Lecture notes for the course: Quantum Information

Transcription

1 Lecture notes for the course: Quantum Informaton gven by Dr. Benn Reznk Sprng 003, Tel-Avv Unversty wrtten by Amr Segner last modfed: March 3rd 005 Caveat: These lecture notes were wrtten whle I was studyng for the fnal exam n the quantum nformaton course. I have gone over some of the parts agan, but not over all. As a result, the notes are both mssng n some places and plan wrong n others (I just don t know where). If you fnd a mstake, or have any comments, please let us know. 1 Note: These notes are based mostly on notes I took durng the lectures gven, but were also supplemented, n some places, by other materal. Most of ths extra materal was taken ether from the book by A. Peres [1], or from the onlne lecture notes of J. Preskll [] (see references at the end). 1 Send emals to reznk@post.tau.ac.l and to asegner@post.tau.ac.l.

2 Contents Part 1. Introducton 3 Chapter 1. Physcs and Informaton Maxwell s demon (classcal) Quantum nformaton 8 Chapter. Bascs of quantum nformaton Bascs of quantum mechancs 17.. Spn 1 and the Paul matrces Open systems, mxtures and the densty matrx 0.4. Entanglement and the Schmdt decomposton 39 Part. Entanglement 49 Chapter 3. Creatng entanglement expermentally 51 Chapter 4. Hdden varables The EPR Paradox Bell nequaltes Contextualty 59 Chapter 5. Uses of Entanglement Encodng nformaton Data hdng Cryptography teleportaton Ramsey spectroscopy Remote operatons State-operators (stators) POVM (Postve Operator Valued Measures) Measure of entanglement (Dstllaton) 75 Chapter 6. Quantum nformaton Data compresson (classcal) Data compresson (Quantum) Schumacher s noseless encodng Communcaton wth nose (classcal) Accessble Informaton Decoherence and the measurement problem Error correcton - Shor s algorthm 89 Bblography 93 Index 95 1

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4 Part 1 Introducton

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6 CHAPTER 1 Physcs and Informaton 1.1. Maxwell s demon (classcal) Assume two adjonng rooms A and B, each flled wth a gas at equal temperature T. Now, magne also a small demon 1 who control s a shutter between the rooms. The shutter s assumed to be deal, so that no work s produced by openng and closng t. The demon opens and closes the shutter so as to allow fast partcle to pass from room B to room A, and allow slow partcles to pass n the opposte drecton, from room A to room B. As a result, at the end of the process, we have faster gas partcles n room A and slower ones n B, and therefore T A > T B wthout any work beng done. Now, from thermodynamcs we know that for quas-statc processes the entropy S obeys ds = d Q T. Snce the system of the two rooms s a closed one, then the heat d Q A gong nto room A must come from room B so that d Q A = d Q B. Thus, the change n entropy S of the whole system (both rooms) s Z ( d QA S = + d Q ) Z ( B d QA = d Q ) A < 0, T A T B T A T B whch s negatve snce d Q A > 0 (the fast energetc partcles are gong nto room A) and T A T B. We have therefore managed to reduce the entropy of a closed system wthout dong any work (the openng and closng of the shutter requres no work), thus contradctng the second law of thermodynamcs The Szlard model. One way out of the paradox s to say that n order to decde when to open and close the shutter, one must know whch partcle s approachng the shutter, and thus we have a connecton between entropy and nformaton (see also the end of ths secton). As an example of ths let us study the Szlard model (199). In the Szlard model one assumes a small cell wth a sngle partcle n t. If the partcle s n the left-hand sde of the cell, we say that the cell s n state 0, and f the partcle s n the rght-hand sde, we say the cell s n state 1. We can thus code nformaton n a sequence of such cells, each cell n one of the bnary states 0 or 1. Now, n order to keep the partcle on the left-hand sde of the cell (or on the rghthand), we must put a barrer n the mddle. However, before we put the barrer the partcle may be anywhere n the cell. Thus, n order to force t to be on one sde, we have to push a pston from one sde of the cell (untl we reach the mddle). Pushng such a pston sothermally requres work to be done, and so we connect work and nformaton. 1 Ths paradox was suggested by Maxwell and s therefore called Maxwell s demon paradox. Recall that ds s an exact dfferental, whle d Q s an nexact one. Inexact dfferental means that the change n the heat Q depends not only on the ntal and fnal physcal (macroscopcal) states of the system (defned by pressure, volume, temperature,... ), but also on the path taken from one state to the other. 5

7 6 1. PHYSICS AND INFORMATION To fnd the work done by the pston when movng (sothermally) to the mddle of the cell, we assume for the moment that the cell s flled wth an deal gas. The force actng on the pston, by the gas, equals the pston s area A tmes the gas pressure P. The change n volume of the cell, when the pston moves a dstance δx s δv = Aδx. Thus the work δw done (on the system, by the pston) when the volume of the cell s changed by δv (due to the pston s movement) s 3 δw = Fδx = PδV. We have assumed that the gas we use s an deal gas, so t obeys PV = Nk B T, where N s the number of partcles n the gas and k B s Boltzmann s constant. Thus the work done by the pston may be wrtten as Z Z V Nk B T W = PdV = V dv = Nk BT ln V 1. V V 1 Now, snce the temperature was unchanged durng the process, the velocty dstrbuton of the gas partcles has not changed ether (the nternal energy of the system has not changed). The queston s therefore where has the work-energy gone to? It has gone to heatng the heat bath surroundng our system. As we just saw, the nternal energy U of our gas has not changed. We know from thermodynamcs that we may wrte the nternal energy U as or n dfferental form but also U = F + T S, δu = δf + T δs, δu = W + δq = W + T δs, where F s the free energy of the system, W s the work done on the system, and δq s the heat whch entered the system. Snce n our case the temperature T s constant, and we have U = 0, then we must have 4 F = W = T S. Thus, from the result we had for the work done, we may wrte F = Nk B T ln V 1 V. S = Nk B ln V 1 V If we now return to the case of our cell, havng a sngle partcle (N = 1) confned to half of the cell (V = 1 V 1), we fnd that F = k B T ln, and S = k B ln. Ths last result should not be surprsng, snce entropy may also be defned as S = k B lnω, 3 Note the mnus sgn. When the pston s pushed (dong work on the system), the volume of the cell s reduced, and so δv s negatve. 4 Snce for δu = 0 0 = δf + T δs = W + T δs.

