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2 votes

How to determine with local measurements which Bell state we have?

If two-qubit unitary transformations are allowed then the problem is straightforward. Consider the unitary $$ U = (|0\rangle \langle 0 | \otimes I + |1\rangle \langle 1|\otimes I) (H \otimes I), $$ i....
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3 votes

How can orthonormal vectors satisfy $\langle i|j\rangle=\delta_{ij}$?

When both vectors are equal and normalized $\langle 1 \vert 1 \rangle = 1$. This normalization is well defined since any inner product space is also a normed space $\Vert \vert i \rangle \Vert := \...
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2 votes
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Equivalence between quantum circuit: CNOT changes control and target qubit

Your computations are correct. The last steps are: $$\begin{align*} &\frac14\sum_{i,j,k,l\in\{0,1\}}(-1)^{xi+yj+li+k(i\oplus j)}|l,k\rangle\\ ={}&\frac14\sum_{k,l\in\{0,1\}}\left(\sum_{i,j\in\{...
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1 vote

Do unitary matrices acting on entangled states always give a quantum state?

Yes, you can absolutely simplify down. The problem you're having isn't one of simplification, rather one of counting all the terms in the normalisation, although you appear to have picked up a stray $\...
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  • 3,457
1 vote

Solution to problem 5.3 Book Quantum Computation and Quantum Information Nielsen Chuang regarding Kitaev's algorithm

Whenever you have an ancilla qubit in a circuit, you can reduce the problem into a 2-dimensional problem by thinking of the circuit in terms of blocking encodings: $$H \otimes I^n = \frac{1}{\sqrt{2}}...
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  • 316
1 vote
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Implementing deutsch_problem(seed=None) and deutsch() from qiskit text

This is how I do it, I tried to append gate directly with to_gate, remember to transpile or decompose the label gate before run simulator. if you want to get unitary of the gate, you can use Operator ...
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  • 551
4 votes
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Understanding Controlled Operation

The best way to see this is through an example. Consider the circuit below: qubit that is in an eigenstate of the unitary Note that $|-\rangle$ is an eigenstate ...
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3 votes

Probability of success proof for Shor's algorithm

TL;DR This is a consequence of the Chinese Remainder theorem and the fact that in any group if $g^k$ is the identity then the order of $g$ divides $k$. Setup We are factoring an odd composite ...
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0 votes

How to calculate the sum of $\sum\limits_{j}{\langle A|{{B}_{j}}\rangle |{{C}_{j}}\rangle }$ with quantum circuits?

Here is one way to do it: You can compute the inner product $\langle A |B_j\rangle$ with inner() method of Statevector. Just ...
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