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### Author Topic: Can a particle collide with itself using the principle of quantum indeterminacy?  (Read 1003 times)

#### Atomic-S

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##### Can a particle collide with itself using the principle of quantum indeterminacy?
« on: 13/08/2016 06:14:48 »
Can a particle collide with itself using the principle of quantum indeterminacy?  The following experiment, based on the Stern-Gerlach experiment, seeks to find out (see diagram).

Silver atoms are ejected from a source toward a magnet M1.  This magnet is designed to create a nonuniform field between its highly dissimilar pole faces (see detail).  The silver atom is characterized by having completely paired electrons except for one, whose spin therefore determines the magnetic behavior of the atom.  The unpaired electron, having a spin number 1/2, admits of two possibilities: If the spin is one way with respect to the field, it will be attracted toward the one magnetic pole; if the other way with respect to the field, it will be attracted toward the other pole.  If the direction of its unambiguous spin is at some other angle, the principle of quantum eigenstates requires that whatever state it is in will split into a combination of the fore/against states just discussed, and that one of them will end up determining which way the atom, if observed, will go -- toward the one pole or the other.  But that unles it is observed, the atom remains in the combination of both states that corresponds to the prior state.

Atoms proceed from the source through M1, which causes each to have the possibility of going in two different directions: Either the axis-up (counterclockwise) spin, or the axis-down (clockwise) spin, as shown. The apparatus is so arranged that the clockwise state will send the atom off toward the trash bin, but the counterclockwise state will send it on to the next magnet M2.  M2 is another magnet of the same type as M1, but arranged at a 90 degree angle from it. It will spit atoms into states axis-left or axis-right.  Because all the atoms that reach M2 have been polarized into a counterclockwise spin by M1, they are in a state that resolves equally into axis-left and axis-right states in M2.  therefore each atom exting M2 has a 50% chance of being found in the axis-left state and a 50% chance of being found in the axis-right state.  From M2, a left-flying atom proceeds to the left ionizer.  The left ionizer strips the unpaired electron off and discards it. This gives the atom an electric charge of +1 and a spin of 0 (because the spin was in the electron).  From there, the atom is propelled by means of its unbalanced charge into a cyclotron ring, where it is accelerated to high speed.

Meanwhile, the exact same action happens in the other side of the system, the atom going through the right ionizer, becoming likewise propelled into the cyclotron in the opposite direction.

If a swarm of atoms enter this process, they can be expected to collide, when the beams are switched together, at the collision point, sending a shower of particles into the detectors.

However, let the experiment be conducted in the following unusual way: Let the rate of emission of atoms from the source be reduced to a very low rate so that there is virtually no chance that more than one atom will be in the system at a time.  Under this condition, the atom goes through the magnets as already discussed, and in M2 is placed in a mixed quantum state corresponding to axis-left and axis-right.  The direction it takes after M2 depends on which of these states apply. However, the path of the atom remains unobserved, so it must still be in the mixed state.  That being so, each state is still "live" and able to take a path, much as a particle going through the double slit experiment appears to go through both slits if it is not detected while doing so.  So we have the axis-left state going left into the left leg, and the axis-right state going right into the right leg. Because both states remain equally viable, they both are eligible to participate in the subsequent acceleration in the cyclotron ring.  Once they have reached a very high energy, the two states are allowed to collide.  And the critical question is, what will be observed?  I opine that the two states of the atom will collide and produce a shower of particles, but because each state is only 50% present, only 50% of the particles that would ordinarily be detected actually will be. If that happens, it will demonstrate that the atom existed in two different places and velocities at the same time.

Thoughts?

#### Blame

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #1 on: 13/08/2016 07:15:05 »
No. For the same reason as your cat question. Just because the particle's position is indeterminate doesn't mean that there is more than one particle. It just means you don't know where it is. It will be one particle wherever it arrives and not two.

Just another case of "you cant have it both ways". LOL.
« Last Edit: 13/08/2016 07:36:18 by Blame »

#### alancalverd

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #2 on: 13/08/2016 09:59:15 »
Too-literal interpretation of "observed", followed by an "observation"!

Self-interference has, however, been observed with much simpler apparatus, and much larger particles. To quote Wikipedia

Quote
In 1999, the double-slit experiment was successfully performed with buckyball molecules (each of which comprises 60 carbon atoms).

Enjoy!

#### Blame

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #3 on: 13/08/2016 10:27:28 »
Self-interference has, however, been observed with much simpler apparatus, and much larger particles.

Yes,  but given the original question, one should be clear that self-interference is not the same as self-interaction. Diffraction may affect the probability of a particle arriving at a particular point but it doesn't change what happens if it does. Still only a single particle.

#### Atomic-S

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #4 on: 18/08/2016 01:58:45 »
Quote
Diffraction may affect the probability of a particle arriving at a particular point but it doesn't change what happens if it does. Still only a single particle.
Let us consider a modification of the experiment.  Instead of trying to cause the particle to collide with itself,  a target will be placed at the site of collision.  When the particle collides, it will be colliding with a target other than itself.  This would be expected to produce a shower of particles, inasmuch as, regardless of which way the atom goes, the collision of an atom with a target of many atoms is a process that is well understood to work. However, in this case the following question arises:  In what way will the resulting shower of particles be distributed?  If the atom arrives from the left, the shower would tend toward the right. If it arrives from the right, it would tend toward the left.  If it arrives from both directions at the same time, then things get interesting.  For it to arrive from both directions at the same time does not contradict the existence of only one flying atom.  It is one atom, but in a state of indefinite momentum.  If the resulting particles scatter in both directions, that does not violate the conservation of momentum, because the momentum of the particle was indefinite from the moment it exited M2.  So what can we expect as the outcome of colliding the particle with a target different than itself?

