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Accepted

### 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 $$S|x\rangle|0\rangle=|x\rangle |\sqrt{x}\rangle,$$ ...
• 2,048

### 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, ...
• 2,551
Accepted

• 364
Accepted

### 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 ...
• 878

### 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, ...
• 666

### 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 ...
• 14.5k
Accepted

### 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$\subsetneq$...
• 12.5k

### 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, ...
• 14.5k
Accepted

• 22.9k

### 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 ...
• 1,127

### 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 ...
• 1,127
Accepted

### 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....
• 10.6k

### 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 ...
• 25.4k
Accepted

### 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.
• 1,517
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 ...