Qubit channels that achieve capacity with two states

Transcription

1 Qubit channels that achieve capacity with two states Dominic W. Berry Department of Physics, The University of Queenslan, Brisbane, Queenslan 4072, Australia Receive 22 December 2004; publishe 22 March 2005 This paper consiers a class of qubit channels for which three states are always sufficient to achieve the Holevo capacity. For these channels, it is known that there are cases where two orthogonal states are sufficient, two nonorthogonal states are require, or three states are necessary. Here a systematic theory is given which provies criteria to istinguish cases where two states are sufficient, an etermine whether these two states shoul be orthogonal or nonorthogonal. In aition, we prove a theorem on the form of the optimal ensemble when three states are require, an present efficient methos of calculating the Holevo capacity. DOI: /PhysRevA PACS number s : Hk I. INTRODUCTION A quantum channel is a completely positive an tracepreserving CPTP map on quantum states. The conition that it is completely positive means that the result of the map is a positive operator, an therefore may represent the state of a system, even if the map acts on one part of an entangle system. The conition that it is trace-preserving ensures that the final state is normalize. In contrast to unitary operations, quantum channels can increase the entropy of a state. A quantum channel arises if an ancilla space is ae, a unitary operation is performe between the system an the ancilla, an then the ancilla is trace over to obtain the reuce ensity operator for the system. Quantum channels are use to moel communication channels, an therefore an important quantity to consier for these channels is the amount of classical communication that may be performe. This is often quantifie by the Holevo capacity. The Holevo capacity of a quantum channel is given by C = sup S p i S i, p i, i i where = i p i i, an S = Tr log 2 is the von Neumann entropy. The p i are probabilities, an therefore must be nonnegative an sum to 1. The Holevo capacity is the asymptotic classical communication that may be achieve using joint measurements on output states, but unentangle inputs 1,2. In general, etermining the Holevo capacity of a channel is a nontrivial task. For the class of channels consiere here, it will be shown that the capacity may be etermine in a straightforwar way. An important issue is the number of states i that must be consiere in the maximization. It is well known that, for quantum channels that act upon a Hilbert space of imension, the number of states in the ensemble nee not excee 2 3. In particular, for a qubit channel no more than four states are require. For the very simple case of unital qubit channels, where 1 =1, the capacity is achieve for two orthogonal input states 4. For more general qubit channels, the capacity may be achieve for two nonorthogonal inputs 5 ; three states 6 or four states may be require 7. 1 With the exception of the channels consiere in Ref. 7, these results are all for a class of channels that can require at most three states. Here we give simple criteria for these channels that, when satisfie, mean that two states are sufficient. These criteria are not satisfie by the channels that require three states given in 6, but are satisfie by examples given in Refs. 4 6,8 where two states are sufficient. In aition, we give criteria to etermine when the input states shoul be orthogonal or nonorthogonal. This paper is organize as follows. We present the proof of the criteria in Sec. II. Then, in Sec. III we give applications of the result to results presente in previous work. We consier the form of the optimal ensembles for those cases where three states are require in Sec. IV. In Sec. V, we show how our results may be applie to the calculation of the Holevo capacity. Conclusions are given in Sec. VI. II. TWO-STATE ENSEMBLES To obtain the results, we use the representation of the qubit channel on the Bloch sphere. A general qubit ensity operator may be expresse as = r, 2 2 where is the vector of Pauli operators x, y, z T. The length of the vector r oes not excee 1, an its components give the position of the state in the Bloch sphere. A qubit channel maps the sphere of possible input states to an ellipsoi, an may be expresse as = r + t. That is, the channel prouces the mapping r r +t.via local unitary operations before an after the map, the transformation matrices an t may be brought to the form 4 = = t1, t 0 0 t 2 t 3. That is, an arbitrary qubit channel may be expresse as = U t, V, where U an V are unitary channels, an /2005/71 3 / /$ The American Physical Society

