1 More General von Neumann Measurements [10 Points]

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1 Ph5c Sring 07 Prof. Sean Carroll Homework - 3 Solutions Assigned TA: Thom Bohdanowicz thom@caltech.edu Due: 5:00m, 5/03/07 More General von Neumann Measurements [0 Points] Recall that the Bloch shere reresentations of orthogonal qubit states whose Bloch vectors are the antiodal air ±ˆn are ±n = ±ˆnih±ˆn = (I ± ˆn ~ ). We have seen in class how, using the von Neumann model, we can erform the PVM { +ˆnih+ˆn, ˆnih ˆn } by letting the system interact with a single ancilla system (reresenting the measurement aaratus) using the interaction H I = g(ˆn ~ ) x. This von Neumann model of measurement is useful for many reasons. One is that it allows us to measure a quantum system non-destructively (for examle, allowing us to measure a hoton without absorbing it into a hotodetector). Recall that for a qubit, one can not simultaneously know the system s eigenvalues for more than one of the Pauli matrices, because they do not commute. Thus, for examle, it would seem that x and z can not be measure simultaneously. But using the von Neumann measurement model, it would seem that maybe we can! What if we instead let our system interact with two ancilla systems reresenting two measurement aarati, where the two observables that are measured are non-commuting? For examle, we could consider the following interaction Hamiltonian between the system S that we want to measure, and the two ancillary measurement systems M and M: H I = g ( z x I + x I x ). () Suose that we let the joint system S M M evolve under this Hamiltonian (with the corresonding evolution oerator U) for time t = 4g. If we then measure some PVM M = {M 00,M 0,M 0,M } (where the subscrits reresent the basis states in M M). This will corresond to doing a simultaneous joint measurement of x and z on the system S, each of which gives incomlete information about the observables x and z. Our goal here will be to understand how this works using POVMs on the single qubit system. Note that in this roblem, all systems are qubits: M = M = S = C. (a.) [ oints] Denote the initial state of system S, which is initially uncorrelated from M and M, as i S. Assume that M and M are each initially in their ready states 0i / h0 /. Using the generalized Born rule for calculating robabilities of measurement outcomes (recall, it is imlemented via a trace over all three subsystems), find an exression for the POVM elements {E ij } that should corresond to the desired measurement on S as a artial trace over subsystems M and M. This artial trace should involve the interaction evolution oerator U discussed above, and the PVM elements {M 00,M 0,M 0,M }. Solution: The Born rule tells is that if we have a POVM (which includes the secial case of a PVM) {E a } a on Hilbert sace H, then the robability of obtaining measurement outcome a when you measure this POVM on state L(H) ispr(a) =Tr( E a ). The state we would like to simultaneously measure z and x on is the state S = ih S L(S). We are imagining two ways of erforming this measurement: as an indirect measurement (von Neumann) on

2 an auxiliary system, or as a direct measurement on the system itself. The first (indirect/von Neumann) way is to coule the system of interest S to two distinct measurement devices with Hilbert saces M and M resectively. To do this, we take the initially uncorrelated state on the entire system, = ih S 0ih0 M 0ih0 M, evolve the system with interaction unitary U that should entangle them, giving state U U, and then measuring the PVM {M ij } ij on systems M and M. Because the state that you re measuring is a state on S M M, your PVM oerators should also be oerators on S M M, and we can accomlish this by tensoring on an identity oerator for system S (because the measurement oerators are acting on M M and not touching S). So, our PVM that we will measure on U U is {I S M ij } ij, and the outcome ij will be obtained with robability given by the Born rule: Pr (ij) =Tr U U I S M ij. () Now, the whole oint is that because our state on S has become entangled with the state on M M viau, each di erent measurement outcome ij for our measurement erformed on M M gives us di erent information about the state of the system S. To understand what this information is, we will find the equivalent POVM {E ij } ij that we could just measure on S L(S) that would give us the same information. For the measurement on M M tobe equivalent to a measurement on S, the outcome statistics for each outcome ij should match. That is, we should demand that if measuring PVM {I S M ij } ij on state U U gives us the same information about S as measuring {E ij } ij on S, then we must have, by the Born rule, Pr (ij) =Tr( S E ij )=Tr(U U I S M ij ). (3) Note that the first trace is a trace over Hilbert sace S because the argument of the trace is a linear oerator on S, whereas the second trace is a trace over Hilbert sace S M M because its argument is a linear oerator on S M M. The goal here will be to work the second trace into a form that looks like the first trace, allowing us to identify a formula for E ij in terms of U and M ij. The first thing that we can do is use the cyclic roerty of the trace to move one of the Us in the second trace, and relace with its exlicit exression: Tr (U U I S M ij )=Tr( S 0ih0 M 0ih0 M U I S M ij U) (4) The trace over a comosite Hilbert sace S M M can be imlemented as searate artial traces over each of S, M, and M searately, in any order you would like. Taking the artial trace over M M and leaving the trace over S incomlete, one finds Tr S M M ( S 0ih0 M 0ih0 M U I S M ij U)=Tr S ( S Tr M M (I S 0ih0 M 0ih0 M U I S M ij U)). (5) We can now comare this to the POVM Born rule for outcome ij and find that Tr S ( S Tr M M (I S 0ih0 M 0ih0 M U I S M ij U)) = Tr S ( S E ij ). (6) Hence, we can identify that the POVM {E ij } ij = {Tr M M (I S 0ih0 M 0ih0 M U I S M ij U)} ij, (7) which is a measurement imlemented directly on state S, is equivalent to the PVM {I S M ij } ij imlemented indirectly by measuring the state U U.