8 1.1. MAXWELL S DEMON (CLASSICAL) 7 where Ω s the total number of possble states of the system. In our case the system can be n one of two states ( 0 left-hand sde, or 1 - rght-hand sde), whch gves the above result The Landauer prncple. The Szlard model has shown us that there s connecton between nformaton and energy/work. Landauer used ths connecton to gve a lower bound on the energy expendture needed for performng a computaton. The Landauer prncple says that n order to erase nformaton we must expend energy whch then goes nto heatng the envronment. We shall show that ths leads to a lower bound on the energy expendture for performng a computaton. We shall frst examne why t requres energy to erase nformaton. For ths we start wth a Szlard cell. Assume that we are gven a cell whch has a partcle ether on the left or on the rght (we don t know where). We shall say that the cell s erased, f the partcle s (for certan) on the left-hand sde. 5 A method of achevng ths, s to take the barrer out of the cell and then push our pston half way from the rght, thus confnng our partcle to the left half. As we have already seen, the work done n pushng the pston, when done sothermally, goes to heatng the envronment by Q = k B T ln. Thus, we see that the process of erasng nformaton causes the heatng of the surroundngs. Now, f we look at logcal gates n a computer, they are schematcally descrbed as rreversble process n whch two bts of nformaton go n, whle only one comes out. Thus, n the rreversble process of a logc gate we have necessarly erased one bt, whch requres an energy of at least k B T ln. 6 We have therefore found a lower bound for the energy expendture for dong a calculaton. Note, that n today s computers the energy expendture ( 10 8 k B T per bt) s much hgher than Landauer s lower bound Bennett s reversble computer. It has been emphaszed that the lower bound gven by Landauer s only good for rreversble gates. Bennett (1973) has shown that f one uses reversble gates, one may construct a computer whch requres no energy expendture at all. In Bennett s computer a gate stll accepts two bts as nput, however (unlke before), the output s also two bts: One bt, s the logcal result we wanted (from the gate) whle the second bt (together wth the frst) allows us to fnd the ntal nput bts. 7 Although, such a gate gves us superfluous nformaton for the calculaton, t does allow us to reverse the process. Now, durng the computaton, usng reversble gates, we shall not erase any cells and therefore no energy wll be wasted. However, n order to make a dfferent calculaton (after the frst) we must reuse our cells whch means erasng them, and thus seemngly returnng to Landauer s prncple. But, as you recall we used reversble gates, therefore we can wrte down the result at the end of the frst computaton and then reverse the process of computaton. Ths reverse computaton wll brng us back to the ntal condtons wth no net energy expendture. The cells n ther ntal condton can then be used for our next calculaton, and so we have bult a computer whch requres no energy. 8 Havng found a connecton between storage of nformaton and entropy/energy, we can now return to the Maxwell s demon paradox. Bennett (198) suggested a resoluton between the demon paradox and the second law of thermodynamcs. The resoluton s that 5 We use ths defnton of erasure snce we are assumng that the computer has a lmted amount of memory. It therefore has to recycle ts bts, whch means erasng them as we defned here. To manpulate a cell we must frst know n whch state t s, and we know ths for the erased cells. 6 After passng through the gate, we no longer know the state of one of the cells. In order to reuse ths cell (our computer has a fnte amount of memory cells), we must erase t and thus waste energy. 7 Snce we have two bts of nput, and two bts of output, then we can code the nput n the output (for logc operators such as and and or ). 8 Note that the cells n the ntal condtons are all n known states, ether 0 or 1, but known to us. Wth ths nformaton we can construct any other ntal condtons by flppng the necessary cells. The process of flppng requres no work; we don t push a pston we smply flp the whole cell.

9 8 1. PHYSICS AND INFORMATION every tme the demon opens and shuts the shutter he s actually performng a computaton (he s performng an f statement whch can be broken nto logcal gates). a computaton means that he needs memory bts. Assumng that the number of bts s fnte, the demon wll have to erase them and thus the demon wll gve rse to work and entropy (although the shutter tself requres no work to operate t). Ths entropy wll ensure that the second law of thermodynamcs s upheld. 1.. Quantum nformaton In the prevous secton we studed nformaton usng classcal objects. We now wsh to ntroduce nformaton theory usng quantum objects. The followng table compares the classcal and quantum manfestatons of the man ponts of mportance n nformaton theory: Classcal Quantum basc nformaton unt bt: {0, 1} qubt: α 0 + β 1 (superposton prncple) dynamcs determnstc (causal) determnstc (untary evoluton) measurements do not nfluence system effect the system (uncertanty prncple + collapse) We shall see that the superposton prncple and the dfferent effects of measurements wll cause the quantum theory of nformaton to dsplay very dfferent trats from those of the classcal theory the qubt. In the classcal case, the basc unt of nformaton we used was the bt, whch could accept ether the value 0 or the value 1. In the quantum case, the basc unt we use s a two state system. 9 We shall generally denote the two states as 0 and 1, 10 however, due to the superposton prncple, the general state of such a system s ψ = α 0 + β 1 ( ψ ψ = 1 α + β = 1). Snce α and β are complex numbers, they are each descrbed by two parameters (real and magnary parts) whch gves us four parameters descrbng the state ψ. However, we also have the requrement α + β = 1 (due to the normalzaton ψ ψ = 1), whch reduces us to just three contnuous parameters. Of these parameters, one s the global phase of the system whch has no physcal mportance. Thus we are left wth just two (physcal) contnuous parameters for descrbng ψ. One method of wrtng ψ wth two parameters s ψ = cos θ e φ 0 + sn θ e+ φ 1. Snce we have two contnuous parameters, one mght thnk that we can use a sngle qubt to store an nfnte amount of nformaton (unlke the classcal bt whch can store only 0 or 1). Ths s ndeed true, we can store n a qubt an nfnte amount of nformaton, however Holevo (1961) has shown that we can extract from a qubt (wth 100% certanty) a maxmum of only one bt of nformaton. Thus, for all practcal reasons we can store n a qubt only a sngle bt of nformaton. 9 The smplest non-trval Hlbert space s a two dmensonal one. 10 The two state system can be any knd of system wth two orthonormal states. For example, t can be a spn 1 system wth the two states and, or a system wth two energy states E 0 and E 1.

