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Wave particle duality & double slit

  1. Nov 17, 2004 #1

    DaveC426913

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    I've been puzzling over the two-slit experiment. There's an alternate setup where you use lasers and prisms.
    [A laser beam is split with a half-silvered mirror/prism. Photons take one path or the other and hit one of two detectors. We arbitrarily insert a converging mirror/prism near the end, and instead of the photons choosing a path, we get an interference pattern - the photons must have taken both paths. Thus, wave-article duality.]

    It seems to me that the mystery lies - not in the outcome at the detectors - but at the first mirror. How do photons bounce off the mirror at the correct angle?

    A rubber ball bounces off an angled surface because it impacts as an area. The ball "knows" what the surface is like and elastic distortion takes care of the rest.

    But in the subatomic world, the finest of mirrors is a collection of atoms that are alomost entirely empty space. As a particle, the photon is going to encounter - well - most likely nothing. I can see it encountering one atom at most, but never more than one. It cannot possibly "know" the arrangements of other atoms, and therefore "know" it should bounce at the right angle.

    So, its got to be the wave property that encounters the wall of atoms. OK. So the waveform of a single photon must be, at a minimum, wider than two atoms' distance. It must be much wider than merely two, or small perturbations in each atom would send photons reflecting at a range of angles.

    So, how "wide" is a photon wave, what's its amplitude? Wait a minute, amplitude is associated with strength. but when you brighten a beam of light, you don't increase the amplitude, you merely send more photons. Do individual photons have a property of amplitude? Is there a way a photon's amplitude can be changed? Or is amplitude is direct function of frequency? Higher freq photons will interact with the mirror better because they have a wider wavefront (or worse because its narrower)?

    Sure - that's why glass mirrors don't deflect X-rays. The X-rays have a narrow wavefront, too narrow to notice the wall of giant atoms. Wait, wouldn't that suggest that lower frequency photons would be *more* easily stopped by glass? Shouldn't glass deflect radio waves? But I digress...

    Do I have it right that the *amplitude* of a photon's wave is the property that allows it to reflect off a mirror/prism at an angle?

    Can someone enlighten me?
     
  2. jcsd
  3. Nov 17, 2004 #2

    ZapperZ

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    There is a flaw in your understanding here. The "best" mirrors (at least within the visible range) are metals. This means that the photon is NOT encountering "atoms", but rather conduction electrons. The conduction electrons have a very efficient reaction to the E-field of the photons - they react and re-radiate the identical E-field (but with a 180 degree phase difference). I can go through all the solid state physics explanation for the conduction electrons dispersion curves on the requirement of such transition, but it will make this a long, boring explanation of conduction electrons and optical conductivity/reflectivity.

    This process is different than refraction or Bragg scattering due to the lattice spacing of the material that you described.

    Zz.
     
  4. Nov 17, 2004 #3

    DaveC426913

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    "This process is different than refraction or Bragg scattering due to the lattice spacing of the material that you described."

    Let me address this for a moment. I know that the laser/beam splitter experiment does not use refraction. It uses either a half-silvered mirror or an optical prism, either way it's deflecting the beam or allowing it to pass unhindered.

    I was not suggesting there was a lattice, I was suggesting that the rows of atoms themselves (in the surface of the glass/mirror) that the photon encounters would act like a picket fence to an ant. The photon simply doesn't see the rows of atoms, it sees at most, one atom. So, how could it 'bounce' at an angle? This is rhetorical, merely clarifying my initial question.

    On to the meat of your answer:
    "The conduction electrons have a very efficient reaction to the E-field of the photons - they react and re-radiate the identical E-field (but with a 180 degree phase difference)."

    So the 180 degree phase difference results in the emitted photon having a trajectory at 90 degrees to the absorbed photon? (or 0 degrees - I assume it is a quantum mechnical effect that causes the photon to randomly pick whether its going to be deflected or pass straight through.)


    When you say E-field, you mean electromagnetic field, right? So, that is the 'wavefront' I was talking about. The photon approaches the angled mirror and interacts with it over an area (where the photon field and the field from the array of atoms intersect). How it interacts over that area is the cause of the deflection, like a wave of water deflecting off a breaker. Right?

    I think this is what I was trying to say, but I'm still unsure about the actual mechanism of deflection.
     
  5. Nov 17, 2004 #4

    ZapperZ

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    But in any case, this point is moot since the atoms are not playing a signifcant role in the reflectivity of a metallic surface. Unless a photon can get beyond the conduction electron (and the skin depth of a visible range of EM radiation is only a few angstrom) and actually gets to excite the lattice phonons, interaction with the atoms of the material doesn't come into this process.

