Recent Questions - Quantum Computing Stack Exchange most recent 30 from quantumcomputing.stackexchange.com 2020-01-25T08:36:34Z https://quantumcomputing.stackexchange.com/feeds https://creativecommons.org/licenses/by-sa/4.0/rdf https://quantumcomputing.stackexchange.com/q/9635 4 What is the difference between the "Fubini-Study distances" $\arccos|\langle\psi|\phi\rangle|$ and $\sqrt{1-|\langle\psi|\phi\rangle|}$? glS https://quantumcomputing.stackexchange.com/users/55 2020-01-24T13:20:46Z 2020-01-25T00:23:26Z <p>I sometimes see the "Fubini-Study distance" between two (pure) states <span class="math-container">$|\psi\rangle,|\phi\rangle$</span> written as <span class="math-container">$$d(\psi,\phi)_1=\arccos(|\langle\psi|\phi\rangle|),$$</span> for example <a href="https://en.wikipedia.org/wiki/Fubini%E2%80%93Study_metric#In_bra-ket_coordinate_notation" rel="nofollow noreferrer">in the Wikipedia page</a>. Other sources (e.g. <a href="https://arxiv.org/abs/1611.03449" rel="nofollow noreferrer">this paper</a> in pag. 16), use the definition <span class="math-container">$$d(\psi,\phi)_2=\sqrt{1-|\langle\psi|\phi\rangle|}.$$</span></p> <p>What is the difference between these two definitions? Is one preferred over the other?</p> https://quantumcomputing.stackexchange.com/q/9634 1 How will quantum computers access large amounts of storage? vy32 https://quantumcomputing.stackexchange.com/users/9482 2020-01-24T12:42:35Z 2020-01-25T05:28:35Z <p>I've seen many articles in the popular press saying that quantum computers will enable searching through huge amounts of data in an instant. But I can't figure out how the current architectures can do that at all, things like Google's Sycamore architecture don't even <em>have</em> storage. There is literally nothing to search except for the state space that results from the configuration of the gates (and that seems to be an RF signal that is spread out over time). </p> <p>So how will quantum computers search anything, other than tuning parameters?</p> https://quantumcomputing.stackexchange.com/q/9631 0 Right way to use Quantum Phase Estimation using aqua NikPapadopoulos https://quantumcomputing.stackexchange.com/users/9877 2020-01-24T05:42:00Z 2020-01-24T14:19:00Z <p>I have been experimenting with Qiskit lately, but I have found the implementation of algorithms in Aqua extremely confusing. Currently I am trying to implement a very simple circuit that will return the eigenvalues of a 2x2 unitary operator, for example Pauli X. The problem is that I don't know which class I should use: is it PhaseEstimationCircuit from qiskit.aqua.circuits.phase_estimation_circuit or EigsQPE from qiskit.aqua.components.eigs.eigs_qpe or any of the many similarly named modules. An example tentative is shown below.</p> <p><code>pauli = Pauli( z=, x=, label=None) weighted_pauli = WeightedPauliOperator(paulis=[(1,pauli)]) a = qpc.construct_circuit()</code></p> <p>The circuit <code>a</code> does not seem to be the correct one. More objectively, is there anyway to reproduce a QPE circuit as the one shown in <a href="https://en.wikipedia.org/wiki/Quantum_phase_estimation_algorithm#/media/File:PhaseCircuit-crop.svg" rel="nofollow noreferrer">Wikipedia</a> through Aqua?</p> https://quantumcomputing.stackexchange.com/q/9627 -2 Quantum Circuit teleportation Ba. Taj https://quantumcomputing.stackexchange.com/users/9700 2020-01-23T13:18:14Z 2020-01-24T16:05:11Z <p>Sender and receiver use teleportation protocol,where sender teleport a quantum state <span class="math-container">$\left| \varphi \right&gt;=\alpha\left| 0 \right&gt; + \beta \left|1\right&gt;$</span> to receiver . I want to implement that and then find the output of quantum Circuit in the right side <span class="math-container">$\left| ABC \right&gt;$</span> when the measurement of the teleportation protocol in the left side is <span class="math-container">$\left| 11\right&gt;$</span> <a href="https://i.stack.imgur.com/ZxLls.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZxLls.png" alt="enter image description here"></a></p> https://quantumcomputing.stackexchange.com/q/9626 4 Global phase and single qubit gate: does it actually matter for two (or more) qubit gates? StarBucK https://quantumcomputing.stackexchange.com/users/5008 2020-01-23T11:07:34Z 2020-01-23T21:13:56Z <p>Consider the <span class="math-container">$X$</span> gate. Mathematically, we have <span class="math-container">$X=i e^{-i\frac{\pi}{2} X}$</span></p> <p>But as global phase of unitaries don't matter because they will simply act a global phase to the wavefunction, we can consider implementing <span class="math-container">$X$</span> by <span class="math-container">$e^{-i\frac{\pi}{2} X}$</span>, allright.