10 1.. QUANTUM INFORMATION no-clonng theorem. As we noted above, one cannot extract more then one bt of nformaton from a qubt. In spte of ths let us now try. Assume two qubts: the frst we shall denote as θ θ = cos θ e φ 0 + sn θ e+ θ 1 and the second wll smply be the spn up qubt These two states together gve = 0. θ = e φ cos θ, so that the probablty of measurng spn-up for a state θ s cos θ. If we could now make many such measurements, then accordng to the statstcs of our measurement we could deduce θ up to any accuracy. Thus, apparently we can encode n a qubt a contnuous parameter and then extract t (to any desred precson). The problem wth the prevous scheme, s that n order to perform a multple number of measurements, we must frst replcate, or clone, our ntal state θ whle we do not know what t s. Only then (after clonng) can we do the measurements and determne θ. The problem s that n quantum mechancs we cannot clone (unknown states). Ths s called the no-clonng theorem. PROOF. The proof of the no-clonng theorem rests on the fact that the evoluton of a quantum state must be descrbed by a untary operator. 11 In order to clone our partcle N tmes we must start wth N partcles n a known state, whch we shall denote as 0. Thus, our ntal state before clonng starts, s Ψ = ψ. At the end of the process we want to have a state Ψ f = U ψ = ψ ψ ψ. Now, assume that we have found such an operator U, whch we use on two states ψ (1) and ψ () : Ψ (1) f = U Ψ (1) = U ψ (1) = ψ (1) ψ (1) ψ (1), Ψ () f = U Ψ () = U ψ () = ψ () ψ () ψ (). Snce the operator U t s untary (U = U 1 ) then necessarly However, by defnton Ψ (1) Ψ () = whle Ψ (1) f Ψ () f = Ψ (1) U U Ψ () = Ψ (1) Ψ (). ( ψ (1) 0 0 )( ) 0 0 ψ () = ( 0 0 ) N ψ (1) ψ () = ψ (1) ψ (), Ψ (1) f Ψ () f = ( ψ (1) ψ (1) ψ (1) )( ψ () ψ () ψ () ) = ( ψ (1) ψ () ) N Recall that the Hamltonan n quantum mechancs must be Hermtan (H = H). The (tme) evoluton operator s then U(t) = e Ht : ψ(t) = e Ht ψ(t = 0) = U(t) ψ(t = 0), whch s necessarly untaran (U = U 1 ). Note, that ths s all true, assumng that the Hamltonan s tme ndependent. If the Hamltonan s tme dependent, then the evoluton operator s e R Hdt, whch s also Untary.

11 and 13 σ z x = x PHYSICS AND INFORMATION Thus, f there exsts a untary clonng operator U, then we must have (snce we need Ψ (1) f Ψ () f = Ψ (1) Ψ () ) that for any two states ( ψ (1) ψ () ) N+1 = ψ (1) ψ (). Ths s certanly not true for any two states, and therefore there cannot exst a clonng operator. Please note, however, that f we choose an orthonormal bass, we can create a untary operator whch clones the elements of the bass, but not ther lnear combnatons Bt vs. qubt. Although we can extract from a qubt only one bt of nformaton, the qubt s not equvalent to a classcal bt. For example, assume that we are gven the ntegral Z 1 f (t)dt = nα, 0 where we know f (t) and α, and we know that n (an nteger) s ether even or odd. Now, n order to fnd whether n s even or odd, classcally we requre an nfnte number of bts, snce t s contnuous, and we need an nfnte number of bts to descrbe a contnuum (to calculate the ntegral numercally). However, f we use qubts, t suffces to use just a sngle qubt to fnd whether n s even or not. To solve the problem quantum mechancally we take a spn up n the x drecton x, and construct a Hamltonan where σ z s one of the Paul matrces H(t) = λ f (t)s z = λ f (t) 1 σ z, σ z = ( ), 1 Such an operator for two partcles could be U = 0 +, j j,0 j j + 0, whch gves and 13 Recall that n the z bass U 0 = UU = j j = 1. x = 1 ( z + z ), x = 1 ( z z ).

12 The evoluton of the spn x s then gven by QUANTUM INFORMATION 11 U(t) x = e λ R t t =0 Hdtσ z x = e λ nασ z x = (cos λnα σ z sn λnα ) x = cos λnα x sn λnα x. Now f we choose λ so that λα = π, then we have U(t) x = cos πn x sn πn x, and thus, f n s odd, we get x (up to a multplcatve factor), and f n s even, we get x (agan, up to a multplcatve factor). Therefore, by measurng the spn n the x drecton at the end, we can determne whether n s even or odd. We have thus been able, wth just one qubt, to fnd somethng that we couldn t do classcally at all. Note however, that the nformaton we got was just a sngle bt ( even or odd ) smulatng a quantum computer wth a classcal one. As we saw above, we can use qubts to get results whch are much harder, or even mpossble to reach usng just smple classcal bts. However, when we consder a computer, t s smply some black box whch accepts some vectors as nput, operates on them, and returns a new vector as an output. All the operatons whch we do quantum mechancally we can also smulate classcally (manpulate vectors, take ther projectons,... ). The queston that should be asked s how much resources does ths requre? Assume N qubts. The state descrbng them s N ψ = (α 0 + β 1 ) = c j ϕ j, j=1 where ϕ j are N-partcle states, whch gve all N possble combnatons of N partcles beng n ether state 0 or state 1. For example for the case of N = 3 we have ψ = 3 =1 (α 0 + β 1 ) = c c c c c c c c The number of parameters descrbng such a state s N : We have N coeffcents c, each one of those s actually two number snce these are complex numbers, however f we requre that ψ be normalzed (one constrant) and don t mnd f t s multpled by a global phase e θ, then two parameters may be dropped gvng us N. If we assume that we need at least one bt for every such parameter, 15 ths means that for an N qubt system we need at least N bts for the classcal smulaton. Such a fast ncrease makes smulatons mpossble very quckly examples. 14 snce σ z = 1, then σ m z = 1 and σ m+1 z = σ z. Therefore the Taylor seres for e θσz can be wrtten as 1 e θσz = n n! (θσ z) n 1 1 = σ z n odd n! (θ)n + n even n! (θ)n = σ z snθ + cosθ. 15 Snce the c are contnuous parameters we need an nfnte number of bts to descrbe each parameter. However, f we settle for a fnte precson for the c s, then a fnte number of bts wll suffce to descrbe each one of them.

