It’s not a very good name, maybe we should call it “sending qubits over a classical channel using a quantum one-time pad”. The idea is that Alice and Bob are far apart from each other and have a prepared pair of qubits, and Alice also has some qubit in unknown state. She can measure it in a clever way and send some classical bits to Bob, which allows him to recreate the unknown qubit. Both qubits of the prepared pair are spent in the process.
Let’s say the unknown qubit is a|0>+b|1>, and the prepared pair is (|00> + |11>)/√2. So the whole state is (a|0> + b|1>) ⊗ (|00> + |11>)/√2 = (a|000> + a|011> + b|100> + b|111>)/√2. Alice can access the first two qubits and Bob can access the third.
Alice flips the second qubit depending on the first, leading to (a|000> + a|011> + b|110> + b|101>)/√2.
Alice applies a combined reflection and rotation to the first qubit, which is known as Hadamard gate: |0> becomes (|0> + |1>)/√2, while |1> becomes (|0> - |1>)/√2. So the whole state becomes (a|000> + a|100> + a|011> + a|111> + b|010> - b|110> + b|001> - b|101>)/2.
We can rewrite that as ((|00> ⊗ (a|0> + b|1>)) + (|01> ⊗ (a|1> + b|0>)) + (|10> ⊗ (a|0> - b|1>)) + (|11> ⊗ (a|1> - b|0>)))/2.
Notice something funny? In the above state, measuring the first two qubits gives you exactly enough information to know which operation to apply to the third qubit to get a|0> + b|1>. So if Alice measures both her qubits and sends the two classical bits to Bob, he can reconstruct the unknown qubit on his end. And if Carol eavesdrops on the message without having access to Bob’s qubit, she won’t learn anything, because the four possible messages have probability 1⁄4 each, regardless of a and b.
If the unknown qubit was entangled with something else, that also survives the transfer, though proving that requires a bit more calculation.
(10/?)
Quantum teleportation.
It’s not a very good name, maybe we should call it “sending qubits over a classical channel using a quantum one-time pad”. The idea is that Alice and Bob are far apart from each other and have a prepared pair of qubits, and Alice also has some qubit in unknown state. She can measure it in a clever way and send some classical bits to Bob, which allows him to recreate the unknown qubit. Both qubits of the prepared pair are spent in the process.
Let’s say the unknown qubit is a|0>+b|1>, and the prepared pair is (|00> + |11>)/√2. So the whole state is (a|0> + b|1>) ⊗ (|00> + |11>)/√2 = (a|000> + a|011> + b|100> + b|111>)/√2. Alice can access the first two qubits and Bob can access the third.
Alice flips the second qubit depending on the first, leading to (a|000> + a|011> + b|110> + b|101>)/√2.
Alice applies a combined reflection and rotation to the first qubit, which is known as Hadamard gate: |0> becomes (|0> + |1>)/√2, while |1> becomes (|0> - |1>)/√2. So the whole state becomes (a|000> + a|100> + a|011> + a|111> + b|010> - b|110> + b|001> - b|101>)/2.
We can rewrite that as ((|00> ⊗ (a|0> + b|1>)) + (|01> ⊗ (a|1> + b|0>)) + (|10> ⊗ (a|0> - b|1>)) + (|11> ⊗ (a|1> - b|0>)))/2.
Notice something funny? In the above state, measuring the first two qubits gives you exactly enough information to know which operation to apply to the third qubit to get a|0> + b|1>. So if Alice measures both her qubits and sends the two classical bits to Bob, he can reconstruct the unknown qubit on his end. And if Carol eavesdrops on the message without having access to Bob’s qubit, she won’t learn anything, because the four possible messages have probability 1⁄4 each, regardless of a and b.
If the unknown qubit was entangled with something else, that also survives the transfer, though proving that requires a bit more calculation.