</p> <p>Now, consider a CNOT. Formally, we have:</p> <p><span class="math-container">$$CNOT=|0\rangle \langle 0| \otimes \mathbb{I} + |1\rangle \langle 1| \otimes X$$</span></p> <p>If at this point I say "well, a <span class="math-container">$X$</span> gate or a <span class="math-container">$\pi$</span> rotation around <span class="math-container">$x$</span> is the same, up to global phase", I could say:</p> <p><span class="math-container">$$CNOT=|0\rangle \langle 0| \otimes \mathbb{I} + |1\rangle \langle 1| \otimes e^{-i\frac{\pi}{2} X}$$</span></p> <p><strong>But the two expressions of the CNOT do not differ from a global phase.</strong></p> <p>My question is the following. </p> <p>Let's assume we want to implement an algorithm. Is it that we have <strong>at the beginning</strong> to define once for all how we implement an <span class="math-container">$X$</span> gate, and be consistant all along.</p> <p>For example, if as soon as there is an <span class="math-container">$X$</span> in the algorithm and that I replace it by <span class="math-container">$e^{-i\frac{\pi}{2} X}$</span>, then I will be fine.</p> <p>But, if sometime I replace it by <span class="math-container">$ie^{-i\frac{\pi}{2} X}$</span> and sometime by <span class="math-container">$e^{-i\frac{\pi}{2} X}$</span> then I will have problems.</p> <p>So here, indeed my two definitions of CNOT do not implement the same unitary, but if they were inside of an algorithm and that I had chosen a fixed convention for <span class="math-container">$X$</span>, then I will be safe ?</p> <hr> <p>Other question (more important for my purpose).</p> <p>Let's assume I can only do single qubit rotations on which I might have a quantum control on it (I can do controlled rotation in the end).</p> <p>How is it possible from this to implement a CNOT operation ? Indeed this example shows that a CNOT is <strong>not</strong> a controlled <span class="math-container">$\pi$</span>-pulse around <span class="math-container">$x$</span>. How could I add the <span class="math-container">$i$</span> that is missing in practice then ? Because from respect to the target qubit this <span class="math-container">$i$</span> is a global phase. This confuses me.</p> https://quantumcomputing.stackexchange.com/q/9623 3 Constructing a circuit for $C^1(U)$ for rotation operators with TWO single qubit gates and CNOT gate Mao LIN https://quantumcomputing.stackexchange.com/users/9816 2020-01-22T23:02:54Z 2020-01-24T17:05:57Z <p>This is the exercise 4.23 from Nielsen and Chuang, asking that if it is possible to construct <span class="math-container">$C^1(U)$</span> for <span class="math-container">$U=R_{x,y}(\theta)$</span> with TWO single qubit gates and CNOT gate. My answer is no, and I would like to argue in the following way. </p> <p>First, we do have such a construction for <span class="math-container">$U=R_z(\theta)$</span> which is the following (sorry that I have to draw it by hand)</p> <p><img src="https://i.stack.imgur.com/cwVTX.png" width="320"></p> <p>where the (reverse) CNOT has the matrix representation</p> <p><span class="math-container">$$\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ \end{bmatrix}$$</span> such that we have the matrix representation of the circuit as <span class="math-container">$$\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ \end{bmatrix} \big( \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; e^{-i\theta/2} \end{bmatrix} \otimes \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; e^{i\theta/2} \end{bmatrix} \big) \begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ \end{bmatrix} = \begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; e^{-i\theta/2} &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; e^{i\theta/2} \\ \end{bmatrix}$$</span></p> <p>which is the desired matrix for <span class="math-container">$C^1(R_z(\theta))$</span>. Thus we do a have a circuit with only two single qubit gates and CNOT for <span class="math-container">$U=R_z(\theta)$</span>. Now for the rotation <span class="math-container">$U=R_x(\theta) = VR_z(\theta)V^\dagger$</span>, which differ from <span class="math-container">$R_z$</span> by a unitary transformation V, we will need to "sandwich" the above circuit by some other gate. However, that is not possible by just one single qubit gate as we need unitary transformation on both sides of the circuit, despite we may merge two gates into one. Thus I feel that it is not possible for <span class="math-container">$U=R_{x,y}$</span>. However, on the other hand, physically there is no difference between various rotation operators, it feels not right that only <span class="math-container">$R_z$</span> has such circuit construction but not for others. So I may have missed sth here, and any help and clarification is appreciated. </p> https://quantumcomputing.stackexchange.com/q/9620 1 IBM quantum computers gate delay values Josu Etxezarreta Martinez https://quantumcomputing.stackexchange.com/users/2371 2020-01-22T14:25:54Z 2020-01-22T15:36:11Z <p>I am doing an state of the art study about the existing quantum computer technologies and one of the parameters of interest is the gate delay value, as it determines the time expected from a gate from such computer to make its operation, and consequently, it is important to see the coherence time that such computer presents.