to boundaries, we have \f (x)\ = 1 = \f (—x)\. Then

\f (x)\2 = (Mx + b,Mx + b) = (Mx — b, Mx — b) = \ f (—x)\2.

The equation in the middle reduces to (Mx, b) = 0.

Now we show b = 0. Since M is invertible, it is surjective and thus contains the standard basis e1,e2,e3 in its image, so we can find an x £ R3 such that Mx = e1. Writing x as a linear combination of the ei, we have

Similarly, (ei,b) = 0 for all i. So b = 0 and any affine homeomorphism of the unit ball onto itself will have to take the center to the center. However, this means any such qubit channel must be unital, and as shown by elementary means in [8], M must then be an orthogonal matrix with determinant +1, i.e., it must belong to

For those who find elementary proofs distasteful, it is also possible to use Theorem 2.1 of [9] and then appeal to the fact that all unitary channels induce rotations of the Bloch sphere (a complete proof of the latter is given in [8]).

Either way, the image of p is isomorphic to a convex subset of {SO(3)), where SO(3) is the group of orthogonal matrices with determinant +1. □

3 Cyclic groups

By the result of the last section, we know that any group whose convex closure may be embedded into Q has to be a finite subgroup of SO(3). Our first lemma, which we shall refer back to often, tells us that any rotation is conjugate to a principal rotation about the x-axis:

Definition 3.1 A principal rotation about the x-axis is denoted

(ei,b) = (Mx,b) = ^ Xi(Mei,b) = 0.

SO(3).

rx(d) :=

for some angle 0 £ (0, 2n]. Notice that rx(a)rx(() = rx(a + (). Lemma 3.2 If f £ SO(3) then there is r £ SO(3) such that

rfr1 = rx(0)

for some angle 6 e (0, 2n].

Proof. Each 3 x 3 rotation f is a normal matrix, so by Theorem 3.3 of [12], we can find an orthogonal matrix r such that rfr1 is block diagonal, each block being

either a 1 x 1 matrix consisting of a real eigenvalue of f or a 2 x 2 matrix of the form [ . If det(r) = —1, we replace r by —r since det(-r) = (—1)3 det(r) and

—b ay

thus we assume that r £ SO(3). Again because we are in dimension three, only two cases are possible: either rfrf is a diagonal matrix or it has the form

(c 0 0

rfr1 = 0 a b \0 —b a #

where c = ±1 is a real eigenvalue of f. Using det(r) = 1/ det(r4) since rf = r-1,

det(rfr4) = det(r)det(f )det(r4) = det(f) = 1 = c(a2 + b2)

so we see that c = 1 and that (a, b) is a point on the unit circle. In the case that rfrf is diagonal, we see that rfrf must be either the identity or one of the matrices

'1 0 0 \ /—1 0 0 \ /—10 0^ 0 —10 , Sy = 010 \ , sz = 0 —10

^00 —1) \ 0 0 —1/ \ 0 0 1,

The identity and sx have the desired form. In the case rfr1 = sy, we conjugate by the rotation

'0 1 0^ 0 0 1

to obtain arf (ar) = sx which again has the desired form. In the case rfr1 = sz, we conjugate by a1.

Finally, because (a, b) is a point on the unit circle, there is an angle 6 £ (0, 2^] such that a = cos6 and b = sin6. Thus, in all cases, we have rfr1 = rx(6). □

Recall that an element x in a group G is said to have order n if n is the smallest positive integer for which xn = I.

Lemma 3.3 Let f £ SO(3) have order n > 2 and let G = {f : 1 < i < n} ~ Zn be the cyclic group that it generates. Then there is a nonzero integer x £ Z and an r £ SO(3) such that

rf xr* = rx(2n/n).

Thus, G is also generated by an element fx £ G that is conjugate to rx(2n/n).

Proof. Let a = rx(2n/n). It is routine to show that a has order n and thus generates Zn. By Lemma 3.2, take r £ SO(3) with rfr1 = rx(6). Since f has order n, so does rx(6) and thus rx(6)n = rx(n6) = I. Then 6 £ (0, 2n] has the form 6 = 2nk/n for an integer 0 < k < n. But

rx(6)= rx( 2-n) = rx( v) " = ak

and by standard group theory, since a has order n, the order of ak is n/(k, n), where (k, n) is the greatest common divisor of k and n. But we know that ak = rx(0) has order n, so (k, n) = 1. By the Euclidean algorithm, there are integers x and y, not both zero, such that

xk + yn = (k, n) = 1. In addition, notice that x = 0 since n > 2. Now we calculate as follows:

rf xrt = (ak )x = akx = a1-yn = a1a-yn = a(an)-y = a(I)-y = a (1)

and since a has order n, so does fx. Then the fact that (fx)n = I implies

{(fx)i : 1 < i < n}C{fi : 1 < i < n},

and because n is the least such positive integer, both of these sets have exactly n elements and so are equal. Then G contains a generator conjugate to rx(2n/n). □ In particular, a finite cyclic subgroup of SO(3) is conjugate to the one generated by rx(2n/n), so any two finite cyclic subgroups of SO(3) with the same order are conjugate. We now classify the cyclic subgroups of Q that have classical representations. This requires a few lemmas in advance.

Lemma 3.4 For all n > 4, there is an integer k > 1 such that n < 2k < 31.

Proof. First, log2 (3n) — log2(n) = log2(3) > log2(2) = 1, so the interval [log2 n, log2 3n] contains at least one integer, call it m. Then log2(n) < m < log2(3n) where the strict inequality on the right holds because 3n is never a power of two.

