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What happen to a single photon entering a PBS.

  1. Feb 20, 2009 #1
    Hi.
    I have a question regarding what happen to a single photon when it hit a polarization beam splitter (PBS) that splits photon based on Horizontal and Vertical polarization.
    If the input single photon is Horizontal or Vertical, i'm sure it will go out through the appropriate Horizontal output port or Vertical output port (assuming nearly ideal PBS).

    But i'm confused to the case of a single photon linearly polarized at +45 degree. Will the X component and Y component split to the Horizontal or Vertical output port ? This may be in a situation such as the Sagnac Interferometer.
    And also what would happen to a circularly polarized single photon entering a PBS ?

    (sorry, I'm new to this field).

    --islahna
     
  2. jcsd
  3. Feb 21, 2009 #2
    No, you can't split a photon up. Either the photon will go through or it will not. The probability for 45 degrees is 50-50, but it turns out that the probability with respect to angle follows the same rule as the intensity with respect to angle derived from classical physics.

    Your questions are essentially regarding base states. Either you pick your base states as |R> and |L>, corresponding to right and left handed polarisation (circularly polarized states) or you pick them as |x> and |y>, corresponding to linear polarisation. As you may have learned, if you have done any quantum mechanics, any state may be formed from a linear superposition of base states.

    |phi> = a|R> + b|L>
    or
    |phi> = c|x> + d|y>
    In one set of base states, let's pick the linear, it may be that c is 1 and d is 0. i.e. the photon is fully polarized in the |x> state. i.e.
    |phi> = |x>
    This state must be able to be represented in the circular base states also.
    |phi> = |R><R|x> + |L><L|x>

    <R|x> and <L|x> are just two numbers and their values are 1/√2. These values can be worked out and there's a good description in Feynman's Lectures on Physics 3 Volume.

    Thus, if you send linearly polarised light through a right handed polarizer, 1/2 (amplitude squared) of the photons will pass through. These photons are now right circularly polarized, which can also be expressed using the linear base states.

    Picking the correct base states often simplifies a problem enormously, thus working with linear base states and using circular polarizers is a bit awkward, but it can of course be done and will result in the correct probabilities!
     
  4. Feb 23, 2009 #3
    Hi DeShark,
    thank you for responding. Yes, I agree with you that we can't split a photon. However,
    I came across a paper that puzzle me up recently. Or, did I understand it different from what they were trying to tell ..

    The following are snippet from a paper of Chinese Physic Letter Vol 25, No 11 (2008) entitled "Quantum Key Distribution System with Six Polarization States Encoded by Phase Modulation" that puzzled me up recently.


    ----------starts ------------------
    Single photon pulses in 45 degree linearly polarization state
    |in⟩ = |45⟩ = √2/2(|⟩ + |⟩),
    are emitted by the semiconductor laser and delivered to the coding system at Alice side. Each of the pulses splits into two orthogonal component, Px and Py. The transmitted Px is directly reflected back by the FM1 with its state rotated to |V> so that it will be reflected by PBS1 to go anticlockwise through the loop before recombined with the other component at PBS1; While the reflected Py goes clockwise through the loop, reflected again by PBS1 and then reflected by FM1 with rotation to the state |H> so that it can pass through PBS1 where Px and Py reunite. The polarizing interference between Px and Py results in a new polarization state as the output of the Sagnac interferometer.
    -------------- end -----------------------------

    thanks,
    --islahna
     
  5. Feb 23, 2009 #4
    Ahh, it's talking about the interference of two wavefunctions with different polarisations. I'm not sure what FM1 (full mirror?) means. I also assume that |H> is for horizontal and |V> for vertical. All in all, the paper is quite confusing without a diagram. Or maybe it's a little late in the day for me.

