# Tag Info

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The types of interferometers are associated with two different types of interference. In the Mach-Zehnder interferometer, the interference is just pure classical interference. The decoherence in this case is caused by the variation in relative phase for the different frequencies within the bandwidth. If you want to think of this in terms of photons, then you ...

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Is a qutrit considered to be in a state of tripartite entanglement? A qutrit is a (single) quantum system. As such, it may be isolated, or it may be interacting with one or more other systems, and it may (or may not) be entangled with those other systems, depending on the situation. In other words, it is certainly possible to bring in two other systems (...

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If you have two entanged qubits, each placed in different area, you can use them for sending another qubit from one place to the other one. This is called quantum teleportation. However, to do so, you firstly have to measure some properties on one side, then send this information classically and on other side to change qubits placed there with this ...

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Good answer from Penguino. Here is an example. Several items of the same type are for sale in a store with many kinds of merchandise. Customers come by at some average rate. On the average, they have a constant probability of wanting one of the items. The lifetime of an item has an exponential distribution. Examples more along the lines of a light bulb are ...

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The assumptions you're making in your first sentence are incorrect. Since quantum entangled particles can transfer the data of their spin properties faster than the speed of light There is no "data transfer" when one of two entangled particles is measured. The no-communication theorem states that you can't use a measurement of entangled particles ...

3

Because quantum entanglement is not a communication technology. Your question is similar to asking "Why don't we build a communication technology based on the principle of randomly shuffling a deck of cards and distributing it to everyone?" Randomly shuffling a deck of cards is an example of a classical correlation -- if you check your card and see ...

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It might be hard to find a realistic example of exponentially distributed lifetimes (EDL) in macroscopic systems. An EDL requires each member of the population to have a constant probability of failure for each time interval - and if failure doesn't occur during that interval the member's state is 'reset' to its state at the beginning of the time interval. ...

0

If one measures an observable of some system, the wave function collapses to an eigenstate of the Hamiltonian of this system. The Hamiltonian does not change under the measurement. The resulting wave function is an eigenfunction of the Hamiltonian. Thus, for your question, $\phi$ solves $H$ after the measurement since $\phi$ is an eigenfunction of $H$. In ...

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Simply put a quantum entangled system will always "beat" a classical one with the goal of correlating experiment outcomes. Regardless of the type of correlation or strategy taken by the classical system, assuming no communication is allowed once the two objects are separated. How does this "enchanced" correlation work precisely? We aren't ...

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The term "quantum entanglement exists" is a subset of the term "there exists a single quantum mechanical wavefunction describing the system". If you know the wavefunction, i.e. have a mathematical description of it, the theory of quantum mechanics has constraints on quantum numbers and their conservation, so quantum numbers are "...

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This is better suited for a comment, but I cannot post one. Have you found this resource? It appears to contain theoretical background, but I cannot judge if it is appropriate for your situation.

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At first sight the consequences for physics would be earth-shattering: they have showed that there exists a Bell inequality for which the commuting Tsirelson bound is strictly larger than the tensor-product Tsirelson bound, so the commuting bound cannot even be approximated by finite-dimensional systems. If we could reach the commuting Tsirelson bound ...

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This "Bell Game" is just the famous CHSH game. It was rather unfortunate that Gisin didn't use the regular name, as it makes it much harder to search information about it. It is explained here, for example, or here, or here. There is no easy way to calculate the quantum "score", you need to use quantum mechanics. Let me first rephrase the ...

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If we have two entangled particles $A$ and $B$ [...] then we measure a property of particle $A$ The sequel of your question suggests that to the "property" (value of measured quantity) asserted of the state of partcle $A$ (or perhaps of the state of particle $A$ along with attending setup instrumentation, such as an analyzer) there also exists an &...

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Yes, partial trace would suffice to get the density matrix for one wire. Lat A,B,C, and D be the four wires, and $\rho_{ABCD}$ be the density matrix of the 4 qubits together. Then in order to get the density matrix of wire A, we have to take partial trace over systems B,C, and D, i.e., $\rho_{A}=Tr_{BCD}(\rho_{ABCD})$ This will always work irrespective of ...

2

An unstated assumption you're making here is that the combined entangled state is with the two particles that are in opposite states, e.g. spin system $\frac{1}{\sqrt{2}}\left(\lvert+,-\rangle + \lvert-,+\rangle\right)$. This is a specific kind of an entangled state—the general meaning of it is much broader (and has multiple equivalent definitions, which I ...

0

The Toric Code in four space dimensions is thermally stable. (The excitations are loop-like, rather than point-like, and possess a tension, and thus tend to stay small.) Some references: https://arxiv.org/abs/0811.0033 https://arxiv.org/abs/1411.6643 : Section V.C and references therein.

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then it will take some time for the wave function of particle B to collapse after the measurement of A This is your problem. It doesn't take any time- wave function collapse is instantaneous, in every reference frame. That is to say, once a measurement has been taken, all further measurements are taken on the collapsed wavefunction. There is no violation of ...

