Recent Questions - Quantum Computing Stack Exchange most recent 30 from quantumcomputing.stackexchange.com 2022-10-04T03:56:51Z https://quantumcomputing.stackexchange.com/feeds https://creativecommons.org/licenses/by-sa/4.0/rdf https://quantumcomputing.stackexchange.com/q/28391 2 Is there an efficient algorithm for decomposing an arbitrary Hamiltonian into Pauli strings? djs2596 https://quantumcomputing.stackexchange.com/users/18034 2022-10-03T18:22:56Z 2022-10-03T18:43:00Z <p>Basically the title. If I have a <span class="math-container">$2^N\times 2^N$</span> Hamiltonian <span class="math-container">$H$</span> of random numbers (we can take the Hamiltonian as normalized if we want) and <span class="math-container">$N$</span> is an integer, is there an efficient way of writing <span class="math-container">$$H = \sum_{i}{\beta_iP_i}$$</span> where <span class="math-container">$\beta_i \in \mathbb{C}$</span> and <span class="math-container">$P_i$</span> is a Pauli string and the sum ranges over all such possible tensor product combinations of the Pauli group, <span class="math-container">$\{I,X,Y,Z\}$</span>. Apologies if my notation is non-standard; let me know if any clarification is needed.</p> <p>Also, I say efficient because I am aware of such solutions as <a href="https://quantumcomputing.stackexchange.com/questions/8725/can-arbitrary-matrices-be-decomposed-using-the-pauli-basis">this</a> and <a href="https://quantumcomputing.stackexchange.com/questions/2703/is-the-pauli-group-for-n-qubits-a-basis-for-mathbbc2n-times-2n">this</a>, but these seem to become exponentially hard in <span class="math-container">$N$</span>.</p> https://quantumcomputing.stackexchange.com/q/28390 0 How do I get correct measurement probabilities in ZX calculus? jjgoings https://quantumcomputing.stackexchange.com/users/12855 2022-10-03T16:38:58Z 2022-10-03T16:38:58Z <p>I'm learning ZX-calculus, but I'm getting confused when trying to obtain some simple results to compute probabilities for different outcomes.</p> <p>Here's a simple example where I'm getting lost. Here, <code>a</code> is a Boolean / binary variable.</p> <p><a href="https://i.stack.imgur.com/lr8hJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lr8hJ.png" alt="enter image description here" /></a> Given the circuit above, I would expect to measure the qubit to be <code>0</code> with zero probability and <code>1</code> with probability one. But according to the <a href="https://arxiv.org/pdf/2012.13966.pdf" rel="nofollow noreferrer">scalar rules here</a>, I compute probability of 2? I can always renormalize, but this seems (to me) tricky when I want to compute more complex measurement probabilities. What is the best way to go forward in the ZX-calculus?</p> <p>For context, what I <em>really</em> want to show is how to obtain the measurement probability in Figure 2 of <a href="https://arxiv.org/pdf/2206.02171.pdf" rel="nofollow noreferrer">this paper</a> with the ZX-calculus.</p> <p><a href="https://i.stack.imgur.com/T5Lrw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T5Lrw.png" alt="enter image description here" /></a></p> <p>where it is claimed</p> <p><a href="https://i.stack.imgur.com/Zi3a5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zi3a5.png" alt="enter image description here" /></a></p> <p>But I keep getting lost with global phase differences and not being totally sure how to compute probabilities with correct normalization from the get-go.</p> https://quantumcomputing.stackexchange.com/q/28389 0 Are indistinguishable bosons and fermions computationally equivalent to distinguishable qubits? Mark S https://quantumcomputing.stackexchange.com/users/2927 2022-10-03T16:25:20Z 2022-10-03T16:25:20Z <p>Physical qubits as we understand and intuit nowadays are fundamentally <em>distinguishable</em> &quot;particles&quot; or &quot;atoms&quot;. For example, at a high level we need to separate and index the wires in our circuit diagrams, but also more concretely we can point to individual ions in a trap or individual transmons and identify and label them as separate and distinguished.</p> <p>From the middle-late decades of the last century, Feynman and Deutsch, and perhaps Wiesner and Bennett among others, had some intuition that qubits are computationally more powerful or more interesting than classical bits.</p> <p>But, quantum mechanics is also concerned with two classes of <em>indistinguishable</em> particles - namely fermions and bosons.</p> <p>Going back all the way to the 1920's, <a href="https://en.wikipedia.org/wiki/Jordan%E2%80%93Wigner_transformation" rel="nofollow noreferrer">Jordan and Wigner</a> provided a mapping between the wavefunction in Hilbert space spanned by a number of qubits and the wavefunction in a Fock space spanned by fermions. Feynman also hinted that a qubit could be defined by the presence or absence of such a particle. But, for fermions at least the <a href="https://en.wikipedia.org/wiki/Numerical_sign_problem" rel="nofollow noreferrer">sign problem</a> is a separate issue that needs to be carefully addressed - e.g. whenever two fermions are swapped, the wavefunction picks up a negative phase.</p> <p>Furthermore regarding bosons, Aaronson and Arkhipov noted that <a href="https://en.wikipedia.org/wiki/Boson_sampling" rel="nofollow noreferrer">sampling</a> bosons such as photons from a network of mirrors and beam splitters is related to calculation of the <a href="https://en.wikipedia.org/wiki/Permanent_(mathematics)" rel="nofollow noreferrer">permanent</a>, while sampling the same for fermions is related to the determinant. It follows from <a href="https://en.wikipedia.org/wiki/%E2%99%AFP-completeness_of_01-permanent" rel="nofollow noreferrer">Valiant's theorem</a> that boson sampling is likely much (much) more difficult than fermion sampling.</p> <p>On the one hand we have fermions, which naively have the sign problem making simulation more difficult; however, fermion sampling is in <strong>P</strong>. On the other hand boson sampling is most likely not in <strong>P</strong> - nor is it likely to <em>contain</em> <strong>P</strong>.</p> <blockquote> <p><strong>But, can we translate between fermions, bosons, and qubits efficiently? For example could we have created a quantum computer out of indistinguishable bosons instead of distinguishable qubits?</strong></p> </blockquote> <hr /> <p><sup>Because of the <a href="https://en.wikipedia.org/wiki/Pauli_exclusion_principle" rel="nofollow noreferrer">Pauli exclusion principle</a>, a chain can be either occupied or not with precisely one fermion, while bosonic occupancy is unlimited. BosonSampling experiments try to mitigate this by having at least quadratically more modes than bosons</sup>.</p> https://quantumcomputing.stackexchange.com/q/28384 1 Parallel run of qiskit circuits noobier https://quantumcomputing.stackexchange.com/users/18427 2022-10-03T11:20:53Z 2022-10-03T11:20:53Z <p>I am trying to run simulation instances of a parametrized circuit in parallel but my algorithm is extremely slow. Excuse my ignorance, but I just want to be sure that the circuits are indeed running in parallel.</p> <p>Here is my execution workflow:</p> <pre><code> backend = Aer.get_backend('statevector_simulator', max_parallel_threads=N_threads, max_parallel_experiments=N_threads, optimization_level=3, precision=&quot;single&quot;) parametrized_circuits = CreateCircuits() transpiled = transpile(parametrized_circuits, backend=backend) experiments = [] for i in range(N): experiments.append(transpiled[i].assign_parameters(a)) job = execute(experiments, backend=backend, optimization_level=0) </code></pre> <p>So, my first question</p> <ol> <li>Is the structure efficient? E.g. Are the circuits transpiled only once? Or does <code>execute</code> re-transpile them? (I have read this function is just a wrapper for <code>transpile</code> and <code>run</code>)</li> </ol> <p>My parametrized circuit has a constant part, let's say <span class="math-container">$R(a)$</span>, which doesn't change between circuits, and a variable part <span class="math-container">$A_i=$</span><code>UnitaryGate(arbitrary_matrix)</code>, with every circuit <span class="math-container">$Q_i$</span> using a different <span class="math-container">$A_i$</span>.