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A Simple way to create entangled photon pairs?

  1. Jun 13, 2017 #1
    Two photons arrive at a hypothetical 50:50 Beam-Splitter with no phase shift between reflected and transmitted modes. One enters the Left side and the other the Bottom side of the BS as shown in Fig.1 of the link below:

    https://drive.google.com/open?id=0B5JsDLKoUSA5emk5Qk9nUHVIelE

    Each photon is in one of two possible quantum states, |0> and |1>, characterized by a phase of 0 or 180 degrees, respectively. The two photons leave the BS, emerging from the Top or Right sides, being detected by Photo-Detectors, and having their phases measured by PDTop and PDRight. Assuming that the BS is lossless, the two photons are detected simultaneously by PDTop, PDRight, or each one is detected by a different Photo-Detector at simultaneous time-stamps. With regard to this third scenario, if the two photons are allowed to interfere at the BS, it would seem the quantum state of the photon pair would be a Bell State of form C(|00> + |11>) (Fig.2), or, in other words, due to destructive interference, measuring the phase at the photon arriving at PDTop would tell us what is the measured phase of the photon at PDRight, at any time, no matter the distance between the Photo-Detectors. In this case, the two photons would appear to be entangled. Is it possible to create entangled photon pairs in this way, only by allowing them to interfere? I would very much appreciate your opinions. Thank you very much!
     
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  3. Jun 13, 2017 #2

    hilbert2

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    I'm not quite sure about this... If I had to think of a hypothetical way to create an entangled photon pair, my first idea would be to create two electronically excited hydrogen atoms that are in different excited states, and then make the atoms interact in some way (long range van der Waals, or something) before they have time for spontaneous emission. That way the entanglement between the two hydrogen atoms (caused by the interaction) would be somehow transfered to the photons that are emitted a bit later.
     
  4. Jun 13, 2017 #3

    Simon Phoenix

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    It's late here, I'm tired - so I haven't thought about your proposal a great deal but there are a number of issues here.
    1. You want a BS for which we don't get a phase shift between the reflected and transmitted parts? You're not going to be able to preserve the field commutation relations in the output modes in this case, nor are you going to conserve photon number. So that's a non-starter
    2. What do you mean by the phase of a single photon? A photon is a number state of the field and doesn't have a well-defined phase - in fact you'll get any value of phase uniformly at random
    3. Phase and number are complementary variables (the issues with defining a phase operator notwithstanding) - so you can't measure number and phase just like you can't measure position and momentum

    Hope that helps
     
  5. Jun 14, 2017 #4

    vanhees71

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    Concerning 3. You mean you cannot prepare number and phase (or position and momentum of a particle) accurately at the same time. You can always measure any quantity as precisely as you like (modulo technical problems in building an accurate measurement device).
     
  6. Jun 14, 2017 #5

    Simon Phoenix

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    Yes - thanks for correcting my sloppiness :sorry:

    In my defence I was nearly asleep when I responded - not much of a defence as I really ought to know better, asleep or not!
     
  7. Jun 14, 2017 #6

    DrChinese

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    Some (like me) would say that parametric down conversion is a simple method to create entangled photon pairs. :smile:

    As Simon has pointed out, there are a number of issues. The timing would need to be such that there are exactly 2 photons arriving within a narrow time window, and they must be indistinguishable as to source. You might post-select those pairs that meet this criteria, and get some pairs that are entangled. But I doubt many would, regardless of how well you set things up. Even when you get coincidences, you'd get lots of unentangled pairs, or triplets, etc.
     
  8. Jun 15, 2017 #7
    Thank you for your reply. Actually there are some proposals to use semiconductors to generate entangled photon pairs that seem promissing (https://arxiv.org/abs/1702.08823). I am not familiar with the physical process involved but hope to understand it a little better after reviewing the literature :-)
     
  9. Jun 15, 2017 #8
    Hi Simon, thank you! I agree with you entirely, such a BS doesn´t exist. I am new to Quantum Photonics, so forgive my lack of understanding :-). I meant the proposal as a kind of thought experiment, but since came across a very interesting quantum effect related to quantum interference of photon pairs that might work, the Hong–Ou–Mandel Effect. It seems to use the phase difference of 180 degrees between reflected and transmitted modes in a BS to cause a quantum destructive interference. I will try to work out the math and post it here. If I understood your point correctly, though it is possible to define a phase difference between two identical photons, it doesn´t make sense to define phase for individual photons. And, in addition, it is not possible to prepare phase and position of incoming photons due to the uncertainty principle, right? Thanks again.
     
