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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. ...
James Wootton's user avatar
6 votes
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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 ...
JSdJ's user avatar
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5 votes
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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 ...
Craig Gidney's user avatar
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5 votes
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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 ...
Craig Gidney's user avatar
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4 votes
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Does the preparation of magic states need magic gates?

An example of universal quantum computation scheme using magic states can be found in this paper. What is shown there is that if you have access to several copies of noisy magic states, you can purify ...
AG47's user avatar
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4 votes

Magic state distillation of physical vs logical qubit

You would distill at the physical level whenever the fidelity of the output state is lower than the fidelity of your physical operations and storage. For magic states, my guess is you would never do ...
Craig Gidney's user avatar
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4 votes
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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's user avatar
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3 votes

How are these T-distillation circuits derived?

I think there are generally two ingredients you see in state distillation circuits: Compression. The circuit should use T gates in some complicated pattern, but do something simple. For example, a ...
Craig Gidney's user avatar
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3 votes
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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 ...
DaftWullie's user avatar
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3 votes
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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 ...
Norbert Schuch's user avatar
3 votes
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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 ...
DaftWullie's user avatar
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3 votes
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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 \...
Norbert Schuch's user avatar
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 ...
Craig Gidney's user avatar
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2 votes
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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 ...
Craig Gidney's user avatar
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2 votes
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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,...
Craig Gidney's user avatar
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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 ...
Craig Gidney's user avatar
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1 vote

Two Quantum State Density Matrices with Unequal Off-Diagonal Elements but Equal Magnitudes

Generally speaking, two density matrices with equal diagonal correspond to states which are indistinguishable when measured in the computational basis, but which result in different outcome ...
glS's user avatar
  • 25.6k
1 vote

Does the preparation of magic states need magic gates?

Yes, preparing a non-Clifford state (a "magic state") requires a non-Clifford gate (a "magic gate"). I've not heard the term "magic gate" before so it's possible that you ...
Craig Gidney's user avatar
  • 38.8k
1 vote
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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 ...
Craig Gidney's user avatar
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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 ...
Devjyoti Tripathy's user avatar
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's user avatar
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1 vote
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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 ...
Norbert Schuch's user avatar

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