</p> <p>When I run my experiments in a <code>12-core</code> node, most cores seem to be under-utilized (atmost 10%, some even lower), while the experiment taking ages to complete. I have also read this <a href="https://quantumcomputing.stackexchange.com/questions/23353/how-to-order-results-after-multi-circuit-qiskit-execute-parallel-run">question</a> regarding reordering. My results come out with the same order as if the circuits were simulated serially, which makes me even more suspicious. So,</p> <ol start="2"> <li>Are my circuits really running in parallel? Is there something I miss?</li> </ol> https://quantumcomputing.stackexchange.com/q/28383 0 What is SqueezingEmbedding in Pennylane? R-X Zhao https://quantumcomputing.stackexchange.com/users/19201 2022-10-03T09:57:02Z 2022-10-03T09:57:02Z <p><a href="https://docs.pennylane.ai/en/stable/code/api/pennylane.SqueezingEmbedding.html" rel="nofollow noreferrer">enter link description here</a></p> <p>This question comes mainly from the above link. Firstly, I don't know the reference for this coding method. Second, assuming there are N features, how many qubits are needed to use this encoding method?</p> https://quantumcomputing.stackexchange.com/q/28381 3 Projective measurement operation in Qiskit stopper https://quantumcomputing.stackexchange.com/users/18259 2022-10-03T09:15:06Z 2022-10-03T12:35:55Z <p>I would like to implement the operation <span class="math-container">$\pi = \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 0 \end{bmatrix}$</span> on qiskit but I don't know how to do that.</p> <p>If I use the reset gate, but I use it for example on the first qubit of a Bell state <span class="math-container">$|\psi\rangle = \frac{1}{\sqrt 2}(|00\rangle + |11\rangle)$</span>, then the result will be <span class="math-container">$|00\rangle$</span> half of the times and <span class="math-container">$|01\rangle$</span> the other half (randomly). Instead I should obtain always <span class="math-container">$|00\rangle$</span>.</p> <p>If I use a Swap gate with an auxiliary qubit, my final state will be <span class="math-container">$|\psi\rangle = \frac{1}{\sqrt 2}(|00\rangle + |01\rangle)$</span> instead of <span class="math-container">$|00\rangle$</span>.</p> <p>The Bell state is just an example, in general I need to work on an arbitrary state.</p> <p>Thank you!</p> https://quantumcomputing.stackexchange.com/q/28380 1 unexpected keyword argument in qiskit vqe Lord Nexprex https://quantumcomputing.stackexchange.com/users/20328 2022-10-03T07:28:57Z 2022-10-03T12:29:06Z <p>I want to find the estimate of the ground state energy of my Hamiltonian <code>H</code> that is implemented as <code>PauliSumOp</code> in my variable <code>H_op</code>. I then prepare an ansatz circuit <code>circ_ansatz</code> with 5 parameters. Then I try to run the thing with <code>my_vqe = VQE(ansatz = circ_ansatz, optimizer=SPSA, quantum_instance=Aer.get_backend('aer_simulator'), initial_point=[0.5]*N_iters)</code> and then <code>print(my_vqe.compute_minimum_eigenvalue(H_op))</code> but I get the following error: <code> print(my_vqe.compute_minimum_eigenvalue(H_op)) File &quot;/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/qiskit/algorithms/minimum_eigen_solvers/vqe.py&quot;, line 526, in compute_minimum_eigenvalue opt_result = self.optimizer( # pylint: disable=not-callable TypeError: __init__() got an unexpected keyword argument 'fun'</code></p> <p>Any help is highly appreciated!</p> https://quantumcomputing.stackexchange.com/q/28378 1 Why does the quantum walk operator only have two eigenvectors? Loic Stoic https://quantumcomputing.stackexchange.com/users/21418 2022-10-03T04:56:18Z 2022-10-03T04:56:18Z <p>In <a href="https://arxiv.org/abs/0810.0312" rel="nofollow noreferrer">Child's paper on the relationship between discrete and continuous quantum walks</a>, he makes the following claim.</p> <p><a href="https://i.stack.imgur.com/ibL4N.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ibL4N.png" alt="enter image description here" /></a></p> <p>Although he provides a proof after this:</p> <p><a href="https://i.stack.imgur.com/uEHsc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uEHsc.png" alt="enter image description here" /></a></p> <p>I don't understand how the walk operator must have only two eigenvectors. To compute <span class="math-container">$\left | \mu \right &gt;$</span> using <span class="math-container">$\left | \mu_\pm \right &gt; = \frac{1 - e^{\pm i acos \lambda} S}{\sqrt{2(1 - \lambda^2)}}T \left | \lambda \right &gt;$</span>, which eigenvectors, <span class="math-container">$\left | \lambda \right &gt;$</span> of <span class="math-container">$\hat{H}$</span> would we use?</p> <p>Thanks for any help.</p> https://quantumcomputing.stackexchange.com/q/28376 0 Quantum Signal Operator and the unitary state preparation oracle? Loic Stoic https://quantumcomputing.stackexchange.com/users/21418 2022-10-02T22:40:42Z 2022-10-03T04:17:26Z <p>I am looking into IL Chuang and GH Low's <a href="https://arxiv.org/abs/1610.06546" rel="nofollow noreferrer">Hamiltonian Simulation with Qubitization paper</a>.</p> <p><a href="https://i.stack.imgur.com/izkZg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/izkZg.png" alt="signal operator and unitary state preparation oracle" /></a></p> <p>I am very confused on the terminology and motivation behind definition 1.</p> <p>I do not understand what the unitary oracle, <span class="math-container">$\hat{U}$</span> or the unitary state preparation oracle <span class="math-container">$\hat{G}$</span> are meant to represent here. Is it simply that <span class="math-container">$\hat{H}$</span> can be decomposed into them? Also what is the <span class="math-container">$\mathcal{I}_s$</span> meant to represent?</p> <p>I think that a subscript, <span class="math-container">$A_x$</span> is meant to represent that <span class="math-container">$A$</span> is in an <span class="math-container">$x \times x$</span> Hilbert space.</p> <p>However, previously in the paper, they say that <span class="math-container">$\hat{H} \in \mathbb{C}^{N \times N}$</span>, <span class="math-container">$\hat{U} \in \mathbb{C}^{Nd \times Nd}$</span>, and <span class="math-container">$\left | G \right &gt; \in \mathbb{C}^{d}$</span> If this is true, then <span class="math-container">$(\left &lt; G \right | \otimes \mathcal{I}_s)\hat{U}(\left | \hat{G} \right &gt; \otimes \mathcal{I}_s) \in \mathbb{C}$</span>, as it is essentially an inner product, not <span class="math-container">$\mathbb{C}^{N \times N}$</span>, right?</p> <p><a href="https://i.stack.imgur.com/YCmRG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YCmRG.png" alt="" /></a></p> <p>Any help is massively appreciated. This is all so confusing. Thank you.</p> https://quantumcomputing.stackexchange.com/q/28373 2 Algorithm for finding Pauli stabilizers of a code Ian Gershon Teixeira https://quantumcomputing.stackexchange.com/users/19675 2022-10-02T14:15:52Z 2022-10-03T14:50:49Z <p>Given the zero logical <span class="math-container">$|0_L\rangle$</span> and one logical <span class="math-container">$|1_L\rangle$</span> for an <span class="math-container">$[[n,1,d]]$</span> code is there a well known/ efficient algorithm for determining which Pauli operators stabilize the code?