13 1 1. PHYSICS AND INFORMATION Deutsch s problem. Assume a black box whch accepts a sngle bt as nput and gves a sngle bt as output. We shall denote the effect of the box as f (x) [f the nput bt s x then we get f (x) as output]. There are of course 4 dfferent possble functons f (x) whch may descrbe the black box (each of the two possble nputs has two possble outcomes). We would lke to know whether f (x) s a constant functon,.e. f (0) = f (1), or whether t s a balanced functon,.e. f (0) f (1). 16 Classcally, to determne the type of functon, we must make two runs of the system. Frst we enter a 0 nput and see the result, and then we enter 1 as nput and see what the outcome s. Such a test would gve f (x) exactly and wll therefore also tell us f f (x) s constant or balanced. However, as we shall see, usng quantum mechancs and the superposton prncple we can fnd the type of functon (constant or balanced) wth just a sngle run. Now, n order to use quantum mechancs, the effect of our black box must be descrbable by a untary operator. If f (x) s balanced there s no problem, however f f (x) s constant, then we do have a problem: A untary operator cannot transform two orthogonal states nto the same state (a untary transformaton, sends a bass to a new bass, and a constant f (x) lowers the dmenson of the bass). We therefore need a slghtly dfferent box. Instead of f (x) we shall use a untary operator U D. Ths operator wll both accept and gve as output two qubts of nformaton accordng to the rule U x 1 y D x 1 y f (x), where means addng and then takng the modulo of the result: 1 0 = 1, 1 1 = 0 0 = 0. Before usng ths new operator let us frst check that t s ndeed untary. Clearly by the defnton of U D we have and U D x 1 0 U D x 1 1 U D 0 1 y U D 1 1 y (any y,y ), where n the second relaton y and y may be the same or dfferent. Therefore (f x = 0 or 1, and y = 0 or 1) we must have (snce 0 1 = 0) 17 and or smply U D x 1 0 U D x 1 1, U D 0 1 y U D 1 1 y, U D x 1 y U D x 1 y ( x x and/or y y x,x,y,y = 0,1 By ths last result we see that applyng U D to the orthogonal bass { 0 1 0, 0 1 1, 1 1 0, } ). 16 Note that we don t care what f (x) s exactly. If f (0) = f (1) = 0 or f (0) = f (1) = 1 doesn t matter to us. In both cases the functon s constant. 17 If x = 0 or 1 and y = 0 or 1, then applyng U D on x 1 y wll gve x 1 y wth x = 0 or 1 and y = 0 or 1. Thus f we know that two states ( ψ 1 = U D x 1 1 y 1 and ψ = U D x 1 y ) are dfferent, t necessarly means that ther nner product ( ψ 1 ψ ) must nclude 0 1 (or 1 0 ), and snce 0 1 = 0, then they must be orthogonal.

14 1.. QUANTUM INFORMATION 13 gves us a new set of four mutually orthogonal states. 18 Snce the four new states are mutually orthogonal, they must consttute a bass. Thus, the transformaton U D took us from one orthonormal bass to another, whch means that U D must be untary, as clamed. 19 To fnd, usng our new quantum black box, whether f (x) s a constant functon or a balanced one, we can of course run t twce (once puttng x = 0 and once x = 1) and see. However, we can also use the superposton prncple to determne ths wth just a sngle run. To see ths let us frst try as nput x = 0 1 and 1 ( y = 0 y = 1 ). By applyng U D we have [ ] 1 U D 0 1 ( 0 1 ) = 1 ( 1) f (0) 0 1 ( 0 f (0) 1 f (0) ) = 0 1 ( 0 1 ), where the last equalty s due to the fact that and 0 f (0) 1 f (0) = 0 1 for f (0) = 0, 0 f (0) 1 f (0) = 1 0 for f (0) = 1. By the same logc, f we nput x = 1 1 and 1 ( y = 0 y = 1 ) we get [ ] 1 ( 1) f (1) U D 1 1 ( 0 1 ) = 1 1 ( 0 1 ). Takng a super poston 1 ( )( 0 1 ) of the two nputs wll therefore gve us [ ] 1 U D ( )( 0 1 ) = 1 ) (( 1) f (0) ( 1) f (1) 1 1 ( 0 1 ) ( 1) f (0) ) = ( ( 1) f (1) f (0) 1 1 ( 0 1 ). If we now examne partcle 1 after applyng U D, we see that we get (up to a global multplcatve factor) { f f (0) = f (1) f f (0) f (1). These two new states are orthogonal to one another, and so may be dstngushed by a sngle measurement [smply measure partcle 1 n the bass + = 1 ( ) and = 1 ( 0 1 )]. Thus, by applyng U D to a sngle state 1 ( )(( 0 1 ) and performng a sngle quantum measurement we can dstngush whether f (x) s constant or balanced, a feat we could not accomplsh classcally (we had to apply f (x) twce, to two dfferent nputs). Note, that once agan we manged to extract by our measurement just a sngle bt of nformaton ( f s constant or not). The power of quantum mechancs entered n the fact that we can use superposton whch cannot be used classcally Beam spltters and the Mach-Zender nterferometer. 18 The new states are all dfferent, snce two dentcal states would not be mutually orthogonal to each other (unless they are dentcally zero, whch U D cannot produce). 19 If e s an orthonormal bass and h s a second orthonormal bass then the transformaton from e to h s whch s clearly untary. U = h e,

15 14 1. PHYSICS AND INFORMATION dense codng. As we saw earler one can store a lot of nformaton n a qubt, however only 1 bt may be extracted wth certanty. We shall now see how, wth the use of entanglement, we can communcate two bts of nformaton by transferrng a sngle qubt. Our system wll nclude two qubts (A and B), whch we wll descrbe usng a specal bass known as Bell states ψ = 1 ( 0 A 1 B 1 A 0 B ) ψ + = 1 ( 0 A 1 B + 1 A 0 B ) φ = 1 ( 0 A 0 B 1 A 1 B ) φ + = 1 ( 0 A 0 B + 1 A 1 B ), where the subscrpts A,B tell us to whch qubt/partcle the ket belongs (n many cases we shall drop the subscrpts and keep the order of the kets constant). Ths specal bass has the convenent trats that t s orthonormal and that all four states are entangled (see secton.4). The bass s also the set of mutual egenvectors of the complete set of commutng operators 01 σ xa σ xb and σ za σ zb [σ xa σ xb,σ za σ zb ] = 0. We now defne (usng the Paul matrces) a new set of untary operators U (A) j U (A) 00 = 1 A, such that U (A) 01 = σ x A, U (A) 10 = σ y A, U (A) 11 = σ z A. From the trats of the Paul matrces, t s easy to see that applyng these operators on the Bell state ψ gves 3 U (A) 00 ψ = ψ, U (A) 01 ψ = φ, U (A) 01 ψ = φ +, 0 Note that the operators commute snce there are two partcles. For just one partcle we have σ x σ z = σ z σ x = σ y [σ x,σ z ] = σ y. However, when we have two partcles the mnuses cancel and we get σ xa σ za σ xb σ zb = σ za σ xa σ zb σ xb = σ ya σ yb [σ xa σ xb,σ za σ zb ] = 0 1 Note that σ ya σ yb also commutes wth σ xa σ xb and σ za σ zb. However, t suffces to look for egenvectors common to σ xa σ xb and σ za σ zb n order to unquely defne the four Bell states (see secton 5.1). Thus σ xa σ xb and σ za σ zb (or any other par of operators from σ xa σ xb, σ za σ zb and σ ya σ yb ) consttute a complete set of commutng operators. Recall that σ z = ; σ z =, σ x = ; σ x =, σ y = ; σ z =. 3 Note that the operators U (A) operate only on the sngle partcle A. Thus to be rgorous, the operator operatng on ψ s actually U (A) 1 B. That s, t s an operator whch apples U (A) on partcle A and does nothng to partcle B.