If m = log2(n), then because the length of [log2 n, log2 3n] is strictly greater than 1, it must also contain m + 1, so set p := m +1. If not, set p := m. Either way,

log2(n) < p < log2(3n) n < 2p < 3n.

Dividing by 4, we get n/4 < 2p-2 < 3n/4. Because p > log2(n) > log2(4) = 2, p > 3 so the desired integer is k := p — 2 > 1. □

Recall that the element in position (i,j) of a matrix a is denoted by aij.

Lemma 3.5 Let a £ (n, n) and let m £ N be divisible by 2. If am = 0 then

m/2 n aii =°.

Proof. If A and B are two n x n matrices, their product AB satisfies (AB)ii = Yn=l AirBri. Since am = am/2am/2, we have

m/2 m/2 ai,r ar,i .

The entries in a classical channel are nonnegative, so this means we have written zero as the sum of n nonnegative products. Then each product must be zero. In particular, for r = i, the product ar'm/'1 a/m^2 is zero, which means ^2 =0. □

Proposition 3.6 The convex closure of Zn C (m,m) cannot be embedded into Q when n > 4.

Proof. Let G = [gz : 1 < i < n} C SO(3) be a cyclic group of order n with generator g. By Lemma 3.3, we can assume that there is r £ SO(3) with rgr1 = rx(d) and 0 = 2n/n. For the remainder of this proof, rx(d) will be denoted by rx. The average of all elements in rGr1,

± :.= Y '-X ,

• 1 n

is the unique algebraic zero of the monoid {rGr1): the only element L £ {rGr1) such that L ■ x = x ■ -L for all x £ {rGr1). We will now show that there are x,y £ [0,1] with x + y = 1 and

rx+rl) + yl^j^j = (2)

Using Lemma 3.4, let j be a power of two such that ^ < j < X. Then because n/2 < jd < 3n/2, we have cos(j^) < 0, while 0 <6 < n/2 gives cos 6 > 0. Defining

cos j6

x :=-IT--~n £ (0,1]

cos j6 — cos 6 and setting y = 1 — x £ [0,1) we have

rx + +Jr-x±j\ = | 0 x cos 6 + y cos j6 0 ] = (000

V J \0 0 x cos 6 + y cos j6j \0 0 0;

This projection belongs to {rGrf) since we have written it as a convex sum of elements belonging to {rGr1). To see that equation (2) holds in {rGr1), notice that

/1 0 0\ /10 0\ /1 0 0^ rx(a)\ 0 0 0 = 0 0 N rx(a) =000

\0 0 0J \0 0 0/ \0 0 0y

for any a; thus, this projection is the algebraic zero for {rGr1) and thus equal to L.

Multiplying equation (2) on the left by r1 and on the right by r, the analogous equation holds in {G):

g+gt ^ + y (^^j) = £ n = L- (3)

We will now show that this equation is impossible for classical (m, m) channels. Let

T. Crowder, K. Martin /Electronic Notes in Theoretical Computer Science 270 (2) (2011) 37—58 43

a £ (m, m) be a generator for a copy of Zn. If

a + at \ ( aj + (aj )t —)+

is the algebraic zero then multiplying it by a leaves it unchanged:

a2 + I \ ( aj+1 + an-j+1 \ ( a + at \ ( aj + (aj )t

2 )+ H—2—)= n~J+yv 2

Rearranging terms we can see that

xl = xa + xat + yaj + y(aj )t — xa2 — yai+1 — yan-i+1. (4)

Since a generates Zn and j < 3n/4 < n, aj = I. Because a and hence aj is a permutation, there is an entry aJii on the diagonal of aj which is equal to zero. Since j = 2k with k > 1, j is divisible by 2, so by Lemma 3.5, a2i = 0 and thus aii = 0 as well. Since transposes do not alter diagonals, atii = 0 and (aj)tii = 0. For the (i,i) entry of the matrix in equation (4), we have

x = —y(aji+1 + ari+1).

Since the powers of a are permutations, o?i+l+arn-i+l is either 0, 1 or 2, contradicting x > 0.

Since equation (3) holds for any quantum generator of Zn but never holds for a classical generator of Zn, the convex closure of Zn C (m, m) can never be embedded into Q when n > 4. □

An interesting consequence of the last proof is that for an integer n > 4,

^cos(2n/n • k) =0 y^ sin(2n/n • k). k=1 k=1

The last result permits a solution of the classification problem for cyclic groups: let (0 A i0 1 0N

flip =(JoJ £ (2, 2) & c = 00^ £ (3, 3)

denote generators for Z2 and Z3 respectively. Theorem 3.7

• If G C (m,m) is a cyclic group and p : {G) ^ Q is an embedding, then G ~ Z2 or G ~ Z3.

• The convex closure of the groups Z2 ~ {I, flip} and Z3 ~ {I,c,c2} can be embedded into Q.

44 T. Crowder, K. Martin /Electronic Notes in Theoretical Computer Science 270 (2) (2011) 37—58

Proof. For Z2, we map I, flip £ (2, 2) onto the spin channels

10 0 0 1 0 0 0 1

10 0 0 -10 0 0-1

and then extend linearly to obtain an embedding

x1 + x2 0 0

0 x1 — x2 0 0 0 x1 — x2

For Z3, c is a classical channel that generates Z3. However, it is also a qubit channel: as a permutation, it is orthogonal, and direct calculation shows that det(c) = 1, so that c £ SO(3). Thus, its convex closure embeds into Q via the identity map! □

4 Finite abelian groups

Lemma 4.1 If Q contains a copy of the group Zn, then n = 1 or n = 2.

