Problem with the two slit experiment, Observing later

TRoc,

Well, now you've gone and upset the applecart.

No detector = no mixing = no beat frequency. The detector becomes the
point of quantum interaction and frequency mixing.

I think it is fairly obvious that photon EM energy affects the "steady state" EM
fields of matter (electrons) by displacing them from their ground states. This
causes new frequencies to be emitted as the result of those EM enegy fields
interacting, as they mix, by altering the vibrational frequency of the electrons and
atoms.

The resultant frequency is determined by the electrons in the atomic structure
vibrating in "harmony", additively, with the electrical energy of the applied EM
fields and forming a composite (mixed) EM field output frequency.

Can the EM energy of a sinusoidal wavefunction displace or modify
the sinusoidal EM energy of another wavefunction in open space? No way to
prove that it can happen unless the result can be measured. We can, however,
isolate each wavefunction just prior to the detection point to verify if it still
maintains its discrete wavefunction characteristics. This being the case, I
say there is no signal mixing prior to the point of detection.


LL

This post has been edited by Laserlight on Dec 27 2006, 07:58 AM

LL,


Just a note, regarding a past conversation, re. color monitors, etc.

Interpolation and Gamma Correction
http://home.no.net/dmaurer/~dersch/gamma/gamma.html

Mathematically dark, .. but not "real world". hmmm

http://en.wikipedia.org/wiki/Anti-aliasing

All,

Back to the "black and white" version of explaining the S/DSE.

http://en.wikipedia.org/wiki/Anti-aliasing

Our wall is sampling the laser frequency, after having gone through a slit (or 2). Nothing has been "removed".

Hey, good news! Our screen/wall has an finite amount of electrons. Only a finite number of "dots" on the screen.

ouch! amplitude = intensity ; not corresponding to the different energies, of the different frequencies

How about trying an "Energy Transform"?

Great! Someone else agrees that we always need AT LEAST 2n (or 2f) to "sample" (measure). Since the Fourier method HAS j & k, then we are up to 4 frequencies. Methods CAUSE results; theories PREDICT them. Separate the method from the theory to unravel the truth.

That truth applies to filters too, as I have said numerous times. We are dealing with "approximations" already. Do we want to improve this?

mmm... wavelets... g o o d

But if the SOURCE is rectangular, back to the "wavelets". Notice the presence of "destructive/constructive" interference crests.




How would that look if it were also spinning? Can we get a look at the "counter rotational", dualistic vibration, that would enter BOTH slits with an opposing, symmetrically balanced, angular momentum? Yes, we CAN! And, we can use a REAL example of a light source. None other than the Sun itself, with its' entangled E and B fields.


(all images from Wikipedia)




ciao,

T.Roc



This post has been edited by TRoc on Dec 27 2006, 09:08 AM

LL,

Well, you asked the question!


Again, I am not arguing that the electron positions (lattice) are not a point of mixing. You will have to put a detector there, if you want to see the interaction. Just like with the super-colliders.


But, now you've gone and upset the applecart too!

I can use your statement to "disprove" the "photon" transaction altogether! Anywhere there is NOT a detector, then there is NOT a "photon", correct? We could NOT put a detector at any unit of length along the path, and NOT measure anything!

But if we DID, we WOULD. So, what is the difference?

It seems you have erected a fence between GE's "dimensions", and my "Resonant Interactions", and you're standing on it.

Is there "something traveling" between source and detector, or not? It (a "photon") either exists between events, or not, IMO.

Does Doppler redshift change the wave? It shows that way when we measure it. Does Gravity, from a large mass (large energy), bend the path? It shows that when we do the measurements. Do AM stations "fight each other" for the space their amplitude is modulating in? The signal coming from the speaker says they do. What about potential energies (mass) of free particles interacting (colliding) directly? What about watching interference happen through a prism? The rainbow?

http://arxiv.org/PS_cache/quant-ph/pdf/0603/0603048.pdf
Experimental interference of independent photons

50% visibility = light/dark bands

QM predicted this would NOT happen. In 1956, it did; stellar diameters were measured with beat-frequencies.

(emphasis added)

Here! Here! We need a "conceptual advantage".


regards,

T.Roc


BTW: My model is "classical" and undiscovered. It is the WHY for quanta, and already has a 2500 year (written) track record.