    You are mixing things. There is no "wavefronts" when you consider the interacton at one photon scale. A "wavefront" only makes sense when a bunch of photons (as in a beam of light) are being considered. You are causing a confusion by wanting to consider what is going on when a photon interacts, but using the language of "waves" that requires LARGE number of photons.

    The change in the phase of light being reflected has nothing to do with the angle of incidence-reflection. It has everything to do with (i) conservation of momentum and (ii) optical transition made by the conduction electrons.

    Zz.
     
  6. Nov 18, 2004 #5
    Ok, let's try to help a little DaveC426913 for fun :tongue2: : can we replace the half silvered mirror with a thin crystal material with a n allowing light reflexion (with a correct angle). I am not an optical expert so I do not know if it can work or if it is feasible (e.g epitaxial growth or whatever else) or if it is correct :biggrin: . (I am replacing the metal electronic structure by an electrical dipole structure that I think it has not the freedom of the electronic field in metals). Thus If we have a very thin material with this n, I think we can get now a light beam that can be transmitted or reflected by this very "thin mirror".

    Seratend.
     
  7. Nov 18, 2004 #6

    ZapperZ

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    But then, why not go all the way? Why not replace the half-silvered mirror with a single-crystal insulator? Then the "reflection" that you get is nothing more than the Bragg reflection off the lattice planes. This is what we do in x-ray diffraction experiments to determine lattice structure, lattice constant, etc... Rosalind Franklin even used this to do crystallography of DNA that she showed to Watson and Crick.

    However, anyone who has studied Solids State Physics can immediately tell you that this isn't your normal "reflection", because the amount that is reflected depends heavily the angle of incidence, AND, this can occur in discrete values! This certainly isn't the same type of reflection off a mirrored surface, and certainly not one used in the type of experiment cited in the beginning of this thread.

    Zz.
     
  8. Nov 18, 2004 #7

    DaveC426913

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    "There is no "wavefronts" when you consider the interacton at one photon scale."

    OK, wavefront is the wrong term. I figure a photon-as-wave would have an amplitude. With an amplitude, it would interact as a non-point object. It's merely my conception of how a photon could interact with the angled surface so as to "know" it's angled.

    But OK, let's stop with my wrong ideas, and concentrate on the right ones.


    Conduction electrons are the freefloating ones that bop around within a metal, the ones that cause it to be electrically conductive, right?

    1] They absorb incoming photons and reradiate photons with a 180 degree phase diff (I get phase diffs if we're talking about sine curves). This 180 degree phase diff causes the photons to leave at a 90 degree angle to the absorbed ones? So, if the mirror is angled at 44 degrees, the radiated photons will have a <180d phase shift, and thus exit at 88 degrees?

    2] If the reflection is due to conduction electrons in a metal, then why does a glass prism work similarly?


    I guess I"m going to need the full explanation. I can handle it if you can.
     
  9. Nov 18, 2004 #8

    ZapperZ

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    This requires correction. There is no "amplitude" of a single photon. The "amplitude", as described within the wave picture, requires a stream of photons. This will correspond to the intensity of that light, which is the number of photons hitting a unit area per unit time. So you will see how, if you mix things wrong, you will get a scenario that simply makes no sense.

    I think I have explained this already. The phase shift has NOTHING to do with photons leaving at whatever angle it has to. It is the phase shift of the the beam of light upon reflection, so this is also described within classical optics. The requirement of the law of reflection (i.e. angle of incidence = angle of reflection) is due to conservation of momentum during an optical transition in the conduction band.

    What property is "similar"? A glass prism works via refraction!

    Zz.
     
  10. Nov 18, 2004 #9
    You really deserve the title of science advisor o:) in this forum!
    I agree with the different process of the reflection/transmission it is why I have tried to propose it (choosing the angle to be in the 100% reflection domain).
    I am not saying that we will get the same behaviour in all cases but if we concentrate on a given angle (to get the full reflection) at least I think that may be we can get the same behaviour as the previous one (half silvered mirror). Now we can imagine reducing the thickness of this material to get a transmission while keeping the same angle of the 100% reflection.
    The advantage is that now we have a material with a coarser lattice structure parameter (compared with the electronic field of the metal) and we can have a different/analogue? explanation on how the light can be transmitted and reflected (scattering).


    Seratend.