#### William McC

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #5 on: 21/08/2016 03:25:29 »
Self-interference has, however, been observed with much simpler apparatus, and much larger particles.

Yes,  but given the original question, one should be clear that self-interference is not the same as self-interaction. Diffraction may affect the probability of a particle arriving at a particular point but it doesn't change what happens if it does. Still only a single particle.

If refraction is the result of a particle of light making an almost 90 degree turn, a right angle turn?

According to old school no way. Although old school said the particle of electricity the particle of light has no mass, only its repulsive nature, there are observations that can be me made with objects that have mass and weight, but behave and display behavior independently of mass weight. From that we could perhaps explain refraction or at least understand what is actually happening.

According to old school we live in a giant cathode ray tube for lack of a better understand and analogy. Matter, light, heat and everything else is just the effect of crisscrossing rays of particles of electricity through matter. All the rays paths are rather straight. When we see things like refraction we are just watching the effect one ray, say a ray of white light on a Plexiglass prism, has on the already existing undetected rays that are slowed to light speed, by the bottle neck at the surface of the prism, by the ray of white light. These rays are then slowed to the red, yellow and blue rays that the prism outputs.

Very much like the red, yellow and blue rays created by the prism in our first experiment, beamed upon another prism, cause yet, a third set of undetected rays to slow to create white light once again from the red, yellow and blue rays.

I uploaded three attachments one is the single prism the other is the double prism and the third is the white light created by both prisms.

Sincerely,

William McCormick

#### CPT ArkAngel

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #6 on: 21/08/2016 09:28:37 »
Too-literal interpretation of "observed", followed by an "observation"!

Self-interference has, however, been observed with much simpler apparatus, and much larger particles. To quote Wikipedia

Quote
In 1999, the double-slit experiment was successfully performed with buckyball molecules (each of which comprises 60 carbon atoms).

Enjoy!

It is not self-interference, it is an interference pattern constructed from a beam of many large molecules.

http://arstechnica.com/science/2012/03/quantum-interference-with-big-molecules-approaches-the-macroscopic/

http://physicsworld.com/cws/article/news/1999/oct/15/wave-particle-duality-seen-in-carbon-60-molecules

« Last Edit: 21/08/2016 09:33:31 by CPT ArkAngel »

#### Atomic-S

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #7 on: 25/08/2016 06:31:32 »
Quote
It is not self-interference, it is an interference pattern constructed from a beam of many large molecules.
Such interference patterns still occur if the rate of incoming molecules is so low that only one molecule is in the apparatus at a time.

#### Atomic-S

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #8 on: 25/08/2016 06:35:43 »
When the atom proceeds through M2 and enters a state of equal probability in both directions, it has an expectation value of transverse momentum of 0 . (That is, the average value of momentum computed by doing the applicable integral of its state function, which gives the long-term average momentum measured over many experiments, is 0).  When it finally collides with the previously mentioned target, it still has an expectation value of momentum of 0 .  The situation suggests that the expectation value of all the particles scattered as a result of the collision will also be 0.  But there are many particles, and for that to be so, some would have to scatter left and others scatter right.  This is a hypothesis that should be experimentally testable.

#### Atomic-S

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #9 on: 25/08/2016 06:38:07 »
The point of such a finding, assuming it occurs, being this:  That the particle could not have struck from the left only, or from the right only.  It would have had to strike from both directions at once, giving evidence that it had existed in an illocal state.

#### CPT ArkAngel

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #10 on: 25/08/2016 21:15:56 »
If you send only one particle, you will have only one spot on the detector (or screen). This does not constitute an interference pattern. You can infer that there is a wave interfering with the path of the particle though. But you cannot say for sure that this wave is a part of the particle, just that this wave modify the path of the particle. The wave might be an interaction with its surrounding (not an ordinary interaction but via entanglement relations).

What is interesting with the atomic bucky ball experiment, is that it shows that even a macroscopic element like a baseball would exhibit an interference pattern in its path. The problem is that the interference pattern diminishes with the number of particles in the ball so it becomes unmeasurable at a macroscopic scale.
« Last Edit: 25/08/2016 21:33:26 by CPT ArkAngel »

#### William McC

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #11 on: 26/08/2016 06:02:18 »
One atom of electricity, is not going to register in your eye, a sensor, or on anything else. It takes an infinite number of particles of electricity (light) to create or relay an effect.

If someone said to you that they saw one atom on the suns surface was different or causing a strange effect, you would know they were watching to much late night science fiction. Well when someone tells you that they saw one sub matter particle do something I can assure that they are preaching sci-fi. Modern scientists have skewed the size of atoms and the sub matter particle, so much so that they think seeing a particle is different than seeing an atom on the sun. In terms of scale the two events are about the same. You may as well be half a solar system away from an atom on the sun, or a fraction of a micron away from a microscopic event. Both are not going to give you much about either the atom on the sun or the atom of electricity creating light.

Atom means the smallest denominator of a substance.

Sincerely,

William McCormick.

#### Atomic-S

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #12 on: 27/08/2016 05:32:35 »
Quote
One atom of electricity, is not going to register in your eye, a sensor, or on anything else. It takes an infinite number of particles of electricity (light) to create or relay an effect.
Individual subatomic particles are being detected all the time:

https://en.wikipedia.org/wiki/Scintillator
https://en.wikipedia.org/wiki/Geiger_counter
https://en.wikipedia.org/wiki/Photomultiplier

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##### Re: Can a particle collide with itself using the principle of quantum indeterminacy?
« Reply #12 on: 27/08/2016 05:32:35 »