15

There are plenty of different variants, particularly with regards to the conditions on the Hamiltonian. It's a bit of a game, for example, to try and find the simplest possible class of Hamiltonians for which simulation is still BQP-complete. The statement will roughly be along the lines of: let $|\psi\rangle$ be a (normalised) product state, $H$ be a ...


11

A conventional Hamiltonian is Hermitian. Hence, if it contains a non-Hermitian term, it must either also contain its Hermitian conjuagte as another term, or have 0 weight. In this particular case, since $Z\otimes X\otimes Y$ is Hermitian itself, the coefficient would have to be 0. So, if you're talking about conventional Hamiltonians, you've probably made a ...


9

One way order to perform Z rotations by arbitrary angles is to approximate them with a sequence of Hadamard and T gates. If you need the approximation to have maximum error $\epsilon$, there are known constructions that do this using roughly $3 \lg \frac{1}{\epsilon}$ T gates. See "Optimal ancilla-free Clifford+T approximation of z-rotations" by Ross et al. ...


8

An approach for Hamiltonian simulation: Any Hermitian (Hamiltonian) matrix $H$ can be decomposed by the sum of Pauli products with real coefficients (see this thread). An example of 3 qubit Hamiltonian: $$H = 11 \sigma_z \otimes \sigma_z + 7 \sigma_z \otimes \sigma_x - 5\sigma_z \otimes \sigma_x \otimes \sigma_y$$ The final circuit for $e^{iHt}$ can be ...


8

Your question remains very unclear as to what it actually is that you want to calculate. There is no direct correspondence between a system Hamiltonian and the quantum state of the system. No matter what the Hamiltonian, any quantum state is a valid state of the system. Where a Hamiltonian comes in useful is, if you know the state at some time (say, $t=0$),...


8

In each of the examples you mentioned, the task breaks very roughly down into two steps: finding a Hamiltonian that describes the problem in terms of qubits, and finding the ground state energy of that Hamiltonian. From that perspective, the Jordan–Wigner transform is a way to find a qubit Hamiltonian corresponding to a given fermionic Hamiltonian. Once you ...


7

The insight that suggests that sparse matrices are useful goes along the lines of: for any $H$, we can decompose it in terms of a set of $H_i$ whose individual components all commute (making diagonalisation straightforward), $$ H=\sum_{i=1}^mH_i. $$ If the matrix is sparse, then you shouldn't need too many distinct $H_i$. Then you can simulate the ...


7

Reformulating your question: How to perform Hamiltonian Simulation for a generic square matrix $A$? Quick answer: it is not possible. The goal of Hamiltonian Simulation (HS) is to find a quantum circuit (i.e. a succession of gates) that acts like $U(t) = e^{-iAt}$ on a quantum state. Here $U(t)$ needs to be unitary (because of the properties of quantum ...


6

Generally speaking, a realization of a quantum gate involves coherent manipulation of a two-level system (but this is nothing new to you, maybe). For example, you can use two long-lived electronic states in a trapped atom (neutral or ionized in vacuo) and use an applied electric field to implement single-qubit operations (see trapped ions or optical lattices,...


6

Yes, in this special case the circuit will simplify as you suggest. The advantage of the circuit that was given is that it generalises more easily, and works for any $H$ which has $\pm 1$ eigenvalues. Here's a general form of the circuit for your reference: This essentially comes down to an issue of how you can reversibly compute a one-bit function $f:\{0,1\...


6

Your reasoning is correct if your two Hamiltonians commute. But, as you say, it doesn't work if they don't commute. In that case, the trick is to find something that approximates the the thing you want. So, what you should really be thinking about is taking terms in the opposite order: $$ e^{iH_1t/2}e^{iH_2t/2}e^{iH_1t/2}e^{iH_2t/2}\approx(e^{i(H_1+H_2)t/2})...


6

TL;DR: Hamiltonian simulation does not just mean "exponentiating $H$". It means finding a quantum circuit $U$ that approximates the matrix exponentiation $e^{-iHt}$. More importantly, the size of the Hamiltonian matrix $H$ isn't the key concern here. The gate complexity (or query complexity, in case the Hamiltonian is described as an oracle) of matrix ...


5

Note: I'm deliberately leaving a few gaps here. Hopefully I'm saying enough to let you piece te rest together! Let's say that you want to implement $V$ on some state $$ \sum_{x\in\{0,1\}^n}\alpha_x|x\rangle $$ You can fairly easily write down what that state produces. Think about the Hamiltonian $Z^{\otimes n}$. What eigenvalues does it have? $\lambda=\pm 1$...


5

I don't know why/how the authors of that paper do what they do. However, here's how I'd go about it for this special case (and it is a very special case): You can write the Hamiltonian as a Pauli decomposition $$ A=15\mathbb{I}\otimes\mathbb{I}+9Z\otimes X+5X\otimes Z-3Y\otimes Y. $$ Update: It should be $+3Y\otimes Y$. But I don't want to redraw all my ...


5

1. Definitions Names and symbols used in this answer follow the ones defined in Quantum linear systems algorithms: a primer (Dervovic, Herbster, Mountney, Severini, Usher & Wossnig, 2018). A recall is done below. 1.1 Register names Register names are defined in Figure 5. of Quantum linear systems algorithms: a primer (Dervovic, Herbster, Mountney, ...