</p> <p>For a bit more context here is some output from the &quot;dumb&quot; algorithm I'm currently using (kind of misleading to say &quot;I'm&quot; since all the actual code here is being written up and run by an old friend of mine at UCI he deserves 100% of the credit for the actual programming):</p> <hr /> <p>FOUND SOMETHING EXTREMELY HEALTHY HERE...</p> <p>Total number of combos searched so far = 29768</p> <p>Number of '0' coefficients: 120</p> <p>Number of '+1' coefficients: 8</p> <p>Number of '-1' coefficients: 0</p> <p>[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]</p> <p>This meets our distance-three conditions and is indeed HEALTHISSIMO</p> <p>There is/are 64 stabilizer(s) for this ZeroL and OneL:</p> <p>[&quot;[I, I, I, I, I, I, I]&quot;, &quot;[X, I, X, X, X, I, I]&quot;, &quot;[Y, I, Y, Y, Y, I, I]&quot;, &quot;[Z, I, Z, Z, Z, I, I]&quot;, &quot;[X, X, X, I, I, X, I]&quot;, &quot;[I, X, I, X, X, X, I]&quot;, &quot;[Z, X, Z, Y, Y, X, I]&quot;, &quot;[Y, X, Y, Z, Z, X, I]&quot;, &quot;[Y, Y, Y, I, I, Y, I]&quot;, &quot;[Z, Y, Z, X, X, Y, I]&quot;, &quot;[I, Y, I, Y, Y, Y, I]&quot;, &quot;[X, Y, X, Z, Z, Y, I]&quot;, &quot;[Z, Z, Z, I, I, Z, I]&quot;, &quot;[Y, Z, Y, X, X, Z, I]&quot;, &quot;[X, Z, X, Y, Y, Z, I]&quot;, &quot;[I, Z, I, Z, Z, Z, I]&quot;, &quot;[I, X, X, X, I, I, X]&quot;, &quot;[X, X, I, I, X, I, X]&quot;, &quot;[Y, X, Z, Z, Y, I, X]&quot;, &quot;[Z, X, Y, Y, Z, I, X]&quot;, &quot;[X, I, I, X, I, X, X]&quot;, &quot;[I, I, X, I, X, X, X]&quot;, &quot;[Z, I, Y, Z, Y, X, X]&quot;, &quot;[Y, I, Z, Y, Z, X, X]&quot;, &quot;[Y, Z, Z, X, I, Y, X]&quot;, &quot;[Z, Z, Y, I, X, Y, X]&quot;, &quot;[I, Z, X, Z, Y, Y, X]&quot;, &quot;[X, Z, I, Y, Z, Y, X]&quot;, &quot;[Z, Y, Y, X, I, Z, X]&quot;, &quot;[Y, Y, Z, I, X, Z, X]&quot;, &quot;[X, Y, I, Z, Y, Z, X]&quot;, &quot;[I, Y, X, Y, Z, Z, X]&quot;, &quot;[I, Y, Y, Y, I, I, Y]&quot;, &quot;[X, Y, Z, Z, X, I, Y]&quot;, &quot;[Y, Y, I, I, Y, I, Y]&quot;, &quot;[Z, Y, X, X, Z, I, Y]&quot;, &quot;[X, Z, Z, Y, I, X, Y]&quot;, &quot;[I, Z, Y, Z, X, X, Y]&quot;, &quot;[Z, Z, X, I, Y, X, Y]&quot;, &quot;[Y, Z, I, X, Z, X, Y]&quot;, &quot;[Y, I, I, Y, I, Y, Y]&quot;, &quot;[Z, I, X, Z, X, Y, Y]&quot;, &quot;[I, I, Y, I, Y, Y, Y]&quot;, &quot;[X, I, Z, X, Z, Y, Y]&quot;, &quot;[Z, X, X, Y, I, Z, Y]&quot;, &quot;[Y, X, I, Z, X, Z, Y]&quot;, &quot;[X, X, Z, I, Y, Z, Y]&quot;, &quot;[I, X, Y, X, Z, Z, Y]&quot;, &quot;[I, Z, Z, Z, I, I, Z]&quot;, &quot;[X, Z, Y, Y, X, I, Z]&quot;, &quot;[Y, Z, X, X, Y, I, Z]&quot;, &quot;[Z, Z, I, I, Z, I, Z]&quot;, &quot;[X, Y, Y, Z, I, X, Z]&quot;, &quot;[I, Y, Z, Y, X, X, Z]&quot;, &quot;[Z, Y, I, X, Y, X, Z]&quot;, &quot;[Y, Y, X, I, Z, X, Z]&quot;, &quot;[Y, X, X, Z, I, Y, Z]&quot;, &quot;[Z, X, I, Y, X, Y, Z]&quot;, &quot;[I, X, Z, X, Y, Y, Z]&quot;, &quot;[X, X, Y, I, Z, Y, Z]&quot;, &quot;[Z, I, I, Z, I, Z, Z]&quot;, &quot;[Y, I, X, Y, X, Z, Z]&quot;, &quot;[X, I, Y, X, Y, Z, Z]&quot;, &quot;[I, I, Z, I, Z, Z, Z]&quot;]</p> <hr /> <p>FOUND SOMETHING EXTREMELY HEALTHY HERE...</p> <p>Total number of combos searched so far = 31712</p> <p>Number of '0' coefficients: 120</p> <p>Number of '+1' coefficients: 8</p> <p>Number of '-1' coefficients: 0</p> <p>[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]</p> <p>This meets our distance-three conditions and is indeed HEALTHISSIMO</p> <p>There is/are 64 stabilizer(s) for this ZeroL and OneL:</p> <p>[&quot;[I, I, I, I, I, I, I]&quot;, &quot;[X, X, X, I, X, I, I]&quot;, &quot;[Y, Y, Y, I, Y, I, I]&quot;, &quot;[Z, Z, Z, I, Z, I, I]&quot;, &quot;[I, X, X, X, I, X, I]&quot;, &quot;[X, I, I, X, X, X, I]&quot;, &quot;[Y, Z, Z, X, Y, X, I]&quot;, &quot;[Z, Y, Y, X, Z, X, I]&quot;, &quot;[I, Y, Y, Y, I, Y, I]&quot;, &quot;[X, Z, Z, Y, X, Y, I]&quot;, &quot;[Y, I, I, Y, Y, Y, I]&quot;, &quot;[Z, X, X, Y, Z, Y, I]&quot;, &quot;[I, Z, Z, Z, I, Z, I]&quot;, &quot;[X, Y, Y, Z, X, Z, I]&quot;, &quot;[Y, X, X, Z, Y, Z, I]&quot;, &quot;[Z, I, I, Z, Z, Z, I]&quot;, &quot;[X, X, I, X, I, I, X]&quot;, &quot;[I, I, X, X, X, I, X]&quot;, &quot;[Z, Z, Y, X, Y, I, X]&quot;, &quot;[Y, Y, Z, X, Z, I, X]&quot;, &quot;[X, I, X, I, I, X, X]&quot;, &quot;[I, X, I, I, X, X, X]&quot;, &quot;[Z, Y, Z, I, Y, X, X]&quot;, &quot;[Y, Z, Y, I, Z, X, X]&quot;, &quot;[X, Z, Y, Z, I, Y, X]&quot;, &quot;[I, Y, Z, Z, X, Y, X]&quot;, &quot;[Z, X, I, Z, Y, Y, X]&quot;, &quot;[Y, I, X, Z, Z, Y, X]&quot;, &quot;[X, Y, Z, Y, I, Z, X]&quot;, &quot;[I, Z, Y, Y, X, Z, X]&quot;, &quot;[Z, I, X, Y, Y, Z, X]&quot;, &quot;[Y, X, I, Y, Z, Z, X]&quot;, &quot;[Y, Y, I, Y, I, I, Y]&quot;, &quot;[Z, Z, X, Y, X, I, Y]&quot;, &quot;[I, I, Y, Y, Y, I, Y]&quot;, &quot;[X, X, Z, Y, Z, I, Y]&quot;, &quot;[Y, Z, X, Z, I, X, Y]&quot;, &quot;[Z, Y, I, Z, X, X, Y]&quot;, &quot;[I, X, Z, Z, Y, X, Y]&quot;, &quot;[X, I, Y, Z, Z, X, Y]&quot;, &quot;[Y, I, Y, I, I, Y, Y]&quot;, &quot;[Z, X, Z, I, X, Y, Y]&quot;, &quot;[I, Y, I, I, Y, Y, Y]&quot;, &quot;[X, Z, X, I, Z, Y, Y]&quot;, &quot;[Y, X, Z, X, I, Z, Y]&quot;, &quot;[Z, I, Y, X, X, Z, Y]&quot;, &quot;[I, Z, X, X, Y, Z, Y]&quot;, &quot;[X, Y, I, X, Z, Z, Y]&quot;, &quot;[Z, Z, I, Z, I, I, Z]&quot;, &quot;[Y, Y, X, Z, X, I, Z]&quot;, &quot;[X, X, Y, Z, Y, I, Z]&quot;, &quot;[I, I, Z, Z, Z, I, Z]&quot;, &quot;[Z, Y, X, Y, I, X, Z]&quot;, &quot;[Y, Z, I, Y, X, X, Z]&quot;, &quot;[X, I, Z, Y, Y, X, Z]&quot;, &quot;[I, X, Y, Y, Z, X, Z]&quot;, &quot;[Z, X, Y, X, I, Y, Z]&quot;, &quot;[Y, I, Z, X, X, Y, Z]&quot;, &quot;[X, Z, I, X, Y, Y, Z]&quot;, &quot;[I, Y, X, X, Z, Y, Z]&quot;, &quot;[Z, I, Z, I, I, Z, Z]&quot;, &quot;[Y, X, Y, I, X, Z, Z]&quot;, &quot;[X, Y, X, I, Y, Z, Z]&quot;, &quot;[I, Z, I, I, Z, Z, Z]&quot;]</p> https://quantumcomputing.stackexchange.com/q/28371 -1 About swap gate Rayhan https://quantumcomputing.stackexchange.com/users/21849 2022-10-02T11:59:07Z 2022-10-03T13:25:33Z <p><a href="https://i.stack.imgur.com/NFscL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NFscL.png" alt="Swap Gate" /></a></p> <p>As I was reading Qiskit textbook and there is this problem ... How can I swap a <span class="math-container">$|+\rangle$</span> to a <span class="math-container">$|-\rangle$</span></p> https://quantumcomputing.stackexchange.com/q/28366 0 Expectation values of non-local operators in Qiskit Blubsiwubsi https://quantumcomputing.stackexchange.com/users/14430 2022-10-01T14:52:20Z 2022-10-02T14:31:24Z <p>Is there a convenient way in Qiskit to calculate the expectation value for a non-local operator, i.e. I would like to calculate:</p> <p><span class="math-container">$$\langle \Psi|O|\Psi \rangle$$</span></p> <p>More precisely, I would like to calculate the expectation value for an Operator that can be described by the following circuit:</p> <pre><code>operator_circ = QuantumCircuit(4) operator_circ.x(0) operator_circ.cz([1, 1, 2], [2, 3, 3]) </code></pre> <p>I am currently doing the following to calculate the expectation value:</p> <pre><code>operator = Operator(operator_circ) #Where psi is some quantumstate/quantumcircuit psi.save_expectation_value(operator, range(4)) </code></pre> <p>But I am afraid, that this is not what I want since when I look into the decomposition of the Operator it consists of some multiplexers that in turn consists of <span class="math-container">$CCX$</span> gates and a gate called <code>squ_dg</code>. However, what I would like is to calculate the expectation value w.r.t. the decomposition of the operator circuit into summed weighted local Pauli.</p> <p>Ideally some function expresses the <span class="math-container">$CZ$</span> gate like so:</p> <p><span class="math-container">$$CZ_{12} = \frac{1}{2} (I_1 \otimes I_2 + I_1 \otimes Z_2 + Z_1 \otimes I_2 - Z_1 \otimes Z_2)$$</span></p> <p>and then calculates the expectation value over the sums of weighted Paulis on the right hand side.