2 DOMINIC W. BERRY t, is the channel with an t given by Eqs. 4. For this stuy, we consier the restricte case of channels such that the x an y components of t are zero, an use the notation t=t 3. Hence t is given by = 0 t t 0. In orer to evaluate the Holevo capacity, we use an approach similar to that of Ref. 8. The Holevo capacity may be given by the following expression 9,10 : C = min max D 0 0, where D is the relative entropy D =Tr log log. 7 Throughout this paper, we use the convention that log an exp are base 2, an logarithms base e are given as ln. The relative entropy can be evaluate using the following useful result from 8 : where D = 1 2 f r log 1 q2 r cos f q, f x = 1+x log 1+x + 1 x log 1 x, f x = log 1 x. 1+x 10 The Bloch vectors for an are r an q, respectively, an we also efine r= r, q= q, cos =r q /rq. To evaluate the Holevo capacity, we consier the action of the simplifie channel t,. This channel has the same capacity as, because unitary operations o not affect the capacity. The set of possible output states from the channel t, forms an ellipsoi centere on the z axis. The ellipsoi has a raius of 1 in the x irection, an a raius of 2 in the y irection. The nature of the optimal ensemble may be etermine by consiering the states in the minmax formula 6. In the following, we take the states = t, 0 an = t, 0 to be output states from the simplifie channel. If is the average output ensity operator for an optimal ensemble, the operators k that maximize D k are possible output states for this ensemble. It is necessary that there is some set of p k such that k p k k =. The optimal ensemble is not necessarily unique, because there may be ifferent ways of choosing the probabilities such that k p k k =. However, from Ref. 8, the optimal average output state is unique. As we are restricting to operations such that t lies on the z axis, there are many simplifications ue to the symmetry of the system. Many of these simplifications were use in Ref. 8 in the analysis of the amplitue amping channel. We give a general explanation here. First, the optimal state must lie on the z axis. To show this result, for any pair of states an, consier the secon pair an, where r = r x, r y,r z T an q = q x, q y,q z T. Due to symmetry, if an are possible output states, then so are an. From the symmetry of the relative entropy, it is evient that D =D. This immeiately implies that max D =max D. Therefore, if minimizes this quantity, then so oes. However, as the optimal average output state is unique, an must coincie, which implies that lies on the z axis. In the case that 1 2, the k that maximize the relative entropy will lie in the x-z plane if 1 2, an the y -z plane if 1 2. That is because lies on the z axis, so the relative entropy is symmetric uner rotation about the z axis. If 1 2, then the ellipsoi has a raius in the x irection larger than the raius in the y irection. Consier any state that is not in the x-z plane. We can etermine a secon state in the x-z plane with Bloch vector r = 2 rx +r 2 y,0,r z T. This state is in the interior of the ellipsoi, an we may obtain a thir state on the surface of the ellipsoi,, by extening outwars in a straight line from. From Ref. 8 the first lemma in Sec. 5.3, D D = D. 11 This implies that oes not maximize the relative entropy. Hence, all k that maximize the relative entropy must be in the x-z plane. Similarly, if 1 2, the ellipsoi has a raius in the y irection larger than the raius in the x irection, an the optimal k must be in the y-z plane. In the case that 1 = 2, the situation is a little more complicate. For each optimal k, there is a circle of optimal ensity operators aroun the z axis. However, in orer to obtain an optimal ensemble, it is only necessary to use nonzero probabilities such that k p k k =. As lies on the z axis, it is sufficient to take k from a single plane in the Bloch sphere that contains the z axis. This reasoning means that, regarless of the relative values of 1 an 2, we may restrict to consiering k that maximize D k in a single plane in the Bloch sphere. Caratheoory s theorem implies that there nee be no more than three states in the ensemble. This fact was also note in Ref. 6. The examples given by Ref. 7 which neee four states use t that were not on the z axis. In fact, in some cases the number of states require is only two 5, though in some cases three are require. Here we give criteria that can show when only two states are require via the following theorem. Theorem 1. For a CPTP map = U t, V with given by Eq. 4 an t given by Eq. 5, if m = 3 or A 0,1/2, where A = t m 1+ 2 m + t an m =max 1, 2, then there is an ensemble that gives the maximum output Holevo information an has two states. Before we procee to the proof, we give some explanation of the quantity A. Let us consier the output ellipse in the x-z plane if 1 2,orthey-z plane if 1 2. A point on