  10. Jun 15, 2017 #9
    Hello, thank you very much! Ok, I think I understood that :-) In your opinion would be possible to mesure simmultaneously the presence and phase of output photons in that hypothetical setup I mentioned? Thanks again!
     
  11. Jun 15, 2017 #10
    Hello Dr.! Thank you for your feedback. Yes, no doubt about that, the perfect timing of input photons is definitly a big issue. Taking that in consideration parametric down conversion really seems simpler. There are some proposals that use semiconductors to generate entangled photon pairs with higher efficiency, that could be used for integrated quantum photonics. Have you heard? Thanks!
     
  12. Jun 16, 2017 #11

    vanhees71

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    First, one has to understand the meaning of quantum-theoretical states. I'm a proponent of the minimal statistical interpretation, which is what you need to use QT to describe real-world measurements and nothing else. According to this interpretation, the states imply the probabilities for an outcome of the measurement of any observable possible on the system. To verify these probabilities you need to prepare a large ensemble of systems in that state and accurately measure the observable you are interested in. You can measure any observable you like, and thus you can also measure incompatible observables, but of course not simultaneously on the same entity, but you can measure first one observable on a sufficiently large ensemble of equally prepared systems and then the other. The accuracy of the measurements is only limited by your measurement apparati, not by the state the system is prepared in.
     
  13. Jun 17, 2017 #12

    Simon Phoenix

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    The basic math you need is quite straightforward - and it's easier to work in the Heisenberg picture to get an input/output relation for the beamsplitter. I'll use ##a## for the input modes and ##b## for the output modes (refer to the figure)
    Wave_Particle_1.gif
    So the input/output relations for the mode annihilation operators are $$ { \hat b }_1 = r {\hat a }_1 + t { \hat a }_2 ~~~~ \rm{and}~~~~{ \hat b }_2 = r {\hat a }_2 + t { \hat a }_1$$The output operators must satisfy the commutation relation ## \left[ \hat b , { \hat b}^\dagger \right] = 1## and also ## \left[ {\hat b}_1 , { \hat b_2}^\dagger \right] = 0## so that the reflection and transmission coefficients obey $$ |r|^2 + |t|^2 = 1 ~~~~ \rm {and} ~~~~ rt^* + tr^* = 0 $$The inverse relations for the annihilation operators are $$ { \hat a }_1 = r^* {\hat b }_1 + t^* { \hat b }_2 ~~~~ \rm{and}~~~~{ \hat a }_2 = r^* {\hat b }_2 + t^* { \hat b }_1$$So if the input state is a single photon state to input mode 1 and vacuum in mode 2 we have $$ | 1, 0 \rangle = { \hat a_1}^\dagger |0,0 \rangle $$The output state is then determined from $$ | 1, 0 \rangle \rightarrow \left( r { \hat b_1 }^\dagger + t { \hat b_2 }^\dagger \right) |0,0 \rangle = r |1,0 \rangle + t | 0,1 \rangle $$where the arrow here is to be read as "the input state becomes the following output state" - so the state to the LHS of the arrow refers to the input modes, and the state on the RHS refers to the output modes.

    So now we can use this to generate the output states at a beamsplitter - it can get a bit messy but for inputs like ##|1,0 \rangle## and ##|1,1 \rangle ## and so on, it works nicely with not too much tedious algebra.

    Now we have this it's not too difficult (but definitely more tedious) to then work out what happens when we bring those outputs back to interfere at a second beamsplitter, that is the Mach-Zehnder interferometer configuration in the figure above.
     
