# Tag Info

## New answers tagged hamiltonian-simulation

3

Swapping 4-level qudits is equivalent to swapping pairs of qubits. Because you can encode a 4-level qudit into a pair of qubits. Similarly, swapping 8-level qudits will be equivalent to swapping triplets of qubits. The swap gate is convenient enough that this should hold regardless of how you map the 4-level qudit into the qubits (e.g. whether you map qudit |...

1

As hinted in the highlighted text, $E_T$ arises from the normalization factor $A(\tau)$. Specifically, differentiating $(1)$, we get two terms  \begin{align} \frac{\partial|\psi(\tau)\rangle}{\partial\tau} &= \frac{\partial}{\partial\tau}\left(A(\tau)e^{-H\tau}|\psi(0)\rangle\right) \\ &= \frac{\partial A(\tau)}{\partial\tau}e^{-H\tau}|\psi(0)\...

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 ...

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 ...

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 ...

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 = ...

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(...

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