# Tag Info

9

Besides the already given answers note that there is indeed some "mental gymnastics" involved here. As soon as you're getting more acquainted with quantum computing, you know some of your usual gates, including the $\mathsf{SWAP}$ gate that appears in your question: \begin{align} \mathsf{SWAP} = \begin{bmatrix} 1 &0 &0 &0 \\ 0 &0 &1 &...

8

The Toffoli gate is just a permutation. If you start in a known basis state, application of a Toffoli just changes it into another basis state, one that you can easily calculate classically (after all, it’s a decision based on looking at 3 bit values). Repeating that doesn’t change anything. To make it universal, you need to add something like Hadamard ...

7

It's not possible to implement a Toffoli using only Fredkin gates, because Fredkin gates preserve the number of 1s in the state while Toffolis do not.

5

To fix what we are talking about, I think you mean $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \quad S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} \quad T = \begin{pmatrix} 1 & 0 \\ 0 & \exp(i\pi/4) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1+i}{\sqrt{2}} \end{pmatrix}.$$ If these are ...

5

an arbitrary single qubit rotation and a CNOT comprises a universal gate set. Is this true? Yes. If you want to understand why, don't go with any answers about finite gate sets such as CNOT, H, T because the proof of those usually relies on the proof of the set you've stated (so the whole thing becomes horribly circular). Instead, you have to make some of ...

5

Toffoli and Hadamard are computationally universal -- that is, they can be used to carry out any quantum computation. However, they do so by implementing quantum gates in an encoded way. Indeed, this is necessary since both Toffoli and Hadamard have only real entries, so there is no way to obtain quantum gates with complex entries, unless one uses some ...

3

It depends on what you mean by an "actual quantum computer". For arithmetic circuits, and circuits to compute cryptographic operations to act as oracles for Shor's and Grover's algorithm, Toffoli gates are essential and ubiquitous. In fact many papers count Toffoli gates as the main resource. I think part of the reason for this is that a Toffoli gate ...

3

So any universal gate set can replicate any other, since both are universal, but different architectures generally have different physical gates. While Clifford+T is a universal gate set that is very nice to think about theoretically, it isn't generally close to the one used in the lab. In most experimental setups, the physical level universal gate set used ...

3

From the spec, $U_2(\phi, \lambda) = U ( \frac{\pi}{2}, \phi, \lambda)$. We can use Cirq's QasmUGate. import cirq from cirq.circuits.qasm_output import QasmUGate q0 = cirq.NamedQubit('q') u2_gate = QasmUGate(0.5, 0, 1) # The angles are normalized to the range [0, 2) half_turns circuit = cirq.Circuit(u2_gate(q0))

3

I have thoughts on a couple of different approaches, although I'm sure there'll be simpler options. Firstly, imagine you start from a two-qubit state $|00\rangle$, and apply an $R_x$ rotation with an angle equivalent to half that of a Pauli $X$ to the first qubit (I forget which convention N&C is using for their rotation gates). Then apply a controlled-...

3

It’s straightforward to see that a random element $U$ drawn from a design is, with very high probability, far away from any fixed unitary $V$, i.e. you can show that the distance between $U$ and $V$ (in operator norm or in the diamond norm of their channels) is on average very close to maximal. But what you’re asking about is if there exists at least one ...

3

Partial answer (universality): Set of gates consisiting of CNOT, $S$, $T$ and $H$ allows you to approximate any gate arbitrary accuracy. CNOT is part of your set. Sice $S = R_z(\pi/2)$, $T=R_z(\pi/4)$ and $H = R_y(\pi/2)Z = R_y(\pi/2)R_z(\pi)$, your gates are universal. Another perspective: $x$,$y$ and $z$ allow you to build any rotation of a qubit, ...

2

Firstly, I think there's a reason why the bit in the textbook before the question is using $\delta$ instead of $\epsilon$ for the results. So, replacing what you've written, it should really be $$E<\sqrt{2(1-\cos(\delta/2))}$$ Now, start with a double-angle formula: $1-\cos(\delta/2)=2\sin^2(\delta/4)$. Thus, $$E<2\sin(\delta/4)$$ Now if you apply ...

