Lecture 21: Quantum Communication

Transcription

1 CS 880: Quantum Information Processing 0/6/00 Lecture : Quantum Communication Instructor: Dieter van Melkebeek Scribe: Mark Wellons Last lecture, we introduced the EPR airs which we will use in this lecture to erform quantum communication. In the tyical quantum communication setting, two arties, Alice and Bob want to communicate. Beforehand, they create an entangled EPR air and give one qubit to Alice and the other to Bob. Alice and Bob will eloit the entanglement between the two qubits to echange information. In this lecture, we show that quantum communication can outerform its classical counterart and also elore the limits of quantum communication. Teleortation Teleortation is a rocedure that allows Alice to send a qubit to Bob using only two classical bits and one EPR air. Recall that the EPR air we use is Φ + = ( 00 + ). () Suose that Alice has some state Ψ = α b b, () that she wishes to send to Bob, and that she and Bob each hold one qubit of an EPR air. We will denote Alice s qubit as A and Bob s as B. For our first attemt at teleortation, we can try the following circuit, in which Alice entangles A with Ψ, Ψ A B () () (3) At the oint () in our circuit, the system state looks like Ψ Φ + = α b b, c, c (3) b,c {0,} At the oint (), Alice has alied a CNOT gate to her qubit, so the system state becomes α b b, b c, c (4) b,c {0,}

2 Alice then measures her qubit, and gets state d = b c. At (3), she then transmits d to Bob, who will use this to affect his qubit. If d = 0, Bob does nothing, otherwise he flis his qubit, giving us the state α b b, b d α b b, b (5) Note that Alice s EPR qubit is omitted from the state equation, as it has been measured and is no longer useful. At this oint, Bob s qubit is almost where we want it to be. However, it is still entangled with Alice s qubit. Alice could measure her state to remove the entanglement, but this collases Bob s as well, which defeats the urose of sending it to him in the first lace. To resolve this in our second attemt, we will use a similar circuit to the one in the first attemt. Ψ H A B This circuit behaves much as the revious one, ecet that Alice uses a Hadamard gate before taking a measurement. This functionally means that she is measuring in the + basis. Thus, the system state change from the Hadamard is α b ( ) ab a, b (6) α b b, b a,b Alice then takes her second measurement and sends a to Bob. If Alice measured a 0, then the system state would be α 0 ( ) (0)(0) 0 + α ( ) (0)() = α α = Ψ, (7) which means Bob has the state Alice wanted to send him, so we are done. In the case that Alice measured a, we have α 0 ( ) ()(0) 0 + α ( ) ()() = α 0 0 α, (8) which means Bob needs to only aly a hase fli to his qubit, and then he will have Ψ. No Cloning Theorem It is imortant to note that Alice was not able to dulicate the qubit that she wanted to send to Bob during the teleortation rocedure, as her coy was destroyed in the transfer rocess when she measured it. This is not a coincidence, as it is imossible for any quantum rocess to dulicate an arbitrary state. Formally, we can show that there cannot eist a quantum oeration that erforms the transformation ψ ψ 0 ψ ψ. (9) Ψ

3 Proof. Suose that such a Q eists. Let Then ψ = α α. (0) Q ψ ψ 0 = ψ ψ = α α 0 α 0 + α 0 α 0 + α. () Since Q is a quantum oeration, it must be linear, so Q ψ ψ 0 = Q (α α ) ψ 0 = α 0 Q 0 ψ 0 + α Q ψ 0 = α α. () Since equations () and () must be equal, it follows that α 0 α = 0, which imlies that the only states we can ossibly clone are the basis states 0 and. Note that this roof shows that in the {0,} basis, only the states 0 and can be cloned. If we changed into a different basis, the {+, -} basis for eamle, we could reeat the roof to show that we could make a quantum oerator to coy the + and states, but nothing else. This can be generalized to the statement that a quantum oerator can only be constructed to clone basis states for a secific basis. 3 Suerdense Coding In teleortation, we used two classical bits and an EPR air to send a qubit. We can also do the reverse in a rocess known as suerdense coding. In this contet, Alice will use an EPR air and a single qubit to communicate two classical bits to Bob. To start, Alice and Bob jointly reare their EPR air and as before, Alice takes A and Bob takes B. At this oint, the system state is Φ + = ( 00 + ). (3) Later, when Alice wants to send a two-bit message b b to Bob, she transmits A to Bob, but first she alies some transformations to it. If b =, she alies the hase-fli oeration, and if b =, she alies the bit-fli oeration. These two oerations in matri form are hase-fli = [ 0 0 ], bit-fli = [ 0 0 ]. (4) Deending on which oerations Alice alied, the EPR air will be in one of the four states { } Φ ( 00 ± ), ( 0 ± 0 ). (5) These states are called the Bell states, and they are all orthogonal. Thus Bob merely needs to measure in the aroriate basis (the Bell basis in this case), and he can determine the system state with erfect accuracy. From this, he can infer Alice s message. 3

