Back of the Envelope

There's an interesting article on a superconducting quantum computation scheme here, in the EE Times. Be forewarned, while I'm not too familiar with this particular project, I have read about some of the experiments and I'm fairly knowledgeable about superconducting quantum computation in general. I saw a lot of things in this article that seemed fishy or flat out wrong. I don't think it's dishonesty on the part of the researchers, aside from the usual hype about their research, so much as the author's limited knowledge about quantum computation and superconductivity in general. I'll let you know more once I have a chance to read the papers. Maybe I'll do a fisking of the EE Times article.

This is part of the series of posts which are going up automatically while I'm in the middle of my move. All these posts were written before August 29th, in some cases by as much as two weeks.

Old Post: My last post on quantum computation, where I discussed entanglement, is especially relevant to this.

When you think of teleportation, you probably think of Star Trek's transporter, or perhaps the teleportation device in The Fly. I'd hate to disappoint you, but quantum teleportation is something quite different. Quantum teleportation transfers a qubit state from one qubit in an unknown state to another qubit using entanglement.

First let me explain how remarkable this is. You cannot make an independent copy of a qubit state. This is called the no-cloning theorem, and it simply means that if you have a qubit in an unknown state, a|0>+b|1>, there is no operation which will copy that unknown state to another qubit without entangling the two qubits, so that you really get the state a|00>+b|11>. If you could clone qubit states, then you could take a single qubit, make a million copies, and perform measurements on each of those copies in order to determine the exact state of the first qubit while leaving it in its original state. It would be a way to measure the exact state of a qubit. But with the no-cloning theorem, each of those qubits are entangled to the original one, and measuring any one will collapse the whole system, including the first qubit, and every qubit will be measured with the same value. Thus you can only get the 0 or 1 measurement out of the million copies that you can get out of the single qubit, with the same probability, |a|² for 0 and |b|² for 1. You gain no new information by building the entangled system. But while you cannot copy a qubit state, you can transfer it. Now if you could measure the qubit state, converting it into classical information, then you could obviously transfer the state through a classical data channel--except that you could never have enough bits. While you can measure only 0 or 1 from a qubit, its state, a|0>+b|1>, can have any complex value of a and b as long as |a|²+|b|²=1. That's an infinite number of possible values, so even if you had some way of reading out the value of the original qubit, how could you possibly transfer it to another qubit with a finite number of classical bits, much less two?

So let's start with our qubit in its unknown quantum state, a|0>+b|1>. We'll call this qubit A. We also have two entangled qubits, B and C, which are in state (|00>+|11>)/2^1/2. Recall that entangled qubits can be any distance apart and still retain their entanglement. In this case, we'll put qubit B in close proximity to qubit A, and keep qubit C a mile away. A is the qubit from which we're transferring the state, and C is the qubit to which we're transferring the state. The total state of the system of A, B, and C is

1/2^1/2[a|0>(|00>+|11>) + b|1>(|00>+|11>)].
While B and C are entangled, A is not entangled with them yet. Next, we perform the operation A CNOT B, where B is inverted if and only if A is |1>. This gives the result

1/2^1/2[a|0>(|00>+|11>) + b|1>(|10>+|01>)].
Next, we perform a Hadamard gate on A. A Hadamard rotation converts |0> to 1/2^1/2(|0>+|1>), and converts |1> to 1/2^1/2(|0>-|1>). This gives a state of
1/2[a(|0>+|1>)(|00>+|11>)+b(|0>-|1>)(|10>+|01>)]
You can rewrite this as

1/2[a(|000>+|011>+|100>+|111>)+b(|010>+|001>-|110>-|101>)]
and then regroup it as

1/2[|00>(a|0>+b|1>)+|01>(a|1>+b|0>)+|10>(a|0>-b|1>)+|11>(a|1>-b|0>)]
Now let's say we measure the first two qubits and collapse the system. The state which the third qubit collapses into is not a 0 or 1 state--it's still a superposition. (This is distinct from the situation we were discussing when talking about the no-cloning theorem, since in that situation, the system was in the a|00..0>+b|11..1> state, and collapsing one of them to 0 or 1 collapsed them all to 0 or 1.) Moreover, we know which superposition it's in based on the output of the measurement on A and B. If we send those two classical bits to the person sitting a mile away holding onto qubit C, he knows the state of qubit C relative to the original, unknown state of qubit A, and he can perform a qubit rotation on C in order to put it in the same state A was originally in. Specifically, he applies an X rotation if B is measured in the 1 state, and a Z rotation if A is measured in the 1 state (both if they're both in the 1 state, and neither if they're both in the 0 state).

And so the state of qubit A has been transferred to qubit C by performing operations on A and B, which is entangled to C, then sending two bits of classical information to the person holding qubit C. A and B, meanwhile, have been collapsed to the 0 and 1 states.

NOTE: In writing this post, I used Nielsen and Chuang's Quantum Computation and Quantum Information as a reference. This is a very useful book, if a bit dense in places. And I never found it to be very useful while trying to understand Shor's algorithm.