</p> <p>I would like to know where can I find such values for the IBM quantum computers.</p> https://quantumcomputing.stackexchange.com/q/9619 0 Depth of quantum algorithm: definition [duplicate] StarBucK https://quantumcomputing.stackexchange.com/users/5008 2020-01-22T12:27:14Z 2020-01-22T12:27:14Z <p>My question is really simple: what is the depth of a quantum algorithm ?</p> <p>Is it the number of gates that are ran in sequence ? When I am googling it I only find article using the word without defining it (because the definition I guess is pretty standard).</p> https://quantumcomputing.stackexchange.com/q/9614 2 How to interpret a 4 qubit quantum circuit as a matrix? Felipe Rojo Amadeo https://quantumcomputing.stackexchange.com/users/9797 2020-01-21T20:45:19Z 2020-01-21T22:39:48Z <p>This is part of Simon Algorithm (Initial state + some Oracle function) There is a post that explains how to interpret circuits (<a href="https://quantumcomputing.stackexchange.com/questions/2299/how-to-interpret-a-quantum-circuit-as-a-matrix">How to interpret a quantum circuit as a matrix?</a>), but I'm not sure how to apply to the following circuit. </p> <p><a href="https://i.stack.imgur.com/30Bxt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/30Bxt.png" alt="Circuit built with IMB plataform"></a></p> <p>The first part, both Hadamard to first and second qubit:</p> <p><span class="math-container">$M_1 = H \otimes H \otimes I \otimes I$</span> </p> <p>Then, first Controlled NOT:</p> <p>How can I apply a matrix to the first and second qubit if I have <span class="math-container">$M_1$</span> that is a 16x16 matrix. I know I could have applied <span class="math-container">$H$</span> to the first qubit and then do a tensor product with <span class="math-container">$I$</span> (third qubit), and the result, multiply by <span class="math-container">$CX$</span>. But then I have the second <span class="math-container">$CX$</span> which is applied to the first qubit and 4th qubit.</p> <p>Symbol <span class="math-container">$\otimes$</span> is a tensor product.</p> https://quantumcomputing.stackexchange.com/q/9611 2 Partial trace condition in Choi state user1936752 https://quantumcomputing.stackexchange.com/users/4831 2020-01-21T13:41:17Z 2020-01-22T11:43:53Z <p>Consider Hilbert spaces <span class="math-container">$\mathcal{X}, \mathcal{Y}$</span>. For any quantum channel <span class="math-container">$\mathcal{E}_{\mathcal{X}\rightarrow \mathcal{Y}}$</span>, the bipartite Choi state <span class="math-container">$J(\mathcal{E}) \in L(\mathcal{Y}\otimes\mathcal{X})$</span> is given by</p> <p><span class="math-container">$$J(\mathcal{E}) = (\mathcal{E}\otimes I)\sum_{a,b} \vert a\rangle\langle b\vert\otimes\vert a\rangle\langle b\vert$$</span></p> <p>It is also possible to show (see <a href="https://cs.uwaterloo.ca/~watrous/LectureNotes/CS766.Fall2011/05.pdf" rel="nofollow noreferrer">here</a> for example) that the trace preserving condition of the map <span class="math-container">$\mathcal{E}$</span> is equivalent to the following condition on its Choi state</p> <p><span class="math-container">$$\text{Tr}_\mathcal{Y}J(\mathcal{E}) = I_\mathcal{X} \tag{1}$$</span></p> <p>The proof is easy - it relies on noticing that the trace preserving condition implies that <span class="math-container">$\mathcal{E}(\vert a\rangle\langle b\vert) = \delta_{a,b}$</span> due to the trace preserving condition. Meanwhile, positive-semidefiniteness of <span class="math-container">$J(\mathcal{E})$</span> corresponds to complete positivity of <span class="math-container">$\mathcal{E}$</span>. </p> <p>If <span class="math-container">$J(\mathcal{E})$</span> is a density matrix but does not fulfill (1) is it still related to physical quantum channels (i.e. completely positive and trace preserving maps) in some way? Specifically, since completely positive maps correspond to positive semidefinite Choi matrices, does imposing that <span class="math-container">$\text{Tr}(J(\mathcal{E})) = 1$</span> (in addition to positive semidefiniteness) give us any condition on the map <span class="math-container">$\mathcal{E}$</span>? </p> https://quantumcomputing.stackexchange.com/q/9610 4 Why is the size of the top register for Shor's algorithm chosen as it is? David Cian https://quantumcomputing.stackexchange.com/users/9408 2020-01-21T13:29:54Z 2020-01-21T19:25:21Z <p>Let <span class="math-container">$N$</span> be the number we're trying to factor. In Shor's algorithm, the top register then has <span class="math-container">$2 \lceil\log_2(N)\rceil+1$</span> qubits, while the bottom register (the ancilla qubits) has <span class="math-container">$\lceil\log_2(N)\rceil$</span> qubits.