5

When using a simulator, it doesn't really matter what kind of qubit you refer to. You can even mix-and-match the types. The type of qubit only becomes relevant when you intend to run on a device, because devices have qubits at specific locations. For example, if you wanted to run on Bristlecone, you would limit yourself to GridQubit instances that actually ...


5

Both circuits work essentially the same. It's perhaps slightly easier to understand the second because it's being explicit about what it's doing rather than hiding some of it in an oracle. So, take the second diagram. Consider the effect of the middle gates. They basically say if the top register is in the all 0 state, flip the bit of the second register if ...


5

We can't implement $e^{iZ_1 \otimes Z_2 \otimes Z_3 \theta}$ with three separate rotations. In other words: $$e^{iZ_1 \otimes Z_2 \otimes Z_3 \theta} \ne e^{i Z_1 \theta} \otimes e^{i Z_2 \theta} \otimes e^{i Z_3 \theta}$$ The implementation of this gate can be found in this answer. The $e^{-iI \otimes I \otimes I\theta} = e^{-i\theta} I \otimes I \otimes I$ ...


4

Several graph theory problems such as Graph Coloring (which is NP-complete) can be cleverly mapped to finding ground states of some classes of Hamiltonians. Graph Partitioning using Quantum Annealing on the D-Wave System Quantum annealing of the graph coloring problem


4

It's not an area I personally know much about, but I know that many of my physicist friends are excited about being able to investigate the Hubbard Model on larger lattices than we can simulate today. There are known and published algorithms for finding the ground state energy, computing Green's functions, and other important characteristics of the model. ...


4

$\newcommand{\bra}[1]{\left\langle#1\right|}\newcommand{\ket}[1]{\left|#1\right\rangle}\newcommand{\proj}[1]{|#1\rangle\langle#1|}\newcommand{\half}{\frac12}$In answer to your first question, I wrote myself some notes some time ago about my understanding of how it worked. The notation is probably a bit different (I've tried to bring it more into line, but it'...


4

Let's say you have a Hamiltonian of the form $$ H=\sigma_1\otimes\sigma_2\otimes\sigma_2\otimes\ldots\otimes\sigma_n $$ There's a straightforward circuit construction that lets you implement its time evolution $e^{-iHt}$. The trick is basically to decompose the state that you're evolving into the components that are in the $\pm 1$ eigenspaces of $H$. Then, ...


4

You want to start by being careful with the sizes of the operators. $\hat U$ acts on $q$ qubits, and $\hat H$ acts on $n<q$ qubits. I believe that $|G\rangle$ is a state of $q-n$ qubits. So, what we really need to talk about is two distinct sets of qubits. Let me call them sets $A$ and $B$. $A$ contains $n$ qubits, and $B$ contains $q-n$ qubits. I'll use ...


4

how can you compare the result of your algorithm with an ideal evolution? You cannot and you do not need to. As you said, computing $e^{-iHt}$ is intractable for most of the interesting cases. If it was not, chemistry simulations would be easy, solving the Schrödinger equation too. The thing you can do though is to prove that your algorithm will, for a (...


4

There is actually a nice way to do this in Qiskit, since it has decompositions for single-qubit unitaries built in. The QuantumCircuit.squ method takes a unitary 2x2 matrix $U$ and a qubit and computes the decomposition $$ U = R_Z(\alpha) R_Y(\beta) R_Z(\gamma) $$ This is a common decomposition, you can find a proof here https://arxiv.org/pdf/quant-ph/...


4

You want to implement $$ e^{i3\pi/4}e^{iX\pi/4}. $$ I would rewrite this as $$ e^{i3\pi/4}He^{iZ\pi/4}H. $$ This is the same as $$ -HS^\dagger H $$ in standard gate terminology. If you're only implementing the gate $e^{iAt}$, then you can neglect the global phase and just implement $HS^\dagger H$. Both of these gates are readily implemented in qiskit as sdg ...


4

I call this the "Paulinomial decomposition" as you are writing the matrix $H$ as a polynomial of Pauli matrices: $H=a_{XX}X_1X_2 + a_{XY}X_1Y_2 +a_{XZ}X_1Z_2 + a_{XI}X_1 + a_{YY}Y_1Y_2 + \cdots $ (for the 2-qubit case). To get the coefficients, you can use this formula: $a_{ij}=\frac{1}{4}\textrm{tr}\left((\sigma_i\otimes \sigma_{j})H\right)$ For example,...


3

The simplest method to implement $e^{iA\theta}$ for a small, Hermitian matrix $A$ is to: Find the eigenvectors $|\lambda\rangle$ and eigenvalues $\lambda$ of $A$. Construct the unitary $U=\sum_i|i\rangle\langle\lambda_i|$. Implement the gate sequence: $U$ $e^{i\theta\sum_i\lambda_i|i\rangle\langle i|}$ $U^\dagger$ Now, for one qubit, you have the middle ...


3

I don't have a completely general method for doing what you ask. However, there are a few of the steps that I might take: The $4\times 4$ matrix $H$ can always be written in the form $$ H=a\mathbb{I}\otimes\mathbb{I}+\underline{n}_1\cdot\underline{\sigma}\otimes\mathbb{I}+\mathbb{I}\otimes\underline{n}_2\cdot\underline{\sigma}+\underline{\sigma}\cdot M\cdot\...


3

Given a simple quantum system, how do I derive its Hamiltonian? For quantum systems of continuous variables, the most common way to construct the Hamiltonian is to add the kinetic energy and potential energy, as described in this resource. The kinetic energy part is explained here, and various potential energy models are given here: which unfortunately I ...


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