</p> <p>Or is it the same as just using the afformentioned code? What would be the best practice in Qiskit in this case?</p> https://quantumcomputing.stackexchange.com/q/28365 0 Stabilizers Of The H Code esabo https://quantumcomputing.stackexchange.com/users/21576 2022-10-01T03:12:38Z 2022-10-02T17:14:29Z <p>How does one interpret the ... in the third and fourth stabilizers of the <a href="https://errorcorrectionzoo.org/c/quantum_h" rel="nofollow noreferrer">H code</a>? Is this <span class="math-container">$[1, 2, 5, 6, 9, 10, \dots]$</span> or <span class="math-container">$[1, 2, 5, 6, 7, 8,\dots]$</span>? Can someone please write this out for <span class="math-container">$k = 1, 2, 3, 4$</span>? Thanks.</p> https://quantumcomputing.stackexchange.com/q/28357 5 Reasonable Circuit Depth Ahmad Bennakhi https://quantumcomputing.stackexchange.com/users/21661 2022-09-30T16:24:27Z 2022-10-01T21:52:34Z <p>So I'm auto-constructing quantum circuits that are being auto-generated from dimacs files along with the addition of the amplification function. It's for a satisfiability problem where the number of variables is 9. I end up with a quantum circuit depth that is 6k-10k. The results that I'm getting at the end are always wrong after 3 the execution of 3 different circuits(8k shots each) on the Kolkota processor.</p> <p><strong>Getting to the question</strong> : Is it expected to always get wrong answers with such a depth or is the situation salvagable?</p> https://quantumcomputing.stackexchange.com/q/28352 1 State tomography with Pauli basis measurements for a high number of qubits Borja Aizpurua https://quantumcomputing.stackexchange.com/users/21187 2022-09-30T10:58:46Z 2022-10-03T13:53:44Z <p>My end goal is to recover the quantum state in its computational basis or reduced density matrix of a high number qubit circuit in a real QPU. Taking into account that the number of qubits will be high (+16 or +32 qubits) getting the density matrix as it is done in the common quantum state tomography algorithms is unfeasible.</p> <p>My idea is to try tomography with Pauli basis measurements to get the individual reduced density matrix of each qubit, but due to the high number of qubits, the number of measurements required will be huge too.</p> <p>I was wondering if there is a feasible way to get this or in case there isn't which option would be the least unfeasible to continue researching on it. Thanks in advance.</p> <p>Pd: the circuit is a 16 or 32 qubit circuit with some hadamard, cnot and u(rotation) gates.</p> https://quantumcomputing.stackexchange.com/q/28344 0 Understanding Shor algorithm fo Elliptic Curves Demonstration Philip.q.c https://quantumcomputing.stackexchange.com/users/22014 2022-09-30T00:06:41Z 2022-10-01T21:24:01Z <p>I was reading <a href="https://arxiv.org/abs/quant-ph/0301141" rel="nofollow noreferrer">Shor's discrete logarithm quantum algorithm for elliptic curves</a>. And i have two questions.</p> <ol> <li>In page 7 they say that <span class="math-container">$x = (x0 - dy) mod q$</span>, where <span class="math-container">$x0$</span> is between 0 and q-1, but then they replace <span class="math-container">$x = x0-dy$</span> which is not true. <span class="math-container">$x$</span> is a number between 0 and q-1 and the other expression is sometimes negative when <span class="math-container">$dy&gt;x0$</span>. Something similar happens later when they take <span class="math-container">$y' = dx' mod q$</span>. My question is are they replacing for <span class="math-container">$$x = (x0 - dy) mod q$$</span> <span class="math-container">$$y' = dx' mod q$$</span></li> </ol> <p>and they simple don't put the mods, or for <span class="math-container">$$x = x0 - dy$$</span> <span class="math-container">$$y' = dx'$$</span></p> <p>The second ones are not always true because if <span class="math-container">$dy&gt;x0$</span> you get a negative <span class="math-container">$x$</span> and <span class="math-container">$dx'$</span> sometimes is bigger that q while <span class="math-container">$y'$</span> is between 0 and q-1.</p> <p>If is the first one can someone tell me how the sum is done because the paper simple say &quot;The sum over y is easy to calculate&quot;, but you end up with mods in the exponents and i don't know how to solve it.</p> <ol start="2"> <li>In the same page it says that <span class="math-container">$x0 =(x + dy ) mod q$</span>, which is true, but it's also true that for some <span class="math-container">$d1$</span> you have that <span class="math-container">$x1 =(x + d1y ) mod q$</span>, now replace <span class="math-container">$x$</span> for, <span class="math-container">$x = x1- d1y$</span>, and continue with the same logic from the paper and you get that <span class="math-container">$d1 = y'(x')^{-1} mod q$</span>. Why is this wrong?</li> </ol> https://quantumcomputing.stackexchange.com/q/28328 3 How do we compute quantum walks for a graph? Loic Stoic https://quantumcomputing.stackexchange.com/users/21418 2022-09-29T02:11:40Z 2022-10-01T15:58:00Z <p>I am reading <a href="https://arxiv.org/abs/0810.0312" rel="nofollow noreferrer">Childs' paper</a> on discrete and continuous quantum walks. I do not really understand why quantum walks are useful--- as implementing the quantum walk operator requires knowing the principal eigenvector of a Hamiltonian (The eigenvector with eigenvalue equal to the norm of the Hamiltonian).</p> <p>The paper does mention this:</p> <blockquote> <p>However, it is straightforward to implement the walk for many cases of interest, such as for an unweighted regular graph.</p> </blockquote> <p>But it is confusing to me. What is the graph meant to represent in this case? The Hamiltonian? If not the Hamiltonian, how could we simulate Hamiltonian evolution using this unweighted regular graph method. And how would we construct the walk operator from this?</p> https://quantumcomputing.stackexchange.com/q/28327 3 Thermodynamic Speed Limit to Quantum Computing Matthew Cory https://quantumcomputing.stackexchange.com/users/22003 2022-09-28T23:09:47Z 2022-10-01T21:10:55Z <p>There's a lot of mystifying jargon in the field of quantum computation, so I would like to pose a question from elementary physics to maybe help clarify things. Is it not true that the speed of a real-world reversible computer scales linearly with applied force and entropy? The Heisenberg energy-time bound shows that energy release per time step is greater than Planck's constant over the step time. Also, the bound on total entropy over a complete computation is O(Boltzmann's constant) to avoid decoherence. By Boltzmann law, if the entropy release per step is much greater than Boltzmann's constant, then the quantum computer decoheres, and noise will be read out. So the runtime of a general quantum computer seems to be lower bounded by <span class="math-container">$(h*S^2)/(k*T)$</span>, where <span class="math-container">$S$</span> is the number of steps, <span class="math-container">$h$</span> is Planck’s constant, <span class="math-container">$k$</span> is Boltzmann’s constant and <span class="math-container">$T$</span> is the ambient temperature. The runtime bounds on Grover’s and Shor’s algorithms don’t look too impressive under this basic analysis. It seems that MT and ML bounds dramatically overestimate the speed of quantum evolution. What is a straightforward answer to this objection?