</p> <p>This is stated in my <a href="https://www.slideshare.net/DavidCian/week5-ap3421-2019part1" rel="nofollow noreferrer">lecture notes</a> on slide 25.</p> <p>I believe I have an approximate understanding of Shor's algorithm: the end result on the top register, <span class="math-container">$\left| x_{\text{final}} \right&gt;$</span>, is such that <span class="math-container">$\frac{x_{\text{final}}}{T} = \frac{s}{r}$</span> for some integer <span class="math-container">$s$</span>. In the aforementioned equation, <span class="math-container">$T=2^{2 \lceil \log_2(N) \rceil + 1}$</span> and <span class="math-container">$r$</span> is the oracle's (modular exponentiation) period.</p> <p>The end result of the bottom (ancillary) register is the result of the modular exponentiation: <span class="math-container">$\left| f(x) + y \right&gt; = \left| f(x) \right&gt; = \left| a^x \mod N \right&gt;$</span>. The largest result possible is <span class="math-container">$N-1$</span>, for which you require <span class="math-container">$\lceil \log_2(N - 1) \rceil$</span> qubits, so it makes sense that the size of the bottom register is <span class="math-container">$\lceil\log_2(N)\rceil$</span>. </p> <p>I do not however understand the choice of the size of the top register.</p> https://quantumcomputing.stackexchange.com/q/9609 3 Where are the physical gates in the Google processor? vy32 https://quantumcomputing.stackexchange.com/users/9482 2020-01-21T12:21:52Z 2020-01-22T13:26:11Z <p>Google's article <a href="https://www.nature.com/articles/s41586-019-1666-5" rel="nofollow noreferrer">Quantum supremacy using a programmable superconducting processor</a> states that the processor "53 qubits, 1,113 single-qubit gates, 430 two-qubit gates, and a measurement on each qubit, for which we predict a total fidelity of 0.2%."</p> <p>Where are the gates physically located? In the diagram, the qubits are the crosses and the qubits are the couplers. </p> <p><a href="https://i.stack.imgur.com/RjXSG.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RjXSG.jpg" alt="enter image description here"></a></p> https://quantumcomputing.stackexchange.com/q/9608 1 Openfermion to qiskit Andrew Jackson https://quantumcomputing.stackexchange.com/users/9849 2020-01-21T10:45:45Z 2020-01-21T10:45:45Z <p>Is there a direct way to go from an object generated in openfermion to objects usable in Qiskit. I can't find anything about any plugin.</p> <p>It's not too hard to translate into pyQuil and then to Qiskit from there, but a direct way would be much easier.</p> https://quantumcomputing.stackexchange.com/q/9607 1 Measuring T1 and T2 constants on IBM Q Aviv Azran https://quantumcomputing.stackexchange.com/users/9846 2020-01-21T07:22:01Z 2020-01-24T09:06:17Z <p>We have been asked to measure relaxation and dephasing times T1 and T2 on the IBM Q using the composer only, Qiskit not allowed. I am a bit confused about the way to do so. Can someone explain the idea behind and how to implement the measurement in QASM?</p> https://quantumcomputing.stackexchange.com/q/9603 2 Quantum teleportation between two computer chips, how relevant? Max https://quantumcomputing.stackexchange.com/users/308 2020-01-20T18:57:18Z 2020-01-20T19:15:46Z <p>After stumbling upon this recent news article: <a href="https://newatlas.com/quantum-computing/quantum-teleportation-computer-chips" rel="nofollow noreferrer">Information teleported between two computer chips for the first time</a> (which references a "Nature Physics" publication <a href="https://www.nature.com/articles/s41567-019-0727-x" rel="nofollow noreferrer">Chip-to-chip quantum teleportation and multi-photon entanglement in silicon</a>)...</p> <blockquote> <p>...with quantum teleportation, information appears to break that speed limit</p> <p>Harnessing this phenomenon could clearly be beneficial, and the new study helps bring that closer to reality. The team generated pairs of entangled photons on the chips, and then made a quantum measurement of one. This observation changes the state of the photon, and those changes are then instantly applied to the partner photon in the other chip.</p> </blockquote> <p>... I've been wondering what this specific break-through means for the state of quantum teleportation, if anything. </p> <p>The quotes above would appear to imply that information <em>can</em> potentially be transported and used instantly, faster than the speed of light. Is that accurate though? My understanding was that because of its dependency on classical communication, quantum teleportation won't allow for faster-than-light communication in practice. Has that changed?