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Repetition codes are great for helping to demonstrate some understanding of how things work. But they are not good quantum codes. In fact this is part of the reason why they are so good for demonstrating some of the ideas - they help bridge the gap between the classical intuition that we're used to and the quantum world which often feels a bit less ...


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To understand the answer of this question you have to look into no cloning theorem which states that you can't make a copy of an unknown quantum state. e.g. $\alpha|0> + \beta|1>$ where you don't exactly know the values of $\alpha$ and $\beta$. but you can make copy of states like $|0>$, $|1>$ or $\frac{1}{\sqrt2}(|0>+|1>)$. So unlike ...


1

If you run the following code you will see the output statevector plotted in Bloch sphere. qc = QuantumCircuit(1) qc.ry(3 * np.pi/4, 0) sim = Aer.get_backend('statevector_simulator') result = execute(qc, sim).result() output_state = result.get_statevector(qc) plot_bloch_multivector(output_state) Now from the picture you can see that the state lies near ...


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It's impossible to do it in polylogarithmic depth, because for a modular multiplication (or even just an increment!) the output value of the most significant bit is a function of every other input bit. One spot is affected by all the other spots. So your runtime has to be at least as long as the longest path between that spot and any one of the other used ...


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Yes, probabilities shown in a histogram are squares of absolute values of coefficients in a state vector (i.e. probability amplitudes). Note that the amplitudes are generally complex numbers, hence we square the absolute values and not the amplitudes themselves.


2

Applying a reset to a qubit is equivalent to measuring it, and then applying a bit flip to it conditioned on the measurement result. def reset(qubit): if measure(qubit) == ON: X(qubit) For example, in this Quirk circuit, you can see that the post-reset state matches the state you'd get when conditioning on a measurement-via-ancilla+bit-flip of ...


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Bob does not know the outcome of Alice's measurement. All he knows is that Alice would have obtained $|\psi\rangle$ with probability $\frac12$ and $|\psi^\perp\rangle$ with probability $\frac12$ for some orthonormal basis $|\psi\rangle$, $|\psi^\perp\rangle$. Therefore, his state is a mixture $$ \rho_B = \frac12|\overline\psi\rangle\langle\overline\psi|+\...


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The term "non-local states" most likely refers to states which allow to violate a Bell inequality. They give rise to statistics that cannot be explained with a local hidden variable model.


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The claim does not specify what protocols for distinguishing quantum states are acceptable. In particular, it does not state whether we are allowed to err or reserve judgment. Below, we note success probability for protocols allowed to err and compute success probability for an error-free protocol. The success probability of the former is not bounded by $4\...


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An easy (but perhaps cheating) answer: The relative entropy $S(A||B)$diverges when $A$ has support over the kernel of $B$ (e.g., wikipedia). Now, the kernel of $B=|0\rangle^{\otimes n}\langle 0|^{\otimes n}$ is everything other than $|0\rangle^{\otimes n}$, i.e., $\mathbb{I}-|0\rangle^{\otimes n}\langle 0|^{\otimes n}$! For the inequality to be meaningful, ...


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You don't state this explicitly, but I'm guessing this is the crucial part: How do $|A_i\rangle$ relate to $A$? I assume that $|A_i\rangle$ correspond to normalized rows, $i$ of matrix $A$, while $\|A_i\|$ is the weight of the row $i$ such that $\|A_i\||A_i\rangle$ would have all the elements corresponding to the $i^{th}$ row of matrix $A$. In other words, $$...


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There's a very quick proof if you can use the properties of the Choi-Jamiołkowski isomorphism. Define a map that acts on subsystem $B$ as $$\Lambda(\rho_B) = \mathrm{Tr}(\rho_B) |B| I_B - \rho_B.$$ The Choi operator of this map is $J(\Lambda)_{BB'} = |B| I_{BB'} - |B| \Phi^+_{BB}$, where $\Phi^+$ is the maximally entangled state. It follows that $J(\Lambda)\...