Proof. Any involution f £ Q must be a rotation and hence symmetric since f = f-1 = ft. Thus, there is a rotation r £ SO(3) such that rfrt is diagonal. However, rfrt £ SO(3) is also an involution, but the only diagonal involutions in SO(3) are the spin channels:

The reason is: being symmetric, each has three real eigenvalues; being orthogonal, each eigenvalue is ±1; being a rotation, the product of these eigenvalues must be 1.

Now if Q contains a copy of Zn, then it contains 2n commutative involutions. Because all the elements of Zn commute, they diagonalize in a common basis. Thus, there is r £ SO(3) such that rfrt £ SO(3) is a diagonal involution for each f £ Zn. Then r (Zn) rt C {I,sx,sy,sz}. And since the number of elements in the group r (Zn) rt is 2n, we have that either n = 1 or n = 2. □ Let rz (d) denote the principal rotation about the z-axis:

From annoying arithmetic, it follows that rx(0) and rz(a) commute iff they are both diagonal.

Lemma 4.2 If Q contains a copy of the group then n = 1.

Proof. Suppose Q contains a copy of Z2. Let f,g £ Z2 C SO(3) be elements of order 3 such that

{f* : 1 < 3}H[gi : 1 < i < 3} = {I}. Notice that Z2 contains elements of order three that generate distinct subgroups: intuitively, in additive notation, f = (0,1) and g = (1,0) are examples of such elements.

Because f,g £ SO(3) are commutative normal matrices, by Theorem 2.5.15 of [5], there is a single orthogonal matrix r such that rfr1 and rgr1 are block diagonal, each block being either a 1 x 1 matrix consisting of a real eigenvalue or a

2 x 2 matrix of the form

Because f and g have order 3, so do rfrt and rgrt, so neither can have blocks consisting solely of eigenvalues, or they would be diagonal and then have order 1 or 2. Thus, the only possible forms are

c 0 0 \ fab 0 \

0 a b & \—b a 0 . 0 —b a ) \ 0 0 c )

Because c refers to a real eigenvalue of an orthogonal matrix, c = ±1, but since the determinant of each is c(a2 + b2) = 1, being the determinant of the rotations f and g respectively, c =1 and a2 + b2 = 1. Thus, the first matrix has the form rx(6) for 6 £ (0, 2n] and the second has the form rz(a) for some a £ (0, 2n].

However, because f and g commute, rfr1 and rgr1 also commute, and this means that both have the form rx(6) or both have the form rz(a), since rx(6) and rz(a) only commute when they are diagonal (which as we have seen, is impossible because rfr1 and rgr1 have order three). Thus, without loss of generality,

rfr1 = rx(6) & rgr1 = rx(a)

where 6 = 2nk/3 and a = 2nm/3 for integers 0 < k,m < 3. If k = m, then f = g, contradicting the fact that f and g generate distinct subgroups. If k = m, then without loss of generality, k = 1 and m = 2. This gives

fg = (r1 rx(2n/3)r) (r1 rx(4n/3)r) = rVx(6n/3)r = I

which means f= g. However, f has order 3, so g = f= f2, which again contradicts the fact that f and g generate distinct subgroups. Thus, SO(3) does not contain a copy of Z2.

If Q contains a copy of Z3 C SO (3) for n > 1, then it contains a copy of Z2, which as we have seen, is impossible. □

The fundamental theorem on finite abelian groups states that any finite abelian group G is isomorphic to a direct product of cyclic groups whose orders are powers of primes:

G -n i=1

a b —b a

where the pi are primes, not necessarily distinct, and each ri > 1. We will now classify the abelian groups that can be used to represent qubit channels. Let

{I ® I, I ® flip, flip ® I, flip ® flip} C (4,4)

denote the natural copy of the Klein four group studied in [7]. Theorem 4.3

• If G C (m, m) is an abelian group and p : {G) ^ Q is an embedding, then G is isomorphic to Z|, Z3 or Z2.

• The convex closure of the groups Z2 — {I ® I, I ® flip, flip ® I, flip ® flip}, Z3 — {I,c,c2} and Z2 — {I, flip} can be embedded, into Q.

Proof. Let G be an abelian group in (m, m) whose convex closure can be embedded into Q. By the fundamental theorem on finite abelian groups, we can write it as a product of cyclic groups

G — II

where the pi are primes, not necessarily distinct, and each ri > 1. If one of these primes pi > 5, then because pi divides the order of G,

G = n % I = n pr,

i=1 i=1

the converse of Lagrange's theorem for finite abelian groups implies that G must have a subgroup of order pi. But the only finite abelian group of order pi is Zpi, a cyclic group having order greater than 3. Since the convex closure of Zpi cannot be embedded in Q by Theorem 3.7, neither can the convex closure of G. Thus, there are three possibilities for G:

(1) G — nn=i Z2^ with n > 1 and each ri > 1,

(2) G — nn=i Z3^ with n > 1 and each ri > 1, or

(3) G — nn=i Z2ri x[]Z3si with n,m > 1 and each ri,si > 1.

In case (1), each ri = 1 or else G has a cyclic subgroup of order > 4, contradicting Theorem 3.7. Thus, G is simply a product of copies of Z2. By Lemma 4.1, the only possibilities are Z2, whose convex closure can be embedded by Theorem 3.7, and Z2, the latter of which we will show later in this proof has a convex closure that can be embedded.

In case (2), each ri = 1 or else G has a cyclic subgroup of order > 4, contradicting Theorem 3.7. Thus, G is simply a product of copies of Z3. By Lemma 4.2, the only possibility is Z3, whose convex closure can be embedded by Theorem 3.7.