</p> https://quantumcomputing.stackexchange.com/q/9602 1 Can two intrinsic angular momentum directions (spin) be used for future entanglement after collapse? [closed] Steve Freeman https://quantumcomputing.stackexchange.com/users/9826 2020-01-20T18:20:45Z 2020-01-23T22:32:48Z <p>Persuant to all contributors, here I will use a coin toss analogy since the vector direction is either up or down (heads or tails). Two coins will be used for each of two tosses. It's assumed that the process will be one-shot (shots=1).</p> <p>Toss one: Superposition will be applied using Hadamard gates on two qubits. Collapse will result in the following possible binary results: [00,01,10,11] (HH,HT,TH,TT).</p> <p>For toss two: Of the four binary results of toss one, I will choose either 00 or 11.</p> <p><strong>I want to know how to prepare two qubits in an entangled state after either 00 or 11 is measured on those two independent qubits prepared with Hadamards in toss one.</strong></p> https://quantumcomputing.stackexchange.com/q/9599 0 How can I build up an arbitrary quantum circuit given a certain unitary matrix operation? Dani https://quantumcomputing.stackexchange.com/users/9716 2020-01-20T14:15:06Z 2020-01-20T14:38:46Z <p>Suppose I want to put a qubit whose initial state is <span class="math-container">$|0\rangle$</span> to the final state <span class="math-container">$\frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac{2}{3}}|1\rangle$</span>.</p> <p>Well, in that case, the unitary matrix that performs such operation is given by: <span class="math-container">$$U = \frac{1}{\sqrt{3}}\begin{pmatrix}1&amp;-\sqrt{2}\\ \sqrt{2} &amp; 1 \end{pmatrix}$$</span> So the question is, how can I build a quantum circuit with the usual quantum gates (X, Y, Z, etc) which reproduces this behavior?</p> https://quantumcomputing.stackexchange.com/q/9597 5 Is there anything practical that can be done with a single qubit? vy32 https://quantumcomputing.stackexchange.com/users/9482 2020-01-20T13:56:47Z 2020-01-23T01:59:50Z <p>Is there anything practical that can be done with a single qubit? And by "practical," I mean a problem that can be solved or information that can be stored. </p> <p>I realize that one practical thing that you can do with a single qubit is to write a grant application to build a computer that has 9 qubits, but that is a bit too meta for this question.</p> https://quantumcomputing.stackexchange.com/q/9596 0 Code for a simple optimization problem in Criq vy32 https://quantumcomputing.stackexchange.com/users/9482 2020-01-20T13:08:50Z 2020-01-20T13:31:06Z <p>For a demonstration, I would like to code a simple optimization problem in Cirq. I don't care what the problem is, but I want it understandable to someone who has had only basic algebra. </p> <p>One idea is to find the value of <span class="math-container">$x$</span> such that <span class="math-container">$x-(x-3)^2$</span> is optimal. From inspection it is the value of 3.5. How do I code that in Cirq? If that's too complicated, what is not too complicated?</p> <p>But if I can code up any function I want, I'd like to find the minimum value of <span class="math-container">$y=\frac{x^4-8x^2+x}{10}$</span> because the graph is really nice:</p> <p><a href="https://i.stack.imgur.com/Bt0ZV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bt0ZV.png" alt="enter image description here"></a></p> <p>It's obvious from inspection that the minimum value of <span class="math-container">$y$</span> is when <span class="math-container">$x=-2$</span>. And we can do that because we can see all of the points at the same time and pick the minimum. So this is kind of what a quantum computer running a Criq program would do, right?</p> https://quantumcomputing.stackexchange.com/q/9579 3 What does it mean "less than identity" in the operator sum representation? czwang https://quantumcomputing.stackexchange.com/users/8321 2020-01-18T16:02:12Z 2020-01-21T10:11:53Z <p>In Quantum Computation and Quantum Information by Isaac Chuang and Michael Nielsen, section 8.2.3, <span class="math-container">$\mathcal{E}=\sum_{k}E_k\rho E_k^{\dagger}$</span> gives the operator-sum representation. In general, it requires <span class="math-container">$\sum_k E_k E_k^{\dagger}\leq I$</span>. But, what does it mean by the inequation here? Does it mean every entry of the matrix is a nonnegative real value up to 1? </p> <p>Thanks</p> https://quantumcomputing.stackexchange.com/q/9567 0 Problems with Q# installing Jonathcraft https://quantumcomputing.stackexchange.com/users/8746 2020-01-16T20:44:38Z 2020-01-23T19:30:11Z <p>I don't know if this is the right place to ask this question. I just thought it would be a place where people knew what they are doing. If there is a better place to look for answers (stackoverflow for example), feel free to tell me.