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Posted (edited)

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An ironic advantage of spirituality is that to promote relevance to astronomical transcendence or unconsciousness means that you know you’re absolutely not relevant to a specific quantum system. By contrast those promoting materialism might accidentally imply there was a sarcastic risk of themselves being slightly relevant by stating they weren’t relevant in a way that could backfire for atomic nanometers! So when we think of obsessions about semantics in philosophy it might turn out relevant in clarifying the double slit experiment. For example when we observe a classical object like a chair there’s a surplus of photons whereas when we think of the double slit experiment it’s almost like it’s nighttime where the photons are emitted individually.

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So when a camera detects a photon in the double slit experiment the photon is by definition in the camera and no longer on the screen. In other words the surplus of photons for recording a large classical object means that there can be an illusion that there are pairs of photons with one on the camera screen and one on the classical object. Hence when we think of the double slit experiment as being in the dark then we cannot tell how the detected photon travelled from the last black screen to the camera. If the photons were randomly diffracted by the double slits then it’s possible they were randomly reassembled when the photons re-appeared on the camera as if the the last black screen were a lenticular image with a momentary multi-prong interference pattern that appeared like 2 prongs to the camera. We could also risk making a false dichotomy between the observer and a photon if consciousness were itself photonic such that the photon no longer exists after contacting the eye’s retina and is converted to an electrical stimulus in the brain. In other words the conservation of energy states that energy cannot be created or destroyed where energy can only be converted from one form into another. So when we think all that’s needed is a camera to break the interference pattern in the double slit experiment then we could think of a human observer as being just like an eye that captures photons rather than a complex brain. Another metaphor is to think of light as being like alternating current rather than direct current where we cannot tell whether light travelled in a straight line from the last black screen in the diagram above to the camera. A shortcut might be to think of the last black screen as being like a mirror that reflects light back through the double slit such that we can only see two lines where the rest were blocked by the cards used to form the double slit. A last resort were the markings on the last black screen just a hollow vestige of a photon rather than a photon itself might be that if an observer is illogical than the experiment would somehow turn out to be illogical in a random way simply because the material world wouldn’t conspire with our own importance if we’re spiritually self-sufficient! Even though America won the Cold War and put a man on the moon it’s possible welfare is still justifiable if no one can be serious when the standard of capitalism imposed by quantum mechanics is artificial intelligence! 
86CCE549-B041-459C-9E21-B258867FDCD9.jpeg.b21fc0263b37f20db406637920419fa8.jpeg

Darth Vader (Star Wars)

 

Edited by Michael McMahon
Posted
29 minutes ago, Michael McMahon said:

 For example when we observe a classical object like a chair there’s a surplus of photons whereas when we think of the double slit experiment it’s almost like it’s nighttime where the photons are emitted individually.

No, “we” don’t

You can do the double-slit with individual photons, but you don’t have to. 

 

29 minutes ago, Michael McMahon said:

So when a camera detects a photon in the double slit experiment the photon is by definition in the camera and no longer on the screen.

When the camera detects the photon there is no more photon.

The premise on which your conjecture is based is false.

 

Posted
19 minutes ago, swansont said:

When the camera detects the photon there is no more photon.

The premise on which your conjecture is based is false.


Extromission theory about eye beams might appear ridiculous to a materialist but there’s also an inverse problem in intromission theory where if you look at a dark room then the room should somehow get a tiny bit darker to account for the lost light trapped by your eyes that’s no longer capable of reflecting back and forth diffusely off objects in the room. 
“Specular reflection, which occurs with smooth surfaces like mirrors, causes light rays to reflect at the same angle as they hit the surface. Diffuse reflection, which occurs with rougher surfaces, scatters light rays in different directions.”

Posted

So what ? any time a photon has a scattering event (interaction) the original photon is absorbed in inelastic scatterings such as a camera detecting the photon or the screen used in the detector. Elastic scatterings such as a mirror is a distinctively different process.

 

Posted
22 minutes ago, Michael McMahon said:


Extromission theory about eye beams might appear ridiculous to a materialist

Or anyone who relies on empirical evidence.

22 minutes ago, Michael McMahon said:

but there’s also an inverse problem in intromission theory where if you look at a dark room then the room should somehow get a tiny bit darker to account for the lost light trapped by your eyes that’s no longer capable of reflecting back and forth diffusely off objects in the room. 

It will, if the eyes absorbed more than what’s being replaced. The question is if this was a detectable drop. Seeing as how the lifetime of a photon in a room that’s a few meters across is several nanoseconds, a small perturbation wouldn’t be easily detectable. Walls absorb light, too and eyes don’t have a large surface area.

