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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 ...

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 $\alpha |100\rangle + \beta|011\... 5 Let's represent controlled$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 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 ... 4 This course was \$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 it for \$2250 + \$2250 = \$4500. This is 20 times higher for the same course. I see there would be number of other MOOC coming soon. One ... 4 Note that measuring an observable is equivalent to projecting the quantum state into a particular eigenspace of the operator, and the measurement result tells you which eigenspace. So, in the case of measuring an observable on an eigenstate of that observable, you just project the state onto itself, and the outcome tells you the eigenvalue. So, (2)is ... 4 If you're told about an operation on a single qubit, then to convert it into an operation on both qubits, you just include the identity matrix on the other qubit. So, contrast the single qubit state$|\psi\rangle$going through the bit-flip channel $$|\psi\rangle\rightarrow(1-p)|\psi\rangle\langle\psi|+pX|\psi\rangle\langle\psi|X$$ with what happens on two ... 4 This phenomenon is sometimes known as a discretization of errors. It is a property of certain error correcting codes that allows it to work. It is described (somewhat briefly) in Section 10.2 of Nielsen and Chuang. Suppose that we have an arbitrary error that affects just one qubit, and suppose that we represent this error by a channel$\Phi$mapping one ... 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 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 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 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 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 ... 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 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 Seems like you are interested in the theory of applying results from classical linear coding theory to quantum information theory. When you want to understand how to generalize the Hamming code to a quantum code - the so-called Steane code - you should definitely get familiar to the "stabilizer formalism" introduced by Daniel Gottesman in his Ph.D. thesis ... 3 The reason people focus on X and Z errors is because X and Z are super simple. You can propagate the errors through common gates like CNOT and H without any trouble. A product of X and Z errors on various qubits before a series of Clifford gates is equivalent to a (different) product of X and Z errors on various qubits after the gates. And error correcting ... 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 In their case, their finite set of unitary operators is closed under composition. They even have footnote that emphasizes this. You can't approximate infinite set by some finite subset with the error that is less than half the minimum distance between elements in this finite subset. This is like trying to approximate every real number from [0,1] by some ... 3 Describing the Pauli-X matrix as a rotation by \pi about a particular axis is specifically referring to how you would visualise the action of the gate with regards to the Bloch Sphere picture. This most naturally maps to the density matrix, which is why it's most apparent there. But you can apply it to the pure state picture as well. Note that you can ... 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 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, \... 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 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 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$\...

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Most error correction schemes suppose that the transmitted vector state becomes mixed on the receiver end. For example, the transmitted state $|00\rangle$ can turn to $|11\rangle$ with probability $p^2$, to $|10\rangle$ with prob $p(1-p)$, to $|01\rangle$ with prob $(1-p)p$ and to $|00\rangle$ with prob $(1-p)^2$. So, the state in the end is a mixed state ...

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noise should average itself out. Noise doesn't perfectly average itself out. That's the Gambler's Fallacy. Even though noise tends to meander back and forth, it still accumulates over time. For example, if you generate N fair coin flips and sum them up, the expected magnitude of the difference from exactly $N/2$ heads grows like $O(\sqrt N)$. That's ...

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Is it possible for quantum computers to exhibit behavior similar to flip errors in classical computers where a state |0⟩ becomes |1⟩ due to this error? Yes. It would be a very abrupt error if you're talking about errors on physical qubits. Usually, we'd think of an error as being a little bit of an X rotation (for example). However, the effect of performing ...

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The Quantum Volume is a benchmark for near term, noisy quantum systems. Indeed, like other random unitary benchmarks, you need to be able to sample the ideal distribution. This distribution comes from classical simulations, so your limited to about the ~40 or so qubit limit. However, the Quantum Volume itself was designed to benchmark not only the quantum ...

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Quantum volume is a bad metric for this purpose. For example, suppose you have a ten thousand by ten thousand grid of qubits with a gate error rate of 1 in one thousand. The quantum volume of this grid is basically 0, because if you pick two qubits at random they will on average be more than one thousand steps apart. So an error will almost certainly occur ...

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