</p> <p>The problem is that .NET can't install iqsharp. It says it doesn't recognize the file even though it is in the PATH folder (user/.dotnet/tools).</p> <pre><code>&gt; dotnet iqsharp install Could not execute because the specified command or file was not found. Possible reasons for this include: * You misspelled a built-in dotnet command. * You intended to execute a .NET Core program, but dotnet-iqsharp does not exist. * You intended to run a global tool, but a dotnet-prefixed executable with this name could not be found on the PATH. </code></pre> <p>It shows the same behavior when I want to use other .NET packages. I have dotnet version 3.1.101. I can send dotnet --info if you want.</p> <p>I have looked on github and other websites for solutions but most of the problems were caused by Linux Ubuntu and most of the treads are still open.</p> <p>Thanks for reading and I hope you can help me.</p> https://quantumcomputing.stackexchange.com/q/9536 4 Are there Bell-like violations that can be observed without collecting statistics? glS https://quantumcomputing.stackexchange.com/users/55 2020-01-14T20:09:53Z 2020-01-21T14:05:18Z <p>Observing the violation of Bell inequalities, be it <a href="https://journals.aps.org/ppf/abstract/10.1103/PhysicsPhysiqueFizika.1.195" rel="nofollow noreferrer">in their original formulation</a>, or in the nowadays more commonly used <a href="https://en.wikipedia.org/wiki/CHSH_inequality" rel="nofollow noreferrer">CHSH formulation</a>, involves computing averages of specific experimentally measurable quantities. In the CHSH formulation, these are for example the averages of the products of the experimental outcomes.</p> <p>Are there scenarios in which Bell nonlocality can be observed <em>without</em> such averages, that is, in a single-shot scenario? By this I mean a scenario in which a single measurement outcome (rather than a collection of outcomes used to compute averages, as is done in CHSH formulation) is enough to rule out a local hidden variable explanation.</p> <p>I seem to recall having seen this kind of thing in some paper, possibly in a scheme involving three or more parties (or maybe it was with systems with three or more outcomes?). I can't, however, find the reference right now, so I'm not sure whether I'm remembering this correctly or not.</p> https://quantumcomputing.stackexchange.com/q/9281 2 Is it possible to write a Q# teleportation code which returns a qubit? Coder https://quantumcomputing.stackexchange.com/users/8487 2019-12-21T18:54:09Z 2020-01-21T19:01:55Z <p>In the teleportation codes I have found so far from Microsoft, the codes are written based on returning the <code>bool</code> values but I wonder if in teleportation process, the program takes a qubit and then gives that qubit as its outcome?</p> <p>Thanks. </p> https://quantumcomputing.stackexchange.com/q/9257 -1 Can a classical limit of qubit emerge on a classical computer? [closed] Ryoji https://quantumcomputing.stackexchange.com/users/5712 2019-12-19T15:28:39Z 2020-01-21T02:44:03Z <p><img src="https://i.stack.imgur.com/8yohC.jpg" alt="state “0”"></p> <p><img src="https://i.stack.imgur.com/ehkKe.jpg" alt="state “1”"></p> <p><img src="https://i.stack.imgur.com/4FTif.jpg" alt="state of qubit"></p> <p>Hello,</p> <p>I am uncertain; however, today at a toy museum, I might have captured a qubit on a classical computing structure, “ a broken 90-marble abacus,” as shown in the attached photos above.</p> <p>I did not use it correctly, but if we set the first photo to be the state “0”, then the next one is “1”. You may say the third photo shows 0.5, but you can also say that it is a bit closer to 0 or 1. Thus, while we observe this purple marble positioned between 0 and 1, we cannot tell if it moves to 0 or 1 until we measure its displacement. Therefore, I wonder if it can be said that the classical limit of qubit emerged on classical computer in this third photo? Its probability of being 0 or 1 is fifty-fifty.</p> <p>The picture below gives a further explanation with a chart of marble A's location. In abacus computing, the states of A are limited to four states, namely, resting at 0 or 1, or moving toward 0 or 1. A cannot rest between 0 and 1. Abacus computing is observed by taking two photographs during its sequence giving two locations of A. When taking the first shot, we can know A's location and not sure if it is moving or resting. Then we take a second shot and know how it moves. After all, we can tell or predict its outcome 0 or 1. Thus, when we observe the initial location of A, this is the state of the classical limit of qubit, which has fifty-fifty probability of becoming 0 or 1, until secondary observation.