</p> <p>Update: I additionally found <a href="https://www.cambridge.org/core/journals/philosophy-of-science/article/abs/end-of-the-thermodynamics-of-computation-a-nogo-result/345E8E78F44B00D0877285841D660F86" rel="nofollow noreferrer">this</a> article by John D. Norton. Abstract: &quot;The thermodynamics of computation assumes that computational processes at the molecular level can be brought arbitrarily close to thermodynamic reversibility and that thermodynamic entropy creation is unavoidable only in data erasure or the merging of computational paths, in accord with Landauer’s principle. The no-go result shows that fluctuations preclude completion of thermodynamically reversible processes. Completion can be achieved only by irreversible processes that create thermodynamic entropy in excess of the Landauer limit.&quot;</p> https://quantumcomputing.stackexchange.com/q/28290 3 Show that $E_k=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)$ implies $U=\begin{bmatrix}[E_1]&\cdots\\ [E_2]&\cdots\\\vdots&\ddots\end{bmatrix}$ Sooraj S https://quantumcomputing.stackexchange.com/users/18369 2022-09-26T18:05:46Z 2022-10-04T00:49:49Z <p>In <a href="http://mmrc.amss.cas.cn/tlb/201702/W020170224608149940643.pdf" rel="nofollow noreferrer">Page 365, Operator-sum representation, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang</a>, it is given that</p> <p><a href="https://i.stack.imgur.com/Jd9B9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jd9B9.png" alt="mock" /></a></p> <hr /> <blockquote> <p>We have a principal system <span class="math-container">$Q$</span> and an environment <span class="math-container">$E$</span> and <span class="math-container">$U$</span> is a unitary operator acting on the combined system <span class="math-container">$QE$</span>. <span class="math-container">\begin{align} \mathcal{E}(\rho)&amp;=tr_E\bigg(U(\rho\otimes|e_0\rangle\langle e_0|)U^\dagger\bigg)\\ &amp;=\sum_k(I\otimes\langle e_k|)(U(\rho\otimes|e_0\rangle\langle e_0|)U^\dagger)(I\otimes |e_k\rangle)\\ &amp;=\sum_k\langle e_k|(U(\rho\otimes|e_0\rangle\langle e_0|)U^\dagger)|e_k\rangle\\ &amp;=\sum_k(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)\rho(I\otimes\langle e_0|)U^\dagger(I\otimes|e_k\rangle)\\ &amp;=\sum_k\langle e_k|U|e_0\rangle\rho\langle e_0|U^\dagger|e_k\rangle\\ &amp;=\sum_k E_k \rho E_k^\dagger \end{align}</span> where <span class="math-container">$E_k=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)=\langle e_k|U|e_0\rangle$</span> is an operator on the state space of the principal system <span class="math-container">$Q$</span>, <span class="math-container">$|e_k\rangle$</span> are the orthonormal basis vectors of the environment, <span class="math-container">$|e_0\rangle$</span> be the initial state of the environment.</p> </blockquote> <p>Please check <a href="https://math.stackexchange.com/questions/4528926/prove-that-tr-big-sum-k-e-k-dagger-e-k-rho-big-1-for-all-rho-implies-s">Prove that <span class="math-container">$tr\Big(\sum_k E_k^\dagger E_k\rho\Big)=1$</span> for all <span class="math-container">$\rho$</span> implies <span class="math-container">$\sum_k E_k^\dagger E_k=I$</span></a> for the derivation.</p> <p>In the problem, <span class="math-container">$E_k=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)=\langle e_k|U|e_0\rangle$</span> where <span class="math-container">$|e_k\rangle$</span> are some orthonormal basis vectors and <span class="math-container">$|e_0\rangle$</span> be the initial state of the environment <span class="math-container">$E$</span>.</p> <blockquote> <p>How do we say that <span class="math-container">$U$</span> is represented as the block matrix <span class="math-container">$$U=\begin{bmatrix} [E_1] &amp; \cdots &amp; \cdots\\ [E_2] &amp; \cdots &amp; \cdots\\ [E_3] &amp; \cdots &amp; \cdots\\ \vdots &amp; \vdots &amp; \vdots \end{bmatrix}$$</span> in the basis <span class="math-container">$|e_k\rangle$</span> ?</p> </blockquote> <p><strong>My observation:</strong></p> <p>Given an operator <span class="math-container">$A:V\to V$</span> <span class="math-container">$$A=\begin{bmatrix}a_{00}&amp;a_{01}\\a_{10}&amp;a_{11}\end{bmatrix}=\begin{bmatrix}\langle0|A|0\rangle&amp;\langle0|A|1\rangle\\\langle1|A|0\rangle&amp;\langle1|A|1\rangle\end{bmatrix}=\sum_{i,j}a_{j}|i\rangle\langle j|$$</span> If we had an operator <span class="math-container">$A:V\to V$</span> and <span class="math-container">$\{|v_i\rangle\}$</span> constitute a basis of <span class="math-container">$V$</span> then</p> <p>Let <span class="math-container">$\{|v_i\rangle\}$</span> be an orthonormal basis then the change of basis matrix is <span class="math-container">$P_v=V^{-1}=V^\dagger=\begin{bmatrix}\langle v_0|\\\langle v_1|\end{bmatrix}$</span></p> <p>The matrix <span class="math-container">$A$</span> in the orthonormal basis <span class="math-container">$\{|v_i\rangle\}$</span> is, <span class="math-container">$$B=[A]_v=P_vAP_v^{-1}=\begin{bmatrix}\langle v_0|\\\langle v_1|\end{bmatrix}A\begin{bmatrix} |v_0\rangle&amp; |v_1\rangle\end{bmatrix}=\begin{bmatrix}\langle v_0|A|v_0\rangle&amp;\langle v_0|A|v_1\rangle\\\langle v_1|A|v_0\rangle&amp;\langle v_1|A|v_1\rangle\end{bmatrix}$$</span> <span class="math-container">$$A=P_v^\dagger BP_v=\begin{bmatrix} |v_0\rangle&amp; |v_1\rangle\end{bmatrix}\begin{bmatrix}\langle v_0|A|v_0\rangle&amp;\langle v_0|A|v_1\rangle\\\langle v_1|A|v_0\rangle&amp;\langle v_1|A|v_1\rangle\end{bmatrix}\begin{bmatrix}\langle v_0|\\\langle v_1|\end{bmatrix}\\ =\sum_{i,j}\langle v_i|A|v_j\rangle|v_i\rangle\langle v_j|$$</span> That means, <span class="math-container">$$A=\sum_{i,j}a_{ij}|i\rangle\langle j|=\sum_{i,j}b_{ij}|v_i\rangle\langle v_j|$$</span> where <span class="math-container">$a_{ij}=\langle i|A|j\rangle$</span> and <span class="math-container">$b_{ij}=\langle v_i|A|v_j\rangle$</span></p> <p>This is clear!</p> <p>So the <span class="math-container">$(i,j)^{th}$</span> term of <span class="math-container">$A$</span> in the basis <span class="math-container">$|v_i\rangle$</span> is <span class="math-container">$b_{ij}=\langle v_i|A|v_j\rangle$</span> and <span class="math-container">$B=[A]_v=\sum_{i,j}b_{ij}|i\rangle\langle i|$</span></p> <p>Unlike <span class="math-container">$\langle v_i|W|v_j\rangle$</span> which is a number and <span class="math-container">$|v_j\rangle$</span> is a vector, the term <span class="math-container">$E_k=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)=\langle e_k|U|e_0\rangle$</span> is an operator acting on the first system (principal system <span class="math-container">$Q$</span>) and <span class="math-container">$I\otimes|e_0\rangle=|e_0\rangle,I\otimes|e_k\rangle=|e_k\rangle$</span> are matrices.</p> <p><strong>My Attempt</strong></p> <p>Thanks @glS for the hint.</p> <p><span class="math-container">\begin{align} E_k&amp;=(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)=\langle e_k|U|e_0\rangle\\ \end{align}</span> If we interchange the order of systems, i.e., the first system is the environment <span class="math-container">$E$</span> and second is the principal system <span class="math-container">$Q$</span>, then</p> <p>Decomposing <span class="math-container">$U$</span> and <span class="math-container">$(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)$</span> in terms of the orthonormal basis vectors of the respective systems as,</p> <p><span class="math-container">\begin{align} &amp;(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)=\\ &amp;=(\langle e_k|\otimes I)(\sum_{m,n,m',n'}\mu_{m,n,m',n'}(|e_n\rangle\otimes|p_m\rangle)(\langle e_{n'}|\otimes\langle p_{m'}|))(|e_{0}\rangle\otimes I)\\ &amp;=\langle e_k|\otimes I\bigg[\sum_{m,n,m',n'}\mu_{m,n,m',n'}|e_n\rangle\langle e_{n'}|\otimes|p_m\rangle\langle p_{m'}|\bigg]|e_{0}\rangle\otimes I\\ &amp;=\sum_{m,n,m',n'}\mu_{m,n,m',n'}\langle e_k|e_n\rangle\langle e_{n'}|e_{0}\rangle\otimes|p_m\rangle\langle p_{m'}|\\ &amp;=\sum_{m,n,m',n'}\mu_{m,n,m',n'}\delta_{k,n}\delta_{n',0}\otimes|p_m\rangle\langle p_{m'}|\\ &amp;=\sum_{m,m'}\mu_{m,k,m',0}|p_m\rangle\langle p_{m'}|\\ \end{align}</span></p> <p><span