1 W of visible light is of order 10^18 photons.

Posted

This particular article seems most appropriate to this thread

The Double Slit Experiment and Quantum Mechanics

 

https://www.hendrix.edu/uploadedFiles/Departments_and_Programs/Physics/Faculty/The Double Slit Experiment and Quantum Mechanics.pdf

"  Let me say this again to emphasize it. Your eyeball is covered with a large number of photon detectors. When you see something, each detector counts the number of photons it received and transmits that number to the brain. Some of the detectors (the cones) can detect the energy of the photons, and they transmit that value to the brain also (thus providing color vision). Your eyeball works much like the detector portion of a digital camera. You have never observed a light wave in your life, but you have added up the numbers of photons striking different places on your retina to create a diffraction pattern. To me, the most convincing evidence that all particles, including photons, are always detected as individual and whole particles was observing the output of a particle detector on an oscilloscope. The output is a series of pulses. Each pulse represents the passage of one particle (a photon, an electron, or whatever) through the detector. You get the same effect with an old fashioned geiger counter: each click represents the passage of a particle through the detector. If you have never had the opportunity to observe this, you should at least read Wikipedia’s article on particle detectors. "

Posted
40 minutes ago, swansont said:

1 W of visible light is of order 10^18 photons.


True yet the double slit experiment uses very little light. 
 

“The human eye is very sensitive; but can we see a single photon? The answer is yes: sensors in the retina can respond to a single photon. But neural filters only allow a signal to pass to the brain to trigger a conscious response when at least about five to nine arrive within less than 100 ms.“ math.ucr

Posted
1 hour ago, Michael McMahon said:

True yet the double slit experiment uses very little light. 

Not true. It depends on how you do it. You can use a bright source. You aren’t required to do a single-photon experiment. Young used the sun shining through a hole

Posted (edited)
1 hour ago, swansont said:

Not true. It depends on how you do it. You can use a bright source. You aren’t required to do a single-photon experiment. Young used the sun shining through a hole


One way to resolve a paradox is to counterbalance it with another paradox. So the way the double slit experiment is ranked less difficult than the mind-body problem means that the double slit experiment might be metaphysically flawed rather than just mathematically flawed. The quantum uncertainty principle applies not only to the photon interference pattern but also to the imprecise location of the camera detector or the eye. The more you know about movement the less you know about location rhymes with the motion blurs of a hallucination rather than just consciousness(!):
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The Babysitter - Bella Thorne 

An advantage of free thought is that even if you’re wrong there’s no limit to how extreme any other belief system can be in such a way that rejecting my belief system can still anyone to be more extreme about their own scientific beliefs. For example the idea of using a mirror in the detector in my opening post is a bit like time symmetry in a light beam as if the light were sucked out of the black screen back to the electron emitter much like a magnet:

Harry Potter and the Philosopher’s Stone - First scene: street light suction 

In other words if light is fully distinct from atomic matter then light might not obey entropy seeing as any heat that causes chaos is caused by changing electron orbitals but technically not the light emission spectra itself. That way light could theoretically move forward and backward in time. 
 

“By splitting a photon, or packet of light, using a special optical crystal, two independent teams of physicists have achieved what they describe as a "quantum time flip," in which a photon exists in both forward and backward time states.”

https://www.space.com/quantum-time-flipped-photon-first-time

Edited by Michael McMahon
Posted (edited)

Those are some rather lousy articles with regards to CPT which the time symmetry is well known.

2 hours ago, Michael McMahon said:

They are rather misleading. Typical pop media coverage but the articles aren't much better in my opinion.

Really all that's happening is photons have two polarizations left and right polarizations.

 

Edited by Mordred
Posted

I gotta admit I loved this line

9 hours ago, Michael McMahon said:

A last resort were the markings on the last black screen just a hollow vestige of a photon rather than a photon itself might be that if an observer is illogical than the experiment would somehow turn out to be illogical in a random way simply because the material world wouldn’t conspire with our own importance if we’re spiritually self-sufficient! Even though America won the Cold War and put a man on the moon it’s possible welfare is still justifiable if no one can be serious when the standard of capitalism imposed by quantum mechanics is artificial intelligence! 

I haven't a clue what you mean by it ( or maybe you haven't a clue ).

I think you should discuss this with John John in the "Photons and Light" thread.
Apparently there is no light, no double slit, no diffraction pattern and no detector/observer; it is all just electrical signals to the brain.

Posted (edited)
On 8/17/2024 at 9:37 PM, Mordred said:

Please keep in mind the experiment has also been performed by quantum dot detectors/emitter. 