3 Now, consider the two orthogonal Bloch shere axes (careful: the ure states in these directions are not orthogonal!), û = (ˆx +ẑ) and ˆv = ( ˆx +ẑ). Consider the observables U =û ~ and V = ˆv ~, with corresonding eigenvectors ±û/ˆvi. If ±i are the eigenvectors of x, thenit is ossible to show (lease don t) that the evolution oerator U discussed above can be written U = +uih+u ( ++ih++ + i i h ) + i uih u ( ++ih++ i i h ) + +vih+v ( + i h+ + i +ih + ) + i vih v ( + i h+ i +ih + ). (8) Let the PVM oerators {M ij } discussed above be the rojectors onto the states iji, where 00i = ( ++i + i i) 0i = ( + i + i +i) 0i = ( + i i +i) i = ( ++i i i). (9) (b.) [ oints] Calculate the corresonding POVM elements E ij as defined in art a, and show that they indeed form a POVM (that is, they are ositive oerators which sum to the identity). (Hint: think about what density oerator you get when you add antiodal oints on the Bloch shere). Solution: We would now like to lug in our exlicit exressions for U and M ij in to our result from art (a.) to exlicitly calculate the oerators E ij. Because U is such an unwieldy looking oerator, this seems like a lot of work. But there are some observations one can make to drastically simlify the calculation. First, notice that because of the rojector I S 0ih0 M 0ih0 M sitting on the left hand side of the trace, only the art of the trace where you sandwich your oerators with I S h0 M h0 M and I S 0i M 0i M will be non-zero, because all otherse will give orthogonal inner roducts on this rojector. So, E ij = (I S h0 M h0 M ) I S 0ih0 M 0ih0 M U I S M ij U (I S 0i M 0i M (0) ) = (I S h0 M h0 M ) U I S M ij U (I S 0i M 0i M ) () Now, looking at just the art U (I S 0i M 0i M ), we see that we will need to calculate the inner roducts h++ 00i, h+ 00i, h + 00i, and h 00i, which are all equal to. So, we can see that or U (I S 0i M 0i M )= +uih+u ( ++i + i i) h00 + i uih u ( ++i i i) h00 +vih+v ( + i + i +i) h00 + () + i vih v ( + i i +i) h00. (3) U (I S 0i M 0i M )= +uih+u 00ih00 3