class="math-container">\begin{align} U&amp;=\sum_{m,n,m',n'}\mu_{m,n,m',n'}(|e_n\rangle\otimes|p_m\rangle)(\langle e_{n'}|\otimes\langle p_{m'}|)\\ &amp;=\sum_{m,n,m',n'}\mu_{m,n,m',n'}|e_n\rangle\langle e_{n'}|\otimes|p_m\rangle\langle p_{m'}|\\ &amp;=\sum_{n,n'}|e_n\rangle\langle e_{n'}|\otimes\sum_{m,m'}\mu_{m,n,m',n'}|p_m\rangle\langle p_{m'}|\\ \end{align}</span></p> <p>If I define the change of basis matrix <span class="math-container">$P=P_E\otimes P_p$</span> then (extending the ideas in the previous section) the operators in the new basis are, <span class="math-container">$$(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)'=\sum_{m,m'}\mu_{m,k,m',0}|m\rangle\langle {m'}|\\$$</span> <span class="math-container">\begin{align} U'&amp;=\sum_{m,n,m',n'}\mu_{m,n,m',n'}|n\rangle\langle {n'}|\otimes|m\rangle\langle {m'}|\\ &amp;=\sum_{n,n'}|n\rangle\langle {n'}|\otimes \sum_{m,m'}\mu_{m,n,m',n'}|m\rangle\langle {m'}|\\ &amp;=\sum_{n,n'}|n\rangle\langle {n'}|\otimes\sigma_{n,n'} \end{align}</span> where <span class="math-container">$\{|n\rangle\}$</span> and <span class="math-container">$\{|m\rangle\}$</span> are the standard basis such that <span class="math-container">$|n\rangle\langle {n'}|$</span> is the matrix with <span class="math-container">$1$</span> as its <span class="math-container">$(n,n')$</span> entry, and <span class="math-container">$\sigma_{n,n'}=\sum_{m,m'}\mu_{m,n,m',n'}|m\rangle\langle {m'}|$</span> is the <span class="math-container">$(n,n')^{th}$</span> block entry of <span class="math-container">$U'$</span>.</p> <p>Therefore, the <span class="math-container">$(k,0)^{th}$</span> block entry of <span class="math-container">$U'$</span> is</p> <p><span class="math-container">$\sigma_{k,0}=\sum_{m,m'}\mu_{m,k,m',0}|m\rangle\langle {m'}|=(\langle e_k|\otimes I)U(|e_0\rangle\otimes I)'$</span></p> https://quantumcomputing.stackexchange.com/q/28233 0 How to convert a large CQM to a BQM [closed] Roland https://quantumcomputing.stackexchange.com/users/20619 2022-09-20T10:58:04Z 2022-10-02T08:20:08Z <p>My constraints are defined in a generic format, i.e. as an arithmetic expression (something that is accepted by CQM). I am trying to convert this expression to a BQM that can be added as a penalty term to the 'composite' BQM.</p> <p>I tried to set my problem directly as a CQM and then convert the CQM to a BQM to sample it with DWaveSampler, but the process is terribly slow as my problem is a large one (90 x 90000 variables). More specifically, I tried to convert a CQM with many variables to a BQM with the method <code> bqm = dimod.cqm_to_bqm(cqm)</code>, and this operation is taking forever (or even crashing).</p> <p>I am thinking about some possible alternative, as individually converting each constraint (together with the objective) to a BQM and then add it to the 'composite' BQM.</p> <p>Maybe there is a better strategy for this, but I do not know.</p> https://quantumcomputing.stackexchange.com/q/28206 8 Non-entangling two-qubit gates Ian Gershon Teixeira https://quantumcomputing.stackexchange.com/users/19675 2022-09-18T00:32:57Z 2022-10-02T16:20:40Z <p>The non-entangling gates in <span class="math-container">$SU_4$</span> contains the entire group of gates of the form <span class="math-container">$$SU_2 \otimes SU_2.$$</span> It also contains <span class="math-container">$$\zeta_8 SWAP= \zeta_8 \begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ \end{bmatrix}$$</span> where <span class="math-container">$\zeta_8=e^{2\pi i/8}= e^{\pi i/4}$</span> is a primitive eighth root of unity.</p> <p>Are there any other non-entangling two-qubit gates? A related (perhaps equivalent?) question is what is the normalizer of <span class="math-container">$SU_2 \otimes SU_2$</span> in <span class="math-container">$SU_4$</span>? Does the normalizer <span class="math-container">$$N(SU_2 \otimes SU_2)$$</span> just have two connected components (the component of the identity and the component of SWAP)? Does it have more connected components? Do these other components correspond to other non-entangling gates? Also, interesting to note that <span class="math-container">$$(\zeta_8 SWAP)^2=iI \not \in SU_2 \otimes SU_2$$</span> is not in <span class="math-container">$SU_2 \otimes SU_2$</span> even though we wouldn't think of it as an entangling gate since it is just a global phase and moreover it is in <span class="math-container">$U_2 \otimes U_2$</span>.</p> https://quantumcomputing.stackexchange.com/q/28144 1 Show the linearity of $(\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|) U(I_{A}\otimes I_B\otimes |0_{C}\rangle\otimes |0_{D}\rangle)$ Sooraj S https://quantumcomputing.stackexchange.com/users/18369 2022-09-12T13:38:08Z 2022-10-03T01:22:40Z <blockquote> <p>Suppose a composite system <span class="math-container">$AB$</span> initially in an unknown quantum state <span class="math-container">$\rho$</span> is brought into contact with a composite system <span class="math-container">$CD$</span> initially in some standard state <span class="math-container">$|0\rangle$</span>, and the two systems interact according to a unitary interaction <span class="math-container">$U$</span>. After the interaction we discard systems <span class="math-container">$A$</span> and <span class="math-container">$D$</span>, leaving a state <span class="math-container">$\rho\:'$</span> of the system <span class="math-container">$BC$</span>. Show that the map <span class="math-container">$\mathcal{E}(\rho)=\rho\:'$</span> satisfies <span class="math-container">$$\mathcal{E}(\rho)=\sum_k E_k\rho E_k^\dagger$$</span> for some set of linear operators <span class="math-container">$E_k$</span> from the state space of system <span class="math-container">$AB$</span> to the state space of system <span class="math-container">$BC$</span>, and such that <span class="math-container">$\sum_kE_k^\dagger E_k=I$</span>.</p> </blockquote> <p>This is given, in <a href="http://mmrc.amss.cas.cn/tlb/201702/W020170224608149940643.pdf" rel="nofollow noreferrer">Exercise 8.3, Page 361, Operator-sum representation, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang</a>.</p> <p>Overall state of the system after the interaction is <span class="math-container">$U(\rho\otimes|0\rangle\langle 0|)U^\dagger$</span></p> <p>Let <span class="math-container">$|a_m\rangle,|b_n\rangle,|c_p\rangle,|d_q\rangle$</span> be orthonormal basis for the state space of the systems <span class="math-container">$A,B,C,D$</span>, respectively.</p> <p>Discarding systems <span class="math-container">$A$</span> and <span class="math-container">$D$</span>, i.e., tracing out system <span class="math-container">$A$</span> first then tracing out the system <span class="math-container">$D$</span> obtains</p> <p><span class="math-container">\begin{align} \mathcal{E}(&amp;\rho)=tr_D\Big[tr_A\big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big]\Big]\\ &amp;=tr_D\Big[\sum_m(\langle a_m|\otimes I\otimes I\otimes I) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes I)\Big]\\ &amp;=\sum_q(I\otimes I\otimes I\otimes \langle d_q|)\\ &amp;\times\Big[\sum_m(\langle a_m|\otimes I\otimes I\otimes I) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes I)\Big](I\otimes I\otimes I\otimes |d_q\rangle)\\ &amp;=\sum_{m,q}(\langle a_m|\otimes I\otimes I\otimes \langle d_q|) \big[U(\rho\otimes|0\rangle\langle 0|)U^\dagger\big](|a_m\rangle\otimes I\otimes I\otimes |d_q\rangle) \end{align}</span></p> <p>Expanding <span class="math-container">$\rho\otimes|0\rangle\langle 0|$</span> into the products</p> <p><span class="math-container">\begin{align} \rho\otimes|0\rangle\langle 0|&amp;=\rho_{AB}\otimes|0_{CD}\rangle\langle 0_{CD}|=\color{blue}{(\rho_{AB}\otimes I_{CD})(I_{AB}\otimes|0_{CD}\rangle)}(I_{AB}\otimes\langle 0_{CD}|)\\ =&amp;\color{blue}{(\rho_{AB}I_{AB}\otimes I_{CD}|0_{CD}\rangle)}(I_{AB}\otimes\langle 0_{CD}|)=\color{blue}{(I_{AB}\rho_{AB}\otimes |0_{CD}\rangle.1)}(I_{AB}\otimes\langle 0_{CD}|)\\ =&amp;\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)(\rho_{AB}\otimes 1)}(I_{AB}\otimes\langle 0_{CD}|)=\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)\rho_{AB}}(I_{AB}\otimes\langle 0_{CD}|)\\ =&amp;\color{blue}{(I_{AB}\otimes |0_{CD}\rangle)\rho}(I_{AB}\otimes\langle 0_{CD}|) \end{align}</span> Substituting back into the expression, <span class="math-container">\begin{align} \mathcal{E}(&amp;\rho)=\\\\ =&amp;\sum_{m,q}(\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)\rho(I_{AB}\otimes\langle 0_{CD}|)U^\dagger(|a_m\rangle\otimes I\otimes I\otimes |d_q\rangle)\\\\ =&amp;\sum_{m,q} E_{m,q}\rho E_{m,q}^\dagger \end{align}</span> where <span class="math-container">$E_{m,q}=\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)$</span></p> <p><span class="math-container">\begin{align} &amp;\color{red} {\text{How do I show that this is a linear operator from the state space of $AB$ to the state}}\\ &amp;\color{red} {\text{ space of $BC$ ?