If you're not concerned with global phase then the following works using only two rotation gates: \begin{align} R_y\left(-\frac{\pi}{2}\right) R_x\left(\pi\right) &= \exp \left(i\frac{\pi}{4}Y\...

Yes its true. Define another orthogonal projector $\Pi_\perp$ such that $\Pi + \Pi_\perp = I$ and write $\rho$ in terms of a spectral decomposition \begin{equation} \rho = \sum_k \lambda_k(\rho) |\...

The two pages you link are using opposite conventions, the first defines \begin{align} R_z(\theta) &:= \exp(i \theta Z /2)\\ &= \cos (\theta/2) + i \sin (\theta/2)Z\\ &= \begin{pmatrix} \...

No such (orthonormal) basis can exist. An orthonormal basis $\{|\psi_i\rangle\}$ requires $\langle \psi_i | \psi_j \rangle = 0$ for $i\neq j$, and so clearly \begin{align} [\rho_i, \rho_j] &= |\...

As mentioned already, both of those unitaries are the same up to a global phase. It might be useful to think about how you can actually arrive at one of these definitions in terms of the "Not ...

Consider a simple implementation of a Support Vector Machine (SVM) that finds a hyperplane (defined by its normal vector $w$) that maximally separates vectors $\{v_1, \dots, v_m\}$ according to their ...

As the previous answer mentions, how a controlled qudit gate is defined is up to a choice of convention. This paper contains a few examples of intuitively appealing definitions for controlled qudit ...

In addition to the accepted answer and @user1271772's examples, here is a circuit primitive referred to explicitly as a "T-gate gadget" in  (originally appearing in ): where ...

Summary: There is a solution for expressing a tridiagonal matrix of the form you've provided for arbitrary $n$ in terms of Pauli operators using recursion. This procedure is given at the bottom of ...

Summary: The expression you're looking for is: $$\frac{1}{4} \left[ (III + IZZ + ZIZ + ZZI) + (XXX - XYY - YXY - YYX)\right]$$ where Pauli string notation like $XYX$ denotes $\sigma_1 \otimes \... View answer Accepted answer 5 votes Yes, the formula you have shows that applying QFT to a given computational basis state$|j\rangle = |j_1 j_2 \dots j_n\rangle$results in an unentangled output state. However when applied to ... View answer Accepted answer 5 votes Without sacrificing any generality we can define an ebit as a Bell state$\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$shared between two parties$A$and$B$, and since we're concerned with ... View answer 5 votes The spectral norm$\|H\|$(sometimes denoted$^1\|H\|_2$or$\|H\|_\infty$) in this case is the largest eigenvalue of$H$. There's no meaningful bound for this number without having additional ... View answer Accepted answer 5 votes There are two different ancillas floating around, one used in$|\psi\rangleand another to conduct the swap test later on: In the above picture with subsystems explicitly labeled we have \begin{... View answer Accepted answer 5 votes As mentioned in the article, you can rewrite the GHZ state as \begin{align} \frac{1}{\sqrt{2}} (|000\rangle + |111)&= \frac{1}{2\sqrt{2}}(|000\rangle + |111 \rangle + \overbrace{|001\rangle + |... View answer Accepted answer 5 votes That quantity appears to be identical to Holevo information, which turns out to be the upper bound on how much classical information you can transmit using a quantum channel . More generally the ... View answer 5 votes In one sense, the Xmon qubit is a transmon qubit, in that they both operate in theE_J>>E_c$regime of the CPB Hamiltonian and take advantage of the exponentially suppressed charge noise vs. ... View answer Accepted answer 4 votes The teleportation should behave just the same with a mixed state as it does with a pure state. I'm going to assume a bit of familiarity with how teleportation works for pure states, as you can find ... View answer Accepted answer 4 votes Python currently doesn't support an operator for Kronecker products. Note how the @ symbol works: when you write the statement A @ B, Python$^1$checks the objects A and B for a __matmul__ method and ... View answer Accepted answer 4 votes Here is an approach that requires no specific knowledge about$|\psi\rangle$whatsoever. In your description you implied that each$H_i$has the same maximum and minimum eigenvalues$\lambda_m$and$\...

There are a couple of ways reversibility might be coming into play in this context. The first is that the measurement at the end of the circuit will be typically be an irreversible step. For example ...

The basic idea of how the quantum feature map works is that you're using a quantum computer to map each input datapoint $x$ from your training domain $\mathcal{X}$ into a quantum state $|\phi(x)\... View answer Accepted answer 4 votes The mixed state is invariant under unitary operations; no choice of$U$that you might apply to$\rho_2$will change its output statistics from 50/50 distribution of "0" versus "1".... View answer Accepted answer 4 votes It seems like the article you're referencing is defining "maximally entangling" as "capable of producing Bell states from product states". However there are other ways to describe ... View answer Accepted answer 4 votes The model's accuracy is purely empirical observation. The error trend (Fig 4, or 41:50 in the video) demonstrates that the error of the system (cross entropy fidelity with respect to simulated results)... View answer Accepted answer 4 votes You can look up work by Gil Kalai, who is a longstanding and outspoken critic of quantum computing (his most recent essay: Kalai, 2019). He often bases his view on assumptions that I entirely disagree ... View answer 4 votes Another way to think about this: To simulate what goes on in a quantum computer we have to do a lot of matrix math using$(2^N \times 2^N)$matrices$^1\$, and the action of (most) of the clifford gates ...

This is not a complete answer but describes a case where knowing polynomially many Pauli expectation values is not sufficient to solve the same problem. Consider that the set of expectation values for ...