Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.
9

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, $\langle\phi_i|\phi_j\rangle=\delta_{ij}\ \forall\ i,j$. ...


8

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 ...


8

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, $|\...


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

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

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 ...


5

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 ...


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

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 ...


3

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-...


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.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 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.


1

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

"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 ...


Only top voted, non community-wiki answers of a minimum length are eligible