In case (3), G has a subgroup H — Z2ri x Z3s\ where r1,s1 > 1. Because H is a product of cyclic groups whose orders are relatively prime, by Theorem 6.1 of [11], we have H — Z2ri3n, a cyclic group of order IH| = 2n3si > 6 > 4, which is impossible by Theorem 3.7.

T. Crowder, K. Martin /Electronic Notes in Theoretical Computer Science 270 (2) (2011) 37—58 47

Now let us prove that the convex closure of Z2 = {I &I, I & flip, flip & I, flip & flip} can be embedded into Q. Using the spin channels from Lemma 4.1, we extend the natural isomorphism

I & I ^ I, I & flip ^ sx, flip & I ^ sy, flip & flip ^ sz convex linearly to obtain

(x1 x2 x3 x4\

X2 X1 X4 X3 X3 X4 X1 X2 \X4 X3 X2 X1j

X + x2 — x3 — x4 0 0

0 x1 — x2 + x3 — x4 0

0 0 x1 — x2 — x3 + x4/

This clearly defines a function. As a convex linear extension of an isomorphism, it preserves the identity, convex sums and products. We need to prove that it is injective. Let f,g £ (Z2) be defined by a convex sum involving probabilities x* and y* respectively. If p(f) = p(g), then

P(f) — P(g) = c1I + c2Sx + c3Sy + c4Sz = 0

where each c* = x* — y* so that ^4=1 c = 0. Then the c* satisfy the following four equations:

c1 + c2 + c3 + c4 =0 (5)

c1 + c2 — c3 — c4 =0 (6)

c1 — c2 + c3 — c4 =0 (7)

c1 — c2 — c3 + c4 = 0. (8)

Adding (7) and (8) gives c1 = c2. Substituting this into (7) gives c3 = c4. Substituting both of these into (6) gives c1 = c3. Then c1 = c2 = c3 = c4. By (5), each c* = 0 and thus x* = y* for all i. This proves f = g so that p is injective. □

5 Finite groups

In this section, we show that there are at most two nonabelian groups that can be used to represent qubit channels. Let A4 denote the alternating group on four letters and let S3 denote the symmetric group on three letters.

Theorem 5.1 If G C (n,n) is a nonabelian group whose convex closure can be embedded into Q, then G is either isomorphic to A4 or to S3.

Proof. If |G| is divisible by a prime p > 5, then by Theorem 5.2 of [6], G contains an element of order p and thus a subgroup isomorphic to Zp. By Theorem 3.7, (G) cannot be embedded into Q. Then let us factor |G| into a product of its possible prime powers as |G| = 2p3q where p,q > 0.

If q > 2, then G contains a subgroup of order 32 by the first Sylow theorem, Theorem 5.7 of [6], but then by Exercise 13 on page 98 of [6], this subgroup must be abelian, which is impossible by Theorem 3.7. Thus, 0 < q < 1.

Write |G| = 2p3q, where p > 0 and 0 < q < 1. If p > 3, then by the first Sylow theorem, G contains a subgroup of order 23 = 8. By Theorem 4.3, this subgroup cannot be abelian. But if it is nonabelian, it must contain an element of order 4, by the proof of Prop. 6.3 in [6], so G contains a copy of Z4, which is impossible by Theorem 3.7. Thus, 0 < p < 2.

Write |G| = 2p3q with 0 < p < 2 and 0 < q < 1. If p = 0, then G is either the trivial group (q = 0) or Z3 (q = 1), using the table on page 98 of [6]. Similarly, if q = 0, then by the same table, the only nontrivial groups are Z2, Z4 and Z2, all of which are abelian. Thus, q = 1 and 1 < p < 2.

Write |G| = 2p3 with 1 < p < 2. If p = 1, then |G| = 6, and the only nonabelian group of order 6 is S3, using the table on page 98 of [6]. If p = 2, then |G| = 12, and using the same table, the only nonabelian groups of order 12 are D6, T and A4. Both of the groups Da and T contain elements of order 6, by remarks (i) and (ii) on page 98 of [6], and thus contain a copy of Z6, in contrast to Theorem 3.7.

Thus, the only two nonabelian groups capable of being embedded into Q are A4 and S3. □

By Lagrange's theorem, any subgroup of a finite group must divide the order of the group. Because |S3| = 6 and |A4| = 12, any proper subgroup of S3 must have order 2 or 3, while any proper subgroup of A4 must have order 2,3,4 or 6. However, by Exercise 8 on Page 51 of [6], A4 does not have a subgroup of order 6. The only groups of order 2 and 3 are Z2 and Z3, while the only groups of order four are Z4 and Z2. Thus, all the proper subgroups of A4 and S3 have convex closures that can be embedded into Q, so there is no obvious way to rule either case out. In addition, S3 is not a subgroup of A4, so the cases A4 and S3 must be considered separately.