16 1.. QUANTUM INFORMATION 15 U (A) 11 ψ = ψ +. Havng constructed our tools, we can now turn to our orgnal problem: communcatng two bts of nformaton usng only a sngle qubt. To see how ths may be done let Alce and Bob be two dstant persons, Alce holdng partcle A and Bob holdng partcle B. The partcles, as Alce and Bob both know, are gven to be n the Bell state ψ. Now, assume that Alce wshes to communcate to Bob two bts of nformaton: and j ( = 0,1 and j = 0,1). To do ths Alce operates locally on her partcle wth the operator U (A) j we defned. As a result (accordng to the effect of U (A) j on ψ gven above) the two partcles A and B together, are now n one of the orthonormal Bell states (up to a global phase). Next, Alce sends her partcle, whch s a sngle qubt, to Bob. Havng both partcles, Bob can now make a measurement (locally) on the state and determne n whch of the orthogonal states the two partcles are. 4 snce Bob knows, that the partcles were orgnally n state ψ, he can therefore nfer whch operator Alce appled on her partcle and thus fnd, j. We have thus seen that by merely passng a sngle qubt from Alce to Bob, Alce could communcate (to Bob) two bts of nformaton. The extra (second) bt communcated was hdden n the entanglement of the two partcles. Note, that a beneft of ths method s encrypton. If a thrd person tres to ntercept the message, all he gets s a sngle qubt, whch gves hm no nformaton at all. Unlke Bob, any other person who gets the transmtted partcle has no extra nformaton and therefore cannot nfer from t anythng. 4 snce the possble states are orthogonal, Bob can make a measurement whch dstngushes between all four. For example he can measure the operator O = 1 ψ ψ + ψ + ψ φ φ + 4 φ + φ +. If the result we measure s 1, we know the partcles were n state ψ f we measure we know the partcles were n state ψ +, and so on.

17

18 CHAPTER Bascs of quantum nformaton.1. Bascs of quantum mechancs Every physcal theory s defned by the followng: The method of descrbng a system. The dynamcs of a system. The method of measurng a system. In quantum mechancs observable quanttes are descrbed by Hermtan operators (O = O). Such operators have the followng trats: All egenvalues are real: λ a R (O a = λ a a ). Egenvectors of dfferent egenvalues are orthogonal: λ a λ a a a = 0. Every Hermtan operator may be wrtten n a spectral decomposton form O = λ a Π a, a where Π a s the projecton onto the subspace of egenvectors wth egenvalues λ a Π a = Π a, Π a Π a = δ aa Π a, Π a = 1. a In general, projectons are Hermtan operators (.e. they are observables) such that f Π s a projecton then Π = Π (Π = Π). From the spectral decomposton trat, every state may be wrtten as ψ = Π a ψ. a Usng ths decomposton we can defne the effect of measurement n quantum mechancs as follows: A measurement of the quantty A for a state ψ results n a collapse of the state nto one of the egenvalue subspaces of A,.e. ψ measure A Π a ψ. ψ A ψ The probablty of the collapse to the subspace a s gven by prob(π a = 1) = prob(a = a) = ψ Π a ψ. CONCLUSION. If two states are not orthogonal ( ψ 1 ψ = 0), then one cannot dstngush between them wth certanty. In other words, there exsts no projecton Π such that prob(π = 1) = ψ 1 Π ψ 1 = 1 17

19 18. BASICS OF QUANTUM INFORMATION and prob(π = 0) = ψ Π ψ = 0. PROOF. Let us assume that there exsts such a projecton (for ψ 1 ψ = 0). We defne by a Grahm-Schmdt process a new state ϕ orthonormal to ψ 1 : ϕ = ψ ψ 1 ψ ψ 1, ϕ = ϕ 1 ϕ ϕ. Snce ψ 1 and ψ are not orthogonal whle ψ 1 and ϕ are, 1 then there exst α,β 0 such that ψ = α ψ 1 + β ϕ. If we now substtute ths new form of ψ nto the assumpton ψ Π ψ = 0, we get 0 = (α ψ 1 + β ϕ )Π(α ψ 1 + β ϕ ) = α ψ 1 Π ψ 1 + β ϕ Π ϕ + α β ψ 1 Π ϕ + αβ ϕ Π ψ 1. Now, snce ψ 1 Π ψ 1 = 1 and also ψ 1 ψ 1 = 1, then we must have Π ψ 1 = ψ 1 (and ψ 1 Π = ψ 1 ). Thus f ψ 1 and ϕ are orthogonal, then ( ψ 1 Π ϕ = ϕ Π ψ 1 = 0 ψ ) 1 ϕ = 0,, Π ψ 1 = ψ 1 and snce Π s hermtan, then ψ 1 Π ψ 1 0 and ϕ Π ϕ 0. Therefore, together wth α, β > 0 and ψ 1 Π ψ 1 = 1, the expresson we just found for ψ Π ψ = 0 becomes 0 = α + β ψ Π ψ > 0, whch s a contradcton. Hence, there does not exst a projecton Π such that prob(π = 1) = ψ 1 Π ψ 1 = 1 and f ψ 1 ψ = 0 prob(π = 0) = ψ Π ψ = 0, Before gong on t should be mentoned that n quantum mechancs one can dstngush between two types of systems. The frst s that of a closed system: a system for whch all elements are known as well as ther nteractons wth one another. The second type, s that of an open system, where n addton to the elements we are nterested n, there s also an envronment. Ths envronment nteracts wth our system, however, n a way that we do not know exactly. 1 And we assume ψ1 = ψ. We may always wrte Π ψ1 = α ψ1 + β ψ, such that α + β = 1, and ψ 1 ψ = 0. Thus ψ 1 Π ψ 1 = α ψ 1 ψ 1 + β ψ 1 ψ = α ψ 1 ψ 1. Snce we also have ψ 1 Π ψ 1, then we must have α = 1 and β = 0, whch proves that Π ψ 1 = ψ 1.

20 .. SPIN 1 AND THE PAULI MATRICES 19.. Spn 1 and the Paul matrces Snce spn 1 partcles are used very often n quantum nformaton t s worth whle to make a short revew of (some of) ther trats. The observables for measurng spn 1 n the x, y, and z drectons are the Paul operators σ x, σ y, and σ z respectvely. These are also denoted as σ 1, σ, and σ 3 respectvely. The operators obey the commutaton relaton of angular momentum [σ,σ j ] = ε jk σ k, where ε jk s the antsymmetrc tensor: +1 for ε 13 and all cyclc permutatons of, j,k ε jk = 1 for ε 31 and all cyclc permutatons of, j,k. 0 otherwse In other words [σ x,σ y ] = σ z, [σ z,σ x ] = σ y, [σ y,σ z ] = σ x, wth all other cases obvous from these. The Paul operators also have the followng trats: σ σ j + σ j σ = δ j 1, σ σ j = δ j 1 + ε jk σ k, Tr[σ ] = 0, σ x = σ y = σ z = 1, σ = σ (Hermtan), and σ σ = 1 (untarty). Note, that the Paul operators are both untary and Hermtan. The Paul operators are usually represented by the standard Paul matrces: ( ) 0 1 σ x =, 1 0 ( ) 0 σ y =, 0 and ( 1 0 σ z = 0 1 The egenstates of the of the Paul matrces are the up (egenvalue +1) and down (egenvalue 1) states n the approprate drecton. The relatons between these egenstates are as follows: ). x = 1 ( z + z ) x = 1 ( z z ) y = 1 ( z + z ) y = 1 ( z z ). The effect of the Paul operators on the up and down states n the z drecton s as follows: σ x z = z,