3 QUBIT CHANNELS THAT ACHIEVE CAPACITY WITH the surface of this ellipse has a istance from the origin r, which is given by Eq. 16 in the proof below. Taking the erivative of r 2 with respect to gives 2 2 r2 = 2 sin m cos 3 t. This expression is zero if sin =0, m = 3 t=0, or cos = 3 t m The thir case is only possible if the absolute value of the right-han sie RHS oes not excee 1. If it oes not, then substituting this expression for cos into the expression for r gives the extremum r 2 ex = t m + 2 m + t 2 = A Therefore, in this case, A is the ifference between the square of an extremum of r an 1. In the case 2 m 2 3 = 3 t=0, the raius is inepenent of. This possibility will be exclue in the iscussion of A, because m = 3 is an alternative criterion to A 0,1/2, an leas to infinite A. If A were positive, then r 2 ex woul be larger than 1, which is not possible for CPTP maps. Therefore, for any map such that an extremum of r is obtaine for sin 0 an m 3, the conition A 0,1/2 is automatically satisfie ue to the fact that states cannot be mappe outsie the Bloch sphere. However, A 0,1/2 is not satisfie for every possible CPTP map, because for some 3 t/ 2 m Another case where A 0,1/2 is automatically satisfie is when m 3. That is because the conition that the map is CPTP implies that 2 m +t 2 1, an if m 3, then t / 2 m 2 3 is negative. Therefore, from the efinition of A, it is clear that A 0. We now procee to the proof of the theorem. Proof. We begin the analysis by mentioning some trivial cases that woul otherwise complicate the analysis. If t =0, then the channel is unital, an the result in this case was proven in Ref. 4. If all three of the k are zero, then the channel capacity is zero, an the result is trivial. If two of the k are zero, then the possible output states form a line in the Bloch sphere, an the result follows from the fact that there are only two extremal output states. The result is also trivial if 3 =0. In that case, since we may restrict ourselves to consiering states in the x-z or y-z plane, the set of output states that it is sufficient to consier forms a line. The result again follows from the fact that there are only two extremal states. For the remainer of the analysis, we take t 0, 3 0, an assume that no more than one of the k is zero. This thir assumption means that m 0. For the remainer of this proof, we consier the input an output states for the simplifie channel t,. The input an output states for the total channel will simply be rotate from these states. We take the input state to have r = sin,0,cos T for 1 2,orr = 0,sin,cos T for 1 2. The output state will then have r = 1 sin,0, t + 3 cos T or r = 0, 2 sin,t+ 3 cos T. The state has q = 0,0,q z T. In either case, we have for the output r = m 2 sin 2 + t + 3 cos 2, r cos = t + 3 cos sgn q z. 16 To search for the optimal, it is merely necessary to search for the optimal. Because sgn q z f q = f q z, we may write the relative entropy as D = 1 2 f r log 1 q z t + 3 cos f q z. 17 The erivative of D with respect to is D = 2 1 r f r f q z 3cos t + = m 2 3 cos t 3 f r /r + f q z 3 sin. 18 There will be extrema of D for =0 an =, as well as when 2 m 2 3 cos t 3 f r /r = f q z We will consier the solutions of this equation for in the interval 0,. Any solution in 0, will yiel a corresponing solution in,0 ue to symmetry. Taking the erivative of the left-han sie LHS gives 2 m 2 3 cos t 3 f r /r = 2 m 2 3 f r + 2 r m 2 3 cos t r r f r sin. 20 r In the case that m 3, 2 m 2 3 cos t 3 2 = 2 m r 2 + A. We then obtain 2 m 2 3 cos t 3 f r /r where g r = r f r r = = 2 m 2 3 sin h r + Ag r, r r 2 r ln r r2log 23 1 r, h r = 2 r f r 2 r 2 = r ln r r2log 1 r. The functions g r an h r satisfy the inequalities

4 DOMINIC W. BERRY g r 0, h r 0, 2h r + g r 0, 25 for r 0,1. IfA 0, then h r +Ag r is negative for r 0,1. Similarly, if A 1/2, then h r +Ag r is positive for r 0,1. In either case, h r +Ag r has constant sign. We o not nee to consier the possibility that r=0, because this value is only possible when sin =0 for m 0. The case where r=1 is more complicate. It is possible for r to be equal to 1 for 0,. In the case where r has a maximum for 0,, the maximum value of r is A+1. If r is equal to 1 for 0,, this must be a maximum, an therefore A=0 as we are taking m 3. That implies that the expression in square brackets on the LHS of Eq. 19 is proportional to 1 r 2. Hence the LHS of Eq. 19 approaches zero as r approaches 1, an is continuous as a function of for 0,.Ash r +Ag r has constant sign for all values of 0, except where r=1, an the LHS of Eq. 19 is continuous where r=1, the LHS of Eq. 19 is one-to-one in this interval. For the case m = 3, 2 m 2 3 cos t 3 f r /r = t sin g r /r. 26 Therefore, the erivative of the LHS of Eq. 19 is nonzero for 0,. Note that we are assuming that t 0 an 3 0, so the RHS of Eq. 26 is nonzero. Thus we have shown that, regarless of the relative values of m an 3, the LHS of Eq. 19 is a one-to-one function of, an there can be at most one solution of Eq. 19 in 0,. If there is a solution, it must correspon to an extremum, because a point of inflection woul conflict with the fact that the LHS of Eq. 19 is one-to-one. As D is symmetric about =0, there must be two solutions of Eq. 19 with sin 0 or none. In the case where there are no solutions, there are only two extrema for =0 an, an only one of these can be a maximum. This is not consistent with being optimal, because