The set of even permutations in (4,4) comprises a group A4 called the alternating group on four letters. Explicitly, its involutions are

/1 0 0 0\ 0 1 0 0\ 0 0 1 0 0 0 0 1

0 10 0 1 0 0 0 0 0 0 1 0 0 1 0

0 0 10 , X = 0 0 0 1 . y = 1 0 0 0 , ^ = 0 1 0 0

\0 0 0 1/ 0 0 10/ 0 1 0 0 1 0 0 0

its order three elements are

/0 0 0 1\ /01 0 0\ 1 0 0 0 0 0 1 0

0 10 0 h _ 0 0 0 1 , C = 0 0 1 0 , d = 1 0 0 0

10 0 0 ) b = 0 0 1 0 0 0 0 1 0 1 0 0

\0 0 1 0/ \10 0 0/ 0 1 0 0 0 0 0 1

together with their squares:

/0 0 1 0\ 0 10 0 0 0 0 1 \1 0 0 0/

/0 0 0 1\ 10 0 0 0 0 10 \0 1 0 0/

/1 0 0 0\ 0 0 0 1 0 10 0 \0 0 1 0/

/0 1 0 0\ 0 0 10 10 0 0 \0 0 0 1/

A quick proof that the matrices above give a copy of A4 is to first notice that a is even because it can be written as the product of two transpositions a = fg, where f swaps the first and third elements and g swaps the third and fourth. Similarly, x is also even. Thus, xa = b, bx = c and xc = d are even as the product of evens, as are their squares. Finally, y = ac2, z = xy and x2 = I are even. Since we have generated twelve distinct even permutations and a four element set only has 4!/2 = 12 even permutations, this set of matrices forms a copy of A4:

Definition 6.1

{I, x, y, z, a, b, c, d, a2,b2,c2,d2} C (4,4).

Notice that any copy of A4 can be generated from an order three element and an even involution. Let us use this intuition to generate a likely candidate for a copy of A4 within SO(3). Starting with the involutions of Lemma 4.1, we take

'1 0 0^ 0 1 0 | ,sa .0 0 1,

'10 0 0 -1 0 Is ,0 0 -1,

—1 0 0 0 1 0 | ,Sz 0 0-1,

—1 0 0^ 0 -10 0 0 1;

and we choose our order three generator to be

0 —1 0N 0 0 1 — 1 0 0,

Now multiplying sx and f in various ways gives other order three elements

'0 —1 0 \ /0 1 0\ /010

0 0 11 , j = Sxh = 0 0 —1

g = sxf = I 0 0 -1 I , h = gsx ,10 0

J 0 0>

,-1 0 0

along with their squares:

0 0 -1N -10 0 0 1 0

0 0 1N -10 0 0 -1 0>

'0 0 1N 1 0 0 ,0 1 0;

c2 „2 1,2 -2i

'0 0 -1N 1 0 0 ,0-1 0

Definition 6.2 G := {I, Sx,Sy,Sz,f,g, h,j, f2,g2,h2,j2} C SO(3).

The set G forms a group isomorphic to A4 in a natural way. Let us prove this:

Lemma 6.3 The set G C SO(3) is a group isomorphic to A4 C (4,4). An isomorphism p is

• P(I) = I, P(x) = Sx, p{y) = sy, p(z) = Sz,

• P(a) = f, P(b) = g, P(c) = h, p(d) = j,

• p(a2) = f2, p(b2) = g2, p(c2) = h2, p(d2) = j2.

Proof. By page 225 of [1], a presentation of the alternating group on four letters is given by generators a and (3 and relations a3 = (33 = (afî)2 = I. Within G, the elements j and h2 satisfy such relations. Thus, the group they generate is isomorphic to A4. We define p on these generators

p(d)= j & p(c2) = h2

and then extend p homomorphically. By [1], c2 and d generate A4, so p is an isomorphism between A4 and the group generated by j and h2. What we have to prove is that the latter group is G. We do so by explicitly calculating p on the remaining ten elements.

Let us calculate the value of p on the involutions I and x. Using jh2 = sx and dc2 = x,

• p(x) = p(dc2) = p(d)p(c2) = jh2 = sx,

• p(I) = p(x2) = p(x)p(x) = SxSx = I.

Using the equations mentioned before the definition of A4, as well as those mentioned during the definition of G, we can now calculate it on a, b and c:

• p(c) = p(xd) = p(x)p(d) = Sxj = h

• p(b) = p(cx) = p(c)p(x) = hsx = g

• P(a) = P(xb) = P(x)P(b) = sx g = f-

Then the equations p(a2) = f2, p(b2) = g2, p(c2) = h2 and p(d2) = j2 follow. Finally, we calculate the value of p on y and z. Using y = ac2 and fh2 = sy,

p(y) = p(a)p(c2) = fh

while the fact that z = xy gives

p(z) = p(x)p(y) = sxsy = sz

finishing the proof. □

We now show that this isomorphism extends to the convex closures of both groups.

Theorem 6.4 The convex closure of A4 can be embedded into Q.

Proof. Let p : (A4) ^ {G) be defined on group elements in the natural way and then extended convex linearly. If we show that p is a function and that it is injective, then it follows from [7] that it must actually be an embedding. An

T. Crowder, K. Martin /Electronic Notes in Theoretical Computer Science 270 (2) (2011) 37—58 51

arbitrary f E (A4) can be written

( e + c + C2 x + b + d2 y + d + a2 z + a + 62 \ x + d + 62 e + a + a2 z + c + d2 y + b + C2 y + a + d2 z + d + c2 e + b + b2 x + c + a2 \z + b + a2 y + c + b2 x + a + c2 e + d + d2j

where we have written a general element f E (A4) as a convex sum

f = e ■ I + x ■ x + y ■ y + z ■ z + a ■ a + b ■ b + c ■ c + d ■ d + a2 ■ a2 + b2 ■ b2 + c2 ■ c2 + d2 ■ d2

of elements of the group A4 C (4,4). Notice that each probability in a general convex sum f E (A4) has the same name as its associated group element except in cases where this would create clear inconsistencies 3 . The element p(f) is thus

(e + x — y — z —a — b + c + d -a2 + b2 + c2 — d2N —a2 — b2 + c2 + d2 e — x + y — z a — b + c — d —a + b + c — d a2 — b2 + c2 — d2 e — x — y + z

Let fij and p(f )j denote the entries of f and p(f) located at position (i,j). Then f and p(f) are related by:

* Hf)11 = +fll + f21 — f42 — f32, 0(f) 12 = —f 14 — f24 + f42 + f32, 0(f) 13 = —f13 — f23 + f32 + f42;

* Hf)21 = —f13 + f24 — f33 + f44, 0(f )22 = +f11 — f43 + f31 — f23, 0(f )23 = +f 14 — f41 — f21 + f34;

* Hf)31 = —f 14 + f23 + f33 — f44, 0(f )32 = +f13 — f21 + f43 — f31, 0(f )33 = +f11 — f43 — f42 — f 14.

