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7 votes
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Preparing a quantum state from a classical probability distribution

Suppose we have two quantum circuits, the first one $S$ computes (or at least approximates) the classical squareroot function ($\sqrt{\cdot}$) via $$S|x\rangle|0\rangle=|x\rangle |\sqrt{x}\rangle,$$ ...
Condo's user avatar
  • 2,048
6 votes

Complex conjugate state preparation

$\langle \psi|$ is not a quantum state, but a linear functional on the set of quantum states. $|\psi\rangle$ is a quantum state, any gates that you can apply to it can only take it to quantum states, ...
Mateus Araújo's user avatar
6 votes
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How instantaneous is state preparation in a quantum register, if all possible superpositions are to be initialized equally?

We know that giving a single qubit starting in the state $|0\rangle$, which is a state one can initialize very fast with high fidelity, then we can put it in the superposition state $|\psi \rangle = ...
KAJ226's user avatar
  • 13.9k
6 votes
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Forming states of the form $\sqrt{p}\vert 0\rangle+\sqrt{1-p}\vert 1\rangle$

As long as you want to set arbitrary states for a single qubit, like in your example, the solution is straightforward and it makes use of standard 2x2 $R_y$ gate. In there if you set $\theta = 2\...
Gianni's user avatar
  • 364
5 votes
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Is the CNOT in the standard three-qubit circuit for the GHZ state necessary?

If you initialize three qubits to $|0\rangle$, apply a Hadamard gate to each, then measure each in the computational basis, the result will be an independent coin flip for each bit: that is, any of ...
benrg's user avatar
  • 878
4 votes

Forming states of the form $\sqrt{p}\vert 0\rangle+\sqrt{1-p}\vert 1\rangle$

I did not thought much about that issue, so my answer may not be the best one, but it has the advantage of being quite simple to understand, it is exact even if you restrict yourself to Clifford+T, ...
Léo Colisson's user avatar
4 votes

How instantaneous is state preparation in a quantum register, if all possible superpositions are to be initialized equally?

You can prepare equal superposition by application of Hadamard gate on each qubit. The result will be state $$ |q\rangle=\frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n}|i\rangle, $$ i.e. the desired equally ...
Martin Vesely's user avatar
4 votes
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References for quantum state praparation: what states are easy to prepare and which ones aren’t?

That's a pretty broad question. But we can use complexity theory to get some guidance. Below I explore the implications of the standard assumptions that BPP$\subsetneq$BQP$\subsetneq$QCMA$\subsetneq$...
Mark Spinelli's user avatar
4 votes

What exactly does state preparation mean in quantum computing?

Quantum state preparation is similar to initialization of variables in classical computing. At the beginning all qubits are in state $|0\rangle$ (similarly classical numerical variables contain zeros, ...
Martin Vesely's user avatar
3 votes
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Adding phases of two qubits

You've asked for a $2$-qubit quantum circuit, thus we are looking for a $2$-qubit unitary that performs the desired transformation. Note that if we take $\theta=0$, then $\frac{1}{\sqrt{2}}\left(|0\...
Tristan Nemoz's user avatar
  • 6,462
3 votes

How many quantum gates are needed to prepare an arbitrary state?

A canonical reference for gate decompositions is Barenco et al., Elementary gates for quantum computation. In particular, it also contains recipes to decompose an arbitrary $n$-qubit unitary into ...
Norbert Schuch's user avatar
3 votes

Is there an efficient circuit implementing the unitary $U|x\rangle|0\rangle=|x\rangle\Big(\sqrt{1 - x/2^n}\,|0\rangle+\sqrt{x/2^n}|1\rangle\Big)?$

I think that asking for an exact solution is pointless, because quantum computers don't have infinite precision. You are limited, for example, by accuracy of pulses that control the gates. To ...
Egretta.Thula's user avatar
3 votes

How to prove that EPR outcomes have equal probability no matter the basis?

The local state (described by density matrix) of each qubit in EPR state is \begin{equation} \rho=\frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{equation} It does not depend on ...
kludg's user avatar
  • 3,214
3 votes

How to prove that EPR outcomes have equal probability no matter the basis?

Let $\rho_{AB}$ be a quantum state shared between two parties, Alice and Bob. Suppose Alice performs a POVM measurement $\{M_i\}_i$ on her half of the state. Then the probability that Alice obtains ...
Rammus's user avatar
  • 5,853
3 votes

Is there a known classification of quantum states on $n$ qubits preparable with O(1) circuit depth?

Suppose you have a real function $f$ such that the state you want to prepare is: $$\newcommand\ket[1]{\left|#1\right\rangle}\ket{\psi_f}=\sum_xf(x)\ket{x}$$ (up to normalisation). We also assume $f$ ...
Tristan Nemoz's user avatar
  • 6,462
3 votes
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What is the state of the art quantum state preparation algorithm?

