New answers tagged

1

The other answer already uses this, but just to make the general fact more explicit: if $\mathcal E=\mathcal E_A\circ\mathcal E_B$, that is, $\mathcal E(\rho)=\mathcal E_A(\mathcal E_B(\rho))$, and the Kraus decompositions of the single channels read $$\mathcal E_A(\rho)=\sum_a A_a\rho A_a^\dagger, \qquad \mathcal E_B(\rho)=\sum_b B_b\rho B_b^\dagger,$$ then ...


4

You can obtain the Kraus operators of the combined channel by taking products of the Kraus operators of the individual channels (using the notation from the paper you linked): Amplitude damping: $E^{AD}_1 = \begin{bmatrix} 1 & 0 \\ 0 & \sqrt{1-p_{AD}} \end{bmatrix}$, $E^{AD}_2 = \begin{bmatrix} 0 & \sqrt{p_{AD}} \\ 0 & 0 \end{bmatrix}$ Phase ...


2

By part 1, we have that for any $\delta > 0$, then for sufficiently large $n$, $tr( \rho ^ {\otimes n} P(n, \epsilon)) \geq 1 - \delta$. This means that $tr( \rho ^ {\otimes n} P(n, \epsilon)) \rightarrow 1$ as $n \rightarrow \infty$, since it is at most 1.


0

The distance of an error correcting code is the smallest number of single-qubit rotations that you have to apply to map one logical codeword into an orthogonal one. If I express it in terms of the stabilizers of the code, it's the smallest tensor product of single-qubit unitaries (usually Paulis) that commutes with all the generators. Let's see how this ...


0

As a supplement to the previous answer: No need to compute inner products ... It is well known that any $n$-qubit stabiliser state is of the form $|\psi\rangle = |K|^{-1/2} \sum_{x\in K} i^{d\cdot x} (-1)^{q(x)+b\cdot x} |x\rangle$ where $K\subset \mathbb{F}_2^{n}$ is an affine subspace, $b,d\in \mathbb{F}_2^{n}$ are vectors and $q$ is a quadratic form (...


2

Let's make a few observations first: Since an $N$-qubit stabilizer state can be generated starting from $| 0 \rangle^{\otimes N}$ and applying H, CNOT, and S gates, we make the following observations. By simply applying the Hadamard on all qubits, one can generate the state $ | + \rangle^{\otimes N}$ which is maximally coherent (under the free operations ...


1

So the biggest number used when comparing families of QECCs is the threshold, which is the error rate (generally depolarizing noise or XZ noise, depends on the paper) for which increasing the size (and distance) of the code actually increases performance. Look at figure 4 in this paper. On the top plot, you can see that above a certain error rate, the extra ...


1

If you focus solely on stabilizer/additive codes, I believe that the weight of the Paulis in the stabilizer is highly important. The weight of a $n$-qubit Pauli is the number of non-trivial factors in it. The weight of the correctable errors has a close correspondence with the distance of a code, but the weight of the elements of the stabilizer is also ...


1

Note that if you don't keep track of the phase $i$ in $Y = iXZ$, so letting $\hat{Y} = XZ$, then something 'weird' happens: $$ \hat{Y}\hat{Y} = XZXZ = X(-XZ)Z = -XXZZ = -I. $$ This is not just an oddity. Any stabilizer code with a generator $G_{1}$ containing an (odd number of) $Y$ will now not be possible, because: $$ G_{1}G_{1} = -I \in \mathcal{S}. $$ But ...


1

This paper uses $H$, $Z$ and $C_Z$ to generate the real Clifford group. You can replace $C_Z$ by $C_X$, but you will need the same number of generators.


1

You might also want to check out Quaec, documentation here, which is: QuaEC is a library for working with quantum error correction, including support for efficiently maniuplating Pauli and Clifford operators. The use of this would pretty much be as-is, since it is not actively (or even passively I believe) developed anymore.


1

You might find some examples in this two-year-old question :) To the best of my knowledge, the most recent work that implements some code on IBMQ's quantum devices is about the repetition code (see the textbook or the paper). If you only want to do the simulation, there should be no problem to take a further step towards more advanced codes. But if you mean ...


2

You can think of this measurement as an 'error' on the (encoded) state that needs to be corrected. Quantum error correction is all about subspaces of the Hilbert space, and during QECC we are always trying achieve information in what subspace our state lies. The state lies in some subspace, which is either the codespace or some orthogonal space. With every ...


1

For quantum circuits: So the main answer is that far-term quantum computers will implement Quantum Error Correction, where each logical qubit is composed out of a number of physical qubits and the information is stored in a complex entangled state. Errors on these physical qubits can be caught and corrected, and these error correction steps will occur ...


Top 50 recent answers are included