Thus, ^ is a well-defined function. To see that it is injective, suppose that p(f) = ^(g) where f is written as a convex sum using probabilities e, x, y, z, etc, 3 is written using probabilities e',x',y',z', etc and we set q = i — i' for each i E {e, x, y, z, a, b, c, d,a2,b2,c2,d2}. Because p(f) — ^(g) = 0, we get three sets of three different equations: the first set involves only variables e, x, y, z, the second involves only a,b,c,d and the third involves only a2,b2,c2,d2. The argument applied to Z2 in the proof of Theorem 4.3 applies to each of these sets allowing us to conclude that

* ce cx cy cz,

* ca c6 cc cd,

* ca2 c&2 cC2 cd2 .

3 This unorthodox notation is necessitated by the complexity of working with ^

However, because ci = 0, we then have ci + cj + ck = 0 for any i e {e, x, y, z}, j e {a,b,c,d} and k e {a2,b2,c2,d2}. This implies that f — g = 0 so that f = g and hence that p is injective.

Because the convex linear extension of the isomorphism p is a well-defined injective function, it is an embedding by the proof of Theorem 2 in [7]. □

We denote by S3 C (3, 3) the natural copy of the symmetric group on three letters:

( /0 1 0\ /0 0 1\ /1 0 0\ /0 0 1\ /0 10 S3 = I, a = 001 \ , a2 =100 ,b = 001 , c = 010 , d =100 [ \1 0 0/ \0 1 0/ \0 1 0J \1 0 0/ \0 0 1

The elements b, c and d are all involutions and the entire group is determined by a and and any of its nontrivial involutions since c = ab and d = a2b. Notice too that aba = a. Let us point out that this group has the following property:

I + a + a2 = I + a + a2 + b + c + d

3 = 6 '

We will now show that this equation is impossible quantum mechanically: Lemma 7.1 The convex closure of S3 C (3, 3) cannot be embedded into Q. Proof. If a copy {I, a, a2,b, c, d} C SO(3) of S3 satisfies equation (9), then

I + a + a2 = b + c + d.

So taking the trace of both sides gives

3 + tr(a) + tr(a2) = —3

where we note that each nontrivial involution is conjugate to one of the spin channels in Lemma 4.1, and that conjugate matrices always have the same trace: in this case, tr(b) = tr(c) = tr(c) = —1. However, each rotation is conjugate to some rx(d) by Lemma 3.2, so tr(a) > —1 and tr(a2) > —1, contradicting the fact that their sum is -6. □

Thus, the convex closure of the natural copy of S3 cannot be embedded into Q. Let us now look to (4,4). For the sake of simplicity, we introduce some new notation for permutations n e (4,4). Let v = (1,2, 3,4). Then nv = (p,q,r,s) arises from rearranging the elements of v. We can thus denote n by 1234 ^ pqrs. We define the following permutations of order three:

• a1 is the permutation 1234 ^ 1342

• a2 is the permutation 1234 ^ 3241

• a3 is the permutation 1234 ^ 2431

• a4 is the permutation 1234 ^ 2314.

Notice that ai keeps element i fixed. Lemma 7.2

* There are eight permutations in (4,4) of order three: a1; a2, a3, a4, a2, a2,, a3, and a4.

* For each i E {1, 2,3,4}, there is an involution x E (4,4) such that ai — xa1x.

Proof.First, we claim that a permutation of order three must keep one element fixed. To see why, let us enumerate the fixed-point free permutations of (4,4) and notice that none of them have order three:

* The fixed point free permutations that 'begin' in 2 are: 1234 ^ 2413, 1234 ^ 2341 and 1234 ^ 2143. They have orders four, four and two, respectively.

* The fixed point free permutations that 'begin' in 3 are: 1234 ^ 3412, 1234 ^ 3421 and 1234 ^ 3142. They have orders two, four and four, respectively.

* The fixed point free permutations that 'begin' in 4 are: 1234 ^ 4123, 1234 ^ 4321 and 1234 ^ 4312. They have orders four, two and four, respectively.

Thus, any x E (4,4) of order three keeps one element fixed and permutes the other three. By writing out the six permutations on a three element set, we see that there are only two that have order three: a = 123 ^ 231 and a2 = 123 ^ 312. Then for each i E {1, 2,3,4}, there are two permutations of order three that have i as a fixed point: the first acts as a on the other three elements, the second acts as a2 on the other three elements. Thus, there are a total of eight permutations in (4,4) of order three: {ai,a'2 : 1 < i < 4}.

To see that each ai is conjugate to a1 by an involution:

* a1 = xa1x where x is the involution 1234 ^ 1234

* a2 = xa1x where x is the involution 1234 ^ 2134

* a3 = xa1x where x is the involution 1234 ^ 3412

* a4 = xa1x where x is the involution 1234 ^ 4231. This finishes the proof. □

Proposition 7.3 If G C (4,4) is a copy of S3, then its convex closure cannot be embedded into Q.