3

Let $\mathcal{H}_A$ denote the Hilbert space of qubits in partition $A$ and similarly for $\mathcal{H}_\bar{A}$. Define the operator $P:=Q|0^n\rangle\langle 0^n|Q^\dagger$ and write its Schmidt decomposition $$ P = \sum_a R_a\otimes Q_a\tag1 $$ where $R_a$ are operators on $\mathcal{H}_A$ and $Q_a$ are operators on $\mathcal{H}_\bar{A}$. See for example the ...


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Yes, the reasoning is correct. In fact, it can be generalized beyond pure states. By definition, every mixed quantum state $\rho$ is a positive semidefinite operator with unit trace. Since every positive semidefinite operator is Hermitian, we may interpret $\rho$ as an observable. In this case, the expectation of observable $\rho$ in state $\sigma$ $$ \...


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(Case of pure states) Let $\rho=|\psi\rangle\!\langle\psi|$ be a pure bipartite state, suppose the underlying space is $\mathbb C^n\otimes\mathbb C^m$, and write as $|\psi\rangle=\sum_{k=1}^r \sqrt{p_k}(|u_k\rangle\otimes|v_k\rangle)$ the Schmidt decomposition of $|\psi\rangle$, with $r\le \min(m,n)$. Then $\rho$ has a single nonzero eigenvalue equal to $1$, ...


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A single qubit is indeed different from a classical coin in a probabilistic state, for the reasons that the other answers have articulated. But at a deeper level, you are correct: a single qubit can indeed be thought of as a classical system, because there exists a local hidden-variable model that exactly reproduces its statistics. The only important ...


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Summary: The expression you're looking for is: $$ \frac{1}{4} \left[ (III + IZZ + ZIZ + ZZI) + (XXX - XYY - YXY - YYX)\right] $$ where Pauli string notation like $XYX$ denotes $\sigma_1 \otimes \sigma_2 \otimes \sigma_1$, for example. To start, we need to write the $n$-qubit GHZ state as an operator, namely \begin{equation} |\psi^{(n)}\rangle\langle\psi^{(n)...


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A direct while brute-force way is by writing state $|\psi\rangle$ as density matrix $\rho$. Then by noticing that $\rho=\frac{1}{2^3}\sum_{ijk}t_{ijk}\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{k}$ for $i,j,k=1,2,3,4$ stands for Pauli matrices and identity matrix, find all the coefficients $t_{ijk}$ by $Tr(\rho\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{k})$. If ...


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We have, $$\begin{align}\begin{aligned}\newcommand{\th}{\frac{\theta}{2}}\\\begin{split}Ry(\theta) = \exp\Big(-i \th Y\Big) = \begin{pmatrix} \cos{\th} & -\sin{\th} \\ \sin{\th} & \cos{\th} \end{pmatrix}\end{split}\end{aligned}\end{align}$$ If the input state is $|0\rangle$, then the probability of getting $|0\rangle$ as a ...


3

In order to apply Nyquist-Shannon sampling theory, we need to know the maximum frequency that will be present in the signal we intend to measure. We will do this by rewriting a time-dependent observable in terms of the frequencies present in the Hamiltonian $H$. Consider an arbitrary observable $O$ which will time evolve under application of $H$ in the ...


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In Quirk, you look at state vectors of qubits using the amplitude display. You can get conditional state vectors, such as the state vector implied by a particular measurement result, by combining the amplitude display with a control on the measurement bit. Here is an example: And another example:


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Cirq has a function cirq.sub_state_vector which can extract a single qubit's state from a full state vector. It doesn't just do the single qubit case, it can do arbitrary subsets of qubits. It will raise an exception if the subset you pick is entangled with other stuff. It's unfortunately a bit picky about error tolerances and input shape.


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Following up on a comment in @gIS's answer, there is a particular sense in which the computational power of algorithms involving only real amplitudes is very much equivalent, up to a small overhead, as the power of algorithms involving both real and complex amplitudes. I say this because it is known that Toffoli gates and Hadamard gates are sufficient for ...


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To build on the other answer, we can in fact characterise the set of coefficients that can fit into a convex decomposition of a density matrix. Given $\rho$, we can write $$\rho = \sum_k a_k |u_k\rangle\!\langle u_k|$$ for some set of (not necessarily orthogonal) states $\{|u_k\rangle\}$ if and only if $\mathbf a\preceq \boldsymbol{\lambda}(\rho)$, meaning ...