4 + i uih u ih00 + +vih+v 0ih00 We have now simlified enough to quickly comlete the calculation: + i vih v 0ih00. (4) E ij = (I S h0 M h0 M ) U I S ijih ij U (I S 0i M 0i M ) (5) and recognizing that each of the iji are orthogonal to the others (their rojectors form a PVM), and that (I S h0 M h0 M ) U is just the conjugate of what we had for U (I S 0i M 0i M ), we can insert everything and use the orthonormality of the iji to find that (6) E 00 = +uih+u (7) E 0 = +vih+v (8) E 0 = vih v (9) E = uih u. (0) Each of these are roortional to rank- rojectors, with eigenvalues / and 0, so they are all ositive oerators. To see that they sum to the identity, calculate E 00 + E 0 + E 0 + E = ( +uih+u + uih u )+ ( +vih+v + vih v ). () But +uih+u and uih u are each rojectors onto orthogonal qubit states (they reresent qubit states which are antiodal to each other on the Bloch shere), so they must sum to the identity I, likewise for +vih+v and vih v. So, we have that E 00 + E 0 + E 0 + E = I + I = I. () Thus, the oerators {E ij } ij indeed form a POVM on S. Okay, now we will consider the descrition of a new measurement based on the measurement just described above. The new measurement rocedure is as as follows: erform the joint measurement rocedure above, which returns outcome (i, j), and ignore the outcome j of measurement aaratus M. Basic robability tells us that Pr(i) = P j Pr(i j), where Pr(i j) is the conditional robability of i, given that j is true. Our new POVM F will have elements {F i }, i {0, }. (c.) [ Points] Write down an exression for Pr(i) for an arbitrary system state i S as a single trace using the POVM elements E ij. Deduce from this exressions for the oerators F i in terms of the oerators E ij. Solution: Using the Born rule, Pr (i) =Tr( ih S F i ). (3) 4

5 But we also know that Pr(i) = P j Pr(i j), and that Pr(i j) =Tr( ih S E ij). So, we see that Pr (i) = Tr( ih S F i )=Tr( ih S E i0 )+Tr( ih S E i ) (4) (5) = Tr( ih S (E i0 + E i )). (6) So, we conclude that F i = E i0 + E i, which obviously form a POVM because the Es do. (d.) [ Points] Show that we can also write F i = q iihi +( q) I for some aroriate value of q (which you ll find in the rocess), and show that the oerators {F i } form a POVM. Solution: Using the Bloch shere reresentation of the qubit density matrix, we know that So, we see that +uih+u = (I + x + z) (7) uih u = (I + x + z) (8) +vih+v = (I x + z) (9) vih v = (I x + z) (30) F 0 = 4 (I + x + z)+ 4 (I x + z) (3) = 4 (I + z ) (3) F = 4 (I x z)+ 4 (I + x z) (33) = 4 (I z ) (34) Resolving z into its sectral decomosition, z = 0ih0 ih, we can write Similarly, we find that F 0 = 4 (I + ( 0ih0 ih ) 0ih0 + 0ih0 ) (35) = 4 (I ( 0ih0 + ih )+ 0ih0 ) (36) = 4 (I I + 0ih0 ) (37) = 0ih0 +( ) I (38) F = ih +( ) I. (39) As mentioned at the end of (c.), it s clear that these form a POVM. We can nonetheless verify exlicitly here by noting that each F i is ositive because each is a sum of ositive oerators with ositive coe cients, and adding the two obviously gives the identity oerator as required. 5

6 (e.) [ Point] Give an interretation for the measurement rocedure that POVM {F i } reresents that allows us to understand it as a noisy measurement of z. This interretation should be something like with robability, a measurement of insert measurement here is erformed, and with robability, the POVM insert POVM here is measured. (Hint: the POVM should be able to be interreted as just ignoring the system and fliing a fair coin.) Solution: This POVM is a robabilistic mixture of the PVM {P 0 = 0ih0,P = ih } which measures z, and the POVM {V 0 = I,V = I } which does absolutely nothing to the system, and gives each outcome with 50 ercent likelihood. How do we see this? Well, first, let s think about the POVM {V 0 = I,V = I }. For any state, the outcomes are 50/50 because Tr ( I )=, and resolving each V i = I into measurement oerators M i = I such that V i = M i M i, the measurement udate rule tells us the ost-measurement state should be M i M i Tr ( I ) = I I =, (40) the state is comletely unchanged, regardless of what the measurement outcome was! So if instead of actually measuring your state, you just flied a fair coin and used that as your measurement outcome, the POVM {V 0 = I,V = I } accurately describes this oeration. Now, suose that we had some machine that we ut our quantum state that we want to measure in, and we ush a button, it measures, and tells us outcome 0 or outcome. We want it to measure z, but what actually goes on inside the machine is that with robability it measures z as a PVM and tells us the outcome (0 or ), and with robability it does absolutely nothing to the state, flis a fair coin, and tells us it was outcome 0 if it was heads and outcome if it was tails. What should the outcome robabilities be (for either 0 or )? We can take a robabilistically weighted sum of Born rules: Pr (0) = Tr ( 0ih0 )+( )Tr ( I ) (4) I = Tr 0ih0 + (4) = Tr ( F 0 ) (43) Pr () = Tr ( ih )+( )Tr ( I ) (44) I = Tr ih + (45) = Tr ( F ) (46) and so we see that F 0 and F, as a POVM, reroduce the measurement statistics of this interreted imerfect measurement of z. It is unnecessary, but one could also go through the exercise of decomosing F 0 and F into their measurement/kraus oerators and calculate the udated state, and would see that the udated state is always a mixed state that is with robability the state rojected onto ii (which would be what haened if z were actually measured and outcome i attained), and with robability is left as (as it should be if the machine did nothing)! 6