One way to think of the observer effect in the double slit experiment is if the lens of the eye functioned like a prism where the multi-pronged interference pattern on the screen merges into two prongs in your eyes much like white light separating into multiple colours on a prism. 
 

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prism

Edited by Michael McMahon
Posted
7 minutes ago, Michael McMahon said:

One way to think of the observer effect in the double slit experiment is if the lens of the eye functioned like a prism where the multi-pronged interference pattern on the screen merges into two prongs in your eyes much like white light separating into multiple colours on a prism. 

You're going all over the place with this. Young's experiment works better with monochromatic light. And trajectories split at the double-slit piece, not in the observer's eye. Newton's experiment of splitting light by their frequencies (energy of the photons) can be explained classically and does not demonstrate quantum mechanics.

And, btw, I don't know of any single case in the history of science when a paradox was solved by throwing another paradox at it. Do you?

Posted
18 minutes ago, Michael McMahon said:


One way to think of the observer effect in the double slit experiment is if the lens of the eye functioned like a prism where the multi-pronged interference pattern on the screen merges into two prongs in your eyes much like white light separating into multiple colours on a prism. 
 

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prism

How about simply thinking of observer effect as any measurement ?

That is how QM describes observer effect.

Posted
45 minutes ago, Mordred said:

How about simply thinking of observer effect as any measurement ?

That is how QM describes observer effect.

Doesn't the HUP also apply to calculations, whether or not they are for a real experiment ?

Posted
1 hour ago, studiot said:

Doesn't the HUP also apply to calculations, whether or not they are for a real experiment ?

Not necessarily. The solution to the Schrödinger equation for the hydrogen atom gives the energy eigenstates, but does not give the width of the energy states. 

Posted
1 hour ago, Mordred said:

Here is a decent article involving Fourier transform uncertainty

http://math.uchicago.edu/~may/REU2021/REUPapers/Dubey.pdf

Thanks for that.

 

There is a simpler version in Appendix appendix C p465 of Elmore and Heald 'Physics of Waves' but along similar lines.

 

@Michael McMahon

 

ou could profit from viewing things from an energy flow POV.

Here is a short extract from A P French's Vibrations and Waves (MIT)

french1.thumb.jpg.5eaaa83ba06a92c1b3ef9930e07c3498.jpg

2 hours ago, swansont said:

Not necessarily. The solution to the Schrödinger equation for the hydrogen atom gives the energy eigenstates, but does not give the width of the energy states. 

I would be interested in any elaboration on this.

Posted
46 minutes ago, studiot said:

I would be interested in any elaboration on this.

The energy states above the ground state have a finite width owing to their lifetime (energy and time being conjugate variables). But lifetime is not part of the solution, and the energies are single-valued. So while other non-commuting variables have the uncertainty built in, the energy-time pairing do not.

Posted (edited)

It may be best to add some important details as it goes beyond the good example provided by Swansont. As dealing with this gets lengthy I will include an article to supply the details behind eugenstates which has no uncertainty.

The article opens with the following key statement.

"As we know, observables are associated to Hermitian operators. Given one such operator A we can use it to measure some property of the physical system, as represented by a state Ψ. If the state is in an eigenstate of the operator A, we have no uncertainty in the value of the observable, which coincides with the eigenvalue corresponding to the eigenstate. We only have uncertainty in the value of the observable if the physical state is not an eigenstate of A, but rather a superposition of various eigenstates with different eigenvalues.

https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/005979fa741c3ea2e0430456b70caf93_MIT8_05F13_Chap_05.pdf

In essence eugenstates has no uncertainty however this gets into the measurement axiom of QM.

Where the act of measurement produces a state however any further measurement will produce a new state. Better described below

https://www.britannica.com/science/quantum-mechanics-physics/Axiomatic-approach

 

 

 

For the OP this deals specifically with "observation" measurement where the superposition of state's is lost due to observation. 

However I will leave that to a mental exercise with the article Studiot posted.

Side note graph A is a delta function that is localizable you can readily determine the boundaries from graph a. Whereas a sine wave is not localizable.  In terms of a particle the mean lifetime can be determined by graph a) using Breit Wigner distributions.

The outside amplitudes not the primary amplitude would be considered resonance however as the width is equal or greater than the amplitude would not be considered a resonant particle.

(Just a little side note and taking advantage  of the graph provided by Studiot.

Edited by Mordred
Posted (edited)
6 hours ago, swansont said:

The energy states above the ground state have a finite width owing to their lifetime (energy and time being conjugate variables). But lifetime is not part of the solution, and the energies are single-valued. So while other non-commuting variables have the uncertainty built in, the energy-time pairing do not.