21 0. BASICS OF QUANTUM INFORMATION σ x z = z, σ y z = z, σ y z = z. Thus we say that σ x flps n the z drecton, whle σ y not only flps n the z drectons, but also adds a phase (dependent on the orgnal state)..3. Open systems, mxtures and the densty matrx So far we have only dealt wth closed quantum systems, we shall now turn to treat open systems. There are two general cases n whch we encounter open systems: (1) Lack of knowledge of the full system: We mght not know some of the ntal condtons (the ntal state of the system), or some of the parameters of the system, or not know exactly the dynamcs of the system. () We are dealng wth a system of two (or more) subsystems, whch we fully know how to descrbe, however, we are nterested n makng measurements only on part of the full system. In both these cases the treatment s dfferent than that of closed systems. We shall see that we have to use mxtures nstead of regular states, where these mxtures wll be descrbed by densty matrces. Further more, probabltes wll behave slghtly dfferent: Instead of one state evolvng n tme, we shall have several, each wth a dfferent probablty to occur. Ths s dfferent from a lnear of combnaton of states (superposton), snce here each state s treated separately and there s no nterference effect. 3 To see the dfference between open and closed systems let us study an example. Assume two states; state ψ A wth probablty p a to occur; and state ψ B ψ A = a a 1 1 ( a 0 + a 1 = 1), ψ B = b b 1 1 ( b 0 + b 1 = 1), wth probablty p b = 1 p a to occur. What s the probablty to measure 0 n ths case?.e. what s prob(π 0 = 1) =? (Π ). If we make many measurements, n p a of them the measurement wll be of state ψ A and n p b = 1 p a they wll be of state ψ b. Therefore, the probablty to measure 0 wll be p a tmes the probablty to measure 0 n case ψ A plus p b tmes the probablty to measure 0 n case ψ B : prob(π 0 = 1) = p a ψ A Π 0 ψ A + p b ψ B Π 0 ψ B = p a a 0 + p b b 0 = p a a 0 + (1 p a ) b 0. If on the other hand, nstead of havng a probablty for each state ( ψ A and ψ B ), we make a superposton ψ AB = α ψ A + β ψ B = (αa 0 + βb 0 ) 0 + (αa 1 + βb 1 ) 1 ( α + β = 1), then, n ths case, we shall fnd prob(π 0 = 1) = αa 0 + βb 0. If we now compare the two results, we see that they are markedly dfferent. In the mxture (assumng a 0,b 0 0), no matter the value of p a there wll always be a fnte probablty to measure 0. However, n the superposton case, we may choose α and β such that the probablty to measure 0 wll be zero. The dfference, as mentoned above, s that n the latter case we have nterference: all the coeffcents appear wthn one absolute value 3 Recall, that n a lnear combnaton of states, the coeffcents appearng are not the probabltes of each state, but ther ampltude. You must take the absolute value squared to fnd the probablty.

22 .3. OPEN SYSTEMS, MIXTURES AND THE DENSITY MATRIX 1 (squared). However n the mxture case, there s no nterference and we have a sum wth two absolute values (squared) the densty matrx. The mathematcal tool we use to descrbe mxtures s the densty matrx. Assume a set {p, ψ } of possble states ψ (not necessarly orthogonal, but ψ ψ = 1), each wth probablty p to occur. We defne the densty matrx/operator ρ as ρ p ψ ψ (0 p 1, p = 1). If we wrte t n an orthonormal bass n ( n m = δ nm ), then and ρ nm = n ρ m Trρ = n ρ n = ρ nn. n The densty matrx has the followng trats ( n s an orthonormal bass): (1) Its trace s 1: Trρ = n ρ n = ρ nn = 1. n () The densty matrx s Hermtan and the sum of ts egenvalues s 1 ρ = ρ λ k = 1 (ρ ϕ k = λ k ϕ k ). k (3) The densty matrx s a postve operator,.e. for every state ψ n the Hlbert space, we have ψ ρ ψ 0 ( ψ ), whch s equvalent to havng all ts egenvalues nonnegatve ψ ρ ψ 0 λ k 0. Note, that together wth the prevous trat ( k λ k = 1), we must have 0 λ k 1. PROOF. (1) To prove that Trρ = 1 we shall use the defnton of the densty matrx. The trace of an operator s ndependent of the (orthonormal) bass we work n. If n s some orthonormal bass, then ( ) Trρ = ρ nn n p ψ ψ n n n = p n ψ ψ n = p ψ n n ψ n, n, ( = p ψ = p = 1, n n n where we have used the trat of orthonormal bases n n = 1. n ) ψ = p ψ ψ () Provng that ρ s Hermtan s very smple from ts defnton. Snce the p are real (0 p 1), then ( ρ = p ψ ψ ) = p ψ ψ = ρ.

23 . BASICS OF QUANTUM INFORMATION Snce ρ s Hermtan, then t may be dagonalzed. The sum of ts egenvalues, s ts trace, and thus from the prevous trat we must have λ k = Trρ = 1. k (3) To show that the densty matrx ρ s a postve operator ( ψ ρ ψ 0, ψ) we shall calculate ψ ρ ψ usng the defnton of the densty matrx. For any state ψ : ( ) ψ ρ ψ = ψ p ψ ψ ψ = p ψ ψ ψ ψ = p ψ ψ. Snce p 0 and ψ ψ 0, then we necessarly have ψ ρ ψ 0, as requred. Wth ths result we can now show that all egenvalues of ρ are non-negatve. To show ths, we choose ψ = ϕ k, where ϕ k s an egenvector of ρ wth egenvalue λ k (ρ ϕ k = λ k ϕ k ). Usng the last result we fnd 0 ϕ k ρ ϕ k = ϕ k λ k ϕ k = λ k λ k 0, whch s just what we wshed to prove. 4 As we saw, the densty matrx descrbes a mxture of states, however, t may also descrbe a regular state. Ths latter case occurs when the mxture ncludes only a sngle state wth probablty p = 1. We say that such a mxture s a pure state (otherwse t s called a mxed state). In other words, a system s n a pure state f there exsts a state ψ such that ρ = ψ ψ (pure state). The densty matrx of a pure state has the specal trat that ρ = ρ. Note, that ths trat holds only for pure states,.e. ρ = ρ pure state. PROOF. Clearly, f we have a pure-state densty matrx ρ (ρ = ψ ψ, ψ ψ = 1), then ρ = ( ψ ψ )( ψ ψ ) = ψ ψ ψ ψ = ψ ψ = ρ, whch proves one drecton (pure state ρ = ρ). As for the opposte drecton, t s smple to show (see the end of ths proof) that a densty matrx ρ obeys ρ = ρ f and only f ρ has a sngle egenvector ϕ 1 wth egenvalue λ = 1, whle all other egenvectors have an egenvalue λ = 0. If ths s ndeed the case, then by the spectral decomposton we may wrte ρ = λ Π λ = 1 Π Π 0 = ϕ 1 ϕ 1, 4 We showed above that f ψ ρ ψ 0 for any state ψ, then necessarly all egenvalues obey λk 0. To show the opposte drecton (assumng the operator can be dagonalzed), smply wrte ψ n the bass of egenvectors ϕ k ψ = α k ϕ k. k Now ψ ρ ψ wll gve ψ ρ ψ = k,k α k α k ϕ k ρ ϕ k = k,k λ k α k α k ϕ k ϕ k = whch s the desred result ψ ρ ψ 0. λ k α k δ kk = α k λ k 0, k,k k

