Tag Info

New answers tagged

2

Once Alice makes the measurement, her state is collapsed and her spin will not be affected by Bob's subsequent measurement. Bob's spin is now determined in the z-direction, however, which is what allows the usual quantum key-distribution techniques. To be more precise: Before Alice's measurement the state is $$|+ \rangle |-\rangle - |-\rangle |+\rangle.$$ ...


2

No, other properties do not necessarily become entangled. Example: You can take twin photons from type-II parametric down conversion. These two photons are entangled in energy and momentum, but not in polarization. For more information about this example, see: Physical Review A 50, 5122 (free version available here). As described in the article, terms ...


1

Suppose the particles are initially in the (entangled) state $$A\otimes B+C\otimes D$$ where $A$ and $C$ are position eigenstates for particle 1 and $B$ and $D$ are position eigenstates for particle 2. Note that this state is the same as $$X\otimes Y+Z\otimes W$$ where $X=(1/2)(A+C)$ and $Y=(1/2)(A-C)$ are momentum eigenstates for particle 1 and ...


2

I think the key point that you're missing is that as soon as you make a measurement, the entanglement between the two particles is broken. It should also be noted that the original particle also obeyed the uncertainty principle, and that at a quantum level there is no direct relationship between position, momentum and time. Another confusing factor is that ...


1

Ok. Let's suppose that the initial state of the two particle are an eigenstate of the momentum operator (momentum is well defined). Quantum mechanics tell us that the position of the center of mass is not well defined. If we measure the position of the particle 1 (electron), then we do two thing in the system: We apply a measurement in a part of the whole ...


3

Since you're an electronics student, I'll speak your language. Think of momentum and position as parameters in time and frequency domain of a signal rather than classical observable that are well defined. If you do so, you can easily realize that your frequency isn't well defined if you don't do an infinitely long measurement. This is simply due to the wave ...


1

The problem is that you propose to make the two measurements "at time $t$ simultaneously". Measuring the particle's momentum cannot be done instantaneously; the more precisely you want to measure it, the longer the minimum required observation time becomes. (Rougly speaking that's because knowing the particle's momentum is equivalent to measuring its ...


2

What is wrong with this logic is that you are supposing a particle has simultaneously well-defined position and momentum. This is not true - a state localized in real space is delocalized in momentum space, and vice versa. The classical conservation laws hold on the quantum level as operator laws, not as laws on the states.


1

My 1st answer described what is a coherent superposition. Now, in short about entanglements. Taking two quantum systems, e.g. one quantum fowl described by $(\text i) \ |\psi_1 \rangle = \frac {|QT\rangle \ + \ |QD\rangle}{\sqrt {2}}$ and the other quantum fowl described by $(\text {ii}) \ |\psi_2 \rangle = \frac {|QT\rangle \ - \ |QD\rangle}{\sqrt ...


1

Superposition & Measurement The two analyzers measure different properties, or observables, of the beam quanta. Prior to measurement, a quantum state may be in a superposition of many values for an observable, so there is a probability that you will measure any of these values. When you measure an observable, the quantum state "collapses" so that a ...


2

You confuse a couple of things: a pure state exists also without a beam-splitter. For instance, a beam of photons can be prepared in a state of linear polarization $ (\text i) \ |\phi \rangle = \frac {|x\rangle + \sqrt {3} |y \rangle}{2}.$ So it gets out from the apparatus which prepares it. Of course, if the beam is prepared this way, then every photon ...


1

If you are interested in how these sets look like, I can provide their characterization. A state on two $d$-dimensional systems is maximally entangled if and only if it can be written as follows: $$\frac{1}{\sqrt{d}} \sum_{i=1}^d |u_i\rangle |v_i\rangle$$ where $|u_i\rangle$ is the $i$-th column of some $d \times d$ unitary matrix $U$ and similarly ...


3

Following http://arxiv.org/abs/quant-ph/0110082, for two qudits the manifold of all product states is $4(d-1)$ dimensional, while the manifold of maximally entangled states has $d^2-1$ dimensions. Thus, with the possible exception of $d=3$, these two sets cannot be mapped onto each other by any kind of "nice" mapping (and in particular not by a linear map). ...


0

Any 4x4 unitary matrix is onto, so onto is going to happen one you insist on a unitary linear map. Neither the set of maximally entangled, nor the set of separable states is a linear subspace. However, if you insist that every (nonzero) separable state map to a maximally entangled state, then you would have to send $|++\rangle$, $|+-\rangle$, $|-+\rangle$, ...



Top 50 recent answers are included