9
votes
Accepted
How does magic state distillation overhead scale compare to quantum advantages?
In the context of scalable quantum computing, the polylog scaling needed for magic state distillation should not be a problem.
Indeed, it is not the only polylog scaling we need to contend with. ...
- 11.2k
5
votes
Accepted
Injecting arbitrary rotations into the surface code
Generally speaking, you would never do a $\pi/2^{10}$ rotation by physical injection and distillation. It would be horrendously expensive. It's far cheaper to use a series of T and H gates to ...
- 33.3k
5
votes
Accepted
Good references to learn magic state distillation for fault tolerance
Disclaimer: these links do not all fit the criterion recent.
First and foremost, it is important to realize that there is no 'one' magic state, and no 'one' magic state distillation.
The term 'magic ...
- 5,369
4
votes
Accepted
15-to-1 distillation protocol: why is the probability of error of the distilled state $35p^3$ and not $\binom{15}{3} p^3=455p^3$?
It's because not every triplet is a failure. Most are detected.
A basic error is for one of the T gates to instead be Z*T. There are 15 T gates. Each of the 15 possible basic errors has a unique ...
- 33.3k
4
votes
Accepted
Why is $\rho$ NPT if and only if $\rho^{\otimes N}$ is NPT?
The short answer is that $(\rho^{\otimes N})^{T_B}=(\rho^{T_B})^{\otimes N}$.
More explicitly, if $\rho=\sum_{ii'jj'}\rho_{ii',jj'}|i\rangle\!\langle i'|\otimes |j\rangle\!\langle j'|$, then we can ...
glS♦
- 23.4k
3
votes
Accepted
Motivation for the definition of k-distillability
Remember that the partial transpose condition is generally good for detecting entanglement, i.e. a bipartite state $\rho$ is certainly entangled if the partial transpose is not non-negative. In other ...
- 55.7k
3
votes
Accepted
What do entanglement cost and distillable entanglement have to do with measuring entanglement?
The entanglement entropy (what you call "von Neumann entropy") is a good measure for entanglement of pure states in the asymptotic setting, i.e. when one is dealing with many copies. However, it is ...
- 5,682
3
votes
Accepted
Entanglement distillation by local operations and post-selection using one entanglement pair
There are two different contexts where the term "entanglement distillation" is used, and are largely incomparable, even if they are conceptually extremely close (and I'm sure you'll be able to find ...
- 55.7k
2
votes
Accepted
Understanding entanglement distillation via stabilizer codes?
In stabilizer circuits, there's an equivalence between a qubit's worldline just sitting around and a Bell pair preparation or measurement linking two qubits. This is the basis of quantum teleportation,...
- 33.3k
2
votes
Accepted
How to prove that the distillable entanglement satisfies $E_D(|\psi_d\rangle\!\langle\psi_d|)\ge \log d$?
The distillable entanglement of this state is equal $\log d$ by definition. Thus, both $E_D(\lvert \psi_d\rangle\langle\psi_d\rvert)\ge \log d$ and $E_D(\lvert \psi_d\rangle\langle\psi_d\rvert)\le \...
- 5,682
2
votes
Under what conditions does entanglement distillation work?
My intuition says to me that, since pairs may be affected by any noise (X and Z i.i.d.), the protocol ends up in a loop where whenever you try to correct an error, you create another one.
This only ...
- 33.3k
2
votes
15-to-1 distillation protocol implemented with lattice surgery: don't we loose transversality by commuting Clifford toward measurements?
We don't actually care about the fact that the T gates are "transversal". That's just a way of explaining why the circuit works. The underlying thing we care about is the fact that replacing ...
- 33.3k
1
vote
Accepted
15-to-1 distillation protocol in lattice surgery: doesn't the potential rejection of the level-1 magic states makes it last longer than $15$ timestep?
Yes, it takes slightly longer on average due to discards. If the injection error rate is 0.1%, then your first stage factory will discard 1-0.999^15 ~= 1.5% of the time. Accounting for this increases ...
- 33.3k
1
vote
What is the intuition behind the following entanglement distillation protocol for continuous variable systems?
To see intuitively why this protocol increases the entanglement after each iteration, we can work out an example where our initial state is say $\lvert 00\rangle+\lvert 11\rangle$. Upon passing ...
1
vote
Accepted
Necessity for discarding errored magic state distillation
The reason you discard them is because this makes distillation much more efficient.
For example, consider 15-to-1 T state distillation using the Reed-Solomon code. This code has distance 3, so you can ...
- 33.3k
1
vote
How are the singlets distilled in Bennett et al.'s 1995 protocol?
$\newcommand{\ket}[1]{|#1\rangle}$I figured out the answer while writing the question. The gist is
Each batch of $n$ two-qubit states, after the measurement, results in a state that is a balanced ...
glS♦
- 23.4k
1
vote
Accepted
How can one impliment Bennett's partial measurement onto a binomial subspace for state distillation?
Alice has to project onto the subspace with some number $n$ of $1$'s in their state $\lvert x_1,x_2,\dots,x_N\rangle$. This can be done by first running a circuit which adds the value of all $x_i$s ...
- 5,682
Only top scored, non community-wiki answers of a minimum length are eligible