24 .3. OPEN SYSTEMS, MIXTURES AND THE DENSITY MATRIX 3 or smply ρ = ϕ 1 ϕ 1. We have thus proven the second drecton,.e. that ρ = ρ mples that ρ has the form of a pure state. To complete the proof, we stll have to fll n one gap. We must show, as clamed above, that ρ = ρ corresponds to ρ havng a sngle egenvalue 1 wth all others 0. To show ths, smply dagonalze ρ to gve ρ D. The dagonal elements of ρ D are the egenvalues of ρ. Snce ρ s a densty matrx, then these egenvalues obey 0 λ k 1 and λ k = 1. Clearly, n such a case the dagonal matrx ρ D obeys ρ D = ρ D f and only f a sngle element on the dagonal s 1 and all others are 0 (.e. f ρ has a sngle egenvalue of 1 and all the rest 0). 5 Further more, snce the dagonalzaton of ρ to ρ D s just a base change, then ρ = ρ ρ D = ρ D. Consequently, ρ = ρ f and only f ρ has a sngle egenvalue 1, whle all others are 0. The densty matrx (pure or not) has one more mportant trat: For any projecton operator Π, the probablty of t measurng true (.e. of the mxture collapsng, due to the measurement, to the subspace of Π) s prob(π = 1) = Tr(ρΠ) = Tr(Πρ). Ths trat may be further generalzed as follows: the average value O of an observable O, when measured, s O = Tr(ρO) = Tr(Oρ). PROOF. We shall start by provng the smple form of the trat. By defnton (recall ρ p ψ ψ ) prob(π = 1) = p ψ Π ψ. If we now use an orthonormal bass n, we know that n n n = 1, and we can therefore wrte the last relaton as ( ) prob(π = 1) = p ψ Π n n n ψ = p ψ Π n n ψ = p n ψ ψ Π n,n (,n ) [( ) ] = n n = Tr(ρΠ), p ψ ψ Π n = Tr p ψ ψ whch proves the smpler trat. We can now use the last result to prove the more general trat. By defnton O = p ψ O ψ. Π 5 If ρd has more than one non-zero elements on ts dagonal, then these elements must be dfferent from 1 (due to λ = 1, λ 0). As a result ρ D wll not gve ρ D. For example =

25 4. BASICS OF QUANTUM INFORMATION Snce O s an observable we can always wrte t n a spectral decomposton O = λ k ϕ k ϕ k λ k Π k. k k Insertng ths nto the prevous relaton and usng the trat prob(π = 1) = Tr(ρΠ) we get O = λ k p ψ Π k ψ = λ k Tr(ρΠ k ). k k Now, snce Tr s a lnear operator then O = λ k Tr(ρΠ k ) = Tr k ( ) ρ λ k Π k = Tr(ρO). k To prove that Tr(ρΠ) = Tr(Πρ) and Tr(ρO) = Tr(Oρ), we can smply use the trat of the trace that Tr(AB) = Tr(BA). On the other hand, n provng the smpler form, we could have started wth ( ) prob(π = 1) = p ψ n n n Π ψ nstead of ( ) prob(π = 1) = p ψ Π n n ψ, n whch would have led us to prob(π = 1) = Tr(Oρ). The trats we have found for the densty matrx put constrants on ts elements ρ nm. We mght therefore ask how many ndependent (real) parameters descrbe an N N densty matrx. If we had no constrants, then there would be N complex elements n the matrx, whch would therefore gve N ndependent real parameters. However, we have three constrants ρ = ρ, and Trρ = 1, λ 0 where λ are the egenvalues of the densty matrx. The frst constrant (ρ = ρ) s actually N equatons snce on the dagonal ρ = ρ gves ρ nn = ρ nn and off the dagonal (n m) we have ρ nm = ρ mn, (N equatons), (N N equatons). Note, that n the off-dagonal case, the number of equatons takes nto account that we should have both doubled and halved the number of equatons (relatve to the N N off-dagonal elements). The number of equatons should have been doubled snce each equalty gves two equatons: one for the real part and a second for the magnary part. On the other hand, the number of equatons should have been halved snce t suffces to count only the pars n,m above the dagonal [ 1 (N N) pars], as those below wll gve us the same equatons agan. We dd not double the equatons for the elements on the dagonal, snce these equatons only tell us that the magnary part s zero, but do not tell us anythng about the real part (t equals tself, whch s trval). To the above constrants we must also add the one on the trace: Trρ = 1 (1 equaton).