This post has been edited by TRoc on Dec 27 2006, 10:13 AM

Hi TRoc,

Just to make sure we're looking at the same experiment..

Clicker
http://www.teachspin.com/instruments/cricket/index.shtml

Clicks when a photon is detected

Instrument specification
http://www.teachspin.com/instruments/two_s...fications.shtml

Needs a counter of at least 0-1Mhz (supply your own). The counter counts when a photon is detected. After (say) 1 second the the counter displays how many it has counted .. leaves that number (eg 2400) displayed while it resets it's count and counts up again for another second .. then displays the new number.. etc.

http://www.teachspin.com/instruments/two_s..._combiplot2.gif

Looking at the counts when the detector is 6.1mm from the origin..

Slit A only .. 600 photons per second
Slit B only .. 600 photons per second
Both open 2400 photons/second

Of the original (say 1 million/second) that actually pass through the slits ..
We're never seeing 4 photons being counted .. the photomultiplier can only respond to one photon at a time. We're seeing that the probability of 1 photon being found at a particular location has increased by a factor of four at a point of constructive interference.

Clearly there are less photons being counted when the detector is at 5.6mm and 6.5 mm .. this is the reason why we can count 4 times as many when the detector is at 6.1mm and say that both the counter and our (my) theory looks pretty good.

TRoc theory will have to predict the same result if it is to agree with the result of this experiment.

Best wishes,

-C2.

Evidence that a photon can only interfere with itself..

We happen to think of the DSE as a laser effect and we know lasers give a coherent output but the DSE works with thermal sources .. light bulbs and the Sun. (after monochromatic filter)... regardless of intensity. If non-phase matched photons interfered with each other then there would be no possibility of cancellation... the observed interference result requires perfect phase matching at the slits and that only leaves one obvious possibility. For this explanation to 'work' you have to accept that the cancellation (dark bits) are the result of the difference in path length from the slit to the detector... see the DSE Equation . We have two slits .. the only thing that we can be virtually sure has a fixed phase relationship at these two points is a single photon..

IMHO just looking at the DSE equation reveals a great deal about the DSE and interference in general.

-C2.

Hi Good Elf,

You imply that a single expanding wave (of any form) can simulate the effects of the DSE equation which suggests that interference is an effect caused by the path difference between two points (the slits) .. this would bring us back to the result we see in the DSE. If there are expanding spheres then to get the right result I think there must be two of them and they are centered on the slits.

Best wishes,

C2.

Good Day!
Keep going, now we are getting into the inside of things.
Question: I interpret "classical" to mean particle-like. Correct?
jal

Hi jal,

My 'classical' means electromagnetism .. Faraday, ripple tank etc. I'm not sure about anyone else though.

Best wishes,

-C2.

This post has been edited by Confused2 on Dec 27 2006, 04:43 PM

TRoc,

Yep, I am standing in a node.


You might find these of interest.

http://www.csm.astate.edu/music.html


http://id.mind.net/~zona/mstm/physics/wave...eDiagrams1.html

http://id.mind.net/~zona/mstm/physics/wave...dingWaves1.html

http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=35

http://id.mind.net/~zona/mstm/physics/wave...erference2.html

http://lectureonline.cl.msu.edu/~mmp/applist/beats/b.htm

http://webphysics.davidson.edu/applets/Sou...t/SoundOut.html

http://www.walter-fendt.de/ph14e/beats.htm


LL

C2,


I can't believe it, but here we go again. The last time around, I bashed you so hard with truth and logic, you ran crying down the hall, yelling "TRoc's gone amuck in the lab!" Funny once, but NOT twice.

YOU

GUESS WHAT? YOU are NOT looking at the same experiment.

(emphasis added)

I am talking about "1-at-time-photons" C2.

So, all of the first post (Confused2 Posted on Today at 11:16 AM) is pointless. If you are talking about "2400 photons/second", then the STATISTICS model works.

Not "one photon"; No more comment.


Your next post (Posted on Today at 1:22 PM), starts with:

First of all, your "evidence" can ONLY be isolated, "1-at-a-time photons", otherwise, the statement is PLURAL, and has an ENTIRELY different meaning. You then go on to say all of this:

Since my argument is that the COUNTER is flawed, then YOUR argument (QM) falls apart. There is nothing close to a guarantee that "1-at-a-time photons" are being produced. The Sun, a light bulb, and the laser have ONE thing in common: they produce HOARDS of "photons". Get a single atom, excite it by ONE orbital, and THEN you get ONE "photon".

