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## Hot answers tagged superposition

17

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 ...

10

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 $\sqrt{SWAP}$. ...

10

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 his ...

9

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 $\... 9 The short answer is that there is more to quantum information than "uncertainty". This is because there is more than one way to measure a state; and that is because there is more than one basis in which, in principle, you can store and retrieve information. Superpositions allow you to express information in a different basis than the computational basis &... 7 Here is how you might go about designing such a circuit.$\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 ... 7 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 ... 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. 6 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):... 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

$$\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{... 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 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. ... 4 Your question is not quite correctly defined. First of all, |a\rangle + |b\rangle is not a state. You need to normalize it by considering \frac{1}{|||a\rangle + |b\rangle||}(|a\rangle + |b\rangle). Secondly, in fact, you don't have access to the states |a\rangle and |b\rangle but to the states up to some global phase, i.e. you can think that the ... 4 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,... 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+|...

4

Yes. You already have the reasoning for why. Sometimes the basis is implied but not explicitly stated though. So caution when reading something like that.

4

If you know, in advance, that the state you want to deamplify is specifically $|000\rangle$, there are a couple of strategies that you could follow. For example, introduce an ancilla and perform the multi-controlled not, targeting the ancilla, where it is controlled off every qubit in the original state being in the $|0\rangle$ state. So, you'd be doing $$\... 4 Whether or not quantum superposition is a "truth", is a philosophical question. Quantum theory is simply an axiomatic mathematical model of the universe that happens to give correct experimental predictions (at least, for a fairly broad range) for several physical phenomena. It certainly might be possible to come up with a different mathematical model of the ... 3 I believe the question boils down to "How to apply a controlled version of the X gate if the control qubits should be in states other than |11\rangle?" Toffoli gate CCNOT applies an X gate to the target qubit if both control qubits are in the |1\rangle state. If you need to apply an X gate if the control qubits are in some different state, you can do ... 3 I think the major advantages of the Hadamard gate are "usability" stuff, as opposed to fundamental mathematical stuff. It's just easier to remember and simpler to apply. The Hadamard gate's marix is real and symmetric. Makes it easy to remember. Hadamard is its own inverse. Makes it easy to optimize in circuits. Any two Hs that meet cancel out; whereas \... 3 A QFT can't arbitrarily raise the probability of any state you want to any value you want. Once you create a superposition, you need to find some way to make destructive interference occur between the states you don't want, and construct interference between the states you do want. Finding ways to do this is essentially the entire field of quantum algorithms.... 3 By an example with a control qubit in superposition and the target in  |0\rangle  state:$$ \frac{|0\rangle + |1\rangle}{\sqrt{2}} |0\rangle = \frac{|0\rangle|0\rangle + |1\rangle |0\rangle}{\sqrt{2}}$$Applying a CNOT will have the following result:$$ \frac{ CNOT(|0\rangle|0\rangle + |1\rangle |0\rangle)}{\sqrt{2}} = \frac{ CNOT(|0\rangle|0\rangle) + ...

3

Any Hermitian quantum gate $U$ is "self-recovering". This is because $U$ is unitary, and $$UU^{\dagger}=U^{\dagger}U=I$$ If $U$ is also Hermitian, then $U=U^{\dagger}$ and $$UU=I$$ Hadamard gate prepares $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ superposition from $|0\rangle$ state. If you need this superposition, you use Hadamard. If you need a different ...

3

No, using a controlled gate doesn't measure the control. In a sense, the idea that controlled gates would be implemented via measurement is exactly backwards. It's measurement that is implemented in terms of controlled gates, not vice versa. A measurement is just an interaction (i.e a controlled gate) between the computer and the environment that's ...

3

The word superposition had many meanings long before the phrase quantum superposition became popular. A linear super-position of A and B is just "A + cB" where c is a constant. So you have your binary optimization problem that has to be solved, for example: H_problem = b1*b2 - b1*b3 + 2*b2*b4. Then you have a constraint such as "b1*b2 = 1" which can be ...

3

Most probably because your input state was $|0\rangle$. The Hadamard gate has the property that $H^2=I$, thus acting on an arbitrary input $|\psi\rangle$ with the Hadamard gate twice in a row results in the output being identical to the input (that is, $|\psi\rangle$).

3

Much as in CMOS we almost always equate a low voltage with binary zero, and high voltage with binary one, and a classical $\mathsf{AND}$ gate as implementing the truth-table by adjusting voltages to be consistent, we have similar requirements in quantum computing to call $\vert 0\rangle$, $\vert 1\rangle$, and the Hadamard gate acting on it in the manner you ...

3

Classical computers are inherently deterministic, so they either generate pseudorandom numbers, or use an external physical process with statistically random noise to generate random numbers. Quantum computers are inherently probabilistic, so generating true random numbers is very natural for them. Quantum random number generators are already on the market ...

3

[Can't comment and so, writing an answer] As @Sanchayan aptly pointed the theoretical/mathematical arguments which could indicate that superposition is not just an assumption. A couple more thoughts: You're right in observing that the superposition collapses to 0 or 1 when observed and that's expected due to the collapse of the wave function. A quick ...

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