    Basically, since we don't measure the polarization states of the photons after passing them through the polariser, they are free to interfere and show their wave-like properties. I'm not all that familiar with Sagnac interferometers (I've read a bit about them, but never used one), so I wouldn't like to comment too much. However, it would seem that the purpose of the exercise is to replace the half-silvered mirror in a normal sagnac interferometer with a linear polariser. Thus the horizontal portion of the wave function takes a clockwise (for example) path and the vertical portion takes an anti-clockwise path. The two are then brought together and interfere. This interference creates a new polarisation state which can be measured (well, it has a probability, which can be measured if we use many photons) by passing the beam through yet another polariser. If that makes sense and helps any?
     
  6. Feb 23, 2009 #5
    If you're talking about spontaneous parametric down conversion as 'described' by most
    papers involving Bell test experiments (and Wikipedia), I too would like to know...
    ... as every one of them describes a single photon 'splitting' into two.
    What is the actual quantum theoretical description of this process?
     
  7. Feb 24, 2009 #6
    Yes that helped.
    mm..am I totally wrong or almost correct if I assume the following :-

    Based on your explanation on the snippet of the said paper,
    - we can use the wave characteristics of a single photon.
    - where it has two orthogonal components that is |V> (vertical) and |H> (horizontal)
    - we can split the |H> and |V> components of a single photon
    - we can change the phase of the |H> and |V> component
    - we can recombine back the |H> and |V> components to produce interference
    - due to the phase difference between |H> component and |V> component, a new polarization state will emerge from that interference
    - we now have our single photon back but with different polarization state.

    best regards,
    --islahna
     
  8. Feb 24, 2009 #7

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    There are conservation rules that govern quantum processes, and these are evident with PDC. By the way, down conversion may be viewed theoretically as time reversed up conversion. Either way, 1 photon in/out and 2 photons out/in. There are 2 types of PDC crystals usually employed, called Type I and Type II. You need 2 Type I crystals to produce a polarization entangled state. From a paper describing an undergrad Bell test:

    "SPD can be understood as the time-reversed process
    of sum-frequency generation (SFG). In SFG, two beams
    of frequency f1 and f2 meet in a nonlinear crystal that
    lacks inversion symmetry. The crystal acts like a collection
    of ions in anharmonic potentials. When driven at
    both f1 and f2, the ions oscillate with several frequency
    components including the sum frequency f1 + f2. Each
    ion radiates at this frequency (among others). The coherent
    addition of light from each ion in the crystal leads to
    constructive interference only for certain beam directions
    and certain polarizations. The condition for constructive
    interference is called the 'phase matching' requirement:
    inside the crystal the wavevectors of the input beams
    must sum to that of the output beam. In SPD, the
    violet laser drives the crystal at the sum frequency and
    downconverted light at f1 and f2 is produced."


    I hope this helps a little. There are not a lot of good detail theory sources out there on PDC, most is embedded within experiemental papers. Look to texts like "Optical Coherence and Quantum Optics" by Leonard Mandel & Emil Wolf (p. 1074) for a treatment.
     
  9. Feb 24, 2009 #8
    Thanks very much Dr C., it sounds interesting.
     
  10. Feb 26, 2009 #9
    Yes, that's my interpretation anyway. Does that make sense in context? Bear in mind I'm just an undergrad student, so if this is major mission critical work, you might well want a second opinion.
     
  11. Mar 2, 2009 #10
    hmm. Can't get hold of the papers even though they are quite old (without paying!).
    One abstract mentions simply using scattering theory as the basis for correctly calculating the probablility in line with experiment. Couldn't see reading up how that might work. I also read around pertubation theory but I can't get my head around the complete application of this here. It seems there is an amplitude for photons to be emitted for any oscillation that forms part of the anharmonic nature of the ions in the crystal. The probability of absorbtion would equal the probability for emission all things being equal. But why do both photons always get emitted in pairs like this? I.e. why do both down converted photons come out at the same time. I suppose we have to say that it has a probability as a 'single event' somehow. Also the probability of one but not the other is zero, i.e. the ions oscillation is-but-it-isn't comprised of simpler harmonics?
     
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