Proof. Let H C (4,4) be the copy of S3 generated by the elements

a=(J & b=(0b)

where a,b E S3 C (3,3). We claim that G is conjugate to H.

Since G is isomorphic to S3, it contains an element x of order three. By Lemma 7.2, x belongs to {a1,a2,a3,a4,a^,a^,a25,a;j}. If x = a2 E G, then since

G is a group,

x2 = (a2)(a2) = aia3 = ai ■ I = ai E G. Thus, G contains one of the elements a1,a2,a3,a4.

Suppose first that a1 E G and let b1 E G be any nontrivial involution. Then f = a1b1 is an nontrivial involution since G ~ S3. Thinking of b1 as the bijection b1 : {1, 2, 3,4} ^ {1, 2,3,4} given by 1234 ^ b1 (1)b1 (2)b1 (3)b1 (4) and f as the bijection f : {1, 2, 3,4} ^ {1, 2, 3,4} given by 1234 ^ f (1)f (2)f (3)f (4), we see that b1(1) = 1:

* If b1(1) = 2, then f 2(3) = 1, contradicting the fact that f is an involution,

* If b1(1) = 3, then f2(4) = 1, contradicting the fact that f is an involution,

* If b1(1) = 4, then f2(2) = 1, contradicting the fact that f is an involution.

Thus, b1(1) = 1, so b1 is determined by an involutive permutation of the last three elements. Then b1 is one of

b=(00)EH a=(00)EH (0d)EH

Then G contains a1 = a E H and also contains one of the nontrivial involutions of H. Since S3 is determined by an element of order three and a nontrivial involution, G = H.

Now suppose that some ai E G. Then by Lemma 7.2, there is an involution x E (4,4) such that xa1x = ai. Then xGx ~ G ~ S3 is a subgroup that contains a1, so by the above argument, xGx = H. Since H = xGx, {H) = {xGx) ~ (G). But (H) satisfies equation (9), which means that (G) does as well. As seen in Lemma 7.1, no copy of S3 C SO(3) can satisfy equation (9). Thus, (G) cannot be embedded into Q. □ We now look to (5, 5).

Definition 7.4 Let U3 C (5,5) denote the copy of S3 given by

^ = {/, a = (0 0), a2 = (0 02), 5 = (0 2) ,g= (0 2) ,d-= (0d)}

where S3 = {I, a, a2,b, c, d} C (3, 3) is the natural copy of S3 and f = flip E (2,2). U3 is called the unorthodox copy of S3.

A simple proof that U3 is a copy of S3 is that the product of a nontrivial involution (b, c, d) with either a or a2 is a nontrivial involution.

In order to construct a copy of S3 C SO (3), we notice that in S3 = {I,a,a2,b,c,d} C (3, 3), each of the matrices I, a, a2 have a determinant of +1, and so are rotations, while b, c, d have a determinant of —1. Thus, a candidate for S3 C SO(3) is

G = {I, a, a2, —b, —c, —d} C SO(3). Let us prove that this is in fact a copy of S3:

Lemma 7.5 The set G C SO(3) is a group isomorphic to S3. An isomorphism is given by

• p(I) = I, p(a) = a, p(a2) = a2,

• p(b) = —b, p(c) = —c, p(d) = —d.

Proof. By Theorem 6.13 on page 50 of [6], a presentation of the symmetric group on three letters is given by generators a, (3 with relations a3 = (32 = I and a(a = (3. Within G, the elements a and —b satisfy such relations, a property they inherit from S3 C (3, 3). Thus, the group they generate is isomorphic to S3. We define p on these generators

p(a) = a & p(b) = —b

and then extend it homomorphically. Because a and b generate S3, p is an isomorphism between S3 and the group generated by a and —b. What we have to prove is that the latter group is G. We do so by explicitly calculating p:

• p(I) = p(b2) = p(b)p(b) = (—b)(—b) = b2 = I,

• p(c) = p(ab) = p(a)p(b) = a(—b) = —(ab) = c,

• p(a2) = p(a)p(a) = a2,

• p(d) = p(a2b) = p(a2)p(b) = a2(—b) = —(a2b) = d. □

Consequently, the homomorphism specified on generators by p(a) = a and p(b) = —b is an isomorphism between U3 and G. Its convex linear extension is an embedding:

Theorem 7.6 The convex closure of U3 C (5,5) can be embedded into Q.

Proof. Let p denote the natural isomorphism from U3 to G and extend it convex linearly. For f = x0I + x1a + x2a2 + x3b + x4c + x5d e {U3) we have

(x0 + x1 + x2 x3 + x4 + x5 0 0 0 \

x3 + x4 + x5 x0 + x1 + x2 0 0 0

0 0 x0 + x3 x1 + x5 x2 + x4

0 0 x2 + x5 x0 + x4 x1 + x3

\ 0 0 x1 + x4 x2 + x3 x0 + x5)

while the corresponding element p(f) e {G) is then

(x0 — x3 x1 — x5 x2 — x4^ x2 — x5 x0 — x4 x1 — x3

x1 — x4 x2 — x3 x0 — x5i/

Let us first prove that p is a well-defined function. If f = g e {U3), where f is written as a convex sum using (xi) and g is written using (yi), set ci = xi — yi. Since f — g = 0, we have the equations:

• c0 + c3 =0, c0 + c4 = 0, c0 + c5 = 0, c2 + c3 = 0, c1 + c4 = 0, and

• c3 + c4 + c5 = 0.

From these equations, it easily follows that ci = 0 for all i, which means xi = yi for all i and thus that p(f) = p(g).