the optimal ensemble cannot have only one state. Therefore, if is optimal, then there must be two solutions of Eq. 19. As the maxima an minima alternate, the maxima are either at =0 an, or the solutions of Eq. 19. In the case that 1 2, this result immeiately implies that there are only two states in the optimal ensemble. In the case 1 = 2, if the maxima correspon to the solutions of Eq. 19, optimal ensembles may contain any states in a ring about the z axis. However, as iscusse above, it is only necessary to consier k in one plane in the Bloch sphere in this case, so there is again an optimal ensemble with two members. It is also possible to etermine simple criteria for when the optimal states in the ensemble are on the z axis, an when the optimal states in the ensemble correspon to the maxima for sin 0. The result is as follows. Theorem 2. Let t, be a CPTP map with 0 given by Eq. 4 an t given by Eq. 5. The conition that m = 3 or A 0,1/2 may be expresse as two alternative mutually exclusive conitions: Conition 1. m 3 or A 1/2. Conition 2. m 3 an A 0. If Conition 1 is satisfie, the optimal ensemble consists of two states on the z axis. If Conition 2 is satisfie, there is an optimal ensemble consisting of two states equiistant from the z axis an lying on a line perpenicular to an intersecting the z axis. Here we have given the result in terms of the simplifie map t,, rather than expressing it in terms of the arbitrary map. That is because the ellipse of output states will be rotate for the arbitrary map, so it is not possible to express the result in this way. The statement of this theorem also iffers in that is taken to be nonzero. This is to exclue the trivial case where all ensembles give zero Holevo information. Proof. As was shown above, m 3 also implies that A 0. Another consequence of this is that, if A 0, then m 3. Therefore Conition 1 contains three alternatives: i m = 3. ii m 3 an A 0. iii A 1/2 an m 3. It is clear that, for each of these three alternatives, the conitions of Theorem 1 must hol. If none of these alternatives apply, but A 0,1/2, then m 3 an A 0, which is Conition 2 given in the theorem. To etermine which extrema of D are maxima an which are minima, it is sufficient to consier the point =0. At this point, the secon erivative of D is given by 2 2D = m 2 3 t 3 f r /r + f q z We know that the LHS of Eq. 19 is one-to-one, an there must be at least one solution of Eq. 19 if is optimal otherwise there woul be only one possible state for the ensemble. If m = 3, then from Eq. 26, the LHS of Eq. 19 is monotonically increasing for 0,. IfA 1/2 an m 3, then h r +Ag r 0, an from Eq. 22 the LHS of Eq. 19 is monotonically increasing. Similarly, if m 3 an A 0, then h r +Ag r 0, an the LHS of Eq. 19 is again monotonically increasing. Therefore, for all three alternatives for Conition 1, the LHS of Eq. 19 is monotonically increasing for 0,. For Conition 2, m 3 an A 0, so h r +Ag r 0, an the LHS of Eq. 19 is monotonically ecreasing for 0,. If the LHS of Eq. 19 is monotonically increasing for 0,, the LHS of Eq. 19 must be less than the RHS for =0, so m t 3 f r /r + f q z This means that the secon erivative of D is negative for =0, an D is a maximum at this point. Hence, the two maxima are obtaine for =0 an, an these values correspon to the states in the optimal ensemble. Thus we see that, for Conition 1, the LHS of Eq. 19 is monotoni

5 QUBIT CHANNELS THAT ACHIEVE CAPACITY WITH cally increasing an the optimal ensemble consists of two states on the z axis. Alternatively, for Conition 2, the LHS of Eq. 19 is monotonically ecreasing, so the LHS of Eq. 19 is greater than the RHS for =0, an less for =. This implies that the secon erivative of D is positive for =0 an =, an these points are minima. Hence, in this case the states in the optimal ensemble correspon to the extrema of D for sin 0. In the case that 1 2 or 1 2, the optimal ensemble must be in the x-z plane or y-z plane, respectively. In either case, two maxima are obtaine in the appropriate plane for =± 0, where 0 maximizes D. These two solutions are equiistant from the z axis, an on a line perpenicular to an intersecting the z axis. If 1 = 2, then there will be a circle of states about the z axis that maximize the relative entropy. Optimal ensembles may contain any number of these states. However, as iscusse above, we may restrict ourselves to states in one plane. This yiels an ensemble with two members that again lie on a line perpenicular to an intersecting the z axis. Another issue is the position of the optimal average output state. It is possible to use similar techniques as above to show that this state shoul be further from the center of the Bloch sphere than the output for the maximally mixe state. Specifically, q z for the optimal average output state shoul satisfy q z /t 1 for t an 3 both nonzero. The case t=0 means that the map is unital, an it is known in that case that q z =0 is optimal. If 3 =0, then clearly q z =t. To show this result, let