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We can try to geometrically interpret the Knill–Laflamme conditions on a code-space $\mathcal C$, as follows.$\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}$ Images of the code-space under an error operation First, consider any individual error operator $E_j$: this is a Kraus operator of some transformation of the system, and so is not ...

6

There are various ways that you might go about computing the distance. I'll give a fairly general strategy here, although I'm sure here are imprvements that can be made. Your starting point is a set of stabilizers $\{K_n\}$ on $N$ qubits, satisfying $K_n^2=I$ and $[K_n,K_m]=0$. Generically, you want to consider the full set of $4^N$ possible tensor products ...

6

The errors that are described by the Master equation are continuous errors. The action of error correction is to discretize those errors. For example, dephasing noise can be described by the Master equation. The net effect is that an initial state $\rho$ is transformed into $$\rho\mapsto (1-p)\rho+pZ\rho Z,$$ where $p$ is a function of time. However, ...

5

Let's represent controlled $Z_1Z_2$ gate in the projector formalism, as described in this answer: $$C_AZ_1Z_2 = |0\rangle\langle0|_A I_1I_2 + |1\rangle\langle1|_A Z_1Z_2$$ This just tells you to apply identity gates to qubits 1 and 2 if the ancilla is in the $|0\rangle$ state and to apply Z gates to qubits 1 and 2 if the ancilla is in the $|1\rangle$ ...

5

Let me begin by reformulating the Knill-Laflamme Theorem, looking at it from a different angle might help in making the matter a little less abstract. The Knill-Laflamme Theorem for the conditions of quantum error correction can be stated as follows: Let $C$ be a quantum error correcting code defined as a subspace of the n-dimensional Hilbert space, $\... 5 You quoted the first part of the exercise; your question is exactly the second part of the exercise :-) Exercise 10.4 (2) asks you to show that the described recovery procedure (the one which projects the state on one of the 8 basis states) works only for computational basis states. Indeed, if you have a corrupted state$\alpha |100\rangle + \beta|011\...

5

Quantum error correction concerns errors that happen on qubits; it does not provide any protection against errors in operations on those qubits. Note however, that an error on an operation can be seen as the perfect operation plus some error 'on the qubit'. It is, however, the case that, without any precaution, the added operations introduced by error ...

4

If you only have one $E_k$ (i.e., $k=1$ can only take one value), and this $E_k=U$ is unitary, then - as you point out in the comments - the first condition is always satisfied, and the error can be corrected. However, this also means that your "error" is the deterministic application of $U$. So after applying the "error" map, you just have to undo $U$, ...

4

I am wondering how could one implement one of such negative controlled gates by using a standard controlled gate, that is, add some other gates to the controlled unitary so that its operation is reversed. This is addressed in Nielsen and Chuang's section 4.3 (~ p. 184, 10th edition). Basically,$$c_{|0\rangle}U_{AB} \equiv (X_A\otimes I_B)c_{|1\rangle}U_{AB}... 4 In Circuit Composer, you can use an if statement as shown below The circuit measures qubit 1 after a gate operation and, if it is 0, then applies an X gate to qubit 2: OPENQASM 2.0; include "qelib1.inc"; qreg q[3]; creg c[3]; x q[0]; measure q[0] -> c[0]; if(c==0) x q[1]; # here where you can set it to trigger on c == 1 measure q[1] -> c[1]; This ... 4 The model's accuracy is purely empirical observation. The error trend (Fig 4, or 41:50 in the video) demonstrates that the error of the system (cross entropy fidelity with respect to simulated results) is tracked closely by the "high school probability" model he mentions. The way this basic model would work is to assume 1- and 2-qubit gate errors are ... 4 If you write Z^aX^b, there's an implicit "add a phase i to make it Hermitian if necessary", although I guess there are a couple of different conventions you might use the determine the sign used. So long as you're clear about the convention it doesn't really matter because you've got the extra \pm 1 freedom to add in to adjust for it. As for an ... 4 Consulting my local copy of Nielsen & Chuang (10th anniversary edition, p. 457), the complete statement of the exercise is pretty much exactly as you have given it: Exercise 10.34. Let \langle g_1, \ldots, g_l \rangle. Show that -I is not an element of S if and only if g_j^2 = I for all j, and g_j \ne - I for all j. Reading the ... 4 You can actually, most of the time, prepare an encoded qubit state by using the stabilizers; you can also do this for the bit-flip code. I have intentionally left the above sentence a little vague, because there are some very important subtleties. To really understand these subtleties, a good understanding of stabilizer codes is strongly preferred. If you ... 4 Operator applied from t_{0} to t_{1} is CNOT \otimes I, i.e. U_1= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 &... 4 frequency (GHz): The frequency(energy) associated with the transition between the qubit's ground state (|0\rangle) and first excited state (|1\rangle). readout error: The probability of preparing a |0\rangle(|1\rangle) and measuring a |1\rangle(|0\rangle), ie., of having an error in your readout single qubit U2 error rate: The average error per ... 3 This question applies equally to classical codes, and is perhaps easier to understand in that context. Consider, for example, the 5-bit repetition code. The code words are$$ 00000\qquad 11111  Clearly, the distance $d=5$ because it takes 5 bit flips to convert from one code word to the other. Now, let's imagine some errors have happened. We look at our ...