While the term "best" or "state of the art" can be subjective and often evolves (especially in the field of quantum machine learning), a recent protocol that offers an approach to (...
banercat's user avatar
  • 763
3 votes
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How to implement the state $|\psi\rangle = \frac{1}{\sqrt{2}}\left[|0\rangle \otimes |X_i\rangle + |1\rangle \otimes |X_j\rangle\right]$

Suppose that $U_i$ prepares the state $\left|X_i\right\rangle$. Start by applying an Hadamard gate on the first qubit: $$\frac{1}{\sqrt{2}}\left(|00\rangle+|10\rangle\right)$$ Then apply $U_j$ ...
Tristan Nemoz's user avatar
  • 6,462
2 votes

Complex conjugate state preparation

$\left<\psi\right|$ is not a state of a quantum system, it is a linear functional that takes a quantum state and returns a scalar. In terms of basic Linear Algebra, it is a row vector rather than a ...
Joseph Geipel's user avatar
2 votes
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How to prepare a random 1-qubit superposition for data encoding

Theorem: Suppose $U$ is a unitary operation on a single qubit. Then there exist real numbers $\alpha, \beta, \gamma, \delta$ such that $$ U = e^{i\alpha} R_z(\beta) R_y(\gamma)R_z(\delta) $$ This is ...
KAJ226's user avatar
  • 13.9k
2 votes

Forming states of the form $\sqrt{p}\vert 0\rangle+\sqrt{1-p}\vert 1\rangle$

I'm curious about how to form arbitrary-sized uniform superpositions [...] for $N$ that is not a power of 2. In https://arxiv.org/abs/1805.03662 it's explained how to do this by using a single ...
Craig Gidney's user avatar
  • 37.7k
2 votes
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How to create superposition of states with fixed parity with a quantum circuit?

I actually found the solution to my problem - with one caveat: the amplitudes $a,b,c,d$ can not be completely arbitrary. An $n$-qubit superposition state with even parity will have $2^{n-1}-1$ free ...
NaturalLog's user avatar
2 votes
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How many quantum gates are needed to prepare an arbitrary state?

I believe this Q&A answers your question about decomposition in detail: Minimum number of 2 qubit gates to build any unitary In short, you are correct that the lower bound for a number of 2-qubit ...
3yakuya's user avatar
  • 632
2 votes
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Confusion about Rodeo algorithm "spectral weight suppression" argument

I believe the following is the key sentence for understanding their reasoning: we can describe the action of the rodeo algorithm for each individual eigenvector of $H_{obj}$ with energy $E_{obj}.$ ...
MonteNero's user avatar
  • 2,646
2 votes

Given a quantum state, can you generate a uniform superposition over its computational basis vectors with nonzero amplitudes?

TL;DR: No. Assuming we only have a single copy of $|\psi\rangle$, the transformation is prohibited by linearity of quantum mechanics. That said, it can be done if we have multiple copies of $|\psi\...
Adam Zalcman's user avatar
  • 22.9k
2 votes

Is it known whether the Fermi-Hubbard ground state can be prepared efficiently or not?

In this paper, Schuch and Verstraete determined the computational complexity of finding the ground state of the Fermi-Hubbard model, showing that it is among the hardest problems in the complexity ...
bm442's user avatar
  • 1,127
2 votes

How to prepare all the computational basis states by running the same quantum ansatz with distinct $\theta$ values?

The answer to the question is affirmative. Here is an example of a single-parameter two-qubit circuit $U(\theta)$ that allows to prepare all four computational basis states starting from the fiducial ...
bm442's user avatar
  • 1,127
2 votes
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Why don't I receive the output I expect?

If your circuit contains the first operation only: circuit.initialize(initial_state, 0) then the output state would be $$0.577∣000⟩+0.816∣001⟩$$ and you will get 0....
Egretta.Thula's user avatar
2 votes

How does the uncomputation step work in the Grover-Rudolph scheme to prepare $\sum_i\sqrt{p_i}|i\rangle$?

The uncomputation is performed by simply running the same unitary used to generate the $|\theta_i\rangle$ states in reverse. We could describe what's going on more abstractly as follows: Start with ...
glS's user avatar
  • 25.4k
2 votes
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Can we create a W state of n qubits with constant circuit depth using mid circuit measurements?

Section 4.2 of this paper seems to answer your question positively. Section 4.3 of the same paper also shows how to do this for Dicke-states, the generalization of W-states.
nippon's user avatar
  • 1,517
2 votes

Quantum Information Retrieval from Bipartite Mixed States under LOCC: Maximizing Individual State Knowledge

None. What is meaningful is the state $\rho$, not the decomposition $\sum \lvert\psi_i\rangle\langle\psi_i$. There is many different such definitions, and there is no way to distinguish them by any ...
Norbert Schuch's user avatar

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