7 (f.) [ Point] Now, go back to the definition of F given before art (c.), and instead consider the case that when we get measurement outcome (i, j) we instead ignore the outcome i of measurement aaratus M and kee j, calling this POVM G = {G j }. You need not show all of the work for arts (c.), (d.), and (e.) again, but give an analogous interretation to that given in art (e.) but for G. (It should just be a di erence in the observable measured with robability.) Solutions: I won t reeat the work, as it is almost identical to that above, but in this case one would find POVM elements G j = E 0j + E j, which could then be rewritten as G 0 = +ih+ + I and G 0 = i h + I, and the interretation would be that with robability the PVM { +ih+, i h } which measures x is erformed, and with comlementary robability, nothing is done and a coin is flied. One can now conclude that the POVM {E ij } is a noisy joint measurement of x and z on the single qubit system S that gives incomlete information about about the two incomatible observables. Kraus [5 Points] Quantum information theory uses the CNOT or controlled-not gate, a unitary oeration on two qubits that acts as U cnot : 00i! 00i (47) 0i! 0i (48) 0i! i (49) i! 0i. (50) In other words, the first ( control ) qubit stays fixed, while the second ( target ) qubit flis when the first is a, and stays fixed when the first is a 0. (a.) [ Point] Find the Kraus oerators for the quantum channel E CNOT,wherethesystem qubit is the control qubit, and the environment is the target qubit. Assume that the target/environment qubit starts in state 0ih0 Solution: Let the control qubit of the CNOT oeration (which our channel will act on) be described by the state C = ih = 0ih0 + 0ih + ih0 + ih, where i = 0i + i. Then the initial state of the whole system is = C 0ih0. We can evolve the entire system unitarily with the unitary U cnot : U cnot C 0ih0 U cnot = U cnot 00ih ih0 + 0ih00 + 0ih0 U cnot = 00ih ih + ih00 + ih. (5) We can now take the artial trace over the target/environment qubit to obtain E( C ): E( C ) = Tr E U cnot C 0ih0 U cnot (5) = 0ih0 + ih. (53) 7

8 Comaring this to the initial density matrix of our qubit, C = ih = 0ih0 + 0ih + ih0 + ih, we see that this is the same as 0ih0 0ih0 + ih ih. (54) Hence, we can see that Kraus oerators K 0 = 0ih0 and K = ih can be used to imlement the quantum channel E. One imortant examle of a quantum channel is the deolarizing channel. deolarizing channel on a qudit with Hilbert sace C d can be written In general, the E( ) =( ) + I d, (55) giving the interretation that with robability, the inut state is relaced by a comletely unknown random state (which we must reresent with the maximally mixed state I d ), and otherwise does nothing. (b.) [ Points] Consider the deolarizing channel for a qubit (d = ). Find a set of Kraus oerators for this channel. (Hint: Consider the decomosition of the qubit density oerator into Pauli oerators via the Bloch shere. What does conjugating a Pauli oerator by another Pauli oerator do? Using this information, you should be able to write the identity oerator as a linear combination of an arbitrary density oerator, lus conjugations by Paulis.) Solution: Start with the Bloch shere decomosition of the qubit density matrix described by Bloch vector ~v, = (I + v x x + v y y + v z z ). Suose we conjugate by any of the Paulis i, then we obtain x x = (I + v x x v y y v z z ) (56) y y = (I v x x + v y y v z z ) (57) z z = (I v x x v y y + v z z ) (58) which are calculated using the observation that i j i = j for i 6= j. Taking an equally weighted linear combination of all three yields x x + y y + z z = I (I + v x x + v y y + v z z ) (59) = I (60) So, we have then that I = 4 ( + x x + y y + z z ) (6) Plugging this identity into the equation for the deolarizing channel, we get E( ) = ( ) + 4 ( + x x + y y + z z ) (6) r r 3 = x x y y z z + +, (63) q 3 so we can see that an aroriate set of Kraus oerators are {K 0 = 4 I, K = x,k = y,k 3 = z} 8

