7

Hint: Instead of using the BCH formula in the form usually presented, for example at the top of this Wikipedia page, use this consequence of Hadamard's Lemma: $$\tag{1} e^{iHt}\hat{a}e^{-iHt} = \hat{a} + [iHt,\hat{a}] + \frac{1}{2!}[iHt,[iHt,\hat{a}] + \cdots $$ Now substitute $H$ into the right-hand side and evaluate the commutators between $\hat{a}$ and ...


7

Classical Hamiltonians By the spectral theorem, for every Hamiltonian there exists a basis in which it is diagonal. Thus, it is not correct to say that diagonal Hamiltonians are classical since this would apply to all Hamiltonians. A Hamiltonian $H$ which is diagonal in a product basis $\mathcal{B}$ is sometimes described as classical, because the evolution ...


7

You can also check their equivalence using Operator, a class from the Qiskit's quantum_info module as follows. from qiskit import * from qiskit.quantum_info import Operator import numpy as np Build and convert each circuit to an operator.( used your name convention for the circuits ) The code below builds and converts the first circuit qrz to op1 qrz = ...


6

As requested through the comment by the OP. Given a Hermitian matrix $H$, we can always write it as linear combination of Pauli strings. That is, $$ H = \sum_i \alpha_i P_i \hspace{1 cm} P_i \in \{I,X,Y,Z\}^{\otimes n} $$ note this linear combination can have up to $4^n$ terms. Thus, for a general Hamiltonian splitting it into linear combination of Pauli ...


5

TL;DR: The two-qubit gate corresponding to the Hamiltonian is the SWAP gate. For an operator $A$ that squares to identity $A^2=I$, we have $e^{i\theta A} = I\cos\theta +iA\sin\theta$. In our case the Hamiltonian does not square to identity $$ (X\otimes X + Y\otimes Y + Z\otimes Z)^2 \ne I\otimes I. $$ However, we can tweak it by adding a constant term since ...


5

Calculate $$ \begin{align} \hat{U}|00\rangle &= \exp\left(-igt(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)\right)|00\rangle \\ &= \sum_{k=0}^\infty \frac{(-igt)^k}{k!}(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)^k|00\rangle \\ &= |00\rangle + \sum_{k=1}^\infty \frac{(-igt)^k}{k!}(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\...


5

Use the differential form of the time evolution, $$dO/dt=i[H, O]\ .$$


5

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 details about the system. On the other hand if you are indeed working in a basis where $H$ is diagonal than the spectral norm is trivially the largest diagonal ...


5

You can definitely run this on a real quantum computer! In your snippet above you mixed circuits and operators. A circuit is only used for the ansatz of your ground state, not for representing the operators. The website you provided talks about the Hamiltonian in terms of the Pauli X and Z matrices; $\hat\sigma^x$ and $\hat\sigma^z$. If you want to compute ...


5

You can't just replace $R_z$ with $R_x$ - the $R_x$ would commute with the controlled-not, so the two controlled-nots cancel and all you get out is just the $R_x$. The trick that you want to use instead is that $Ue^{-iHt}U^\dagger=e^{-iUHU^\dagger t}$. Now, we know that you can transform $Z\otimes Z$ into $X\otimes X$ using hadamards. So set $U$ to be the ...


5

2-local forms As I have seen the term, 2-local Hamiltonians are those Hamiltonians $\hat H$ which can be written as a sum of independent qubit operators $\hat H = \sum_i \sum_j \hat O_{ij}$, where each $\hat O_{ij}$ acts non-trivially only on two qubits $i$ and $j$. This is as opposed to a global Hamiltonian, whose terms may act on any number of qubits. This ...


4

First, note that we can only measure in the computational basis in quantum computing (at least at the moment). But this is not a problem since \begin{align} \langle \psi | H | \psi \rangle = \langle \psi | 2X + 0.5 Z | \psi \rangle &= 2\langle \psi | X |\psi \rangle + 0.5\langle \psi|Z|\psi\rangle \\ &= 2\langle \psi|HZH|\psi\rangle + 0.5\langle\...


4

There's more than one way, and I'll suggest two of them here: Expand $\hat{U}$ using the formula for the Taylor series of an exponential ($e^\hat{A}$) centered around $\hat{A}=\hat{0}$, and then you will have a sum of terms where each term no longer involves an exponential operator (i.e. you have just pure creation and annihilation operators and products/...


4

Perhaps not surprisingly, the gate is $\exp(-i \theta (X \otimes X + Y \otimes Y + Z \otimes Z))$. More usefully, the terms that are being summed happen to commute so you can decompose the problem into $\exp(-i \theta X \otimes X) \cdot \exp(-i \theta Y \otimes Y) \cdot \exp(-i \theta Z \otimes Z)$. import cirq from cirq import X, Y, Z import math theta = ...


4

$\gamma$ should still go from $[0, 2\pi]$, as $U1$ also has domain on $[0, 2\pi]$. See https://qiskit.org/documentation/stubs/qiskit.circuit.library.U1Gate.html. $U1$ is cyclic mod $2\pi$ so in general one cannot distinguish $U1(x \pm 2\pi)$


4

I do not know what the state-of-the-art for this problem is, but here is my attempt at it. First, I doubt whether the required unitary $U$ can be found for every matrix $A$. The most glaring issue occurs when $b\in\ker A$. In this case, $|Ab\rangle$ is ill-defined, so it is not clear what $U|b\rangle$ should be approximately equal to. A slightly more subtle ...


