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15

To simplify the question consider CNOT gate instead of Toffoli gate; CNOT is also fanout because \begin{align} |0\rangle|0\rangle \rightarrow |0\rangle|0\rangle\\ |1\rangle|0\rangle \rightarrow |1\rangle|1\rangle \end{align} and it looks like cloning for any basis state $x\in\{0,1\}$ \begin{align} |x\rangle|0\rangle \rightarrow |x\rangle|x\rangle \end{...


13

Numerous papers on quantum cloning have been written since 1996, including both theoretical and experimentally focused papers. The following survey paper is a good place to start if you want to learn more: Valerio Scarani, Sofyan Iblisdir, Nicolas Gisin, and Antonio Acin. Quantum cloning. Reviews of Modern Physics 77: 1225-1256, 2005. arXiv:quant-ph/...


10

Regarding the optimality of the results of your linked article [1],$\def\ket#1{\lvert#1\rangle}\def\bra#1{\!\langle#1\rvert}$ we find in Section III A that on input $\ket{\phi}$, the states produced by this imperfect cloning operation are of the form $$ \qquad\qquad\qquad \rho_{\text{out}} \,=\, \tfrac{5}{6}\ket{\phi}\bra{\phi} \,+\, \tfrac{1}{6}\...


8

The no cloning theorem only applies when quantum information is in an unknown superposition. If you know a basis in which the state of some qubits is not under superposition, then you can make all the copies you want. Classical information encoded directly into qubits is going to be in the computational basis state. Therefore you can clone it. You use CNOT ...


8

The no cloning theorem says that there is no circuit which creates independent copies of all quantum states. Mathematically, no cloning states that: $$\forall C: \exists a,b: C \cdot \Big( (a|0\rangle + b|1\rangle)\otimes|0\rangle \Big) \neq (a|0\rangle + b|1\rangle) \otimes (a|0\rangle + b|1\rangle)$$ Fanout circuits don't violate this theorem. They don't ...


7

The answer is that the no-cloning theorem states that you cannot clone an arbitrary unknown state. This circuit does not violate the no-cloning theorem, because let's look at what it does when the input is $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. The output at the third register still has to be a $|0\rangle$ or a $|1\rangle$. Therefore it's impossible ...


6

Given the constraints you impose (why those are appropriate constraints is perhaps another discussion), I think you're over-complicating things. Without loss of generality, you can assume $$ |\alpha_0\rangle=|0\rangle,\qquad |\alpha_1\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle $$ and clearly $$ U|\alpha_0\rangle|0\rangle=a_0|00\rangle+a_1|\alpha_1\...


5

As John Watrous said, the Rev. Mod. Phys. article is an excellent starting point. If you want to know the sort of thing that's been looked at since, then in a shameless bit of self-promotion, you might look at this paper. There have been a couple of follow-up papers as well (including one that closes a small step left open in one of the proofs). What is ...


4

TL;DR: The assumption of non-orthogonality is implicitly used by the linked answer. It is needed due to a "loophole" in the no-cloning theorem that allows cloning of known orthogonal states. Universal cloner is prohibited The no-cloning theorem is the statement that there is no unitary $U$ such that $$ U|\gamma\rangle|b\rangle = |\gamma\rangle|\...


4

Assume this works. Then, nothing prevents Alice from applying the same protocol to a quantum state that is known to her, such as $|0\rangle$ or $|1\rangle$. This way, she could send information to Bob instantaneously. Thus, it violates faster-than-light communication and thus is impossible.


3

You may also want to check for: state dependent deterministic cloners which clone with a better fidelity when input state comes from a known ensemble. Ref: Bruss et al., PRA 57, 2368 (1997) probabilistic cloners which clone with unit fidelity but with less than unity success probability asymmetric cloners where the outputs have cloned with different ...


3

The no-cloning theorem states that an unknown quantum state cannot be copied exactly --- so this rules out any algorithm that attempts to produce perfect copies of an arbitrary quantum state (including squeezed and coherent states). As you note, however, the no-cloning theorem does not rule out the production of approximate quantum state clones. Andersen et ...


2

You seem to be mixing two very different concepts here. Quantum cloning is talking about the absolute limits of what is theoretically possible in a perfect world. In this absolute theoretical limit, yes we can derive how well quantum cloning can work, and we also know that classical cloning is nominally perfect. There is then a separate question of how well ...


2

The fastest way to check is with a simulator: You can tell 00 has twice the amplitude of 11 (and 10) because their probabilities have a $2^2:1$ proportion. You can tell the phases are right because all the blue circles have indicators pointing directly to the right. The circuit being used is a bit inefficient. You can get the same state from one CNOT ...


1

One way to do this is simply working through the quantum circuit one step at a time. Note that your preparation circuit for the state $|\phi\rangle$ can be written as: The initial state is $|00\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} $. The circuit itself has the matrix representation of: $$ \big( R_Y(\theta_3) \otimes I \big) \cdot CNOT_{...


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