9 (c.) [ Point] By studying the action of the deolarizing channel on the Bloch shere reresentation of a qubit, describe what this channel does to the Bloch shere. Solution: We have that E( ) = ( ) + I (64) = ( ) I + v x x + v y y + v z z + I = I +( )v x x +( )v y y +( )v z z So, we see that the deolarizing q channel takes a qubit state with Bloch vector ~v =(v x,v y,v z ) with magnitude k~v k = vx + vy + vz to a qubit state with Bloch vector ~v 0 =(( )v x, ( )v y, ( )v z )=( )~v, whose magnitude is k~v 0 k = k~v k < k~v k. So, alying the deolarizing channel shrinks the Bloch vector of the qubit state until its Bloch vector is zero, in which case the state is the maximally mixed state = I. (d.) [ Point] While the interretation given makes it seem like should be between 0 and, this actually need not be true for the formula given for the deolarizing channel to correctly reresent a quantum channel. Find the most general bounds on which allow for E( ) =( ) + I d to be a quantum channel for d =. Solution: Looking at the Kraus oerators {K 0 = q 3 4 I, K = x,k = (65) (66) y,k 3 = z}, we see that if <0 then the last three are no longer ositive, which they must be. So, 0. But the first Kraus oerator retains ositivity for u to 4/3, after which it is no longer ositive, so we see that we must have [0, 4/3]. 3 Stinesring and Kraus [5 oints] It was discussed in class that the unitary evolution of a density matrix is not the most general way a quantum system can evolve, but rather a quantum channel reresented by the CPTP ma E : L(H)!L(H) is. We exlored the Kraus reresentation of such a quantum channel and saw that this kind of ma bears a similarity to a mixture of unitary evolutions (that is, a density matrix can in general evolve to become a mixture of density matrices under a quantum channel). It was demonstrated in class that in aroriate situations, Kraus oerators describing the evolution of a subsystem can be obtained by considering unitary evolution of a larger system, and then tracing out everything excet the subsystem of interest (via the artial trace). Kraus oerators do not always need to be obtained in this fashion, but there is a theorem from oerator theory called the Stinesring Dilation Theorem that says you always can if you want to. In our language, the Stinesring Dilation Theorem says: Any CPTP ma E : L(H A )! L(H A ) can be exressed as the reduced action of a single unitary oerator acting on an extended Hilbert sace U : H A H B!H A H B. That is, we can write E( ) =Tr B (U U ), (67) 9

