# Tag Info

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The cloning theorem requires that the result of the cloning is two independent copies of the starting qubit, i.e., the state of the system in the end should be $\big(\alpha |0\rangle + \beta |1\rangle \big) \otimes \big(\alpha |0\rangle + \beta |1\rangle \big)$. This is not the state CNOT will give you. The qubits you get after applying CNOT as you ...

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By "copying a quantum state", we mean that we cannot take $$|\psi⟩|0⟩=\alpha|00⟩+\beta|10⟩$$ into $$|\psi⟩|\psi⟩=(\alpha|0⟩+\beta|1⟩)(\alpha|0⟩+\beta|1⟩)=\alpha^2|00⟩+\alpha\beta|01⟩+\beta\alpha|10⟩+\beta^2|11⟩$$ for arbitrary single qubit state $|\psi⟩=\alpha|0⟩+\beta|1⟩$. Notice that this resulting two-qubit state $|\psi⟩|\psi⟩$ is still separable. But in ...

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The no-cloning theorem itself can be stated very precisely. Given an unknown pure state $|\psi\rangle$ that is drawn from a distribution $\{p_i,|\phi_i\rangle\}$ (known to the counterfeiter), it is impossible to create a perfect clone with unit probability unless all the states are orthogonal, $\langle\phi_i|\phi_j\rangle=\delta_{ij}\ \forall\ i,j$. ...