4 4 Bounds on Quantum Communication Suerdense coding allows us to transmit two bits with a single qubit and an EPR air. It immediately follows that any n bit message can be transmitted with n/ qubits if we allow rior entanglement. Naturally, we are interested in whether we can do better. The answer is no, a reduction by a factor of two is the best quantum communication can do over classical communication. Theorem. Given a quantum communication rotocol that allows Alice to send any message where {0, } n to Bob with robability of correctness, let m AB be the number of qubits Alice sends to Bob, and m BA be the number of qubits Bob sends to Alice, then without rior entanglement and with rior entanglement m AB + m BA n log m AB ( ), (6) [ ( )] n log. (7) We now rove the secial case of one-way communication without rior entanglement. Proof. Given message and some otimal rotocol, Alice will send ψ over the channel to Bob, where ψ consists of m AB qubits. Bob then alies some quantum oeration D on ψ, and then erforms a rojective measurement onto P y for y {0, } n. The robability that Bob gets the correct result is Pr[Bob reads the correct ] = P D ψ. (8) The average robability over all ossible messages would then be n P D ψ. (9) Since ψ lives in a subsace of dimension d m AB, then so does D ψ. Let φ i, i =,,... d be an orthonormal basis for the D ψ subsace, then D ψ = d α,i φ i. (0) Substituting this back into equation (9) and choosing our rojection oerators such that all P s have orthogonal ranges gives d n P α,i φ i, () n n d α,i P φ i, () d α,i P φ i. (3) }{{} This is the robability that Bob receives the message that Alice actually sent. Obviously, this needs to be very close to. 4

5 The last equation is true as P φ i = length of the rojection of φ i φ i =. (4) Thus which leads to d n m AB n, (5) m AB n log ( ). (6) 5 Holevo s Theorem The revious roof can be generalized into a more owerful theorem, but first let us review some information theory terminology. Given some random variable X with range R, we define its classical entroy H to be H(X) ( ) log where = Pr[X = ]. (7) R H has the roerty that 0 H(X) log R (8) where H(X) = log R if X is the uniform distribution and H(X) = 0 if X is comletely deterministic. For a mied state ψ with density oerator ρ, we define the Von Neumann entroy to be S(ρ) H(Probability distribution induced by the eigenvalues of ρ). (9) Finally, we define the mutual information of random variables X and Y to be I(X, Y ) = H(X) + H(Y ) H(X, Y ). (30) Informally, the mutual information says how much information X gives about Y, and vise versa. We can now state Holevo s Theorem, Theorem. Suose Alice has a random variable and sends ρ over the channel. Bob receives ρ and alies some oerations on it to obtain y. Then I(X, Y ) S(ρ) ρ Sρ where ρ = ρ. Let s consider the eamle where X is uniform over {0, } n and we want zero error in our channel. Note that zero error is equivalent to = y in Holevo s Theorem. In this eamle, we can see that I(X, Y ) = log n = n, and S(ρ) = m AB. Putting these into Holevo s Theorem gives n m AB ρ Sρ m AB. (3) Thus, classically, Bob cannot learn more bits of information than the number of qubits Alice transmits. Net lecture, we will eend on quantum communication rotocols where Alice and Bob want to jointly comute a boolean function rather than just transmit a string. 5