# Tag Info

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No-cloning theorem says that you cannot "clone" the state. As in there is no way to make to make perfect copies of a quantum state. In the simple teleportation case with two parties the Alice's state is transferred to Bob by using the EPR pair. There is no copying. But if Alice were to try and send her state to two people(Bob and Dina) then that would mean ...

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First, we need to define the interference pattern. It is the pattern formed by the fundamental frequency of the wave properties of the electron, passing simultaneously through two slits with "suitable" width and separation distance. When a "detector" is placed on one slit (A), it takes away some of the energy and allows only a higher harmonic (with lower ...

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I would highly recommend the two seminal papers by E. T. Jaynes, http://journals.aps.org/pr/abstract/10.1103/PhysRev.106.620 and, http://journals.aps.org/pr/abstract/10.1103/PhysRev.108.171 Also check out the book by E. T. Jaynes, which has a focus on the foundations in probability but is rather light on applications in physics: ...

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The faster processor would have less mass near it. Furthermore if Photons and Time are slowed by the back hole's gravity creating a red shift then phonons would also be slowed and pulled in by gravitational pressure making matter colder after passes the photon sphere. Temperature and energy would be slowed near absolute 0 before the horizon. It would be ...

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The OP's confusion seems to stem from the incorrect assumption that if my detector isn't triggered I cannot see how one could argue it interacted [with the electron] Just because the detector sometimes does not click, does not mean that there is no interaction at all. A good way to think about this is in terms of continuous measurement. This and this ...

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The problem is that you are treating quantum objects as both classical waves and classical particles simultaneously. More specifically, you talk about them passing through one slit or the other and sensing which slit an electron goes through. But in order for the interference pattern to emerge, the electrons have to pass through both slits at a time. We can ...

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The correct formula is $$\mathrm{tr}[A^TB]=\langle m \vert A\otimes B\vert m\rangle\ ,$$ so your proof is correct, you're just trying to proof an erroneous formula. (You can easily verify this because with a $\dagger$ the l.h.s. is sesquilinear while the r.h.s. is bilinear.) But I have the feeling this has been asked before. If you have this from ...

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Presumably your vector stands for a three-dimensional quantum state. The Bloch sphere is really only useful for two-dimensional quantum states, as commented by Emilio Pisanty. Theoretically, there is an equivalent description to the Bloch sphere for three dimensional quantum states, but it is not useful for visualization as it is an eight dimensional ...

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If B released energy when A was measured, you could use it for communication. And you know the answer to "Can entanglement be used for communication?" is NO. You mentioned it in your question! Therefore... No, entangled particles don't release energy when their partner is measured. They don't change in any locally determinable way when their partner is ...

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If the particles are nontrivially entangled, then particle $A$ cannot be in an eigenstate of the energy operator (or any other operator that acts just on $A$'s state space) in the first place. If the initial entangled state is, say, $X\otimes Y+Z\otimes W$, where (for example) $X$ and $Z$ are energy eigenstates, then an observation of particle $B$ will ...

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The best way to understand this is to work through an example. Here is how you factor the number 15 using Shor's algorithm. (If you prefer, view the problem not as factoring 15 but as solving $2^r=1 \hbox(mod 15)$, and skip Step Seven). As you work through this, imagine replacing 15 with a much larger composite number, and you'll see the advantage of ...

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There are two kinds of operators which have an intuitive meaning when you apply them to a state: Evolution operators like $e^{-i\hat H\Delta t/\hbar}$ simply take the state at time $t$ and give you the state at $t+\Delta t$ Projectors tell you what the state will be after you take a measurement. For example, supose you measure an observable $\hat A$ on ...

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NOTE: This answer has now been merged into Understanding the quantum eraser from a quantum information stand point (part IV). Let me start by copying the first part of my previous answer which describes the circuit model of a double-slit or other interference experiment; then, I will try to describe the delayed choice setting (the way I understand it). ...

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You are absolutely right with the cards. To lift the entanglement on atomistic level talented physicists created the spontaneous parametric down conversation. They solved the task to create pairs of particles (photons) with one entangled parameter. Often this parameter is the direction of the electric field component of the two involved photons. This ...

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The answer is structured as follows: I will first give the quantum circuit corresponding to a normal double slit (or interferometer), then the circuit where the which-way information has been recorded, a circuit where the which-way information is first recorded and then erased in a unitary way, and finally a circuit where the which-way information is ...

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R.F. Werner, Optimal Cloning of Pure States describes the optimal procedure for cloning multiple copies of the same pure state (which, remarkably, for pure states is independent of the figure of merit). Abstract: We construct the unique optimal quantum device for turning a finite number of $d$-level quantum systems in the same unknown pure state ...

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I have a lot of experience with this particular book, but unfortunately it is in my office right now so I can't reference the exact page you're on. In QIQC, and this book in particular, you're going to be doing a lot of manipulating the Hamiltonian of the QSHO using commutators. Since constants always commute, the constant term of the QSHO falls out of ...

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I have not looked at the book , however the sense in which it is an approximation is that it is neglecting the constant term The Hamiltonian of a SHO is , $H= (a^{\dagger}a + 1/2)\hbar\omega$ This means that the ground state energy of the SHO is $1/2\hbar\omega$. This is what is being neglected since it is only a constant.

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The delayed choice quantum eraser doesn't delete a qubit, it measures an entangled qubit along a parallel axis or else along a perpendicular axis. As an analogy, suppose Alice and Bob share an EPR pair $\left| 01 \right\rangle - \left| 10 \right\rangle$. Alice measures her qubit from the pair along the Z axis. Now, Bob could also measure along the Z axis, ...

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Barium ions are used in ion traps and have a similar energy level structure to what you are referring. The ground state is the $6^2\text{S}_{1/2}$ and the excited state is the $6^2\text{P}_{1/2}$ driven at $493.6\ \text{nm}$. There is a decay down the metastable $5^2\text{D}_{3/2}$ with a relatively high branching ratio though (3:1 ground state to ...

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The Hadamard gate is a 180 degree rotation around the diagonal X+Z axis of the Bloch sphere. Terrible picture (from a blog post): In the diagram the |0> state is at the top (+Z) of the sphere and the |0>+|1> state is at the front (+X). Rotating around the back-bottom-to-front-top axis (X=Z,Y=0) moves the top point (|0>) to the front point ...

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No, you can't tell if someone else has observed an entangled copy of your quantum information. And the YouTube videos about the delayed choice experiment are particularly bad, even for quantum things. You can get evidence that an entangled copy was made, because your qubit's density matrix will correspond to a mixed state instead of a pure state (center of ...

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