Light casts evenly in a cone past the slit. The banding is made by the dual nature of light, and the ELECTRIC AND MAGNETIC fields that MAKE it work. These two things DO NOT ever go together, they remain AT 90 degrees to each other. When you take away 1 degree of freedom (dimension), that is from "free-flight" 3D, to impacting the wall 2D, that is what happens. They can NOT remain at 90 degrees, so they end up in separate areas. That completely explains the light and dark bands. Just break up a bar magnet into several pieces, and line them up. S,N,S,N,S,N ... get it? No math or geometry is necessary, just a basic understanding of EM. My theory does not need "proof", it predicts the outcome from the MOST simple explanation, that is ALREADY accepted.


From YOUR link: http://www.teachspin.com/instruments/two_s...periments.shtml

Here is the first point: a shift in wavelength is a change in frequency. There is now more than 1 frequency in the experiment, just by lowering the intensity. (among other reasons)

Arranging the parameters to SHOW the desired outcome. This is not Science, it is dramatization.

C2, they "MUST BE SET" to "REJECT THE DARK CURRENT". This is direct EVIDENCE for the FACT that there is MORE THAN ONE FREQUENCY to account for.
"Dark current" is code-word for other "photons".

At this level of education, THE STUDENT ALREADY KNOWS "PROPER" COUNTING METHODS. The reality here is that this is INDOCTRINATION into the QM world. This is the "seed" that is going to generate the "desire to purchase"

Directly from THEIR mouth: "single photon" requires "special" interpretation to be valid. Logic is not allowed here, apparently.


Jal, when I say "classical", I mean Maxwell & EM; NOT a particle.


regards,

T.Roc



This post has been edited by TRoc on Dec 27 2006, 07:11 PM

Good Day!
"classical" to me is not particle but particle-like which includes both your interpretations. Okay!

I want to try to improve the particle-like picture of a photon.
In order to get what we want in the DSE we must remove the extra/unwanted/noise that would “blur” the result that we want to achieve.
TRoc has illustrated this point to my satisfaction.

This is what we do to any signal that we want to receive/produce. (radio,laser, DSE etc.)
So what you are doing is looking at how you are manipulating the photon prior it being sent out by the transmitter. (Signal processing)

Due to treating the photon as particle-like we got a lot of technology. For a start look at
http://en.wikipedia.org/wiki/Ray_tracing
What happens in nature
In nature, a light source emits a ray of light which travels, eventually, to a surface that interrupts its progress. One can think of this "ray" as a stream of photons travelling along the same path. In a perfect vacuum this ray will be a straight line. In reality, any combination of three things might happen with this light ray: absorption, reflection, and refraction. A surface may reflect all or part of the light ray, in one or more directions. It might also absorb part of the light ray, resulting in a loss of intensity of the reflected and/or refracted light. If the surface has any transparent or translucent properties, it refracts a portion of the light beam into itself in a different direction while absorbing some (or all) of the spectrum (and possibly altering the color). Between absorption, reflection, and refraction, all of the incoming light must be accounted for, and no more. A surface cannot, for instance, reflect 66% of an incoming light ray, and refract 50%, since the two would add up to be 116%. From here, the reflected and/or refracted rays may strike other surfaces, where their absorptive, refractive, and reflective properties are again calculated based on the incoming rays. Some of these rays travel in such a way that they hit our eye, causing us to see the scene and so contribute to the final rendered image.
For the application of lens design two special cases of interference is important. In a focus rays from a point light source meet again. In a closeup of the focal region all rays are replaced by plane waves. They inherit their direction from the rays. The optical path length from the light source is used for the phase. The derivative of the position of the ray in the focal region on the source position is used to get the width of the ray and from that the amplitude of the plane wave. The result is the point spread function, it Fourier transform is the MTF, and from the former the Strehl ratio can be calculated also. The other case is wavefront calculation of a plane wavefront. Of course when the rays come to close together or even cross the wavefront approximation breaks down. Interference of spherical waves is usually not combined with ray tracing thus diffraction at an aperture cannot be calculated.
This is used to optimize the design of the instrument by minimizing aberrations, for photography, and for longer wavelength applications such as designing microwave or even radio systems, and for shorter wavelengths, such as ultraviolet and X-ray optics.
Before the advent of the computer, ray tracing calculations were performed by hand using trigonometry and logarithmic tables. The optical formulas of many classic photographic lenses were optimized by rooms full of people, each of whom handled a small part of the large calculation. Now they are worked out in optical design software such as OSLO or TracePro from Lambda Research, Code-V or Zemax. A simple version of ray tracing known as ray transfer matrix analysis is often used in the design of optical resonators used in lasers.