}} \end{align}</span> <strong>My Attempt</strong></p> <p>If <span class="math-container">$|a_m\rangle,|b_n\rangle,|c_p\rangle,|d_q\rangle$</span> be orthonormal basis for the state space of the systems <span class="math-container">$\mathbb{C}^{i},\mathbb{C}^{j},\mathbb{C}^{k},\mathbb{C}^{l}$</span> respectively, then <span class="math-container">$|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle$</span> is an orthonormal basis for the combined system <span class="math-container">$\mathbb{C}^{i}\otimes\mathbb{C}^{j}\otimes\mathbb{C}^{k}\otimes\mathbb{C}^{l}=\mathbb{C}^{ijkl}$</span>. <span class="math-container">\begin{align} &amp;\color{gray}{\text{Lemma: If $V$ and $W$ are vector spaces with basis $|v_i\rangle$ and $|w_j\rangle$ respectively. Then }}\\ &amp;\color{gray}{|v_i\rangle\otimes_{outer}|w_j\rangle=|v_i\rangle\langle w_j| \text{form a basis for the vector space $V\otimes_{outer}W$.}} \end{align}</span> Therefore, <span class="math-container">$(|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle)(\langle a_{m'}|\otimes\langle b_{n'}|\otimes\langle c_{p'}|\otimes\langle d_{q'}|)=|a_m\rangle\langle a_{m'}|\otimes|b_n\rangle\langle b_{n'}|\otimes|c_p\rangle\langle c_{p'}|\otimes|d_q\rangle\langle d_{q'}|$</span> for a basis for the space <span class="math-container">$\mathbb{C}^{ijkl}\otimes_{outer}\mathbb{C}^{ijkl}$</span>.</p> <p><span class="math-container">$\therefore U\in\mathbb{C}^{ijkl}\otimes_{outer}\mathbb{C}^{ijkl}$</span> can be written as a linear combination of the basis <span class="math-container">$(|a_m\rangle\otimes|b_n\rangle\otimes|c_p\rangle\otimes|d_q\rangle)(\langle a_{m'}|\otimes\langle b_{n'}|\otimes\langle c_{p'}|\otimes\langle d_{q'}|)=|a_m\rangle\langle a_{m'}|\otimes|b_n\rangle\langle b_{n'}|\otimes|c_p\rangle\langle c_{p'}|\otimes|d_q\rangle\langle d_{q'}|$</span>.</p> <p>i.e., <span class="math-container">$$U=\sum_{m',n',p',q'}\nu_{m',n',p',q'} (|a_{m'}\rangle\otimes|b_{n'}\rangle\otimes|c_{p'}\rangle\otimes|d_{q'}\rangle)(\langle a_{m''}|\otimes\langle b_{n''}|\otimes\langle c_{p''}|\otimes\langle d_{q''}|)\\ =\sum_{m',n',p',q'}\nu_{m',n',p',q'} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}|$$</span></p> <p>And <span class="math-container">$|0_{CD}\rangle$</span> being a standard state of the system <span class="math-container">$CD$</span> can be represented in terms of the orthonormal basis vectors as, <span class="math-container">$$|0_{CD}\rangle=\sum_{p,q}\eta_{p,q} |c_p\rangle\otimes|d_q\rangle$$</span></p> <p>Substituting back into the expression for <span class="math-container">$E_{m,q}$</span> obtains <span class="math-container">\begin{align} &amp;E_{m,q}=\\ &amp;=\langle a_m|\otimes I\otimes I\otimes \langle d_q|) U(I_{AB}\otimes |0_{CD}\rangle)\\ &amp;=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|) U\sum_{p''',q'''}\eta_{p''',q'''}(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ &amp;=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|)\Big[ \sum_{m',n',p',q'\\m'',n'',p'',q''}\nu_{m',n',p',q'\\m'',n'',p'',q''} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}|\Big]\\ &amp;\times\sum_{p''',q'''}\eta_{p''',q'''}(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ &amp;=\langle a_m|\otimes I_B\otimes I_C\otimes \langle d_q|)\Big[ \sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} |a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|\otimes|d_{q'}\rangle\langle d_{q''}|\Big]\\ &amp;\times(I_{A}\otimes I_{B}\otimes|c_{p'''}\rangle\otimes|d_{q'''}\rangle)\\ \\ &amp;=\sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} \langle a_{m}|a_{m'}\rangle\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\langle c_{p''}|c_{p'''}\rangle\otimes\langle d_{q}|d_{q'}\rangle\langle d_{q''}|d_{q'''}\rangle\\ \\ &amp;=\sum_{m',n',p',q'\\m'',n'',p'',q''}\sum_{p''',q'''}\eta_{p''',q'''}\nu_{m',n',p',q'\\m'',n'',p'',q''} \delta_{m,m'}\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\delta_{p'',p'''}\otimes\delta_{q,q'}\delta_{q'',q'''}\\ \\ &amp;=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} \langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle\otimes 1\\ &amp;=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} (1.\langle a_{m''}|\otimes|b_{n'}\rangle\langle b_{n''}|\otimes|c_{p'}\rangle.1\otimes 1.1)\\ &amp;=\sum_{n',p',q,m'',n'',p'',q''}\eta_{p'',q''}\nu_{m,n',p',q\\m'',n'',p''} (1\otimes|b_{n'}\rangle\otimes|c_{p'}\rangle\otimes 1)(\langle a_{m''}|\otimes\langle b_{n''}|\otimes 1\otimes 1)\\\end{align}</span></p> <p><span class="math-container">$\implies E_{m,q}$</span> is a sum of operators from <span class="math-container">$AB$</span> to <span class="math-container">$BC$</span>.</p> https://quantumcomputing.stackexchange.com/q/28114 0 Building Unitaries in TSP Quantum Phase Estimation Qiskit? Nghia Nguyen Huu https://quantumcomputing.stackexchange.com/users/21843 2022-09-10T06:29:20Z 2022-10-03T09:28:50Z <p><a href="https://i.stack.imgur.com/qIOnZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qIOnZ.png" alt="enter image description here" /></a></p> <p>In the paper [https://arxiv.org/pdf/1805.10928.pdf] <a href="https://arxiv.org/pdf/1805.10928.pdf" rel="nofollow noreferrer">2</a> published by IBM, they use Quantum Phase Estimation to solve the TSP problem. I don't understand the reason why they can decompose the diagonal unitary matrix U(j). Is that their own work to find the two multiplying components in the form of P-gates to have the unitary matrix U at the end, or have they followed the lemma or some other form to decompose? Thanks for your help!</p> https://quantumcomputing.stackexchange.com/q/28039 1 Calculating state fidelity and space complexity of Minimum Eigen Optimizers (VQE, QAOA and Grover Optimizer) in qiskit Janani Ananthanarayanan https://quantumcomputing.stackexchange.com/users/21767 2022-09-03T08:51:39Z 2022-10-03T15:08:20Z <p>I'm a beginner in using Qiskit and my Computer Science background is not extensive. But I understand the Quantum Physics aspects of it relatively well.</p> <p>I solved a QUBO problem in Qiskit using VQE (with CVaR), QAOA and Grover Optimizer, and got the correct results with all three and wish to compare the performance of each. I have already done that for execution time using <code>time.process_time()</code> for each and averaging over 15 runs, at relevant intervals.</p> <p>I would also like to compare the state fidelities and memory used up by each (for space complexity), but don't know what to import and which function to execute from the qiskit library. I also am not sure how to call for the final state-vector results/density matrices that are stored in the result.