    P.S. I should have not taken a coffee break, I've lost several posts! :yuck:
     
  11. Nov 18, 2004 #10

    ZapperZ

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    Hahahah... well thanks! That should lay to rest all those accusations that I slept with someone here just to get that title! :)

    I'm not exactly sure how we'd do that. For the 50-50 beam splitter, we characterize it as "50-50" based on the transmission and reflection measurement of the rated frequency of light. In your scheme, we are sitting right at the 100% reflection (this automatically implies we are considering the ideal, zero-temperature state). Considering that the ideal angular spread at this location is a delta function, are you sure that by reducing the thickness of the material that you can achieve the same 50-50 condition of the beam-splitter? This is not something that I'm familiar with nor have I come across.

    Zz.
     
  12. Nov 18, 2004 #11

    DaveC426913

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    But a photon can be considered as a wave or a particle. As a wave, even a single one has a wavelength and thus an amplitude.


    Oh wait. Maybe I mislearned my physics?
    *Light* can be considered wavelike (with a frequency/wavelength).
    Or *light* can be considered a particle (photon).
    But not both at the same time. So photons (QM) do not act like waves(classical).
    Is that more accurate?


    Oh I'm babbling - maybe you should just skip to my last point**...


    Ah, I was thinking of a right-angled prism, where it's face was at 45 degrees to the incident beam. Some of the light would reflect off its face at 90 degrees.

    I was thinking that the rest of the light would pass unimpeded through the glass to emerge in a straight line. But I realize (after sketching it out) that I may have been lazy in examining book diagrams. The beam of light would not pass *directly* through the glass and emerge without changing course, it would of naturally refract as it passed into the prism at a 45 degree angle - which is of no use.

    So, they don't use prisms, they use half-silvered mirrors.


    I could have sworn that's what you meant when you said: "The conduction electrons have a very efficient reaction to the E-field of the photons - they react and re-radiate the identical E-field (but with a 180 degree phase difference)."


    OK. I keep getting off on tangents, running around my own incorrect ideas. (Really, I swear I'm well read in physics. (At least, I *thought* I was... my old brain must be getting rusty.))

    ** Can you quickly walk me through what happens between the moment the photon approaches the mirror and the moment the/a photon leaves at a 90 degree angle?
    .
    .
     
  13. Nov 18, 2004 #12

    ZapperZ

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    There appears to be several problems on top of one another in here. I can only tackle ONE thing at a time before it gets messy.

    So why even bother using a prism since you are not going to use its refractivity? Why not use a plain old glass block?

    Here's the problem with doing that. Transmission and reflection of a piece of dielectric like glass is of a different process than transmission/reflection from a metallic/mirror surface. We can already see this because a glass block will only "reflect" a smaller range of wavelength when compared to a metallic surface (see, for example the "brewster angle"). This is because in a glass block, the reflection and transmission properties are due to the phonon modes of the dielectric, or more precisely, the optical phonon modes that are accessible by the oscillating E-field of an EM radiation.

    In any case, I really do not understand why we are reaching for such a thing. What IS the problem with a beam splitter? And take note that the original reason why I entered into this thread was to correct the idea you had that "atoms" and their valence electron orbits are somehow participating in the optical processes in solids such as a metallic mirror. I think it would be impossible for me to give a lesson on optical conductivity in solids on here since such a thing is covered in whole chapters of books.

    Zz.
     
  14. Nov 19, 2004 #13
    Let's go again if you mind (anybody else in this forum is welcome :tongue: ).
    I am not at all familiar with this possibility but this thread put the question and I think it is interesting to push to the limit common interpretations of reflections/transmission and photon size thanks to the coarser lattice of the material (for fun).
    I know that some high power lasers diodes (array) are now using 45° etched facets coated with SiO2 as a way to have an analogue of a VCSEL laser diode. (they are using, I think, the SiO2/air interface to get the ~100% reflection). This is what is behind my initial though proposition.

    Now the point is the following : if we decrease the thickness of the reflection material (we can assume the zero –temperature state and every simplification) there should exist a thickness “e” where some part of the incident light may cross the material. An engineer point of view :uhh: : if we do not have the material the light is transmitted, if we have some material light is reflected, :confused: ,so there should be an intermediate value where we may have both transmission and reflection :eek: .
    Classic representation:
    n1 < n2 > n1
    | |
    _____| |
    |\ |
    | \ |
    | /|
    _____|/ |
    | |
    | |

    Now, if we assume that the transmission becomes possible how can we explain it? It should be correlated to the thickness of the material.
    Do we have a smooth transition between the 100% reflection and the 100% transmission? Can we use a basic scattering process; do we need to take in account the wavelength of the photon vs size/thickness of lattice to model it in a first approach?