3

I believe that this is actually two separate questions; I'll try to explain the issue concerning errors as channels with multiple Kraus operators instead of unitaries first: You are correct in saying that errors, in general, are not unitary operations. Rather, they are quantum channels that most often have more than 1 Kraus operators. Consider, for instance,...

3

I would say the most important classes are concatenated codes, topological codes and LDPC codes. But it depends on how one defines 'classes', as one could also talk of CSS vs non-CSS, stabilizer vs subsystem. In any case, I'll explain my three chosen 'classes'. Concatenated codes are used in cases where we know a code that has a finite code distance (and ...

3

Both circuits start by turning $|\psi\rangle = a |0\rangle + b|1\rangle$ into $|\psi\rangle = a |000\rangle + b|111\rangle$, copying the 0/1 information onto two additional qubits. Note that, for any pair of qubits in this state, either both are in state $|0\rangle$, or both are in state $|1\rangle$. Any superposition must be expressed only in terms of ...

3

Content wise very similar course but different name Quantum Information Science I (three parts - part 1, 2, and 3) and Quantum Information Science II were $49 per course as verified certified learning outcome - a series of 3 courses + 1 more extra course on edX. Now the course has been taken down; at least no more new enrollments, and MITx Pro is offering ... 3 Here is a way to think about the role of entanglement in quantum codes which I think is complementary to Felix Hubers response. Suppose that we take a maximally entangled state$|\Psi\rangle_{RQ}$and record the$Q$system into some quantum error-correcting code. Suppose the code records$Q$into subsystems$S_1,S_2,S_3$such that erasure of any one ... 3 There are two methods, when you look at the code you'll see their names: pseudo_inverse and least_squares (https://github.com/Qiskit/qiskit-ignis/blob/master/qiskit/ignis/mitigation/measurement/filters.py). pseudo_inverse is the simpler one, it applies the inverse of the calibration matrix on the measurement results. However it has some issues, which require ... 3 1) There are 4 Bell states, namely the ones you listed divided by$\sqrt{2}$. There is no "the bell state". The Bell states are only defined for 2 qubits, so there is no "higher dimensional definition of a Bell state". One of the key features of the Bell states is that they're maximally entangled. If this is what you'd like in a higher dimensional analog of ... 3 I am not following all of the calculations in your post (for one thing because your first displayed equation does not mention$C_1$), but I know why the goal of the exercise is true. In fact it is true for any code at all, not just a CSS code. If$\mathcal{C} \subseteq \mathcal{H}_\text{qubit}^{\otimes n}$is any code, then you get an equivalent code$\...

3

Before starting, I should probably emphasise that, although useful for the practice of working through the maths of quantum error correction on a relatively simple case, amplitude damping combined with the repetition code is a really bad thing to be thinking about. This is because, if there's an error, and they you apply a syndrome measurement, so that you ...

3

AFAIK, what you want can't be run on the hardware right now. See this github issue. However, you can do this in the simulators. If, for example, c[0] and c[1] make up a two-bit classical register c, you can do this: qc.x(q[0]).c_if(c,3) qc.x(q[1]).c_if(c,3)

3

For $|0\rangle$, $|1\rangle$, $|+\rangle$, and $|-\rangle$ you do transversal initialization (initialize all physical qubits to the desired state, then turn on the stabilizers). For $|i\rangle$ and $|-i\rangle$ you do topological initialization using twists. For $T|+\rangle$ states and other states with nice state distillation protocols, you do noisy low ...

3

Each vertex has a physical data qubit. But what exactly do the operators (green circles) represent? Is there an X and Z operator on every vertex site, or only some of them (Figure 1b)? The circles on, e.g. the bottom-left of Fig. 1, show you how to describe each of the stabilizers. So, for every mustard yellow square in the top-left diagram, there is a ...

3

an X error on the top left qubit and the qubit beneath it would be undetectable That error would be detected by the flipping of the four body Z stabilizer adjacent to the lower qubit you operated on: If you or I got X and Z mixed up, then the error is undetectable, but it corresponds to a topologically trivial cycle from a boundary to itself, so it has no ...

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