New answers tagged quantum-gate
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The basic idea is to multiply $U$ on the left with $2\times 2$ unitaries until the identity is obtained. This method provides a sequence of gates $U_k$ such that $U_1\cdots U_n U=I$, which then gives you the decomposition of $U$ in terms of $2\times2$ unitaries: $U=U_1^\dagger \cdots U_n^\dagger$.
For example, suppose you start with
$$U=\begin{pmatrix}1/2&...
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Their approach is more advanced than the simple one, described in the book "Quantum Computation and Quantum Information" by M. Nielsen and I. Chuang, section 4.5.1. It's better to understand it first. Basically we are just making zeros under diagonal step by step, where each step is the multiplication by some two-level unitary. Hence there are only $d(d-1)/2$...
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The metric you're looking for is known as entangling power.
Here are some references:
Entangling power of two-qubit gates on mixed states (Guan et al., 2014)
Entangling power of two-qubit unitary operations (Yi Shen & Lin Chen, 2018)
I will expand on this later.
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$\mathrm{X}$ is not equivalent to a $\mathrm{CNOT}$ gate. The former is a 1-qubit gate whereas the 2nd is a 2-qubit gate (in essence, a controlled-$\mathrm{X}$). The $\mathrm{X}$ basically flips the state of qubit B i.e., $|0\rangle_B\to|1\rangle_B$ and $|1\rangle\to|0\rangle_B$, and does not depend on the state of qubit A.
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How it works depends on the choice of quantum system used for computation. For any choice of quantum system, the common theme is that $\text{CNOT}$ does not collapse the wavefunction, i.e. force a choice between $\vert 0 \rangle$ and $\vert 1 \rangle$, while a measurement does.
A simple example (oversimplified here) uses a non-linear Kerr medium to create ...
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A mechanistic way of looking at it that is sure to infuriate any genuine physicists in the discussion :) is that a CNOT is two supercooled transmons interacting as described in the Open Pulse documentation and that the mystical "observation" does not occur until there is an interaction outside that environment.
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The key to understanding many quantum protocols and circuits is in the following circuit:
This is especially true in the case where $U^2=I$, such that $U$ has eigenvalues $\pm1$. You can readily calculate that if the input, $|\psi\rangle$, of the second qubit has an amplitude $\alpha_+$ for being supported on the $+1$ eigenspace, then at the end of the ...
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What that cSWAP test does (and doesn't) do
The important thing about the controlled-SWAP test is that what it does isn't just to SWAP, or to not SWAP, the two inputs. The controlled-SWAP test involves a control qubit which is in a superposition of $\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\ket{0}$ and $\ket{1}$: that is, we measure the first ...
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In general, "operating on a state with an observable" does not have direct physical meaning (i.e. you cannot think of it as evolving the state doing something to it).
What does have physical meaning, is applying a unitary operation to a state. Every unitary operator corresponds to a physical operation that you can (in principle) implement, transforming (...
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I think I understand where you're getting tripped up. When considering this portion of a circuit,
it seems contradictory that both $\vert x \rangle$ is unchanged by $\text{CNOT}$, and $\text{CNOT}$ maps $\vert \psi_1 \rangle = \vert +- \rangle \rightarrow \vert -- \rangle = \vert \psi_2 \rangle$. It turns out that this is not a contradiction, but it is ...
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You can easily check that $H^1=XHX$, which is another way to say that $H^1$ is the same as $H$ modulo swapping $|0\rangle$ and $|1\rangle$.
This means that it is a different gate.
On the other hand, it is "equivalent" in the sense that, given an arbitrary circuit given as a sequence of gates $\prod_k U_k$, if you swap all the $|0\rangle$s and $|1\rangle$s, ...
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The Hadamard gate has close ties to the discrete Fourier transform. Consider the DFT for an $N$-level system:
$$\vert j \rangle = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} e^{\frac{i2 \pi j k}{N}} \vert k \rangle.$$
For $N=2$ this is simply
$$\vert j \rangle = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \, \vert k \rangle = H \...