Hopefully this will help the readers if it does not help the debaters.
jal

Hi LL,


Thanks for the links; they were interesting.


I noticed in ALL of them, that the "phenomenon" of wave motion, nodes & anti-nodes, superposition, constructive & destructive interference can ONLY be described by an "on the way" process. Something that is happening BETWEEN point A and B (source and detector).

Was that your point?



ciao,

T.Roc

TRoc,

Good observation! The point is that it takes standing waves for
wavefunctions to interact. The actual result of the interaction is at the reflecting
wall. Everything else showing wave interaction in free space is theoretical.

Have you ever used a stereo spectrum analyzer? It is able to separate out
the desired discrete frequency (band) from the musical score. True, all
frequencies within its selected bandwidth are detected and cannot be individually
isolated. The electronic isolation filtering cannot resolve identical frequencies
because they are mixed at the detector....we have been thru this scenario before.

My point being that other frequencies outside of the selected band are not
distorted and can be similarly "isolated" within the accuracy of the filtration to
resolve them.

IMO, it is the same with photons. Different frequencies of photons cannot
interact since they are phased and oriented differently. They cannot mix (blend)
together or else they would become distorted and not transport the qubit
information that each is carrying. Signal mixing distortion would cause
blurring, attenuation, and loss of qubit information.

LL

Hi TRoc,

If my efforts produce a result then hopefully (in time) they might be slightly better understood.

Maybe we have a result......

We can separate the DSE Equation from any theoretical concept .. let it just tell us that (empirically) the bright bits are where the path difference adds up to a whole wavelength and the dark bits are where they add up to half a wavelength. We could (conversely) use the same equation (as Thomas Young did) to tell us the wavelength of the phenomenon that creates it.

Can you extend your explanation to predict how far apart the bright regions will be?

Best wishes,

-C2.

Confused2, all,

Just a review of what I have read and see if this summarizes most of what has been expressed?

I think Laidback and others were saying this if not it is not my intentions to put any interpretation or misconstrue a thought.

Interference can be demonstrated by placing a thin slit in front of a light source, stationing a double slit farther away, and looking at a screen spaced some distance behind the double slit. Instead of showing a uniformly illuminated image of the slits, the screen will show equal spaced light and dark bands. Particles coming from the same source and arriving at the screen via the two slits could not produce different light intensities at different points and could certainly not cancel each other to yield dark spots. Light waves, however, can produce such an effect.

Now I know what they say about 'assuming', but as did Huygens, that each of the double slits acts as a new source, emitting light in all directions, the two wave trains arriving at the screen at the same point will not generally arrive in phase, though they will have left the two slits in phase. Depending on the difference in their paths, “positive” displacements arriving at the same time as “negative” displacements of the other will tend to cancel out and produce darkness, while the simultaneous arrival of either positive or negative displacements from both sources will lead to reinforcement or brightness. Each apparent bright spot undergoes a time wise variation as successive in-phase waves go from maximum positive through zero to maximum negative displacement and back.

This seems to be important to the DSE, neither the eye nor any classical instrument, however, can determine this rapid “flicker,” which in the visible-light range has a frequency from 4 × 1014 to 7.5 × 1014 Hz, or cycles per second. Although it cannot be measured directly, the frequency can be inferred from wavelength and velocity measurements. The wavelength can be determined from a simple measurement of the distance between the two slits, and the distance between adjacent bright bands on the screen; it ranges from 4 × 10-5 cm (1.6 × 10-5 in) for violet light to 7.5 × 10-5 cm (3 × 10-5 in) for red light with intermediate wavelengths for the other colors.

Hope I have it right on the post?
Happy New Year
Duality/Lisa