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I'm assuming you are asking specifically about the need for complex numbers in the context of a quantum algorithm written as a decomposition in terms of quantum gates. If you are instead asking about the need for complex numbers more in general in quantum mechanics, the answer would be a bit different, depend on what precisely you mean with "need", ...


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There is a physical meaning to the complex amplitude. The phase of the state affects what happing to the state. For example, consider photon qubit, which might be located in a superposition of $2$ optic fibers (fiber $A$ is $|0\rangle$, fiber $B$ is $|1\rangle$) they might be in the same phase, they might be mirrors (phase $-1 / \pi / 180^\circ$) or phase $...


2

No. In the ensemble interpretation of a density matrix $$ \rho=\sum_k p_k|\psi_k\rangle\langle \psi_k| $$ the states $|\psi_k\rangle$ are not necessarily orthogonal. Also, the probabilities $p_k$ are not necessarily eigenvalues. Ensembles vs eigendecomposition A density matrix $\rho$ encodes multiple ensembles $\{p_k, |\psi\rangle_k\}$, see theorem $2.3$ on ...


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The following construction works for a pair of qubits and small perturbations with $\|H\|_2\le\frac15$. Remarkably, in this case we can choose operators $A_k$ and $B_k$ independently of $H$. Basis Begin by defining pure states $$ \begin{align} |\psi_0\rangle &= |0\rangle& |\psi_2\rangle &= \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac23}e^{\frac{2\pi ...


1

The question is a bit ambiguous but yes you can think of unentangled states as independent and the process of entanglement as "adding" correlations. Take the following unentangled state for instance: $$\vert 00 \rangle + \vert 10 \rangle$$ Then the probability distribution of each qubit is independent of the other: $$P(q_1|q_0) = P(q_1)\>\> ...


2

In general, the problem of finding a separable decomposition (or even deciding if it exists) is NP-hard. I don't know if it's that hard for operators of the form $S = \mathbb{I}_\mathcal{X} \otimes \mathbb{I}_\mathcal{Y} - H$ where $\lVert H \rVert_2 \leq 1$, but if you want just examples it's easier to go backwards. You can start with a separable $S$, say, ...


1

To some extent, you can think that many implementations of qubits can be interpreted in terms of waves, as any particle can be interpreted as such (known as wave-particle duality.) I do not, however, believe you can generalize the mapping of qubit-related terms like "amplitude" or "frequency" to waves directly. For some implementations ...


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I don't believe that there is an immediate connection between waves and qubits. Most of the terminology is probably due to the historical development of quantum mechanics. For instance, you may have heard about waves when talking about the two slit experiment, where you shoot electrons through a pair of slits and you get an interference pattern on the screen....


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Wave-particle duality is a general feature of quantum mechanics, see e.g. the relevant Wikipedia page, and this post on physics.SE. A "qubit" is an abstraction of a two-dimensional quantum system. When you talk about qubits, you are not making any reference to the actual physical substrate that you are modeling. Any two-dimensional quantum system ...


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Below is a partial analytical solution to a variant of the problem using the trace distance. Solution approach We find an analytical expression for a lower bound on the trace distance $D(\rho_{AB}, \tilde{\rho}_{AB})$ and - under certain additional conditions described below - quantum states that achieve the bound. This gives us a partial solution of the ...


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Here's a direct circuit to produce the logical $\bar{|0\rangle}$ state from https://arxiv.org/abs/1501.02524: I'm not a qiskit expert, but in Cirq, I can check the superposition created by this: import cirq q = cirq.LineQubit.range(7) state = cirq.Circuit( cirq.H(q[0]), cirq.H(q[1]), cirq.H(q[3]), cirq.CNOT(q[0], q[2]), cirq.CNOT(q[3], ...


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It is not true that the increase in von Neumann entropy of entanglement between the two partitions due to a single cross-partition two-qubit gate is at most one. See below for an example where the increase is two. However, we can use log Schmidt rank to justify a different asymptotically linear upper bound. Product states A two-qubit gate cannot produce more ...


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