</p> <p>Can you please help me with this? Would also appreciate if you can tell me what other metrics exist that can help me compare the performance of each QA.</p> https://quantumcomputing.stackexchange.com/q/26582 3 How do I perform an erasure error in stim? Craig Gidney https://quantumcomputing.stackexchange.com/users/119 2022-05-26T18:13:11Z 2022-10-03T15:29:38Z <p>An erasure error is a heralded error that completely destroys a qubit (e.g. resets it or maximally mixes it). The qubit is gone but you <em>are told</em> it's gone. How do I simulate this kind of error using stim?</p> https://quantumcomputing.stackexchange.com/q/26216 2 Is the Clifford group superperfect? Ian Gershon Teixeira https://quantumcomputing.stackexchange.com/users/19675 2022-04-29T17:43:55Z 2022-10-02T16:51:37Z <p>This is a follow up to <a href="https://quantumcomputing.stackexchange.com/questions/26150/is-the-clifford-group-perfect-equals-its-own-commutator-subgroup/26165?noredirect=1#comment34160_26165">Is the Clifford group perfect (equals its own commutator subgroup)?</a></p> <p>Motivation:</p> <p>Since global phase is unphysical in quantum mechanics we often consider projective representations, where the matrices are only well defined modulo a global <span class="math-container">$e^{i\theta}$</span>, instead of true linear representations.</p> <p>It turns out that the projective representations of <span class="math-container">$G$</span> correspond exactly to the linear representations of the universal cover which is a central extension of the original group. For example <span class="math-container">$SO_3$</span> has universal cover <span class="math-container">$SU_2$</span> and projective representations of <span class="math-container">$SO_3$</span> correspond to half integer spin in quantum mechanics.</p> <p>This story for semisimple Lie groups has an analogue in the theory of perfect finite groups. For a perfect group <span class="math-container">$G$</span> there is a universal central extension, sometimes called the universal cover, with the property that the projective representations of <span class="math-container">$G$</span> are in exact correspondence with the linear representations of the universal cover.</p> <p>In the theory of semisimple Lie groups a group which is its own universal cover is called simply connected,this is equivalent to the fundamental group of <span class="math-container">$G$</span> being trivial. For a semisimple Lie group the fundamental group is always a finite abelian group.</p> <p>In the theory of perfect groups a group which is its own universal cover is called superperfect, this is equivalent to the schur multiplier being trivial. For a perfect finite group the schur multiplier is always a finite abelian group.</p> <p>The Clifford group <span class="math-container">$\overline{Cl}_n$</span> (the automorphism group of the Pauli group <span class="math-container">$P_n$</span> ) is a perfect group which is important in quantum computing. I want to know if <span class="math-container">$\overline{Cl}_n$</span> is superperfect or if the there exists some nontrivial perfect central extensions (i.e. the schur multiplier is nontrivial). If the schur multiplier is nontrivial I would certainly be curious which finite abelian group it is.</p> https://quantumcomputing.stackexchange.com/q/18630 2 Qiskit CNOT-gate matrix mixup? Alvo https://quantumcomputing.stackexchange.com/users/16771 2021-07-27T13:03:19Z 2022-09-30T22:45:37Z <p>In the qiskit textbook <a href="https://qiskit.org/textbook/ch-gates/multiple-qubits-entangled-states.html#3.1-The-CNOT-Gate-" rel="nofollow noreferrer">chapter 1.3.1</a> &quot;The CNOT-Gate&quot; it says that the matrix representation on the right is the own corresponding to the circuit shown above, with q_0 being the control and q_1 the target, but shouldn't this matrix representation be for the case of q_1 being the control and q_0 the target? This seems to be presented the other way round...or there seems to be something I am not quite getting yet.</p> <p>Thanks so much :)</p> <p>Quick edit: By &quot;right&quot; I am referring to this: <a href="https://i.stack.imgur.com/ohi7V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ohi7V.png" alt="enter image description here" /></a></p> https://quantumcomputing.stackexchange.com/q/16067 1 How to implement Majority Vote Kiji https://quantumcomputing.stackexchange.com/users/14447 2021-02-16T12:48:51Z 2022-10-02T17:45:16Z <p>I am trying to boost the success probability of standard phase estimation by repeating the procedure enough times and taking a majority vote that will be encoded in a quantum register. My problem is I don't know how to implement a majority vote.</p> <p>For 3 qubits <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> taken in <span class="math-container">$\{|0\rangle, |1\rangle \}$</span> the majority vote circuit returns <span class="math-container">$MV(a, b, c)$</span> and can be implemented as follows:</p> <p><a href="https://i.stack.imgur.com/TxcE0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TxcE0.png" alt="Majority vote 3 qubits" /></a></p> <p>For example, if <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> are all in state <span class="math-container">$|0\rangle$</span> or if <span class="math-container">$b$</span> or <span class="math-container">$c$</span> are in state <span class="math-container">$|1\rangle$</span> then nothing happens and <span class="math-container">$MV(a, b, c) = a$</span>. If <span class="math-container">$b$</span> and <span class="math-container">$c$</span> are in <span class="math-container">$|1\rangle$</span>, the CCNOT gate flips <span class="math-container">$a$</span>.</p> <p>If <span class="math-container">$a$</span>, <span class="math-container">$b$</span> and <span class="math-container">$c$</span> are in state <span class="math-container">$|+\rangle$</span>, then <span class="math-container">$MV(a, b, c) = |+\rangle$</span> and we have 50% probability for each outcome.</p> <p>How can we generalize this circuit to 5 qubits and higher ?</p> <p>(<a href="https://imgur.com/FVQezhb" rel="nofollow noreferrer">My motivation is to implement this</a>)</p> https://quantumcomputing.stackexchange.com/q/10290 3 Installing Q# in jupyter quanity https://quantumcomputing.stackexchange.com/users/10471 2020-03-25T14:36:42Z 2022-10-03T06:06:25Z <p><a href="https://i.stack.imgur.com/oL6DS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oL6DS.png" alt="enter image description here"></a>I have followed all steps described in <a href="https://docs.microsoft.com/en-us/quantum/install-guide/qjupyter" rel="nofollow noreferrer">https://docs.microsoft.com/en-us/quantum/install-guide/qjupyter</a></p> <p>But I am unable to get through part 3. It is showing syntax error in Python terminal in Jupyter notebook. I am not seeing Q# terminal in jupyter notebook (New) folder</p> <p>Now after so many attempts its showing<a href="https://i.stack.imgur.com/NVdSP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NVdSP.png" alt="enter image description here"></a> I am done with this. I think its better to switch to Qiskit.</p> <p><a href="https://i.stack.imgur.com/qUcfE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qUcfE.png" alt="enter image description here"></a> </p> https://quantumcomputing.stackexchange.com/q/6958 7 What design considerations set the frequency bounds for superconducting qubits? psitae https://quantumcomputing.stackexchange.com/users/1867 2019-08-07T04:42:09Z 2022-10-04T03:09:11Z <p>Superconducting qubits generally have frequencies within the range of 4 - 8 GHz. What design considerations give the upper and lower bounds for what is a feasible design. I.e, why can't they be higher or lower in frequency?</p>