26 .3. OPEN SYSTEMS, MIXTURES AND THE DENSITY MATRIX 5 Ths constrant s just a sngle equaton, snce we already know that the trace has no magnary component (the dagonal elements are all real, due to ρ = ρ). Subtractng the number of equatons from the total number of parameters (n the case of no constrants) we fnally get 6 #of ndependent parameters = N 1. Now that we know that densty matrces have N 1 parameters, the queston mght be, how do we parametrze these matrces,.e. how do we wrte the matrces as a functon of N 1 parameters. To do ths we note that not only densty matrces have N 1 parameters, but that the SU(N) group of matrces 7 also has N 1 parameters. Snce both sets have the same number of parameters, we mght try and relate the two somehow. In general, t s mpossble (at least n a smple way) to construct densty matrces usng a lnear combnaton of untary matrces from SU(N). However, we may use ther generators. 8 We shall next see how ths s done for the case of N =. For N =, one possble set of generators, of SU(), s the Paul matrces σ. 9 For convenence, we defne a vector of matrces and an nner product of matrces 10 σ (σ x,σ y,σ z ) (σ 1,σ,σ 3 ), A,B Tr(A B). Usng ths last defnton we fnd that the Paul matrces are orthogonal to one another σ,σ j = δ j. If we also add the unt matrx to the Paul matrces, we now have N = 4 matrces, and these four (usng complex coeffcents) span the space of matrces. To see ths, note that f we defne σ 0 1, 6 Note, that we have not used the constrant that all egenvalues must be non-negatve. Ths constrant does not change the number of parameters, t just reduces the range of the parameters. It reduces the regon of allowed parameters n the N 1 dmensonal space. 7 The SU(N) group (Specal Untary group) s the group of all N N untary matrces wth determnant +1: U SU(N) U U = 1,det(U) = +1. (In general, untary matrces have a determnant of e θ, wth θ R). It s easy to see that SU(N) has N 1 ndependent (real) parameters, snce U U = 1 gves N equatons, and det(u) = +1 s another equaton. Thus we have N + 1 equatons for the N real parameters of U (.e. N 1 real ndependent parameters). The N orgnal real parameters are due to the fact that an N N matrx has N elements, but each has two components: a real part and magnary part. Note, that ths logc does not work for the number of equatons n U U = 1 and det(u) = +1. Ths s because U U mxes real and magnary parts, and thus we cannot splt the N complex equaton nto N equatons (real and magnary). 8 The generators of a group, n ths case, are a set of matrces g such that any element n the group may be wrtten as e θ jg j (or e θ j g j ), where the θ j s are real. For the SU(N) group there are N 1 generators g. 9 Remnder: The Paul matrces are σ 1 = ( ) ; σ = ( 0 0 ) ( 1 0 ; σ 3 = It can easly be checked that Tr(A B) obeys all the requrements of an nner product (α a scalar): A + B,C = A,C + B,C, αa,b = α A,B, B,A = A,B, A,A 0 where A,A = 0 A = 0. ).

27 6. BASICS OF QUANTUM INFORMATION then the above nner product σ,σ j = δ j stll holds, even when, j run from 0 to 3. Snce the four matrces are orthogonal to each other, then they necessarly consttute a bass of all the matrces (a four dmensonal space). Snce the Paul matrces, together wth the unt matrx, consttute a bass, then any densty matrx may be wrtten n the form ρ = a a 1 σ x + a σ + a 3 σ 3 a a σ. To fnd the coeffcents a we apply the constrants we had on the densty matrx. Frst we apply the constrant on the trace: Trρ = 1. Snce Trσ = 0 for all three Paul matrces ( = 1,,3), then the condton Trρ = 1 becomes (recall that here 1 s a matrx) 1 = Trρ = a 0 Tr1 + 0 = a 0 a 0 = 1. Now, the second requrement we had s that ρ be Hermtan (ρ = ρ). Snce the Paul matrces themselves (and 1) are Hermtan, the requrement becomes whch means that 11 ρ = ( a 1σ x + a σ y + a 3σ z ) = ( a 1σ x + a σ y + a 3 σ z ) = ρ, For convenence we defne a = a a R. p a, whch allows us to wrte the densty matrx as ρ = 1 (1 + p σ) ( p R3 ), or n matrx form ρ = 1 [( ) ( ) ( p p = 1 ( ) 1 + p3 p 1 p. p 1 + p 1 p 3 ) + p 3 ( The fnal requrement of the densty matrx, s that t be a postve operator. Snce we are dealng wth a matrx wth a postve trace, then a necessary and suffcent condton s that the determnant be non-negatve 1 det(ρ) 0. From the matrx form we found for ρ, ths means that det(ρ) = 1 + p 3 p 1 p p 1 + p 1 p 3 = 1 (p 1 + p + p 3) = 1 p 0. Therefore, we fnally have the general form of the densty matrx ρ = 1 (1 + p σ) ( p R3, p 1). )] 11 Snce the Paul matrces together wth the unt matrx consttute a bass, then there s only a sngle choce of coeffcents a whch gves a certan matrx (f there were more, the matrces would not be lnearly ndependent, and therefore not a bass). The above condton gves two sets of coeffcents {a } and {a }. For these sets to be the same we must have a = a. 1 Recall, that the trace of a matrx equals the sum of ts egenvalues, and ts determnant s the product of the egenvalues. Snce we are dealng wth a matrx, t has two egenvalues. The trace s 1, whch s postve, and therefore the product of the egenvalues,.e. the determnant, must be non-negatve for both egenvalues to be non-negatve.

28 .3. OPEN SYSTEMS, MIXTURES AND THE DENSITY MATRIX 7 The vector p s called the polarzaton vector. The name s used snce p gves the average drecton (polarzaton) of the spn: σ x = Tr(ρσ x ) = p x, σ y = Tr(ρσ y ) = p y, σ z = Tr(ρσ z ) = p z. Note, that as promsed, we have three parameters (N 1 = 3, for N = ) whch descrbe the densty matrx. These parameters are the three real components of the vector p. To conclude, we see that we can represent all possble densty matrces (of a two dmensonal Hlbert space) by the possble vectors p. The possble vectors p ( p 1) form a ball of radus 1. Ths ball s known as the Bloch sphere. 13 We shall see that the case p = 1 (ponts on the surface of the Bloch sphere) concdes wth pure states. CLAIM. A densty matrx descrbes a pure state f and only f the polarzaton vector p s a unt vector ( p = ˆn),.e. ρ = ψ ψ ρ = 1 (1 + ˆn σ). PROOF. Let us start wth the reverse drecton,.e. that p = ˆn gves a pure state. We know that f we have ρ = ρ then ρ descrbes a pure state, thus t wll suffce to show that p = ˆn mples ρ = ρ. Let us therefore start by calculatng ρ for p = ˆn. By defnton (when p = ˆn) [ ] 1 ρ = (1 + ˆn σ) = 1 4 [1 + ˆn σ + ( ˆn σ) ], and We know that for the Paul matrces and ( ˆn σ) = (n 1 σ 1 + n σ + n 3 σ 3 ) = n σ + 1 (n n j σ σ j + n j n σ j σ ), j j = n σ + 1 n n j (σ σ j + σ j σ )., j j σ = 1 σ σ j = σ j σ ( j). Therefore, by the above relaton for ( ˆn σ), we must have ( ) ( ˆn σ) = n 1 = 1. Thus ρ becomes ρ = 1 4 [ 1 + ˆn σ] = 1 [1 + ˆn σ] = ρ. Ths result (ρ = ρ when p = ˆn) proves that ρ = 1 (1 + ˆn σ) descrbes a pure state (and can thus be wrtten as ρ = ψ ψ for some ψ ) Yes, the Bloch sphere, s actually a ball. However, n some places the term Bloch sphere s ndeed reserved only for the boundary of the ball. 14 If we defne ˆn by the sphercal angles θ and ϕ ˆn = snθcosϕ ˆx + snθsnϕŷ + cosθẑ,