4

When you have $\langle \varphi | I \otimes Z | \varphi \rangle $ It means you are calculating the expectation of the operator $I \otimes Z$ with respect to some state $|\varphi \rangle$. Since $I$ is on the first qubit, we would not need to do anything there, no need to do any rotation or even measurement. The eigenspace is decomposed into two halves ...


4

Ok, here's my attempt: take a time-dependent Hamiltonian $H(t)$ and consider its evolution in the time interval $[0,t]$. Discretize this interval in $k$ steps of length $\Delta \tau \equiv t/k$ $$ \tau_{n} \equiv n \Delta \tau, \qquad n = 0,1,\ldots,k-1. $$ Now consider the piecewise-constant product of the propagators $\exp(-i H(\tau_n)\Delta \tau)$ taken ...


4

Yes, they should be. You can check their statevector to confirm that they indeed generate the same state. First, you can generate some arbitrary initial state and use the following two methods to show that they generate the same statevector. For instance: init_circ = QuantumCircuit(3) init_circ.ry(2.5,0) init_circ.ry(1.2, 1) init_circ.ry(0.5, 2) print(...


4

Yes. In general, a trivial construction of controlled-$U$, given a circuit for $U$ is to make every circuit element controlled off the control qubit. However, this introduces more overhead than necessary. If you have a part of your circuit that looks like $VWV^\dagger$, it is only necessary to make the $W$ controlled because the $VV^\dagger$ cancel in the ...


4

TL;DR: No, in practice quantum computers cannot implement irrational numbers exactly. However, this does not prevent us from realizing quantum gates and algorithms whose properties depend on irrationality of some of their parameters, because for all practical purposes sufficiently good approximations, rational or otherwise, are indistinguishable from the ...


3

The oldest and most commonly known way is the Jordan-Wigner transformation. The qubit operators will be $\mathcal{O}(N)$-local for $N$ occupiable orbitals. A significantly more complicated way is the Bravyi-Kitaev transformation for which the qubit operators will be $\mathcal{O}(\log N)$-local. There's many other ways, but the above two are by far the most ...


3

Stoquastic Hamiltonians have only non-positive off-diagonal terms, see for instance the abstract of this paper by Bravyi et al. The diagonal terms may be zero, but may also be stricly positive. The sign problem is not restricted to only an appearance in quantum computing; it even stems from more general physics - check for instance this question and answer ...


3

There is a lot of proposition how to use quantum computers in finance, see this question. Your are right that you can simulate Hamiltonians on a quantum computer easier than on classical computer. However, in some cases it is difficult to construct respective quantum circuit. On the other hand, circuits for so-called Ising Hamiltonians used for solution of ...


3

The Quantum Approximate Optimization Algorithm is closely related to the Quantum Adiabatic Algorithm. Let's say we have a simple Hamiltonian (in our case $H_B$) with a known ground state and another Hamiltonian $H_C$, whose ground state we want to calculate. Consider the time-dependent Hamiltonian \begin{equation} H(t) = \left(1-\frac{t}{T}\right)H_B(t) + \...


3

Here is an implementation from qiskit.circuit import QuantumCircuit, Parameter theta = Parameter('θ') qc = QuantumCircuit(2) qc.cx(0, 1) qc.crx(-1 * theta, 1, 0) qc.cx(0, 1) print(qc) --- ┌────────┐ q_0: ──■──┤ RX(-θ) ├──■── ┌─┴─┐└───┬────┘┌─┴─┐ q_1: ┤ X ├────■─────┤ X ├ └───┘ └───┘ and to evaluate that it works: from ...


3

Let $H$ be a $2\times 2$ Hermitian matrix. You can always choose a basis with respect to which $H$ is represented as a diagonal matrix, so let us fix this basis. We can also assume, without loss of generality, that the first eigenvalue is zero (because changing the Hamiltonian by a constant factor doesn't affect the physics). Then, $H=\operatorname{diag}(0,\...


3

So, you want to implement a time evolution $$ e^{iA t}. $$The first thing to note is that $$ e^{iAt}=U^{\dagger}e^{iUAU^\dagger t}U. $$ So, we can choose a $U$ that diagonalises $A$. This works particularly well here because all the terms commute and are the stabilizers of the two-qubit cluster state. For those that know the cluster state, you'll know that ...


3

While a general QUBO task has form $$ \sum_{i=1}^n b_ix_i + \sum_{i=1}^n\sum_{j=1}^na_{ij}x_ix_j \rightarrow \min, $$ where $x_i \in \{0;1\}$ and $a_{ij},b_{i} \in \mathbb{R}$, an Ising Hamiltonian has form $$ \sum_{i=1}^n b_iZ_i + \sum_{i=1}^n\sum_{j=1}^na_{ij}Z_i\otimes Z_j, $$ where $Z_i$ is $Z$-gate applied on $i$th qubit while identity operators are ...


3

--Self-answer made CW-- The local circuit labelled $f^2$ above is likely incomplete and does not create the kind of entanglement I envisioned, as it suggests that $\vert\mathrm A\rangle$ can be swapped with $\vert\mathrm B\rangle$, independent of $\vert\mathrm C\rangle$ being swapped with $\vert\mathrm D\rangle$, etc. For example such a circuit corresponds ...


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