10 where is some state on an environment Hilbert sace H B of unsecified dimension. Show that this is true for a quantum channel E : L(H A )!L(H A ) on a d-dimensional quantum system H A = C d, by exlicitly constructing the unitary U in terms of the Kraus oerators {K i } k i=0 of E, assuming initial environment state = 0ih0. (Hint: Similarly to how there is a unitary freedom in choosing Kraus oerators to imlement a quantum channel, the unitary matrix U here is not unique. In fact, you should be able to argue that only d columns of U are fixed by the Kraus oerators. If you choose to reverse the ordering of your Hilbert saces (and hence their bases) to H B H A instead of H A H B, it will be exactly the first d columns of U : H B H A!H B H A that will be fixed by your Kraus oerators, and that art of your unitary will have a nice block form. You can choose to work in this reversed basis for convenience if you wish. The remaining columns of U can be chosen arbitrarily as long as all of the columns end u orthonormal). Solution: For convenience, let s work with the tensor factor ordering H B H A.Ifdim(H B )= b, then we can take orthonormal basis { ii B } b i=0 for H B and orthonormal basis { ii A } d i=0 for H A, so that we have orthonormal basis { ii B ji A } b,d i,j. Note that the convention for ordering of such a tensor basis is such that i and j start at 0 and the index j indexing H A s basis increases until it gets to d, then i indexing H B s basis increases by, j rolls back to zero, and we iterate through j again. That is, the ordered basis can be written { 0i B 0i A, 0i B i A,..., 0i B d i A, i B 0i A, i B i A,..., b i B d i A }. The ordering of this basis determines the ositions of rows/columns of matrices reresenting oerators that act on H B H A. Now, we re given a set of Kraus oerators (let s say that there s k of them) {K i } k i=0 which act on H A, and imlement a quantum channel on H A. That is, for state L(H A ), the action of the channel can be written E( ) = P k i=0 K i K i. Because channels reserve trace (as they must in order to ma quantum states to quantum states again), we know that the Kraus oerators satisfy the relation P k i=0 K i K i = I. Let s start by asking: is there a matrix U L(H B H A ) that we could choose such that Taking the artial trace, we find that Tr B U 0ih0 B U Xk = E( ) = K i K i? (68) i=0 Tr B U 0ih0 B U = = Xb (hr B I A )U( 0ih0 B )U ( ri B I A ) (69) r=0 Xb (hr B I A )U( 0i B I A ) (h0 B I A )U ( ri B I A ). (70) r=0 This suggests that if we took b = k, maybe choosing U such that K r =(hr B I A )U( 0i B I A ) could work. Let s work this exression out exlicitly to see what the relationshi between matrix elements of U and matrix elements of each K r would have to be. To do this, we can write U in terms of the standard basis for L(H B H A ), U = X U ijkl iji BA hkl BA = X U ijkl iihk B jihl A (7) i,j,k,l where U ijkl is the matrix element at the ijth row and klth column (i and k range from 0 to i,j,k,l 0

11 b, where j and l range from 0 to d ). Let s lug this into our exression: (hi B I A )U( 0i B I A ) = X U ijkl (hr B I A ) iihk B jihl A ( 0i B I A ) (7) i,j,k,l = X U ijkl hr iihk 0i jihl A (73) i,j,k,l = X U ijkl r,i k,0 jihl A (74) = i,j,k,l Xd j,l=0 U rj0l jihl A. (75) Then, exanding the matrix K r in the standard basis as K r = P d j,l=0 K r jl jihl A, equating K r =(hi B I A )U( 0i B I A )yields Xd j,l=0 K rjl jihl A = Xd j,l=0 U rj0l jihl A (76) which is true if and only if K rjl = U rj0l. So we see that by choosing matrix element U rj0l of U to be K rjl, we will have that Tr B U 0ih0 B U Xk = K i K i = E( ) (77) as desired. We can see, then, that the 0lth column of U is the lth column of each Kraus oerator stacked on to of each other, and iterating l from 0 to d we see that the Kraus oerators fix exactly the first d columns of the matrix U, and it doesn t matter what any of the remaining unfixed matrix elements of U are, the relation Tr B U 0ih0 B U = E( ) will hold. But, what about U being unitary? The whole oint is that we want to say that what aears to be a non-unitary evolution on quantum system H A is actually a unitary evolution on the larger system H B H A, so that the axiom of quantum mechanics that all time evolution is unitary is reserved. For U to be unitary, we need for the inner roducts of the columns of U are orthonormal to each other. Is this true for the d columns that are fixed by the Kraus oerators? The (comlex) inner roduct of the mth column of U with the nth column is Xd r,j=0 i=0 Xd Xd Urj0mU rj0n = Kr jm K rjn. (78) r=0 j=0 Now, observe that P d j=0 K r jm K rjn is the mn entry of the matrix K rk r. So we see that the inner roduct of the m and nth columns of U is equal to the mn entry of the matrix P d r=0 K rk r.but we know that because K r are Kraus oerators for a quantum channel that P d r=0 K rk r = I A, so we see that the inner roduct of the m and nth columns of U are equal to if n = m, and 0 otherwise. So, the columns are orthonormal, as they would need to be for U to be unitary. The remaining db d columns of U can be anything we want, so in articular, we can use the Gram- Schmidt orthogonalization rodecure to come u with db d columns to ut in the unsecified art of the unitary that are orthonormal to themselves as well as the first d columns, and U will be unitary. Thus we have shown that for any quantum channel, there exists a unitary U acting on a larger Hilbert sace that imlements this quantum channel.

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