</p> <p>This concept can be extended to tossing coin or tumbling dice. If the observer is unable to predict those outcomes by observing those motions, the observer can tell only classical probabilities.</p> <p>Thank you,</p> <p>Ryoji</p> <p><a href="https://i.stack.imgur.com/aWwMH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aWwMH.png" alt="eabacus observation"></a></p> https://quantumcomputing.stackexchange.com/q/9222 3 On the probability of preparing of a uniform superposition by performing a controlled-multiplication and post-selecting $0$ Mark S https://quantumcomputing.stackexchange.com/users/2927 2019-12-15T18:30:31Z 2020-01-23T21:47:07Z <p>I take as a starting point Watrous's celebrated <a href="https://arxiv.org/abs/cs/0009002" rel="nofollow noreferrer">paper</a> defining the Quantum Merlin-Arthur (QMA) class. He provides a protocol for Arthur to test whether an element <span class="math-container">$h$</span> is <em>not</em> in a group <span class="math-container">$\mathcal{H}$</span> with generating set <span class="math-container">$\langle g_1,g_2,\cdots, g_k\rangle$</span>. Merlin gives Arthur a quantum state <span class="math-container">$\vert\mathcal{H'}\rangle$</span> that is alleged to be a pure state corresponding to the uniform superposition over all elements of <span class="math-container">$\mathcal{H}$</span>:</p> <p><span class="math-container">$$\vert\mathcal H'\rangle = \frac{1}{\sqrt{|\mathcal H|}}\sum_{a\in \mathcal{H}} |a\rangle.$$</span></p> <p>However, Arthur must test that Merlin really <em>did</em> provide such a state <span class="math-container">$\vert\mathcal{H'}\rangle$</span>. He does this by initially performing <span class="math-container">$j$</span> iterations of a controlled multiplication of <span class="math-container">$\vert\mathcal{H'}\rangle$</span> by elements known to be in <span class="math-container">$\mathcal{H}$</span>, and testing that the control register reverts back to <span class="math-container">$\vert 0\rangle$</span> after Hadamarding. Watrous states that as long as we can post-select upon measuring <span class="math-container">$0$</span> in the control register, the state <span class="math-container">$\vert \mathcal{H'}\rangle$</span> "will in fact be changed (by quantum magic!) to one that is invariant under right multiplication by elements in <span class="math-container">$\mathcal{H}$</span>."</p> <p>According to these <a href="https://www.cs.cmu.edu/~odonnell/quantum15/" rel="nofollow noreferrer">notes</a> from O'Donnell, it might suffice to uniformly choose <span class="math-container">$j$</span> generators <span class="math-container">$z_i$</span> from <span class="math-container">$\langle g_1,g_2,\cdots, g_k\rangle$</span>. That is, if initially <span class="math-container">$\vert \mathcal{H'}\rangle=\sum_{g\in\mathcal{G}}a_g\vert g\rangle$</span>, then upon multiplying by <span class="math-container">$z_1$</span> and post-selecting <span class="math-container">$\vert 0\rangle$</span> on the control register for the first of the <span class="math-container">$j$</span> iterations, our state is:</p> <p><span class="math-container">$$\vert \mathcal{H'}\rangle=\sum_{g\in\mathcal{G}}a_g\vert g\rangle + a_g\vert gz_1\rangle;$$</span></p> <p>upon multiplying by <span class="math-container">$z_2$</span> and post-selecting <span class="math-container">$\vert 0\rangle$</span> on the control register for the second iteration, our state is:</p> <p><span class="math-container">$$\vert \mathcal{H'}\rangle=\sum_{g\in\mathcal{G}}a_g\vert g\rangle + a_g\vert gz_1\rangle+a_g\vert gz_2\rangle+a_g\vert gz_1z_2\rangle;$$</span></p> <p>etc.</p> <p>Indeed, <span class="math-container">$\vert \mathcal{H'}\rangle$</span> may initially correspond to a <em>singleton</em> state such as the state <span class="math-container">$\vert e \rangle$</span>, where <span class="math-container">$e$</span> is the identity of <span class="math-container">$\mathcal{H}$</span>. Our probability of post-selecting <span class="math-container">$\vert 0\rangle$</span> in the control register appears to increase with each iteration - it may start off low, at <span class="math-container">$\frac{1}{2}$</span>, then with "quantum magic," increase to greater and greater than <span class="math-container">$\frac{1}{2}$</span> when <span class="math-container">$\vert\mathcal{H'}\rangle$</span> includes most elements of <span class="math-container">$\mathcal{H}$</span>, then settle close to <span class="math-container">$1$</span> when <span class="math-container">$\vert\mathcal{H'}\rangle$</span> is finally right-invariant.</p> <p><strong>If we did start off with such a singleton state <span class="math-container">$\vert\mathcal{H'}\rangle=\vert e\rangle$</span>, how many such iterations must we do to be able to post-select a state that is exponentially close to being right-invariant under elements of <span class="math-container">$\mathcal{H}$</span>, and what is our probability of successfully post-selecting such a state?