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The notation is fine. I would probably put a subscript $A$ on the $\langle 0|_A$ state, such as to make clear that this is a state on $A$, and thus, you are left with a state on B. The formula you want to prove is wrong, there should be a transpose, not a dagger. Other than that, the more general statement is that for the maximally entangled state |\Omega \... 0 Convolver, you commented PM 2Rings answer as follows: In that case, I think superposition doesn't really exist... Take the case of two colored marbles. Indeed there isn’t a superposition. Different from two photons, made by Spontaneous parametric down-conversion. The photons directions are entangled, but the directions (together) are random around 360°. ... 2 Can you know the exact time at which the particle B goes from superposition into a known state due to the remote measurement of particle A, only by waiting on particle B without knowledge of A? No, you cannot tell if particles B & A are entangled without measuring both particles and comparing the measurements, and to compare the measurements you need to ... 1 Bell's theorem is a lot more subtle. First replace the coloured balls with gyroscopes, and assume that when you measure the spin direction of a gyroscope you also give it a kick, such that the gyroscope automatically aligns with the apparatus used to measure it, with a probability for aligning either way which depends on the angle between prior gyroscope ... 1 This is an excellent question with a subtle answer. To answer it, let us consider the following two states: \begin{align} |\psi \rangle_{\textrm{bw}} &= \sqrt{p}|bb \rangle + \sqrt{1-p}|ww \rangle \, , \\ \rho_{\textrm{bw}} &= p|bb\rangle\langle bb|+1-p|ww\rangle\langle ww| \end{align} Both states are correlated, but they are fundamentally different. ... 2 There are many ways of creating entangled photons, the mainly used methods are: Spontaneous Parametric Down Conversion In this case they use a special crystal, and input a single (pump) photon, and the output is a pair (or more) photons, whose total energy equals the input photons'. Momentum is conserved as well, the total input photon momentum and the ... 1 imagine that you have two particles that first interact in such a way that their combined spin wave function becomes a superposition of one having spin up and the other spin down, and vice versa. Spin and magnetic dipole of subatomic particle are correlated. The production of superposition means that these two magnetic dipoles are oriented anti-parallel ... 4 You are asking the right kind of question, and the main thing I want to say is to warn you that a huge amount of stuff has been written on this and only about 0.00001 percent of all that stuff is worth reading. These questions go to the heart of what we mean by a 'state' when talking about quantum systems. Most people try to frame their discussion by ... 4 This arises from a fundamental misconception in the early days of quantum mechanics, that the quantum state describes the physical state of a particle. In fact the quantum state describes an observer's knowledge of the particle (well illustrated in Schrodinger's cat and Wigner's friend). So, if Alice measures particle A, she acquires knowledge of particle B, ... 1 Make your measurement at B. You'll get either "up" or "down". Repeat the whole experiment many times. You'll get "up" about half the time. Is this because the wave function is collapsing as you measure, or because it's collapsing shortly before you measure? There's no way to tell. Either description leads to the observed 50/... 0 the concept of what happens before loses meaning when the events have a space-like separation. In fact geometrically speaking, this whole business of saying before for such events is ill-defined. There is no strict time ordering possible between both events precisely because any ordering is observer-dependent What do they really mean when you say that in ... 0 If you have many "copies" of your qubits, you can measure both qubits in different bases and do quantum tomography to "scan" state of the qubits. Finally you can compare whether the qubits are same. Note that they can differ in global phase. See more on quantum tomography for example here. 2 If trying to use the wave function description, one cannot always uniquely determine the wave function of the sub-system qubit 1. This roots from the famous EPR (Einstein-Podolsky-Rosen, 1935) paradox. To understand this, considering the following entangled state between qubit 1 and qubits 2-3 \begin{align} |\psi \rangle_{1, 23} & = \frac{1}{\sqrt{2}...

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The non-signaling principle stems from the fact that in our Universe the past cannot be re-written. In fact, a Universe in which the past can be re-written cannot exist, it is self-contradictory. Faster-than-light communication appears in some frame of coordinates as sending information backward in time.

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I expect that Carrol is referring to cluster decomposition, a principle satisfied by many quantum field theories. This principle says that if two quantities are located in spacelike-separated regions very far from one another, then they are going to be uncorrelated. That is, if operators $A$ and $B$ are localized in two spacelike-separated regions a ...

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I don't think you should pay any attention to this. Entanglement refers to quantum states, not to quantum fields.

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A mixed state is very different than an entangled state, but in this case the entanglement can only be detected by measuring both photons. If you only look at a single photon, the state is effectively a classical mixture, because the entanglement cannot be detected in any way.

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Entanglement describes only the relationship (or rather, what is known of it) between the entangled particles. It describes nothing else,

4

Short answer: as you said the nature of the uncertainty associated with the polarization of each of the photons is not classical since they are prepared in a pure state. The uncertainty after observing each photon alone is purely quantum. Long-and-maybe-boring answer: The most general representation of a quantum system is written in terms of the density ...

2

Maybe you're asking to resolve the tension between the following facts: (1) For finite length $N$, the $W$-state $|W_N\rangle = \frac{1}{\sqrt{N}}(|10...0\rangle + |010...0\rangle + ... + |00...01\rangle)$ is not a product state. (In particular, it has up to one bit of entanglement on sub-regions, and an MPS representation requires bond dimension at least ...

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