    Seratend
     
  15. Nov 19, 2004 #14

    ZapperZ

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    I am using a Ce-doped YAG crystal to reflect a 3.3 eV laser. You can never tell simply by looking at it because the damn thing is completely transparent at almost all angle in most of the visible range. It means that this crystal isn't suitable to be used to reflect almost anything else except for a small range of frequency. This, again, is exactly the example of what I've mentioned earlier.

    I'm not exactly sure I understand where this is going. There have been studies on the "usual" reflection off mirrored surfaces and making the thin-film reflecting surface to be smaller than the skin depth. If the thickness is less than the skin depth, then you no longer have your "100%" reflection. However, this is well-known, and I'm guessing this is not what you are talking about here..... so I don't know.

    Zz.
     
  16. Nov 19, 2004 #15
    In fact that is just what I am talking about, If I have understood what you've said :wink: .
    The skin depth may be viewed in several ways (classically or quantum). Now we are just considering the quantum view of a single photon arriving on such interface (so we can consider a photon and a laser with the adequate wavelenght).
    This was only proposed at the beginning of the thread as an alternate possibility just to replace the electronic field of the metal with a coarser dipole structure and to see how the single photon limit behaves at such interface (interpretation).

    Seratend.
     
  17. Nov 19, 2004 #16

    ZapperZ

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    OK, I will reproduce what I believe, was the central question of the beginning of this thread:

    1. I assume the fallacy surrounding a "reflection" off a metallic surface has been answered.

    2. If #1 is correct, then the ONLY question left is the reflection process off dielectric material. I have discussed this at length already, both in this thread and in other previous threads. It can be either direct Bragg scattering, or one is exciting various phonon modes in the dielectric.

    Therefore, I don't understand what is left to be said. I don't think I want to tackle the "photon is wave or particle" issue for the gazillion'th time on here because I want to save it for the next quack that will drop by any second now. Besides, I've cited ample experiments already (one is even listed in my Journal entry on the coincidence measurement done at the undergraduate level, no less).

    Zz.
     
  18. Nov 19, 2004 #17

    DaveC426913

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    OK, while this thread is at risk of getting hijacked by tangential discussion, I'm going to make an attempt to pursue my original question (ZapperZ, I'm starting to feel like you and I are standing in a room full of people, trying to have a 1-on-1 ... ).


    "...the ONLY question left is the reflection process off dielectric material. I have discussed this at length already..."

    I appreciate your patient attempts at enlightening me, (the quack remark was uncalled for). But to my dismay, a review of all the posts in this thread has turned up little that specifically discusses the actual mechanism by which a photon intreracts with a mirrored surface:

    "...conduction electrons have a very efficient reaction to the E-field of the photons - they react and re-radiate the identical E-field..."
    "...The requirement of the law of reflection (i.e. angle of incidence = angle of reflection) is due to conservation of momentum during an optical transition in the conduction band..."

    I understand how a wave can reflect, but can a single photon reflect? This mechanism is what escapes me.


    Perhaps you could direct me to some URLs where I could read up on it. (layperson preferably. I'm well-read, but the math will be lost on me.)
     
  19. Nov 19, 2004 #18

    ZapperZ

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    Er.. the "quack" reference wasn't meant to refer to you at all. If I ever thought that you were a quack, trust me, you'd never get any serious physics answers from me, because I would already conclude you would be either incapable or do not want such answers.

    You need to make sure you understand when I'm refering to "photon" and when I'm refering to "photonS". ONE single photon hitting a metallic surface will cause an optical transition in the electronic conduction band. What is a conduction band? Read http://euch3i.chem.emory.edu/proposal/www.atip.or.jp/~johan/cth/solid/report.html

    Because of the conservation of momentum parallel to the plane of the conductor (and the recoil momentum of the lattice ions), the decay transition will re-emit that single photon! In an ideal process, that will give you the reflection of a single photon!

    The same occurs when you have a gazillion photons as in a beam of light. However, when you have THAT, it is no longer efficient (or even makes sense) to talk about it at a single-photon level. YOu now have a collection of many photons in a coherent beam and it is perfectly valid to consider this as a classical optics problem. This, however, doesn't mean we have no idea the microscopic mechanism of reflection and transmission. People study the properties of materials using optical processes, so we must know a lot about it to use it as a tool.

    Zz.
     
  20. Nov 20, 2004 #19

    DaveC426913

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    Thanks. I'll read it.
     
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