How are Superposition and Entanglement related?

Question

How are Superposition and Entanglement related? I don't know much of Quantum Mechanics. I am CSE student and got started with this Quantum Computing. It is interesting! If anybody can help me on this, without digging deep into mathematical details.

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"The principle of quantum superposition refers to pure states of a quantum system. One considers a primary beam of quantal entities that passes into a primary beam splitter or quantum analyser that has multiple output channels. A beam has a pure state when every one of its quantal entities passes into one and the same output channel. A primary beam with a pure state is passed into another, different, secondary beam splitter or quantum analyser. Then the emerging quanta are probabilistically in its several output channels. These several emergent, intermediate, beams are respectively pure with respect to the secondary analyser. They are then passed to respective copies of the secondary analyser, arranged so as to bring them together into a single reconstituted beam. That beam is then passed to a copy of the primary analyser. In general, it will be probabilistically split into the several output channels." - Wikipedia

I did not get that! Any help is appreciated.

Answer 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 repeated measurement gives the same result with probability 1.

However these analyzers are measuring two non-commuting observables. What this means is that the quantum states representing definite values of these observables - "eigenstates" - cannot be the same; a eigenstate of observable A is a superposition of eigenstates for observable B, and vice versa.

So if a beam with quanta in eigenstate A=1 is measured by the first analyzer it takes the A=1 channel. But when the second analyzer measures B, the state of each quanta is in a superposition of B-eigenstates and can take multiple channels. Each quanta collapses to different B-eigenstates. Upon measuring A again at the first analyzer, the quanta are now in superpositions of A-eigenstates and can take channels other than A=1. At the end of this process each of the quanta is in one of the A-eigenstates.

(FYI this picture of quantum collapse, while it describes sequences of measurements very well, is probably incorrect. The more complicated picture is that the state doesn't collapse, instead it becomes entangled with the measuring device and all the superposition states decay away thanks to decoherence)

Entanglement Sets of quanta can be in a superposition. They are entangled if that superposition cannot be factored into superpositions of the individual quanta, which results in a correlation between measurements. For instance, if observable A of two quanta can be 0 or 1 with equal probability, you might measure {0,0}, {1,0}, {0,1}, and {1,1} all with equal probability - these quanta are not entangled because the results are factorable into "A1 can be 0 or 1" and "A2 can be 0 or 1", with no relation between them. An entangled example would be that observable A for the two quanta is measured as {1,0} or {0,1} with equal probability. These are entangled because while measuring either quanta can give A=0 or A=1, you can't measure {0,0} or {1,1}. When quanta are entangled, measuring one allows you to better (or even completely) predict the result of the other measurement.

Answer 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 obeys this wave-function. We can check this by sending the beam onto a polarizer with the direction $\pi /3$ ; all the photons will pass, none will be absorbed.

You also say

"A beam has a pure state when every one of its quantum entities passes into one and the same output channel."

It's not necessary. If the polarizer is oriented in the direction $x$, only 1/4 from the number of photons will pass, and 3/4 will be absorbed. And though, the beam is in a pure state.

You say next

"A primary beam with a pure state is passed into another, different, secondary beam splitter or quantum analyzer. Then the emerging particles are spread probabilistically in its several output channels. These several emergent, intermediate, beams are respectively pure with respect to the secondary analyzer."

Let me return to the experiment above, just replace the polarizer with a polarization beam-splitter (PBS). And let the PBS be oriented so as do discern between polarization $x$ and $y$. By sending our beam $(\text i)$ on it, we get at the output with $x$-polarization a beam with 1/4 of the original beam intensity, and at the output with $y$-polarization a beam with 3/4 from the original beam intensity.

Indeed, each beam is in a pure state, i.e. respectively $|x\rangle$ and $|y\rangle$. But the two outputs still form a pure state. If you bring the outputs on a second PBS, the $|x\rangle$ beam landing on one input face and the $|y\rangle$ beam on the other input face, for a suitable orientation of this PBS both beams will emerge through one single output.

This is not a trivial fact, because if you bring on this PBS two beams, one $x$-polarized and one $y$ polarized, they will be split s.t. you'll have beams exiting through both outputs. he rejoining of the two beams obtained from the state $(\text i)$ is due to the fact that they still remained coherent with one another, i.e. they remained in a quantum superposition. In other words, between the $x$-polarized and $y$ polarized beam, a constant phase was preserved.

No doubt, however, if the 2nd PBS has an arbitrary orientation with respect to the 1st one, the input beams won't exit through the same output face but through two faces. The output intensities will depend on the orientation of the 2nd PBS in comparison with the orientation of the 1st PBS.

Answer 3

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

$(\text {ii}) \ |\psi_2 \rangle = \frac {|QT\rangle \ - \ |QD\rangle}{\sqrt {2}}$

where $|QT\rangle$ is the state quantum turkey and a $|QD\rangle$ is the state quantum duck, the joint system of the two fowls should be

$(\text {iii}) \ |\Psi\rangle = |\psi_1\rangle |\psi_2\rangle = \frac {|QT_1\rangle |QT_2\rangle \ - \ |QT_1\rangle |QD_2\rangle \ + \ |QD_1\rangle |QT_2\rangle \ - \ |QD_1\rangle |QD_2\rangle}{2}.$

That means you have four possibilities, you can find that both fowls are turkeys, or the fowl 1 is quantum turkey and fowl 2 quantum duck, etc. each possibility with probability 1/4.

Well, an entanglement is not separable as a product of two independent states, $|\psi_1\rangle |\psi_2\rangle$. E.g. the state of the two systems may be

$(\text {iv}) \ |\Psi \rangle = \frac {|QT_1\rangle |QT_2\rangle \ - \ |QD_1\rangle |QD_2\rangle}{\sqrt {2}}.$

The uncommon situation with this entanglement is that if you check which type is the one fowl and find turkey, the other fowl will be with certainty also turkey. But if you find that the 1st checked fowl is duck, the other fowl will be with certainty duck. So are the entanglements. Checking of which type is one quantum object, you already know something about which is the type of the other object.