us assume some value for q z the other components of q are zero, an take a value of such that t+ 3 cos t 3 cos. We enote the states with r z =t± 3 cos by ±. Determining the ifference in relative entropies gives D + D = f r + f r 2 3 f q z cos f r r + r 2 3 f q z cos, 29 where r ± is the magnitue of the Bloch vector for ±, an r = r + +r /2. In the secon line, we have use the strict convexity of f r an the Hermite-Haamar inequality 11. Now using the fact that r 2 + r 2 =4t 3 cos, we have r + r = 2t 3 cos /r. Therefore, Eq. 29 simplifies to D + D 2t 3 cos f r /r f q z /t. 30 We have chosen such that t 3 is positive, an both f x an f x /x are monotonically increasing functions. Also r t, with equality only if m sin =0. Therefore, q z /t 1 implies that D + D This means that, if t is positive an q z t, then all states that have z component of their Bloch vector less than t o not maximize the relative entropy. In aition, if q z =t, the relative entropy cannot be maximize for r z =t. In the case m =0 this is trivial, because the maxima are for r z =t+ 3 an r z =t 3.If m 0, then f r /r f t /t. As we are also taking 3 0, this inequality means that Eq. 19 cannot be satisfie for = /2. Hence, for q z t 0 an 3 0, all k that maximize the relative entropy must have a z component of their Bloch vector greater than that for, an they cannot give an average equal to. This is not consistent with being the average state for the optimal ensemble, an therefore the average state for the optimal ensemble must satisfy q z t. Similarly, if t is negative an 3 0, then the average state for the optimal ensemble satisfies q z t. With the ai of this result, we can alternatively express Theorem 2 in terms of the orthogonality of the input states. The result is as follows. Corollary 1. Consier a CPTP map = U t, V with 0 given by Eq. 4 an t given by Eq. 5. The conition that m = 3 or A 0,1/2 may be expresse as two alternative mutually exclusive conitions: Conition 1. m 3 or A 1/2. Conition 2. m 3 an A 0. If t 0 an 3 0, the maximum output Holevo information is obtaine for two orthogonal input states if Conition 1 is satisfie, an two nonorthogonal input states if Conition 2 is satisfie. Proof. Note first that unitary operations o not change the orthogonality relations between the states. Therefore, it is sufficient to prove the orthogonality relations for the simplifie map t,. For Conition 1, the result follows immeiately from Theorem 2. The two input states are the extremal states on the z axis, an therefore are 0 an 1, which are orthogonal. To prove the result for Conition 2, we use the result that, for t 0 an 3 0, q z is not equal to t. If the input states for Conition 2 were orthogonal, then that woul lea to q z =t. Therefore, if t 0 an 3 0, the input states must be nonorthogonal if Conition 2 hols. III. APPLICATIONS These results allow us to make sense of the results obtaine in previous work. In particular, 8 foun that only two states in the ensemble were require for the amplitue amping channel, where 1 = 2 =, 3 =, an t=1.we fin that, in this case, A=0, so A 0,1/2 is satisfie an Theorem 1 preicts that the optimal ensemble requires two states. For this channel, m 3 an A 0, which correspons to Conition 2 in Theorem 2. Theorem 2 therefore preicts that, for this channel, the optimal ensemble consists of two states at the same istance from the x-y plane, rather than on the z axis. This is what was foun in Ref. 8. Another channel is the shifte epolarizing channel, which was consiere in Ref. 6. For this channel, k = an t=1. As m = 3, Theorem 1 applies, an the ensemble shoul require only two states. This result is what was foun in 6. Also, because m = 3, Conition 1 in Theorem 2 hols, so Theorem 2 preicts that the states in the optimal ensemble lie on the z axis. This is also consistent with the results of Ref

6 DOMINIC W. BERRY arise naturally when consiering extremal maps 13. Also, it is known that all qubit maps with two Kraus operators are of this form 13. For maps of this form, we fin that A=0, so the conitions of Theorem 1 are satisfie. Therefore, for maps that arise from a unitary interaction with an ancilla qubit, the optimal ensemble requires only two states. This result was also claime in Ref. 14, although the complete proof was not given. In aition, 3 m, so from Theorem 2 the two states for the optimal ensemble are away from the z axis. IV. THREE-STATE ENSEMBLES FIG. 1. The values of A soli line an m ashe line as a function of m for 3 =t=1/2.the shae region shows the region of values of A such that the optimal ensemble may require three states. Results for m 1/ 2 are not shown, because the maps for m 1/ 2 are not CPTP. On the other han, let us consier the examples given in 6 that require three states. For one of these examples, 1 = 2 =0.6 an 3 =t=0.5, so A This is in the interval 0,1/2, so it is not surprising that three states are require. Another example is 1 =t=0.5 an 2 = 3 =0.435; in this case A is about 0.278, which is again in the interval 