  14. Jun 19, 2017 #13
    Thank you so much Simon! Before I saw your math I had worked out some equations for the Hong-Ou-Mandel Effect. I based myself mainly on Wikipedia, so I am not sure if it is 100% correct, though the general idea seems to agree with your equations. The equations and set-up are here (I haven´t figured out yet how to post nice equations or figures directly ;-)):

    https://drive.google.com/open?id=0B5JsDLKoUSA5RWk3dzNGT1g2N05uTjdJYzdWS0d0V3RjUm5z

    As you can see in Eq.1, I define the input of two single photons as the input ports' creation operators acting on two vacuum modes. The input creation operators become output creation operators through the BS transformation matrix (Eq. 2 and 3). Finally, due to destructive interference of all cross-over output creation operators (T.R or R.T), Eq. 3 seems to indicate an equal probability of simmultaneous creation of the two photons at output port T or R, and zero probability for the two photons appearing one at each output port separately. The experimental set-up for the effect is in Fig. 2. Since measuring one photon at a TOP detector determines the second photon will appear at the other TOP detector, and never at a RIGHT detector, and vice-versa, even if the detectors are located in different galaxies, they would seem to be entangled. I just can´t figure out the physical meaning of the 1/2 multiplier in Eq. 3. It does seem to indicate equal probabilities for the two outcomes, but wouldn´t 1/square_root(2) make more sense?
    Do you spot some mistake in the general idea? Thank you!
     
  15. Jun 19, 2017 #14

    DrChinese

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    Yes, there are a number of ways to create entangled photon pairs. Each has various pros and cons. Here is one proposal using a quantum dot, the idea being to create a pair on demand. You might be interested in this.

    https://arxiv.org/abs/1308.4257

    An on-demand source of indistinguishable and entangled photon pairs is a fundamental component for different quantum information applications such as optical quantum computing, quantum repeaters, quantum teleportation and quantum communication. Parametric down-conversion and four-wave mixing sources of entangled photons have shown high degrees of entanglement and indistinguishability but the probabilistic nature of their generation process also creates zero or multiple photon pairs following a Poissonian distribution. This limits their use in complex algorithms where many qubits and gate operations are required. Here we show simultaneously ultra-high purity (g^(2)(0) < 0.004), high entanglement fidelity (0.81 +/- 0.02), high two-photon interference non-post selective visibilities (0.86 +/- 0.03 and 0.71 +/- 0.04) and on-demand generation of polarization-entangled photon pairs from a single semiconductor quantum dot. Through a two-photon resonant excitation scheme, the biexciton population is deterministically prepared by a Pi-pulse. Applied on a quantum dot showing no exciton fine structure splitting, this results in the deterministic generation of indistinguishable entangled photon pairs.
     
    Last edited: Jun 19, 2017
  16. Jun 20, 2017 #15
    Thank you DrChinese, fascinating subject! I will take a look at the paper.
     
  17. Jun 20, 2017 #16
    Hello vanhess71, thank you for the feedback! What is the minimal statistical interpretation? Concerning observables, do you think it would be possible to say that, in the case of transformation of a photon by a 50:50 BS, the observable hermitian operator would take the form of a sigma-3 Pauli matrix (like the case of the spin of an electron?), with eigen-states |1,0> and |0,1>, referring to the two possible outcomes, creation of the photon at the TOP and RIGHT output ports, respectively, and eigen-values of +1/sq_root(2) and -1/sq_root(2), implying equal respective probabilities? Would that be a correct interpretation? Thank you!
     
  18. Jun 20, 2017 #17

    vanhees71

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    States are states and represented by statistical operators and obserables are observables represented by self-adjoint propagators. So please define your experimental setup, i.e., what goes into your optical apparatus and what do you want to measure.

    @Simon Phoenix explained the state preparation for a single photon running to a beam splitter very well in #12. So what's not clear with this?
     
  19. Jun 20, 2017 #18
    Well, I am trying to figure out what is the observable matrix for the BS. First I thought it would be the transformation matrix:
    |1 1|
    |1 -1|

    But then I realized it doesn´t have eigen-states and corresponding eigen-values. Or, at least, I can not find them :-)
    I also don´t understand why the transformation matrix and the observable operator don´t seem to match. Sorry for my lack of understanding of basics, I am new to the field. The experimental apparatus would be two photons arriving at the BS and I would like to measure their position, that is, if they come out at the TOP or RIGHT output ports. Thanks.
     
  20. Jun 20, 2017 #19
    I just found out another thread called "Photon position again", where the subject of photons' observables is discussed. Or, rather, the lack of observables for photons positions. That may explain partly my confusion, or, at least, tells me I am not the only one confused by this :-)
     
  21. Jun 20, 2017 #20

    vanhees71

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    Don't worry, photons confuse me too ;-).
     
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