</strong></p> <p>In his wonderful <a href="https://www.youtube.com/watch?v=D1cZbRvBMqY" rel="nofollow noreferrer">lecture</a> from 2012 at about the 25 minute mark, Watrous proposes that <span class="math-container">$j$</span> should be linear or at best quadratic in <span class="math-container">$\log \Vert \mathcal{H}\Vert$</span> and mentions that the generators should be "cube generators" having nice properties according to computational group theory.</p> <p>But in the case of starting off in a singleton state such as <span class="math-container">$\vert e\rangle$</span>, can we learn anything about the properties of the specific set of generators <span class="math-container">$\langle g_1,g_2,\cdots, g_k\rangle$</span> by post-selecting to try and build our own uniform superposition?</p> https://quantumcomputing.stackexchange.com/q/9046 5 Self reducibility of QCMA problems BlackHat18 https://quantumcomputing.stackexchange.com/users/1351 2019-12-03T07:30:07Z 2020-01-20T14:34:13Z <p>Self reducibility is when search version of the problems in a language reduce to decision versions of the same problems. NP-complete problems are self reducible. Are QCMA complete problems self reducible?</p> https://quantumcomputing.stackexchange.com/q/8799 0 Atom magnetic moment caused by orbiting electron czwang https://quantumcomputing.stackexchange.com/users/8321 2019-11-15T15:48:38Z 2020-01-20T21:01:50Z <p>In Nielsen and Chuang Quantum Computation and Quantum Information book section 1.5.1 describing the Stern-Garlach experiment, it says: "Hydrogen atoms contain a proton and an orbiting electron. You can think of this electron as a little 'electric current' around the proton. This electric current causes the atom to have a magnetic field;" I thought the view of an electron orbiting the nucleus is outdated, replaced by the model of a probabilistic cloud around the nucleus in modern quantum physics view.</p> <p>Can anyone help me to understand the relationship between the two views today? (I majored in computer science, without much formal education in modern physics.) Thanks.</p> https://quantumcomputing.stackexchange.com/q/8458 2 IBM 53 qubit cloud access sandualuclopotaru https://quantumcomputing.stackexchange.com/users/8592 2019-10-10T21:05:24Z 2020-01-20T17:13:35Z <p>IBM announced a 53 qubit chip.</p> <p>Will the advertised chip be available for non-paying users, or only to IBM clients?</p> https://quantumcomputing.stackexchange.com/q/6261 2 How does quantum contextuality relate to mutually commuting observables? glS https://quantumcomputing.stackexchange.com/users/55 2019-05-28T16:42:32Z 2020-01-20T16:56:19Z <p>I am trying to get a better understanding of what is the idea behind quantum contextuality. Quoting from the <a href="https://en.wikipedia.org/wiki/Quantum_contextuality" rel="nofollow noreferrer">wikipedia page</a> (emphasis mine):</p> <blockquote> <p>Quantum contextuality is a foundational concept in quantum mechanics stating that the outcome one observes in a measurement is dependent upon what other measurements one is trying to make. <strong><em>More formally, the measurement result of a quantum observable is dependent upon which other commuting observables are within the same measurement set</em></strong>.</p> </blockquote> <p>I am a bit confused by the phrasing here. If I have two commuting observables, <span class="math-container">$[A,B]=0$</span>, that means that they do not interfere with one another, or, quoting from the <a href="https://en.wikipedia.org/wiki/Complete_set_of_commuting_observables" rel="nofollow noreferrer">relevant Wikipedia page</a>, that "<em>the measurement of one observable has no effect on the result of measuring another observable in the set</em>".</p> <p>Is this just bad phrasing (or a typo) on Wikipedia's side, or am I missing something?</p> https://quantumcomputing.stackexchange.com/q/5462 4 Constructing a circuit which performs the transformation $|x,y\rangle \to |x, x + y \bmod 4\rangle$ QCQCQC https://quantumcomputing.stackexchange.com/users/5632 2019-02-11T18:37:37Z 2020-01-24T22:06:39Z <p>When faced with exercises like these, I find it hard to know how to construct the circuits due to the amount of input one needs to account for. I have seen the solution provided <a href="http://web.mit.edu/2.111/www/2010/ps4solution.pdf" rel="noreferrer">here</a> however, I don't think I would have been able to solve this exercise my self.</p> <p>Does anyone have any tips on a systematic way to construct a circuit like this? I start out with a circuit which solves one specific input (for example for <span class="math-container">$x = |00\rangle, y = |01\rangle$</span>), but after this I get stuck. I appreciate any help!</p>