Quantum Computing Asked by Dina Abdelhadi on March 17, 2021
This paper discusses strictly contractive channels, i.e. channels that strictly decrease the trace distance between any two input quantum states.
It is shown that if a quantum circuit is composed of rounds of gates followed by strictly contractive channels then the trace distance between any two input states would decay exponentially with circuit depth, which means we would not be able to distinguish outputs corresponding to any two different inputs.
In light of this, how is it even possible to build any kind of fault-tolerant circuit which is capable of "arbitrarily long" computations, given that the standard noise model, the depolarizing channel, is strictly contractive?
This is a very interesting question. Indeed, CP maps - and this includes the operations used in the error correction (measurement and subsequent unitaries) - will always decrease the trace norm.
The answer is that if you take a (strictly) contractive map on, say, a qubit, and consider how it acts if you apply it to many qubits, there will always be some subspace where the map is much less contractive - still contractive, but supressed exponentially. The art of (quantum) error correction is to encode the information in those subspaces, and to "re-focus" it into those subspaces (which is the actual error correction procedure).
Let me give a (slightly oversimplified) example.
Consider a noise which flips a bit with probability $epsilon=0.01$ (per unit time, if you wish).
Now encode a (classical) bit in $N$ zeros or $N$ ones, $0cdots 0$ and $1cdots 1$. Then, these states (seen as quantum states, if you wish, or as probability distributions) will keep a trace distance on the order of at most $epsilon^{-N/2}$ -- after all, you have to flip half the bits in either to get any overlap between them (and if they are orthogonal, the trace distance stays 1).
So what happened? Before, the noise was $epsilon=0.01$. Now, the noise is $epsilon=0.01^{-N/2}$. So for $N=10$, you might be able to go about $10^{10}$ time steps, rather than $100$.
Of course, this will break down if you let more time pass -- so what you have to do after a short interval of time is to "re-focus" your information, that is, move it back to that subspace which is best protected (like all zeros and all ones). This is what error correction does. This is a CP map and does not increase distinguishability, but it will allow you to stay with the best error rate of $1$ error in $10^{10}$.
(Note: Clearly, this is not a way to safely encode quantum information - this is not what this example is supposed to illustrate.)
Correct answer by Norbert Schuch on March 17, 2021
So that's where quantum error correction comes in. By measuring the stabilizers, the code is projected back into a pure state. I'll give an example using a Bell state:
Imagine the state $$frac{1}{sqrt{2}}(|00rangle + |11rangle),$$
which is stabilized by XX and ZZ (meaning we can catch errors by verifying that these two operators have eigenvalue +1 on our state). Now imagine that the state undergoes a stochastic X channel on the second qubit, described by:
$$E(rho) = (1-p)Irho I + p X_2rho X_2.$$
The overall state will now become a mixed state described by: $$rho = frac{1}{2}[(1-p)(|00rangle + |11rangle)(langle 00| + langle 11|) + p(|01rangle + |10rangle)(langle 01| + langle 10|)].$$ This state is a mixed state with trace less than 1, since the error channel is non-unitary.
When we measure the stabilizers XX and ZZ, we either get [+1, +1] with probability $1+p$, meaning that we have projected ourselves back into the state $frac{1}{sqrt{2}}(|00rangle + |11rangle)$, or with probability $p$ we get [+1, -1], meaning the state is $frac{1}{sqrt{2}}(|01rangle + |10rangle)$. Both of these new states are again pure states, and the goal of error correction is to be able to figure out what errors we have projected onto the state, so we can undo it and recover our state. The key point is that by measuring stabilizers, error correction naturally relies on a non-unitary process in order to deal with these trace reduction concerns.
Answered by Dripto Debroy on March 17, 2021
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