 # Superdense coding. How much classical information in n qubits? Observe that 2 n  1 complex numbers apparently needed to describe an arbitrary n -qubit.

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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  2 n 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 qubitsone cannot convey more than n classical bits of information in n qubits

Holevo’s Theorem U ψψ n qubits b1b1 b2b2 b3b3 bnbn Easy case: b 1 b 2... b n certainly cannot convey more than n bits! Hard case (the general case): ψψ n qubits b1b1 b2b2 b3b3 bnbn U 00 00 00 00 00 m qubits bn+1bn+1 bn+2bn+2 bn+3bn+3 bn+4bn+4 bn+mbn+m The difficult proof is beyond the scope of this course measurement We can use only n classical bits Info only here

Superdense coding

Superdense coding (prelude) By Holevo’s Theorem, this is impossible AliceBob ab Suppose that Alice wants to convey two classical bits to Bob sending just one qubit ab Can we convey two classical bits by sending just one qubit?

Superdense coding How can this help? AliceBob ab In superdense coding, Bob is allowed to send a qubit to Alice first ab The idea is to use entanglement!

H H XZ Alice Bob measures in Bell Base Bob Creates EPR entanglement 0 00 0 In this scheme the measuremen t is only here General idea of Superdense Coding can be explained by this distributed quantum circuit

How superdense coding works? the state  00  +  11  first qubit 1.Bob creates the state  00  +  11  and sends the first qubit to Alice Alice: if a = 1 then apply X to qubit if b = 1 then apply Z to qubit send the qubit back to Bob ab state 00  00  +  11  01  00  −  11  10  01  +  10  11  01  −  10  3.Bob measures the two qubits in the Bell basis Bell basis 2. Alice wants to send two bits a and b Alice sends back to Bob So let us analyze what Alice sends back to Bob? No change Alice applies X to first qubit To analyze this communication scheme we need to use our known methods for quantum circuit analysis See last slide

the state  00  +  11  first qubit Bob creates the state  00  +  11  and sends the first qubit to Alice ab state 00  00  +  11  01  00  −  11  10  01  +  10  11  01  −  10  Bob  00  +  11  X Alice changes 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 10011001 = 01100110 Similarly other cases can be calculated Bob does measurement In Bell basis ( ) Three stages of communication Example of calculation the operator executed by Alice We can say that Alice changes one entanglement to another entanglement

Measurement in the Bell basis H Specifically, Bob applies to his two qubits... inputoutput  00  +  11   00   01  +  10   01   00  −  11   10   01  −  10   11  and then measures them, yielding ab Homework or exam

H to his two qubits... inputoutput  00  +  11   00   01  +  10   01   00  −  11   10   01  −  10   11  and then measures them, yielding ab This concludes superdense coding H XZ Alice Bob measures in Bell Base Bob Creates EPR entanglement 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. 0 00 0 In this scheme the measurement is only here Results of Bell Base measurement

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  00  0101  10  span of  00  and  01   10 

Incomplete measurements Measurements up until now are with respect to orthogonal one-dimensional subspaces: 00 11 22 can have other dimensions: The orthogonal subspaces can have other dimensions: span of  0  and  1  22 (qutrit) Such a measurement on  0  0  +  1  1  +  2  2  results in  0  0  +  1  1  with prob  0  2 +  1  2  2  with prob  2  2 (renormalized)

Teleportation

Measuring the first qubit of a two-qubit system Result is the mixture  00  00  +  01  01  with prob  00  2 +  01  2  10  10  +  11  11  with prob  10  2 +  11  2 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)  00  00  +  01  01  +  10  10  +  11  11 

Easy exercise: show that measuring the first qubit and then measuring the second qubit gives the same result as measuring both qubits at once Homework or exam Hint: continue calculations from last slide

Teleportation (prelude) Suppose Alice wishes to convey a qubit to Bob by sending just classical bits  0  +  1  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  0  +  1  ( 1 /  2 )(  00  +  11  ) In teleportation, Alice and Bob also start with a Bell state 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 H XZ  0  +  1   00  +  11   0  +  1  b a Alice Bob Measurement by Alice in Bell Basis These two qubits are entangled so Alice changing her private qubit changes the qubit in Bob’s possession Bob inputoutput  00  +  11   00   01  +  10   01   00  −  11   10   01  −  10   11  Results of Bell Base measurement

(  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  ….. = 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  ) Initial state: Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state) 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 16 terms after multiplication. Some cancel You can multiply the lower formula to get the upper formula

How teleportation works (  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  ) Initial state: Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state) 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

What Alice does specifically Alice appliesto her two qubits, yielding: Then Alice sends her two classical bits to Bob, who then adjusts his qubit to be  0  +  1  whatever case occurs ½  00  (  0  +  1  ) + ½  01  (  1  +  0  ) + ½  10  (  0  −  1  ) + ½  11  (  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 Before measurement After measurement This is what Alice sends to Bob

Bob’s adjustment procedure if b = 1 he applies X to qubit if a = 1 he applies Z to qubit Bob receives two classical bits a, b from Alice, and: 00,  0  +  1  01, X (  1  +  0  ) =  0  +  1  10, Z (  0  −  1  ) =  0  +  1  11, ZX (  1  −  0  ) =  0  +  1  yielding: Note that Bob acquires the correct state in each case

Summary of teleportation: circuit H XZ  0  +  1   00  +  11   0  +  1  b a Alice Bob Quantum circuit exercise: try to work through the details of the analysis of this teleportation protocol Homework or exam measurement This circuit works correctly regardless the randomness of measurements

No-cloning theorem

Classical information can be copied 0 aa a 0 aa a What about quantum information? ψψ 00 ψψ ψψ ?

works fine for  ψ  =  0  and  ψ  =  1 ... but it fails for  ψ  = ( 1 /  2 )(  0  +  1  )...... where it yields output ( 1 /  2 )(  00  +  11  ) instead of  ψ  ψ  = ( 1 / 4 )(  00  +  01  +  10  +  11  ) Candidate:

No-cloning theorem Theorem: there is no valid quantum operation that maps an arbitrary state  ψ  to  ψ  ψ  Proof: Let  ψ  and  ψ′  be two input states, yielding outputs  ψ  ψ  g  and  ψ ′  ψ ′  g ′  respectively Since U preserves inner products:  ψ  ψ ′  =  ψ  ψ ′  ψ  ψ ′  g  g ′  so  ψ  ψ ′  ( 1 −  ψ  ψ ′  g  g ′  ) = 0 so  ψ  ψ ′  = 0 or 1 ψψ 00 00 ψψ ψψ gg U Homework or exam

Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 3524 cleve@cs.uwaterloo.ca Lecture 2 (2005) Source of slides

Used in 2007, easy to explain, students can do detailed calculations for circuits to analyze them. Good for similar problems.

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