2

A classical proof involves Lie algebra to show that the Deutsch's three Qubits gate is universal for three qubits gates and then showing that we can construct a $n$ qubits Deutsch's gate using Toffoli gates and three qubits Deutsch's gate only, so the results extends to $n$ qubits. You can find the proof in the caltech's course on quantum computation, ...

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.')

2

There's a known theorem known as the Solovay-Kitaev theorem which can be used to deduce the order of the number of gates required to approximate any unitary operator of arbitrary dimension quantum systems. As an example, suppose we have a quantum system with states in $\mathbb C^n$. We take a universal family of gates (the OP wants 1 and 2 qubit ones, so...)...

2

In Elementary gates for quantum computation it is conjectured that based on dimension counting, ... a lower bound on the number of two-bit gates required to produce an arbitrary $n$-bit unitary transformation [is]: $\Omega(n) =\frac19 4^n−\frac13n−\frac19$.

2

For a 0/1 matrix, the sort of protocol that I use is to write out the basis in the standard order. For three qubits, this is: 000,001,010,011,100,101,110,111 (the numbers 0 to 7 written in order in binary). You can list these next to the rows and the columns. Now, a given 1 entry in the matrix has a specific row and column (and there is only one of these for ...

2

You can see from here, nmr.physics.ox.ac.uk/oxonly/C2/QIP2answers.pdf, that Toffoli can't be constructed from Fredkin without the use of ancilla qubits (which in practice qubits are a valuable resource), whilst Fredkin from Toffili doesn't require the use an additional ancilla qubit. Simulating gates with Toffoli is just a more compact way of constructing a ...

1

There are two statements in your question: If $\theta$ is a rational number of $2\,\pi$, then $\left(\theta_k\right)_{k\in\mathbf{Z}}$ does not reach every $x\in[0\,;\,2\,\pi)$ If $\theta$ is an irrational number of $2\,\pi$, then $\left(\theta_k\right)_{k\in\mathbf{Z}}$ does reach every $x\in[0\,;\,2\,\pi)$ We can prove that actually, none of these ...

1

I guess you assume that you can implement any quantum circuit using Toffoli gates only; this is not true. Toffoli gate is classically universal, but not quantum universal. The importance of Toffoli gate in quantum information science is based on 2 facts: Toffoli gate is classically universal, that is you can implement any classical circuit using Toffoli ...

1

I use the ideas from these slides, specifically slide 8,9,10. We can decompose any $U_{3}(\theta,\phi,\lambda)$ into a rotation around the $Z,Y$ & again $Z$ axis, because for any $U \in SU(2)$ we can write: \begin{equation} U = \begin{bmatrix} e^{i(\alpha-\frac{\beta}{2}-\frac{\delta}{2})}\cos(\frac{\gamma}{2}) & e^{i(\alpha-\frac{\beta}{2}+\frac{\...

1

Here is a construction of $CU3$ gate on IBM Q: u1((lambda+phi)/2) c; u1((lambda-phi)/2) t; cx c,t; u3(-theta/2,0,-(phi+lambda)/2) t; cx c,t; u3(theta/2,phi,0) t; Where t is a target qubit and c is control qubit. Note that $U1$ gate is a special case of $U3$, it holds that $U1(\lambda)=U3(0,0,\lambda)$.

1

We care about $F$'s action on the basis states, where basis states are represented by \begin{equation}\begin{pmatrix} 1\\0\\0\\0\\0\\0\\0\\0 \end{pmatrix}, \begin{pmatrix} 0\\1\\0\\0\\0\\0\\0\\0 \end{pmatrix}, \ldots, \begin{pmatrix} 0\\0\\0\\0\\0\\0\\0\\1 \end{pmatrix}\end{equation} It should be fairly easy to see that on most basis states, this gate will ...

Only top voted, non community-wiki answers of a minimum length are eligible