0,1/2. In Ref. 6, a strategy use to fin channels that require three states was to vary the parameters from a channel such that the optimal states are on the z axis to one where the optimal states are away from the z axis. This strategy can alternatively be explaine in terms of Theorem 2. The channel parameters cannot be continuously varie from Conition 1 to Conition 2 without A passing through the interval 0,1/2. That is because it is not possible to continuously vary the channel parameters from m 3 to m 3 while maintaining the same sign for A. To take an example from 6, let 3 =t=1/2, an vary m. Then the variation of A an m are as in Fig. 1. It can be seen from this figure that as m passes through zero, A switches from negative to positive. In fact, the only point where Conition 2 is satisfie is for m =1/ 2. In passing from m =0.5, where m = 3,to m =1/ 2, the value of A passes through 0,1/2. A case of particular interest is that in which an t are given by = cos cos cos cos, t = 0 0 sin sin. 32 This type of channel arises naturally when consiering qubit interactions. If one introuces an ancilla qubit, performs a unitary operation, then traces over this ancilla qubit, the resulting operation is of this form 12. Maps of this form also In the case where three states are require for the optimal ensemble, it is possible to show that one of the states nees to be on the z axis. The result is as follows. Theorem 3. Consier a CPTP map t, with given by Eq. 4 an t given by Eq. 5. If the Holevo capacity cannot be achieve with a two-state ensemble, then any optimal ensemble with three states consists of one state on the z axis, an two states equiistant from the z axis an on a line perpenicular to an intersecting the z axis. The optimal input state on the z axis is 0 if t+ 3 t 3, an 1 if t+ 3 t 3. Proof. In orer to prove the result, we start by consiering the expression in square brackets in Eq. 22. Although h r +Ag r can change sign, it is only zero for one value of r. To show this result, we use the following facts: h r 0, g r 0, g r 0, 33 h r g r g r h r These inequalities are all for r 0,1, an are easily checke by plotting the functions. If h r +Ag r 0 for r =r 0, then A h r 0 /g r 0,soh r 0 +Ag r 0 h r 0 g r 0 g r 0 h r 0 /g r 0 0. Therefore, if h r +Ag r 0 for r =r 0, then h r +Ag r is increasing for r=r 0. This implies that, if there is a value of r for which h r +Ag r =0, then h r +Ag r 0 for all larger values of r. Hence h r +Ag r can be zero for only one value of r in 0, 1. Recall that, if there is an extremum of r for sin 0, then the conition A 0,1/2 is satisfie, an therefore the optimal ensemble requires no more than two states. In the conitions for Theorem 3, the optimal ensemble requires more than two states, so r has no extremum for sin 0. Hence r is a one-to-one function for in the interval 0,. Combining this result with the above reasoning, the RHS of Eq. 22 can be zero for only one value of in the interval 0,. These results imply that the LHS of Eq. 19 can have a turning point for only one value of in 0,, an therefore there are at most two solutions of Eq. 19 for 0,. In turn, this implies that there are no more than two extrema of D for 0,. In fact, there must be exactly two if is optimal, because if there were only one, then the optimal ensemble woul require only two states, which violates the conitions of Theorem 3. Thus there will be two extrema of D for 0,, two symmetric extrema for,0, an ex

7 QUBIT CHANNELS THAT ACHIEVE CAPACITY WITH trema at =0 an. These extrema must alternate between minima an maxima, an so one of the extrema at =0 an will be a maximum, an the other will be a minimum. To etermine which points are minima an which are maxima, consier the secon erivative of D at a solution of Eq. 19, 2 2D = 2 m 2 3 sin 2 h r + Ag r. 35 2r Recall that, if h r +Ag r is positive for r=r 0, it must also be positive for r r 0. Therefore, for the solution of Eq. 19 with smaller r, h r +Ag r is negative, an for the solution with larger r, h r +Ag r is positive. For maps that require three states to achieve the Holevo capacity, A 0. As iscusse above, this implies that m 3,so 2 m 2 3 is positive. Thus multiplication by 2 2 m 3 oes not change the sign, so the solution of Eq. 19 with smaller r is a maximum, an the solution with larger r is a minimum. As the extrema alternate between maxima an minima, the extremum on the z axis that is closer to the origin must be a minimum. Therefore, if t+ 3 is greater than t 3, then the optimal output state on the z axis will be at t+ 3. This correspons to an input state of 0. Similarly, if t 3 is greater than t+ 3, then the optimal output state on the z axis is at t 3, which correspons to the input state 1. The two remaining states in the optimal ensemble will correspon to solutions =± 0 of Eq. 19. In the case that 1 2, these states are in the x-z or y-z plane of the Bloch sphere, epening on whether 1 2 or 1 2.Ineither case, the states are equiistant from the z axis, on a line that