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6

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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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 ...


3

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 ...


3

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 ...


3

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: ...


2

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 ...


2

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 ...


2

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 ...


2

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 ...


2

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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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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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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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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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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