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0

@ahelwer has provided a smart construction, and I will provide a more brute-force method as a complementary for beginners like me. It actually starts out by listing all the possible 16 moves. Let x,y=0,1,2,3, then all the possible moves are $ x=0, \quad y=(0123)\rightarrow(0123) \\ x=1, \quad y=(0123)\rightarrow(1230) \\ x=2, \quad y=(0123)\rightarrow(2301) ...


4

What is this business of global phase? Unfortunately, most people encounter this when seeing their first calculation and the teacher says, "we can just get rid of that because overall phases don't matter." But where does this come from? First, the math answer: States of quantum systems are properly modelled as complex projective spaces. (More loosely: 1 ...


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Note: $I_k$ is unit matrix of order $k$ in the following text. First step of the algorithm is $H \otimes H \otimes I_2 \otimes I_2$ as you mentioned. A controlled gate $U$ with $n$ qubits between the control qubit and the target qubit can expressed as a matrix $$ CU_{n} = \begin{pmatrix} I_{\frac{N}{2}} & O_{\frac{N}{2}} \\ O_{\frac{N}{2}} & I_{\...


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It seems that $\mathrm{CNOT}$ gates should not be in your circuit. Here is Grover algorithm for 3 qubits: Put your Oracle instead of dashed line. The Oracle should have three inputs $q_0$, $q_1$ and $q_2$, output should be on qubit $q_3$ after $\mathrm{H}$ gate.


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An application of the reflection operator change each amplitude of a basis state $ |i\rangle$ by $$ \alpha_i \rightarrow- \alpha_i + 2 \langle\alpha\rangle\ $$ where $ \langle\alpha\rangle\ $ is the average of all amplitudes. It follows the oracle which is use to "mark" the seeked elements $ |i\rangle$. Say for instance that you have 8 elements, so you ...


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If you have one parameter (one $\theta$) for both circuits then I think the first one is better... they are doing the same job, but the second one is creating extra gates. So the first one will be faster and will have fewer errors because there are fewer gates in the first circuit. But if you are obtaining the second circuit with two parameters (two ...


2

It seems that there is a missing $\mathrm{X}$ and $\mathrm{H}$ gates on qubit $q_2$. I used this Grover algoritm shape: Note 1: controlled $\mathrm{Z}$ is replaced by $\mathrm{CNOT}$ and Hadamards on both sides. Note 2: put your Oracle instead of dashed line. Note 3: you do not have to measure $q_2$.


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I don't think a hello world really exists here. You can have different points of view or goals here. I will give references. The first one is speeding up parts of the algorithm with a quantum version (here is an example reference). But here, we assume a perfect hardware. Another one is to apply it to quantum many-body systems. The interesting point here is ...


5

This is a purely hypothetical answer - I don't know if anybody has ever studied it, and have not attempted to find out - but think about public key cryptography. Current public key systems are based on the idea that some problems in the complexity class NP are probably hard to solve directly, but there exists a "proof" that lets you verify the solution ...


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You should put measurement after all quantum gates. According to the theory, there is no difference whether gates (in this case CNOTs) are controlled by qubits or classical bits. I encountered the same problem, see solution here.


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The error is caused by appending gates onto qubits following a measurement. On qubit 1 and qubit 0, you attach a cx gate after a measurementhas already been placed. This will compile on the simulator, but it is not something that is supported on the real hardware.


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As far as I know, there are four possibilities for having a quantum advantage in Bayesian machine learning: Gaussian processes: there is a known quantum speed-up for Gaussian processes that you can easily test on IBM Q [1,2]. The idea is to use HHL (quantum algorithm for matrix inversion) in order to compute the inverse of the kernel matrix, which is used ...


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