# Tag Info

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Regarding your first question, you are essentially asking about the validity of a position taken by David Deutsch - a founder of quantum computing! For example, in his book 'The Fabric of Reality', Deutsch states: When Shor’s algorithm has factorized a number, using $10^{500}$ or so times the computational resources that can be seen to be present, where ...

14

No, a superposition of two different states is a completely different beast than a mixture of the same states. While it may appear from your example that $\rho_1$ and $\rho_2$ produce the same measurement outcomes (and that is indeed the case), as soon as you measure in a different basis they will give measurably different results. A "superposition" like $\... 14 Question 1 This description lies somewhere between the two extremes of a theory and mysticism, depending on how amiable one is to the concept. David Deutsch is vocal proponent of the former, Lee Smolin of the latter (he categorizes it as "Mystical Realism"). The general idea was initiated by one of John Wheeler's PhD students, Hugh Everett III, in ... 12 Here is how you might go about designing such a circuit.$\def\ket#1{\lvert#1\rangle}$Suppose that you would like to produce the state$\ket{\psi} = \tfrac{1}{\sqrt 3} \bigl( \ket{00} + \ket{01} + \ket{10} \bigr)$. Note the normalisation of${\small 1}/\small \sqrt 3$, which is necessary for$\ket{\psi}$to be a unit vector. If we want to consider a ... 11 Break the problem in parts. Say we have already sent$\mid 00 \rangle$to$\frac{1}{\sqrt{3}} \mid 00 \rangle + \frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle$. We can send that to$\frac{1}{\sqrt{3}} \mid 00 \rangle + (\frac{1}{2} (1+i))\frac{\sqrt{2}}{\sqrt{3}}\mid 01 \rangle + (\frac{1}{2} (1-i))\frac{\sqrt{2}}{\sqrt{3}}\mid 10 \rangle$by a$\sqrt{SWAP}$. ... 11 The short answer is that there is more to quantum information than "uncertainty". This is because there is more than one way to measure a state; and that is because there is more than one basis in which, in principle, you can store and retrieve information. Superpositions allow you to express information in a different basis than the computational basis &... 10 I'll tell you how to create any two qubit pure state you might ever be interested in. Hopefully you can use it to generate the state you want. Using a single qubit rotation followed by a cnot, it is possible to create states of the form $$\alpha \, |0\rangle \otimes |0\rangle + \beta \, |1\rangle \otimes |1\rangle .$$ Then you can apply an arbitrary ... 9 Do you mean mapping the state$|0\rangle^{\otimes n} \to \dfrac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1} |i\rangle $? If that is the case then you can just apply$H^{\otimes n}$to the state$|0\rangle^{\otimes n}$. That is, you apply a Hadamard gate to each of the qubit. The reason for this is$H |0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$and so \begin{... 9 Consider a quantum circuit on one qubit initialized in the state$|0\rangle$and consisting of two Hadamard gates where we can insert a measurement between the Hadamards. The input state is$|0\rangle$, so the state after the first Hadamard gate is $$H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$ which is an equal superposition of$|0\rangle$and ... 7 Depending on what precisely your assumptions are about a, b, I think this is essentially impossible, and is something called the "no superposing theorem". Please see this paper. 7 In the many worlds interpretation (MWI) reality consists of a structure called the multiverse that looks like a collection of slightly interacting parallel universes in some circumstances: Deutsch, David. "The structure of the multiverse." Proceedings of the Royal Society of London. Series A: Mathematical, Physical and Engineering Sciences 458.2028 (2002):... 7 Control qubit can be in superposition, why not? This is how entangled qubits are created. But the circuit you have drawn does not entangle qubits, it actually does not change qubit states at all, and$|p\rangle = \alpha| 0 \rangle + \beta| 1 \rangle$,$|q\rangle=|0\rangle$,$|r\rangle=|0\rangle$. It would be more interesting if the input value of the 2nd ... 7 Superposition is a basis dependent concept. Namely the$|0\rangle$and$|1\rangle$states are commonly said not to be superposition states exactly because one of the two coefficients in the$\{|0\rangle,|1\rangle\}$basis expansion is zero. However, using the x-basis representation,$\{|+\rangle, |-\rangle\}$one finds$|0\rangle = (|+\rangle + |-\rangle)/\...

7

As Davit Khachatryan's answer points out, the task is impossible / ill-defined, since the desired target state is generally not normalized and it depends on the relative global phases of the two initial states. However, it is possible to rephrase the question so that is meaningful and has an interesting answer. The two problems -- sensitivity to the global ...

6

The dirac notation itself, the $|$ and $\rangle$ parts is simply a notation to remind you that you're dealing with quantum states. What you write inside this `ket' is completely arbitrary. In the examples you give, somebody has (probably) chosen to depict a quantum system of two quantum bits. The up arrow and the down arrow depict two distinguishable states,...

