Hot answers tagged

13

The statement is either false or unproven, depending on how you take it. Scenario 1: "Quantum Computers Cannot Be Simulated, Even Given Infinite Time and Space" This statement would be one in which we wish to know the outcome of a quantum computation, but we lack a quantum computer. In this case, is there anyway we can make a classical computer do the ...


6

If you go through the proof of the no-cloning theorem (see e.g. wikipedia) you will notice that only the following properties are used: Unitarity of the evolution operator Hilbert space axioms for a composite state $|\phi \rangle | 0 \rangle$ (which imply the Cauchy-Schwartz inequality) The no-cloning theorem than says that no unitary operator exists ...


5

No, not really. The amount of entanglement and the amount of energy in a state are completely independent: entanglement (together with discord) is a property of the state itself and its relationship to a bipartite (or multipartite) structure of the Hilbert space in which it lives, whereas energy is a joint property of the state and the system's Hamiltonian....


4

Quantum computations can be simulated in a classical computer, at least theoretically. Quantum computers (if they end up existing), it is thought (see comments below) that at most a speed advantage over classical computers, such as being capable of speeding up a problem from, an exponential speed on the input length to a polynomial time in input length (if $...


4

Suppose you have a black box, which takes an arbitrary pure quantum state as input, and returns a new pure quantum state. This is not an unreasonable definition of a quantum computer for our purposes. This black box can be expressed as a unitary matrix. If our goal is to determine what that unitary operation is, then we are interested in performing what is ...


4

Aaronson's claims are true, but your statement about what he means is not correct. If cloning were possible, the HUP would still exist but would pose a non-absolute limit on how much you could learn about a single copy of an unknown quantum state- that is, any limit on how much you learned about a state could be attributed to not doing a very good ...


3

Wrote this to address the 2nd question in the OP: "Is it possible to derive a generalized HUP from a nonlinear modification of quantum theory (Invent some nonlinear observables that is neither position nor momentum)?", following the discussion on the topic in comments. Although it is not a clear-cut answer, perhaps it may help. But first a couple of ...


3

Quantum states are rays in the Hilbert space. i.e. $e^{i\alpha}|\psi\rangle$ is the same as $|\psi\rangle$. The set of all $e^{i\alpha}|\psi\rangle$'s with $\alpha\in[0,2\pi)$ is said to form a ray in the Hilbert space. A point on the Bloch sphere represents a ray and not a state. If you want you can think of the point $(\theta,\phi)$ on the Bloch sphere as ...


3

Both. If quantum mechanics is correct, then an unpredictable RNG circuit is extremely simple: (Note: even if you're a proponent of a deterministic interpretation, like many worlds, this circuit remains fundamentally unpredictable. Philosophers may debate whether or not it's "really" random, but predictability is ultimately what matters in practice for ...


2

Simulating quantum computers may be intractable, but it's not impossible. What quantum computers do is still computable. They don't violate the Church-Turing thesis. But they probably violate the extended variant of that thesis (where 'compute' is strengthened to 'efficiently compute'). Proving the computability of a quantum computer is easy. Just ...


2

Can quantum computers compute non-computable elements? No. Why? Quantum computers can be simulated by classical computers. If quantum computers could compute uncomputable things, so could classical computers. But if a classical computer can compute it, it's computable. Not uncomputable. (The simulation is very slow, but 'computable' doesn't require '...


2

That's... weird notation. $M^r$ means "apply operation $M$ controlled by qubit $r$"? (Please avoid calling your states $x$, $y$, or $z$ when you also have operations using those letters.) Anyways, the mistake you made was passing a Hadamard gate over the control. I agree with Peter Shor about this being a mistake attributable to the notation being poor. ...


2

But how we can say that photon pass through the second filter if we know that after passing the first one it's polarized vertically? For me there is probability 0 to passing second filter if that filter isn't polarized vertically. Polarizing filters don't just discard photons, they change the polarization of photons. This is simply true, by experiment. If ...


2

Take a look online at the "Dirac three polarizer experiment" for a more comprehensive answer.The key points are well explained by the following illustrations from "http://alienryderflex.com/polarizer/" keep in mind that these illustrations are based upon an initial horizontal polarizer, not a vertical one as stated above, but the concept is the same either ...