2
I found out the number of ancillas is minimum $n-2$.
I found this line in the Qiskit source code of mct:
if len(ancillary_qubits) < len(control_qubits) - 2:
raise AquaError('Insufficient number of ancillary qubits.')
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Your understanding is correct.
In the theory of photon polarization, the parametrization of the Bloch sphere (or its surfave) has traditionally another name. On the wikipedia page for the Jones calculus (the parametrization of the Bloch sphere surface), you'll find a table for the correspondence between kets and polarizations.
To summarize, eigenstates ...
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In principle, yes, you can always do it. The Bloch representation can be generalised to arbitrary dimensions, and you can always parametrise states in it by their "angle coordinates".
For example, you can write an arbitrary 3-modes pure state as
$$|\psi\rangle=\cos\alpha|0\rangle + e^{i\theta}\sin\alpha\cos\beta|1\rangle+e^{i\phi}\sin\alpha\sin\beta|2\...
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I see the heart of your question. I'd like to clarify a bit, before answering your question though. Matrices (aka operators) do not measure quantum states--they operate on them. Specifically, they project the state into the matrix's eigenvectors. We can then measure that projected state in a particular basis that may be the same or different than what the ...
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Initially $|x\rangle|x\oplus y\rangle$ is a perfectly valid state. The first qubit is in $|x\rangle$ and the second qubit is in $|x\oplus y\rangle$ - that is, the second qubit is entangled with the first qubit. As you study other algorithms you might grok that we can have a state $|x\rangle|f(x)\rangle$ for some function $f(x)$ - the first register might ...
0
Short answer: Yes, except for measurement.
There two postulates in play here:
1- the evolution of a closed QM system can always be described by a unitary matrix.
2- measurement operators (observable operator) are always hermitian
Hermitian operators can't (always) be described as a rotation in any space.
Unitary operators can always be considered as ...
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The state of single qubit can be described as a point on the Bloch sphere. All the allowed transformations of a single qubit can then be described as rotations on the Bloch sphere. Unfortunately, bigger quantum systems can no longer be described as fitting on a sphere like geometry. As a result, this idea of considering transformations as rotations does not ...
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I don't think you can factor. There are many cases where the result of the gate will produce a combined CNOT output (4 numbers) that cannot be factored. I think most states between two qbits is entangled from our perspective. So you cannot trust in factoring. Instead, the four products remain in memory while your program writes the data to both of your ...
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While a follow-up question asks for the motivation behind the two-qubit gates used in Sycamore, this question focuses on the random nature of the single qubit operations used in Sycamore, that is, the gates $\{\sqrt{X},\sqrt{Y},\sqrt{W}=(X+Y)/\sqrt{2}\}$ applied to each of the $53$ qubits between each of the two-qubit gates.
Although I agree with @Marsl ...
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This answer only addresses the part about the necessity of the randomness of the circuit because I am by no means familiar with the physical implementation of the qubits at Google and what kind of constraints these impose on the implementation of certain gates.
Now, for the randomness: Consider the problem of sampling from the output distribution of a ...
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TL/DR: The two-qubit gates are going by the moniker "Sycamore gates" in the paper, and it appears that they would ideally want to explore more of the $(\phi, \theta)$ phase-space but for their purposes (of quantum supremacy) their current Sycamore gate is sufficient. The pattern of gates $\mathrm{ABCDCDAB}$ was chosen to avoid "wedges" and maximize/optimize ...
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What does "obtaining samples" mean in this context?
The same thing it means in a more classical context. Consider the probability distribution of the possible outcomes of a (possibly biased) coin flip. Sampling from this probability distributions means to flip the coin once and record the result (head or tail). If you sample many times, you can retrieve ...
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Use u3 function and rotate by an angle of 90 degrees.
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We should generally think of QASM as agnostic to timing. Time is not represented anywhere in a quantum circuit diagram. The wires between circuit operations simply represent a quantum state. I could move the CNOTs as far to the right as I want in your diagram, or generally modify any spacing between any operations however I like, and the circuits would be ...
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