6

Cloning means you have a unitary transforming $U$ that takes $|\psi\rangle|0\rangle$ to the $|\psi\rangle|\psi \rangle$ for all quantum states $|\psi\rangle$. That is, $$U\big(|\psi\rangle|0\rangle \big) = |\psi\rangle|\psi \rangle \hspace{0.5 cm} \forall \ |\psi \rangle \in \mathcal{H}$$ where $\mathcal{H}$ is a Hilbert space. Thus, what you shown is ...

5

The idea is to apply a unitary transformation which will map these states to a different pair of orthogonal states which are easy to distinguish by measuring in computational basis. Let's construct a unitary $U$ which prepares $|\psi_0\rangle$ starting from $|000\rangle$ state. The first step is to prepare a $W_3 = \frac{1}{\sqrt{3}}\left(|100\rangle + |... 5 A QFT can't arbitrarily raise the probability of any state you want to any value you want. Once you create a superposition, you need to find some way to make destructive interference occur between the states you don't want, and construct interference between the states you do want. Finding ways to do this is essentially the entire field of quantum algorithms.... 5 As long as you want to set arbitrary states for a single qubit, like in your example, the solution is straightforward and it makes use of standard 2x2$R_y$gate. In there if you set$\theta = 2\arctan(\sqrt{1-p}/\sqrt{p})$and apply$R_y(\theta)$in the form $$\begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2)\\ \sin(\theta/2) & \cos(\theta/2) \... 5 It's mostly about simplicity and adopted convention. In the end, this is basically the same question as "why should I pick a universal set of gates A rather than a universal set B?" (see here). Experimentalists would pick the universal set they have available. Theorists just pick something that they like to work with, and eventually a convention is adopted. ... 5 Along with glS' post: A mixed state would be if you had a can of paint, but you weren't sure if it was blue or yellow. You know it is either one of the two, and once you pop the top and measure it, you'd know, but until you do it is in one of those two pure states. If you picked it up from a stack of cans where you knew there were equally many cans of blue ... 5$$ \begin{eqnarray*} \mid 0 0 \rangle &\to& \frac{1}{\sqrt{2}} \mid 0 0 \rangle + \frac{1}{\sqrt{2}} \mid 1 0 \rangle\\ &\to& \frac{1}{\sqrt{2}} \mid 0 0 \rangle + \frac{1}{\sqrt{2}} \mid 1 1 \rangle\\ &\to& \frac{1}{\sqrt{4}} \mid 0 0 \rangle + \frac{1}{\sqrt{4}} \mid 1 1 \rangle + \frac{1}{\sqrt{4}} \mid 1 0 \rangle + \frac{1}{\sqrt{... 5 If you apply the same gate twice on a state$|\psi\rangle$you will get out the state$|\psi\rangle$if the hermitian of the gate is the same as the gate. In your case, you applied the Hadamard gate,$H$, twice on (presumably) the$|0\rangle$state, and so you got$|0\rangle$as your result. This happens because it is a requirement for quantum computing ... 5 That's not too far from the truth. You can always describe the probabilities of a two-outcome event like that: you have$p_1+p_2=1$and thus defining$c_i\equiv\sqrt{p_i}$you observe that$(c_1,c_2)$are distributed on a circle (more precisely, in the upper-right sector of one). More generally, if an event has$n$possible outcomes, you can describe the ... 5 For any problem which is described by an infinite dimensional Hilbert space, you can regard any state as a superposition of an infinite number of states. The only real question is thus whether infinite-dimensional Hilbert space are "used in practice". 5 To talk about entanglement, you have to first identify subsystems. In your$d=4$example, you defined an isomorphism$\mathbb{C}^4\simeq \mathbb{C}^2\otimes\mathbb{C}^2$via the identification of basis states. Whether this is meaningful, depends on the context/the physical scenario you have in mind. But it definitely can be. For$d=3$, this is never possible.... 5 A counterexample that shows that this is not possible in the general case (here I am neglecting post-selection possibility discussed in the comments of the question and in the accepted answer): $$C_1 = X \qquad C_2 = -X$$ Or one can take$C_2 = R_y(- \pi)\$ and all mentioned below equations will reamin true. So: $$C_1 |0\rangle = |1\rangle = |\psi \rangle \... 5 We know that giving a single qubit starting in the state |0\rangle, which is a state one can initialize very fast with high fidelity, then we can put it in the superposition state |\psi \rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}} by applying a Hadamard gate. That is,$$H |0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$$And we can do for N-... 4 According to this paper, A significant conclusion from this solution is that generically the generalized algorithm also has O(\sqrt{N/r}) running time Where 'r' is the number of marked states. By generalized, the authors meant a distribution with arbitrary complex amplitudes. So it seems to answer your question. That the modified initialization would ... 4 Do you want to know about this specific sequence, or more generally? Specifically, this is essentially a W state, about which there have been several questions in the past. You just need one extra term (effectively |32\rangle) that you could later change into |11\rangle, leaving the others untouched. Let's say I've got a state$$ (|1\rangle+|2\rangle+|...

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