2

I think you've been fooled by notation. The value inside the $|\rangle$ is a state; a label. The value outside the $|\rangle$ is a weighting; an amplitude. You're negating the label instead of the weight. The Z gate doesn't produce a $|-1\rangle$, but it can produce a $-|1\rangle$. Said another way, the Z gate doesn't produce a $|-True\rangle$ but it can ...


2

A PSPACE machine can simulate any BQP machine by using path integrals. Since quantum computers are like BQP machines, and BQP is a subset of PSPACE, quantum computers can't efficiently solve problems outside of PSPACE.


1

From what I understand, you want to characterize the solution space of those $a,b\in\mathbb C$ such that \begin{align} a|00⟩+b|11⟩ & = a|{++}⟩+b|{--}⟩ \\ & = \frac{a}{\sqrt 2}\left(|00⟩+|01⟩+|10⟩+|11⟩\right) \\ & \quad + \frac{b}{\sqrt 2}\left(|00⟩-|01⟩-|10⟩+|11⟩\right) \\ & = \frac{a+b}{\sqrt{2}}\left(|00⟩+|11⟩\right) +\frac{a-b}{\sqrt{2}}\...


1

We don't know any simple ways to store a qubit, which is what you're asking for. Definitely nothing you could hold in your hand, unless you like hefting dilution refrigerators. Actually, it's worse than that. We don't have any good complicated ways to store a qubit either. Not for very long, anyways. Especially if you want to be able to apply operations to ...


1

Write $$ U = \frac{1}{2}(I_1 + Z_1) \otimes U'_{00} + \frac{1}{2}(I_1 - Z_1) \otimes U'_{11} + (X_1 + iY_1) \otimes U'_{01} + (X_1 - iY_1) \otimes U'_{10} = \left(\begin{array}{cc} U'_{00} &U'_{01} \\U'_{10}& U'_{11}\end{array}\right) $$ where $U'_{ij} \in G_n$. In terms of the latter, $$ U' |\Psi\rangle = \sqrt{2} \;\langle 0 | U | 0\otimes \Psi \...


1

A qubit is in a mixture of the state "0" and the state "1." We write this mixture as $\alpha | 0 \rangle + \beta | 1 \rangle$. Here, $\alpha$ and $\beta$ are two complex numbers, subject to some constraints. It is not the case that a $Z$ gate will suddenly give us a new state $| - 1 \rangle$ -- the qubit does not possess such a state (it's a qubit, it has ...


1

I will turn my comment to an answer, in reality answering your statement in the comments: Thus I am asking "what everyday processes require quantum entanglement rather than just classical correlation?" Classical probabilistic analysis depends on classical dynamics, the emergence of entropy from statistical mechanics is a good example. There have ...


1

The question assumes that quantum entanglement actually happens the way it has been described by quantum mathematics. Which may or may not be true. All is not set yet. Leaving that aside - In any case, entanglement is not a law in itself, it is a phenomenon. It is a consequence of basic underlying laws at quantum level. For example, anti correlation of spin ...


1

Using the expansion into Fock states: $$|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle $$ and the identity $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$ It is straightforward to work out that: $$\langle \beta |\alpha \rangle=e^{-|\alpha|^2/2}e^{-|\beta|^2/2}e^{\alpha\beta^*}$$ As you can see, if $\beta=\alpha$ this ...


1

Someone please correct me if I'm wrong, but I suspect it might just be a bit ambiguous, like "beam splitter". The $\pi/2$ part of the name refers to the amount of rotation in the Bloch sphere, but single-qubit quantum operations have an extra degree of freedom on top of the amount of rotation and the axis of rotation: the global phase factor. For example, ...


1

You've incorrectly found that the matrix for $H$ is given by $$ \left(\begin{array}{cc} 1&0\\0&H \end{array}\right) $$ where $H$ and $1$ are $2\times 2$ matrices, and we're using the following ordering of basis elements: $(|00\rangle,|01\rangle,|10\rangle,|11\rangle)$. This isn't true. An easy way to see that is to compute the matrix element $\...


1

Coherent states are classical in a precise way which hasn't been stated explicitly yet, although Rod suggests at it. Suppose you want to time-evolve the interaction between a coherent EM state and matter. This amounts to solving the Schroedinger equation for: $$ i\hbar\frac{d}{dt} |\psi \rangle= H |\psi \rangle$$ with $$\hat{H}= \hat{H}_{0A}+\sum_k\omega_k ...



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