To see that p is injective, if p(f) = p(g), we again set c = xi — yi and from p(f) — p(g) = 0 derive equations that are even easier to solve:

co — c3 = 0, co — c4 = 0, co — c5 = 0, c2 — c4 = 0, c1 — c4 = 0.

So ci = c0 for all i. But ci = 0 since both (xi) and (yi) sum to one. Thus, c = 0 for all i, which means f = g, and so p is injective. By the proof of Theorem 2 in [7], p is thus an embedding. □

8 Classification

We now state in final form the collections of qubit channels that have a classical representation: there are five of them, each one being either a subgroup of A4 or a subgroup of S3:

Theorem 8.1 Let A4 C (4,4) denote the alternating group on four letters and S3 C (5, 5) denote the unorthodox copy of the symmetric group on three letters. Then

(i) For each subgroup G of A4, there is an embedding p : (G) ^ Q.

(ii) For each subgroup G of S3, there is an embedding p : (G) ^ Q.

(iii) If G C (m, n) is a group for which such an embedding exists, then G is either a subgroup of A4 or a subgroup of S3. That is, G must be isomorphic to either Z2, Z3, Z2, S3 or A4.

Proof. (i) and (ii) follow from the fact that an embedding restricts to the convex closure of a subgroup. (iii) If G is nonabelian, then it is either A4 or S3. If G is abelian, then it is either Z2, Z3 or Z2, each of which is a subgroup of A4. Thus, G is either a subgroup of A4 or a subgroup of S3. □

As S3 shows, it is not true that the convex closure of any copy of the five groups above can be embedded into Q. However, the converse does hold: if G C SO(3) is one of the five groups above, then its convex closure is isomorphic to a subset of either (A4) C (4,4) or (S3) C (5,5). The reason is that all finite isomorphic subgroups of SO(3) are conjugate [1]. Thus, the particular representations of A4 and S3 considered in this paper capture a certain type of quantum structure.

Let us consider an example that will help to clarify the precise nature of the theorem above:

Example 8.2 A convex linear injection that is not an embedding. Define p :

This is a convex linear injection that does not preserve products. Notice that there is no convex linear injection of (3, 3) that preserves products and the identity: if there were, it would restrict to an embedding of the convex closure of S3 in (3,3),

(3, 3) ^Q by

which is impossible by Lemma 7.1. Similarly, there is no embedding of (n,n) into Q when n > 3.

Thus, Theorem 8.1 is about the interaction between convexity and group theoretic structure.

9 Interpretation

Each time we rule out a group in the classification, we are identifying a difference between classical and quantum, by writing down an equation that holds quantum mechanically but not classically or vice-versa. The physical significance of the former is that they express multiple experiments for achieving the same task. For instance, if f e SO(3) is a generator for Z4, we can have an equation like

which is never true classically. This tells us that generators of Z4 and coin flips can be used to perform projective measurements in at least two different ways -depending on the context, one of these ways may be better than the other. This equation also makes the randomness in a projective measurement explicit.

Another interesting equation arises with the Klein four group V = {e, x, y, z} C SO(3), where we can write

This gives us multiple methods for completely disrupting the correlation between sender and receiver, with the second expression leaving the problem of randomness to nature. This is the quantum analogue of randomly flipping a classical bit [7].

It's natural now to wonder about the classical representation of quantum channels in higher dimensions. Other than the fact that involution groups of all orders can now be used and the fact that all such classes of channels arise as a convex subset of some {SO(n)), we do not know much else about the question.

However, even in the case of qubit channels, there is still much to be understood. For instance, while we have classified the groups that can be used to classically represent qubit channels, what are the classes of qubit channels which result? In the case of the Klein four group, the class represented is any maximal commutative collection of symmetric channels (for instance, the diagonal matrices). In the case of Z2, the channels described are bit flipping channels with respect to a particular spin matrix. For Z3, we know the channels described are those that are both stochastic and circulant, though we are not sure of what this means physically yet. Finally,

10 Closing

and most importantly, we have no idea which classes of channels are represented by A4 and S3, be we suspect the answer is interesting.

Acknowledgement

We thank Alan Aspuru-Guzik for a helpful discussion about this work when it was still ongoing, Chris Fuchs for helping us navigate the literature and Sanjeevi Krishnan for help with algebraic topology (which is needed in the proof of (ii)(a)).

References

[1] M. Artin. Algebra. Prentice Hall, New Jersey, 1991.

[2] T.M. Cover and J.A. Thomas. Elements of information theory. Wiley, 1991.

[3] H. S. M. Coxeter. Introduction to Geometry. John Wiley and Sons, Inc. (1961), New York.

[4] H. S. M. Coxeter and W. O. J. Moser. Generators and Relations for Discrete Groups. Springer-Verlag (1965), Germany.

[5] R.A. Horn and C.R. Johnson. Matrix Analysis. Cambridge University Press, New York, 1985.

[6] T. Hungerford. Algebra. Springer-Verlag, New York, 1974.

[7] K. Martin. How to randomly flip a quantum bit. Proceedings of Quantum Physics and Logic 2008, Electronic Notes in Theoretical Computer Science, to appear.

[8] K. Martin. The scope of a quantum channel. Mathematical Structures in Computer Science, Cambridge University Press, to appear.

[9] A. Nayak and P. Sen. Invertible quantum operations and perfect encryption of quantum states. http://arxiv.org/abs/quant-ph/0605041v4

[10] M. Nielsen and I. Chuang, Quantum computation and quantum information. Cambridge University Press, 2000.

[11] D. Saracino. Abstract algebra: a first course. Addison-Wesley, 1980.

[12] D. Serre. Matrices: Theory and applications. Springer-Verlag, Graduate Texts in Mathematics, 2000.