Linear algebra 18 Quantum teleportation © Tim Byrnes What is quantum teleportation? ψ The aim of quantum teleportation is to send an unknown quantum state held by Alice to Bob At the end of the proces.
18 Quantum teleportation © Tim Byrnes What is quantum teleportation? The aim of quantum teleportation is to send an unknown quantum state held by Alice to Bob At the end of the process Bob should have the unknown quantum state ψ α +β = Alice and Bob never need to know what the state is Initial state ψ © Tim Byrnes What is quantum teleportation? The aim of quantum teleportation is to send an unknown quantum state held by Alice to Bob At the end of the process Bob should have the unknown quantum state ψ α +β = Alice and Bob never need to know what the state is Final state ψ © Tim Byrnes What quantum teleportation is and is not The atom itself doesn’t disintegrate and get sent somewhere So its not quite like the Star Trek-like teleporter But the quantum state is the most fundamental representation of the state, and if the qubits are atoms, then they are fundamentally identical So if you say the quantum state is the most basic reality, then it is quite close to teleportation From the no cloning theorem, we can never make two copies of ψ But teleportation allows one to destroy a state on one side and create it elsewhere ψ © Tim Byrnes Why is teleportation amazing? There is an emphasis on the fact that Alice and Bob not know the state Why is this important? If Alice knew what the state was, she could just tell Bob what the state is, and Bob could make that state No need for teleportation Hi Bob My state is = α i= / 2, β Sure thing, I’ll it right away 3/2 Can you make it there? = ψ α +β © Tim Byrnes Measuring an unknown state Even if Alice didn’t know her state initially, couldn’t she just figure this out and tell Bob? With only one copy of the state, it is impossible to figure out an unknown state There are an infinity of basis choices that could be made, and since the measurement outcome is random, this doesn’t give much information e.g Measuring = ψ α +β gives the outcome All this says for sure is that I measured it and I got Not sure what the state is but why don’t you just make that? β ≠0 Errrrmm Okay sure I guess ψ © Tim Byrnes Theoretical limit to estimate state It was shown that even with the fanciest possible measurements, the best estimate you can get of N unknown copies of a state is fidelity = F N +1 = ψ ψ est N +2 For one qubit N=1, the best fidelity is F=2/3=0.666 If Alice doesn’t know her state its impossible to just tell Bob exactly what it is with one copy Massar & Popescu Phys Rev Lett 74, 1259 (1995) My best guess of the state is ψ Ok let’s go with that est ψ ψ ψ © Tim Byrnes The teleportation circuit The teleportation circuit allows you to overcome this and (under ideal conditions) get perfect transfer of the state! Alice Bob ψ ψ H H X 0 Z ψ © Tim Byrnes The teleportation circuit The main steps of the teleportation circuit are Measurement in Bell basis Alice Bob ψ H H X Entanglement generation Z ψ Classical correction © Tim Byrnes Part 1: Entanglement generation The first part of the circuit produces an entangled state T0 Working through the circuit step by step we have Time T0: T0 = ψ 0 © Tim Byrnes Part 1: Entanglement generation T1 Apply a Hadamard on qubit 2: Time T1: (0 +1 T H= = ψ T0 ψ ( 00 + 10 )0 ) 1 H= 1 −1 H = + H = − © Tim Byrnes Part 1: Entanglement generation Apply a CNOT between qubits and T2 Time T2: ( ctrl q= 2) ( ctrl q 2) = CNOT CNOT T2 ψ U= T1 U ( 00 + 11 ψ ( 00 + 10 ) ) This is a Bell state (entangled state on qubits & 3) ψ ( 00 + 11 ) © Tim Byrnes Part 2: Measurement in Bell basis The next part of the circuit is an example of what was seen in Lecture 14 A unitary in front of a measurement makes a measurement in a different basis M gen n = n bn The basis is in this case 1 bn =( 00 ± 11 ) , ( 01 ± 10 2 ) © Tim Byrnes Part 2: Measurement in Bell basis We can also just work this out step by step as before = ( ctrl q= 1) ( ctrl q1) CNOTq CNOTq T3 U= T2 U U ( ctrl = q1) CNOTq = =U ( ctrl = q1) CNOTq (α +β ) ψ ( 00 + 11 ( 00 + 11 ) ) (α 000 + α 011 + β 100 + β 111 (α 000 + α 011 + β 110 + β 101 ) 1 00 11 + + α β ( 10 + 01 ( ) 2 ) T3 ) © Tim Byrnes Part 2: Measurement in Bell basis Applying the Hadmard T = H1 T 1 H1 α ( 00 + 11 ) + β ( 10 + 01 ) 2 = = α ( α +1 000 ( )( 00 + 11 ) + β ( −1 + 011 + 100 + 111 ) + β )( 10 010 ( + 01 ) + 001 − 110 − 101 T4 ) © Tim Byrnes Part 2: Measurement in Bell basis Now we measure the qubits in the , basis Here it is easiest to write the state before the measurement in terms of the four measurement outcomes 00 , 01 , 10 , 11 T4 T 4= α 000 ( + 011 + 100 + 111 ) + β 010 ( + 001 − 110 − 101 ) 1 1 00 (α + β ) + 01 (α + β ) + 10 (α − β ) + 11 (α − β 2 2 ) © Tim Byrnes Part 2: Measurement in Bell basis 1 00 (α + β ) + 01 (α + β 2 1 + 10 (α − β ) + 11 (α − β ) 2 T4 = Measuring the state randomly collapses the state to T4 ) T5 T5 00 (α + β ) p00 = 1/ 01 (α + β ) p01 = 1/ 10 (α − β ) p10 = 1/ Can work out probabilities using the measurement operators 11 (α − β ) p11 = 1/ M n = 00 00 , 01 01 , 10 10 , 11 11 © Tim Byrnes Part 3: Classical correction After the measurement we have T5 00 (α + β ) 01 (α + β ) T = 10 (α − β ) 11 (α − β ) p00 = 1/ p01 = 1/ p10 = 1/ p11 = 1/ We can see already that the teleportation is nearly there For the 00 outcome it actually is already working, but for the other cases Bob gets a state that is almost the state, but with an extra bit flip (01 case) or a phase flip (10 case), or both (11 case) We can “fix up” the state by removing the extra bit flip or phase flip! Since Alice has in her possession the outcomes, she just needs to tell Bob which measurement she got, and the teleportation will work for all cases! © Tim Byrnes Part 3: Classical correction So for example, if Alice gets the state 01 00 (α + β ) 01 (α + β ) T = 10 (α − β ) 11 (α − β ) Oh ok, so I’ll apply a Z gate and then I’ll have your state Hi Bob I just measured my qubits and I got 10 p00 = 1/ p01 = 1/ p10 = 1/ p11 = 1/ α −β © Tim Byrnes Part 3: Classical correction After the measurement we have 00 (α + β ) 01 (α + β ) T = 10 (α − β ) 11 (α − β ) T5 p00 = 1/ p01 = 1/ p10 = 1/ p11 = 1/ Alice’s outcome qubit Alice’s outcome qubit Bob’s state Classical correction Bob’s state after correction α +β 0 α +β I (do nothing) α +β X α +β 1 α −β Z α +β 1 α −β ZX α +β Bob gets Alice’s state for every case! © Tim Byrnes