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All operations on quantum states are unitary operations. We don't make the rules, this is just how nature seems to work. So if you want to define an operation that copies a qbit, it has to be a unitary operation. That unitary operation would look like this: $U|\psi\rangle_A|0\rangle_B=|\psi\rangle_A|\psi\rangle_B$ So you have the qbit you want to copy, $|\... 8 The no cloning theorem only applies when quantum information is in an unknown superposition. If you know a basis in which the state of some qubits is not under superposition, then you can make all the copies you want. Classical information encoded directly into qubits is going to be in the computational basis state. Therefore you can clone it. You use CNOT ... 7 As already mentioned in the other answers, the crucial point is that copying means implicitly that the state of the original qubit is unknown, i.e. given a qubit in an unknown state, you want to prepare a second qubit to be in exactly the same state. To make it more intelligible, there is a less mathematical argument that this should not be possible: By the ... 7 The proof does not seem to rule out the case that there exists a specific U that can clone only the specific state |ψ⟩. That's because you can clone specific states. Cloning is only impossible if the set of possible input states includes a pair of states that are not orthogonal. For example, here is a circuit that performs$|\psi⟩ \to |\psi⟩|\psi⟩$as long ... 6 To keep the problem small, let's say 1 qubit. In the original statement$| \psi \rangle$could be any state$\alpha | 0 \rangle + \beta | 1 \rangle$for whatever$\alpha$and$\beta$produce a well defined state. Quantifying over all so not a probabilistic statement. Say you uniformly draw over all possible states. It is impossible to guarantee that you ... 6 That's the way that I would initially go about answering the question. There are, however, a few tweaks you could make. Definitive Answer As you point out, the annoying feature is that you can never be definitive about having the state$|\psi\rangle$. There are a couple of ways that you might avoid that pitfall. The first option is to have two ... 5 One (obvious) application is the generation as True Random Number Generators, e.g. IDQ, or you can download some here Free True Random Numbers (please do not use these for security relevant application). In order to build such a TRNG, from a quantum circuit perspective, all you need is a single qubit, a Hadamard gate and a measurement. Although there might ... 4 Certainly not exhaustive, but to get the ball rolling... One possible application is blind quantum computation. In this, there is a user who wants to complete a computation, but only has the capability of producing single-qubit (non-entangled) states. These are sent to a server who can (locally) entangle them for the purposes of performing a measurement-... 4 To answer the first part of the question (whether unitary matrix$U$operates on$|\psi_A \rangle$only): A unitary matrix can operate on an arbitrary number of qubits. Single-qubit gates, like Pauli X, Y and Z gates, operate on one qubit and are represented by 2x2 matrices; CNOT gate operates on two qubits and is represented by a 4x4 matrix, etc. In this ... 3 Firstly, carefully read through the formal presentation of the protocol as described on Wikipedia. Secondly, there's nothing to prove as such here. It is evident from the teleportation protocol itself. Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e.$|q\rangle$. ... 3 For step (116), the equivalence between both of them is proved by \begin{equation} (\langle\psi_1|\otimes\langle0|)C^\dagger C(|\psi_2\rangle\otimes|0\rangle) = (\langle\psi_1|\otimes\langle0|)(|\psi_2\rangle\otimes|0\rangle) = \langle\psi_1|\psi_2\rangle\otimes\langle0|0\rangle=\langle\psi_1|\psi_2\rangle\langle0|0\rangle, \end{equation} where in the ... 3 The no cloning theorem can be stated in the form: If you are given an unknown state$|\psi\rangle$, which is promised to be one of a set of distinct states$\{|\phi_i\rangle\}$, it is impossible to create a second copy of$|\psi\rangle$if there is a pair of$|\phi_i\rangle$which are not orthogonal. Computational basis states are orthogonal, so they're ... 3 Cloning means the generation of$|q\rangle|q\rangle$from$|q\rangle|0\rangle$. This is not what happens in teleportation. Teleportation is kind of a swap operation, i.e. something like$|q\rangle|0\rangle \to |0\rangle|q\rangle$. 2 I think you agree that if you start with the state$(a|0\rangle+b|1\rangle)|0\rangle$, the cnot produces$a|00\rangle+b|11\rangle$. The issue is why is the state of the first qubit not the same as$a|0\rangle+b|1\rangle$. The answer is if you only look at that one qubit and you only look in the standard,$Z$basis, then they do look the same. But those are ... 2 I'm a little confused about which gates operate on which qubits and how, but following the linked question, I think I understand that you are wondering why, given a single qubit in the state in$a|0\rangle+b|1\rangle$and preparing two qubits in a state$a|00\rangle+b|11\rangle$does not qualify as cloning the first bit, especially because the probabilities ... 2 You seem to be mixing two very different concepts here. Quantum cloning is talking about the absolute limits of what is theoretically possible in a perfect world. In this absolute theoretical limit, yes we can derive how well quantum cloning can work, and we also know that classical cloning is nominally perfect. There is then a separate question of how well ... 2 Forget that you know that non-orthogonal states cannot be distinguished, as this is what you're trying to prove. It's like a proof by contradiction. The question is talking about a hypothetical device that could, if it existed, distinguish between$|\psi\rangle$and$|\phi\rangle$. Assume it exists, and prove that it gives you cloning (I know which state I ... 2 No,$|\psi\rangle$and$|\phi\rangle$are non-orthogonal by assumption, they can't be effectively (whatever this means) orthogonal. The device just gives us the answer$\phi$or$\psi$(you can think it returns label and destroys the input state), so we can prepare$|\psi\rangle$or$|\phi\rangle$separately (or many copies of it). It is that trivial. 2 I.e., if we have the entire initial state is written as follows $$| q \rangle \otimes | \beta_{00} \rangle =(\alpha | 0\rangle+\beta | 1 \rangle ) \otimes \frac{1}{\sqrt{2}}( |00\rangle+| 11 \rangle)$$ $$= \frac{1}{\sqrt{2}}(\alpha | 000\rangle+\alpha | 011 \rangle+\beta | 100 \rangle+\beta | 111 \rangle ),$$ then, in step three, we obtain the state (the ... 2 I propose the following advantages to quantum money, over and above blockchain-based cryptocurrencies. The security of Nakomoto-style cryptocurrencies (read, Bitcoin) is based on computational assumptions, at least in that there is a assumption that inverting SHA256 hashes is likely computationally difficult. However, Wiesner-style private-key based quantum ... 2 There is not, generally, a universal result for which you simply plug numbers into a formula. However, there is a good strategy that results in an upper bound for the achievable fidelity, with conditions on when that maximum fidelity can be achieved. I use it quite extensively in these two papers: paper1, paper2. First, be aware that there are several ... 1 Please be careful with your notation; don't get confused with the number of qubits that are in$|0_{Eve}\rangle$(I'm not necessarily saying that you are - just mentioning this as a precaution). Since the state$|\psi\rangle = \frac{1}{\sqrt{N}}\sum_{j \in 0,1,2....N-1}|j\rangle$that Alice wants to send is a$n$-qubit state with$N=2^{n}\$, Alice needs to ...

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It does exist. It's basically the controlled-not gate generalised to higher dimensional systems. The important thing to realise is that means that the bit that Bob ends up with will be highly entangled with what Eve has, and that will have significant observable consequences. As part of a cryptographic protocol, for exampe, Bob could detect that Eve is ...

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"Optimal Cloning of Pure States" by Werner, 1998 Although it really should be called "optimal heralded cloning", since it uses a post-selection step: Basically it works by appending maximally mixed states and projecting the qubit+mixed system into the symmetric state. Note how the state on the left (bloch sphere pointing down-and-left) is turned into two ...

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In lecture 4 of O'Donnell's series on quantum computing, he introduces the Elitzur-Vaidman bomb tester, which is an interesting application of the quantum Zeno effect. O'Donnell introduces the bomb tester prior to discussing multi-qubit entanglement in lecture 5. In the Elitzur-Vaidman tester, a single qubit in a superposition can be used to probe and ...

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