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Superdense coding. How much classical information in n qubits?. Observe that 2 n 1 complex numbers apparently needed to describe an arbitrary n -qubit pure quantum state: because 000 000 + 001 001 + 010 010 + + 111 111
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How much classical information in n qubits? • Observe that 2n1 complex numbers apparently needed to describe an arbitrary n-qubit pure quantum state: because 000000 + 001001 + 010010 + + 111111 • 2n is exponential so does this mean that an exponential amount of classical information is somehow stored in n qubits? • Not in an operational sense ... • For example, Holevo’s Theorem (from 1973) implies: • one cannot convey more than n classical bits of information in n qubits
b1 b1 U U ψ b2 ψ b2 b3 b3 n qubits n qubits bn bn 0 bn+1 0 bn+2 m qubits 0 bn+3 0 bn+4 0 bn+m Holevo’s Theorem Easy case: Hard case (the general case): measurement b1b2 ... bncertainly cannot convey more than n bits! We can use only n classical bits The difficult proof is beyond the scope of this course
Superdense coding (prelude) Can we convey two classical bits by sending just one qubit? Suppose that Alice wants to convey two classical bits to Bob sending just one qubit ab Alice Bob ab By Holevo’s Theorem, this is impossible
Superdense coding In superdense coding, Bob is allowed to send a qubit to Alice first ab Alice Bob ab How can this help? The idea is to use entanglement!
if a=1then apply X to qubit if b=1then apply Z to qubit send the qubit back to Bob Alice: How superdense coding works? • Bob creates the state 00+11and sends the first qubit to Alice 2. Alice wants to send two bits a and b No change So let us analyze what Alice sends back to Bob? Alice applies X to first qubit Bell basis • Bob measures the two qubits in the Bell basis To analyze this communication scheme we need to use our known methods for quantum circuit analysis
Bob creates the state 00+11and sends the first qubit to Alice 00 + 11 X Bob does measurement In Bell basis Bob Alice Similarly other cases can be calculated 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 1 1 0 1 0 0 1 x = 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 1 01 10 =
H Measurement in the Bell basis Specifically, Bob applies to his two qubits ... and then measures them, yielding ab Homework or exam
H H Alice 0 0 X Z In this scheme the measurement is only here Bob Creates EPR entanglement Bob measures in Bell Base to his two qubits ... Results of Bell Base measurement and then measures them, yielding ab This concludes superdense coding Observe that Bob knows what Alice has done. This is used in Quantum Games. Bob knows more than Alice if he sends her quantum entangled info.
10 10 01 00 span of 00 and 01 Incomplete measurements • von Neumann Measurement: associated with a partition of the space into mutually orthogonal subspaces When the measurement is performed, the state collapses to each subspace with probability the square of the length of its projection on that subspace
2 2 span of 0 and 1 0 1 Incomplete measurements Measurements up until now are with respect to orthogonal one-dimensional subspaces: The orthogonal subspaces can have other dimensions: (qutrit) Such a measurement on 0 0 +1 1+2 2 (renormalized) results in 00+11 with prob 02+12 2with prob 22
Result is the mixture0000+0101 with prob 002+012 1010+1111 with prob 102+112 Measuring the first qubitof a two-qubit system 0000+0101 +1010 +1111 • Measuring this first qubit is defined as the incomplete measurement with respect to the two dimensional subspaces: • span of 00 & 01 (all states with first qubit 0), and • span of 10 & 11 (all states with first qubit 1)
Easy exercise: show that measuring the first qubit and then measuring the second qubit gives the same result as measuring both qubits at once Hint: continue calculations from last slide Homework or exam
0+1 0+1 Teleportation (prelude) Suppose Alice wishes to convey a qubit to Bob by sending just classical bits If Alice knows and , she can send approximations of them ―but this still requires infinitely many bits for perfect precision Moreover, if Alice does not know or , she can at best acquire one bit about them by a measurement
Teleportation scenario In teleportation, Alice and Bob also start with a Bell state (1/2)(00+11) 0+1 and Alice can send two classical bits to Bob Note that the initial state of the three qubit system is: (1/2)(0+1)(00+11) = (1/2)(000+ 011+ 100+ 111)
teleportation circuit Measurement by Alice in Bell Basis 0+1 Bob H Alice a 00+11 b Bob 0+1 X Z These two qubits are entangled so Alice changing her private qubitchanges the qubit in Bob’s possession Results of Bell Base measurement
Initial state: (0+1)(00+11)(omitting the 1/2 factor) = 000+ 011+ 100+ 111 = = ½ 000 + ½ 000 + ½ 011 +½ 011+½ 100 +½ 100+ ½ 111 +½ 111+ ½ 110 + ½ 001+ ½ 011 + ½ 100….. etc factors = after factorization = = ½(00+ 11)(0 +1) + ½(01+ 10)(1 +0) + ½(00−11)(0 −1) + ½(01−10)(1 −0) You can multiply the lower formula to get the upper formula 16 terms after multiplication. Some cancel This is state of the qubit in Bob possession which is changed by entanglement with changes in Alice circuit These qubits are measured by Alice Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state)
Initial state: (0+1)(00+11)(omitting the 1/2 factor) = 000+ 011+ 100+ 111 = ½(00+ 11)(0 +1) + ½(01+ 10)(1 +0) + ½(00−11)(0 −1) + ½(01−10)(1 −0) How teleportation works This is state of the qubit in Bob possession which is changed by entanglement with changes in Alice circuit These are classical bits that Alice sends to Bob Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state)
½00(0 +1) + ½01(1 +0) +½10(0 − 1) + ½11(1 −0) (00, 0 +1) with prob. ¼ (01, 1 +0) with prob. ¼ (10, 0 −1) with prob. ¼ (11, 1 −0) with prob. ¼ H What Alice does specifically Alice applies to her two qubits, yielding: Before measurement After measurement This is what Alice sends to Bob Then Alice sends her two classical bits to Bob, who then adjusts his qubit to be 0 +1 whatever case occurs
00, 0 +1 01, X(1 +0)=0 +1 10, Z(0 −1) =0 +1 11, ZX(1 −0)= 0 +1 Bob’s adjustment procedure Bob receives two classical bits a, b from Alice, and: if b=1he applies X to qubit if a=1he applies Z to qubit yielding: Note that Bob acquires the correct state in each case
Summary of teleportation: circuit This circuit works correctly regardless the randomness of measurements measurement 0+1 H Alice a 00+11 b Bob 0+1 X Z Homework or exam Quantum circuit exercise: try to work through the details of the analysis of this teleportation protocol
a a a 0 ψ ψ ψ 0 Classical information can be copied a a a 0 What about quantum information? ?
Candidate: works fine for ψ =0and ψ =1 ... but it fails for ψ =(1/2)(0+1) ... ... where it yields output (1/2)(00+11) instead of ψψ =(1/4)(00+01+10+11)
ψ ψ U 0 ψ 0 g No-cloning theorem Theorem: there is no valid quantum operation that maps an arbitrary state ψ to ψψ Proof: Let ψ and ψ′ be two input states, yielding outputs ψψg and ψ′ψ′g′ respectively Since U preserves inner products: ψψ′ = ψψ′ψψ′gg′ so ψψ′(1− ψψ′gg′) = 0 so ψψ′ =0or1 Homework or exam
Source of slides Introduction to Quantum Information ProcessingCS 467 / CS 667Phys 667 / Phys 767C&O 481 / C&O 681 Richard Cleve DC 3524 cleve@cs.uwaterloo.ca Lecture 2 (2005)
Used in 2007, easy to explain, students can do detailed calculations for circuits to analyze them. Good for similar problems.