is perpenicular to an intersecting the z axis. If 1 = 2, then optimal ensembles may contain any states from a circle about the z axis. However, for optimal ensembles with three states, the conition that the mean state is on the z axis restricts the remaining two states to be equiistant from the z axis, an on a line perpenicular to an intersecting the z axis. V. CALCULATING CAPACITIES These results enable us to etermine numerically efficient ways of calculating capacities. In the case that the channel satisfies the conitions of Theorem 1, the problem becomes particularly simple. First it is necessary to check whether it is Conition 1 or Conition 2 in Theorem 2 that is satisfie. For Conition 1, the optimal ensemble consists of the two extremal states on the z axis. The probabilities may be etermine by the fact that D 1 =D 2. The expression for the relative entropy 8 simplifies to D = 1 2 f r z log 1 q z 2 r z f q z. The conition that D 1 =D 2 then becomes 36 f t + 3 t + 3 f q z = f t 3 t 3 f q z. 37 This may be solve for q z, yieling where q z = X 1 X +1, 38 X = exp f t + 3 f t Recall that we are using notation where exp means 2 to the power of the argument. The channel capacity is obtaine by substituting Eq. 38 into Eq. 36. Thus the channel capacity may be obtaine analytically. The optimal ensemble may also be etermine analytically. The optimal states correspon to points on the z axis at t± 3, an the probabilities are given by p ± = 1 2 ± q z t For Conition 2 in Theorem 2, the optimal states are away from the z axis. Because must be the average of the two k, an the z components of the two r k are equal, the z component of q must also be equal. If is optimal, for the solution of Eq. 19 the z component of r shoul be equal to the z component of q. Therefore, the optimal ensemble may be foun by fining the solution of Eq. 19 with q z =r z. Thus fining the capacity in this case reuces to fining the zero of a function of a single real variable, which is easily performe numerically. As an alternative interpretation of this result, consier the ensemble consisting of two states corresponing to =± 0. The Holevo information of this ensemble is given by D ± = 1 2 f r f r z, 41 where is the average state. If the optimal ensemble is of this form, then the maximum of this quantity gives the Holevo capacity for the channel. Taking the erivative with respect to, we fin that the maximum will be for a solution of Eq. 19 with q z =r z. For the case where an t are as given in Eq. 32, the problem of calculating the capacity has been consiere in Ref. 15. For this case, this reference gives an analytic metho for calculating the Holevo capacity for a given mean state. Although this metho was erive in quite a ifferent way from the metho given here, it is equivalent. In those cases where A 0,1/2, it is still possible that two states may be sufficient for the optimal ensemble. In those cases, the ensemble must still consist of either two states on the z axis of the Bloch sphere, or two states corresponing to =± 0, where 0 is a root of Eq. 19. This result may be shown by consiering D as a function of. As was shown in the previous section, there can be at most three maxima of D. If there are only two, then these are at =0 an or =± 0. In either case, the form of the optimal ensemble is the same as for channels satisfying the conitions of Theorem 1. If there are three maxima, then one of these is on the z axis, an the other two are for =± 0. If two states are sufficient for the optimal ensemble, these states must corre

8 DOMINIC W. BERRY spon to =± 0, because otherwise woul not be on the z axis. Therefore, regarless of whether there are two maxima or three, if two states are sufficient for the optimal ensemble, then these consist of either two states on the z axis, or two states corresponing to =± 0. These results can be use to etermine if the optimal ensemble requires three states in cases where A 0,1/2. From the sufficiency of maximal istance property in 9, we know that the ensemble is optimal if there are no values of that give values of D greater than the k in the ensemble. Therefore, in orer to etermine if the ensemble requires more than two states, etermine via the two ifferent methos above. If, for one of them, D is maximize for the corresponing k, then the optimal ensemble requires only two states. If neither of these methos gives the optimal ensemble, then we have eliminate all possibilities for optimal two-state ensembles, an the optimal ensemble must require three states. It is also possible to efficiently etermine the Holevo capacity in those cases where the ensemble requires three states. The reason for this is that the only unknowns for the three-state ensemble are the value of 0 such that =± 0 for the two off-axis states, an the probabilities for the three states. Given the value of 0, there is an analytic metho to etermine the probabilities. Therefore, the problem reuces to a numerical maximization in a single real variable, which