 

Using what I mentioned above and the graph A of the article posted by Studiot. on the Y axis assign \(\sigma^\ast\) excited quantum state of an atom. The Y axis on the graph assign \(\phi(E)\) on the x axis E. The vertical center line of the amplitude peak assign \(E_R\) for peak energy of the amplitude. 

The width of the amplitude 2/3 up determines the the lifetime of the resonance. Resonance is used for all particles under Breit-Wigner.

further details here. ( a simplified treatment for ease of understanding )

https://web2.ph.utexas.edu/~vadim/Classes/2019f/resonances.pdf

a cross section being the entire graph rather than the localized highest peak resonance. In terms of sound this would apply to the phonon.

(keep in mind by the statement extremely simplified even though the later parts gets complex a full cross section of an interaction would look like below (as I already have this in latex in another thread my Nucleosynthesis thread) I will simply copy and paste from there. It includes further details on Breit Wigner 

Breit Wigner cross section

\[\sigma(E)=\frac{2J+1}{2s_1+1)(2S_2+1)}\frac{4\pi}{k^2}[\frac{\Gamma^2/4}{(E-E_0)^2+\Gamma/4)}]B_{in}B_{out}\]

E=c.m energy, J is spin of resonance, (2S_1+1)(2s_2+1) is the #of polarization states of the two incident particles, the c.m., initial momentum k E_0 is the energy c.m. at resonance, \Gamma is full width at half max amplitude, B_[in} B_{out] are the initial and final state for narrow resonance the [] can be replaced by

\[\pi\Gamma\delta(E-E_0)^2/2\]

The production of point-like, spin-1/2 fermions in e+e− annihilation through a virtual photon at c.m.

\[e^+,e^-\longrightarrow\gamma^\ast\longrightarrow f\bar{f}\]

\[\frac{d\sigma}{d\Omega}=N_c{\alpha^2}{4S}\beta[1+\cos^2\theta+(1-\beta^2)\sin^2\theta]Q^2_f\]

where

\[\beta=v/c\]

c/m frame scattering angle

\[\theta\] 

fermion charge

\[Q_f\]

if factor [N_c=1=charged leptons if N_c=3 for quarks.

if v=c then (ultrarelativistic particles)

\[\sigma=N_cQ^2_f\frac{4\pi\alpha^2}{3s}=N_cQ^2_f\frac{86.8 nb}{s (GeV^2)}\]

2 pair quark to 2 pair quark

\[\frac{d\sigma}{d\Omega}(q\bar{q}\rightarrow \acute{q}\acute{\bar{q}})=\frac{\alpha^2_s}{9s}\frac{t^2+u^2}{s^2}\]

cross pair symmetry gives

\[\frac{d\sigma}{d\Omega}(q\bar{q}\rightarrow \acute{q}\acute{\bar{q}})=\frac{\alpha^2_s}{9s}\frac{t^2+u^2}{t^2}\]

 

 

Edited by Mordred
Posted

A counterargument against materialism might be how the photon seems to require a second photon to be a wave in the double slit experiment rather than just one photon being a wave in itself. So in the double slit experiment not only could you fire electrons individually but for emphasis you could delay each electron for 5 minutes. Then how would an electron be connected temporally to an electron fired 5 minutes prior to be diffracted with where they ought to have been two separate waves. In other words if an electron travels randomly then why would its motion become more random after the double slit towards the interference pattern on the last screen rather than just crashing into the first screen beside the double slit? Could the electron quantum tunnel through the first screen to appear to be deflected on the multi-prong interference pattern?! To paraphrase an argument I made in the opening post if you attempt to emit an electron straight in way that immediately becomes a slightly random direction then maybe it has to be detected on the camera from a random direction as if the electron was always moving randomly rather than temporarily being random after the double slit. 

Posted (edited)

I learned that the camera detector is actually pointing at the double slit in certain versions of the double slit experiments where the top slit is tagged as if with a polaroid instead of the camera pointing towards the final screen. So an alternative solution might be that the final screen that showed the interference pattern is itself also a quantum system like glass. Hence in this inverted solution there might only ever be two-prongs with the remaining prongs being an illusion of double vision where the slight translucency and transparency of a glass screen wouldn’t fully capture a photon to describe its precise location. When you hold your finger in front of your eyes then it’s the light the produces a double image rather than a cloning of objects in the background only that in the double split experiment light or electrons are themselves the objects rather than the “glass” medium. 
AAF2E586-9728-43E8-9D1D-7FC17299A969.thumb.jpeg.a9b41b7886cab816eb2e401d493bcab3.jpeg
Double image blurring of flower vase shadow on double pane windows 

Edited by Michael McMahon
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