is easily performe. From Theorem 3, the state on the z axis will be at t+ 3 if t+ 3 t 3, an t 3 if t+ 3 t 3. Taking the other two states to correspon to =± 0, the conition that the relative entropy D k is inepenent of k becomes f t ± 3 t ± 3 f q z = f r 0 t + 3 cos 0 f q z, 42 where r 2 0 = 2 m sin t+ 3 cos 0 2. We take the plus sign if t+ 3 t 3, an the minus sign if t+ 3 t 3. Solving for q z gives q z = X 1 X +1, 43 where X = exp f t ± 3 f r ±1 cos 0. Note that this solution is reasonable only if the value of q z obtaine is between t± 3 an t+ 3 cos 0 ; otherwise, negative probabilities woul be require for the ensemble. Given this solution for q z, the common value of the relative entropy is given by D k = 1 2 f t ± 3 log 1 q z 2 t ± 3 log X. 45 By fining the maximum of this with q z between t± 3 an t+ 3 cos 0, the Holevo capacity may be etermine. This metho was use to etermine the ifference between the two-state capacity an the three-state capacity for FIG. 2. The ifference between the two-state capacity an the three-state capacity vs the value of A. Ranom samples are shown as gray points, an the numerically obtaine upper boun is shown as the soli line. The cross an plus are examples from Ref. 6. The cross is for 1 = 2 =0.6 an 3 =t=0.5, an the plus is for 1 =t =0.5 an 2 = 3 = a range of ifferent maps. This ifference is plotte as a function of A in Fig. 2. In aition, the states that maximize this ifference were searche for numerically for given values of A; these results are also shown in Fig. 2. It can be seen that the maximum ifference in the capacities is still quite small, less than Also, the ifference can be nonzero in the entire interval 0,1/2. The ifference approaches zero quite rapily as A approaches 1/2, but is still nonzero. For comparison, two of the examples from Ref. 6 are shown in Fig. 2. It was also foun that, regarless of the value of A, there were cases where two states were sufficient for the optimal ensemble. VI. CONCLUSIONS We have shown a number of results on the form of optimal ensembles for qubit channels. The class of channels consiere inclues those that can be simplifie, via unitary operations before an after the channel, to a form that is symmetric uner reflections in the x-z an y-z planes. This class inclues extremal channels, an most examples of channels consiere in previously publishe work. For these channels, we have introuce the parameter A, which can be interprete in some cases in terms of the istance between the output ellipsoi an the unit sphere. The main result is that if A is not in the interval 0,1/2, then two states are sufficient for the ensemble that maximizes the Holevo capacity. In aition, optimal two-state ensembles must consist of either two states on the z axis of the Bloch sphere, or two states on a line that is perpenicular to an intersecting the z axis. For cases where A 0,1/2, we have presente a simple metho to etermine which form the optimal ensemble takes. This result also enables us to etermine if the input states shoul be orthogonal or nonorthogonal. Even in cases where A 0,1/2, if two states are suf

9 QUBIT CHANNELS THAT ACHIEVE CAPACITY WITH ficient for the optimal ensemble, then the ensemble must take one of these two forms. For cases where three states are necessary for the optimal ensemble, our results show that the optimal three-state ensemble consists of one state on the z axis at the maximum istance from the origin, an two states on a line perpenicular to an intersecting the z axis. This emonstrates that the form of the optimal three-state ensembles foun in Ref. 6 is universal. Last, we have provie a computationally efficient metho of etermining the Holevo capacity. For cases where the optimal ensemble consists of two states on the z axis, the capacity may be etermine analytically. For other cases, the calculation is a numerical maximization of a function of a single real variable, which is easily performe. For the specific case of extremal channels, this metho is equivalent to that given in Ref. 15. ACKNOWLEDGMENTS This project has been supporte by the Australian Research Council an the University of Queenslan. The author is grateful for helpful comments from Barry Saners. 1 A. S. Holevo, IEEE Trans. Inf. Theory 44, B. Schumacher an M. D. Westmorelan, Phys. Rev. A 56, E. B. Davies, IEEE Trans. Inf. Theory 24, C. King an M. B. Ruskai, IEEE Trans. Inf. Theory 47, C. A. Fuchs, Phys. Rev. Lett. 79, C. King, M. Nathanson, an M. B. Ruskai, Phys. Rev. Lett. 88, M. Hayashi, H. Imai, K. Matsumoto, M. B. Ruskai, an T. Shimono, Quantum Inf. Comput. 5, J. Cortese, e-print quant-ph/ B. Schumacher an M. D. Westmorelan, Phys. Rev. A 63, M. Ohya, D. Petz, an N. Watanabe, Probab. Math. Stat. 17, J. Haamar, J. Math. Pures Appl. 58, ; D.S.Mitrinović an I. B. Lacković, Aequ. Math. 28, C.-S. Niu an R. B. Griffiths, Phys. Rev. A 60, M. B. Ruskai, S. Szarek, an E. Werner, Linear Algebr. Appl. 347, F. Verstraete an H. Verschele, e-print quant-ph/ A. Uhlmann, J. Phys. A 34,