Jump to content

An Observer


AbstractDreamer

Recommended Posts

What precisely is an observer, in the way they are referred to in Quantum Physics?

 

After agreeing on a definition, can we try to answer this question by listing things that:

 

Almost definitely observers,

Maybe observers,

Almost definitely not observers.

 

We can start with the definition from wiki, a quote from Werner Heisenberg, Physics and Philosophy, p. 137

 

"Of course the introduction of the observer must not be misunderstood to imply that some kind of subjective features are to be brought into the description of nature. The observer has, rather, only the function of registering decisions, i.e., processes in space and time, and it does not matter whether the observer is an apparatus or a human being; but the registration, i.e., the transition from the "possible" to the "actual," is absolutely necessary here and cannot be omitted from the interpretation of quantum theory."

 

Some initial thoughts:

 

Is the definition of an observer relative to that which is being observed?

 

For example, to observe or register the presence of an object photon P1. Could an observer be another photon P2? If the P1 collides with P2, then can it be said that P2 has registered something about P1, therefore has an observation has been made?

 

What if P1 collides with a mirror M2? Can the mirror have registered P1 even in the smallest conceivable sense at the instant of reflection.

 

Is this about preserving information?

How long and how far can information be preserved and does it need to be retrievable to remain observed.

 

If P1 passed the event horizon of a blackhole BH2, can it ever be retrieved?

 

Link to comment
Share on other sites

Anything that measures or interacts with a quantum mechanical system such that any superposition is/would be lost

 

If the quantum mechanical system can maintain a state of superposition / entanglement then no - but if the interaction is such that it the system moves from a state of superposition to a classical defined state of particles then yes

Link to comment
Share on other sites

  • 2 weeks later...

Is there anything that doesn't interact with a quantum mechanic system such superposition is/would be lost?

 

Isn't the quantum universe in perpetual interaction with all the layers of fields around it and all the particles whizzing around doing their thang?

 

Lets take a photon as an excitation in the EM field. I cant imagine anywhere on Earth this photon could be moving in that has no other excitation. I mean there's EM radiation everywhere right? Radio signals, light, gamma rays from space, charged particles, Earth's magnetic field, CMB thermal radiation, all sorts of excitations.

 

Or a quark in the gluon field, or higgs field. Are these fields generally at base excitation level such that most excitations are NOT interacting?

 

Even if the fields themselves do not count as observers, surely other excitations do?

 

Would these other excitations need to be removed or at least absent from Schrodingers box?

 

In the double slit experiment when electrons are used; Does the Earth's magnetic field not observe the electron's path infinitesimally small that might be?

Edited by AbstractDreamer
Link to comment
Share on other sites

No responses. Either my question is totally stupid and meaningless, or my reputation is not worth responding to.

 

Let me simplify. Which of the following are quantum observers:

  • The entire universe
  • My future event horizon
  • My particle horizon
  • A super cluster of galaxies
  • A galaxy
  • A nebula
  • A super massive black hole
  • A star
  • A planet
  • A continent
  • A mountain
  • A human
  • A chair
  • A monkey
  • A cat
  • An insect
  • A mushroom
  • some Algae
  • A plankter
  • A bacterium
  • An amoeba
  • A virus
  • A complex molecule
  • A simple molecule
  • An atom
  • An ion
  • A proton
  • A neutron
  • An electron
  • A photon
  • Quantum stuff.
Edited by AbstractDreamer
Link to comment
Share on other sites

So you're telling me, in the double slit experiment using electrons, that while a single electron is in transit, it encounters NOTHING before it hits the detector?

 

NOTHING that interacts such that superposition is lost?

 

So no magnetic fields, no electric fields, no gas molecules, no thermal radiation, no gravity field, ... NOTHING that could detect its presence other than the detector?

Edited by AbstractDreamer
Link to comment
Share on other sites

So you're telling me, in the double slit experiment using electrons, that while a single electron is in transit, it encounters NOTHING before it hits the detector?

 

Are you asking me if the experiment is conducted within a vacuum? The nozzle that releases the electron wiggles around on purpose to give an electron a random start time and path. As far as I know, it only encounters the slits before the detector. If you have not looked into the quantum eraser experiment yet ..do it now.

 

NOTHING that interacts such that superposition is lost?

 

 

That's right, otherwise, why would we get the interference pattern without the detectors?

 

So no magnetic fields, no electric fields, no gas molecules, no thermal radiation, no gravity field, ... NOTHING that could detect its presence other than the detector?

 

 

Could be, but none of that stuff is recording information at the time. The divulgence of information to seems to be key here.

Link to comment
Share on other sites

Divulging information to who/what? Why does it need to be passed on beyond the first observation?

 

An electron has charge and mass. Mass has gravity. The Earth's gravitational and magnetic field should be able to interact with the electron the moment it is emitted and vice versa.

 

Clearly the experiments don't nullify the Earth's gravity or magnetism. So how is superposition not lost immediately?

Edited by AbstractDreamer
Link to comment
Share on other sites

There is a theory that earths gravity is responsible for wave collapse for anything more massive than a micron. Anything that can be used in the double slit would still be too small.

 

Divulging information to who/what?

 

 

To anyone that will listen

 

So how is superposition not lost immediately?

 

 

I call it an error in the code to our simulation

Link to comment
Share on other sites

So it IS a question of relative scale, as i asked in my OP?

 

And on that basis, if I were to find a location in space far enough away from everything so that i am only RELATIVELY undetectable by anything in the universe, then my waveform could be reformed and my superposition regained. I would change from being matter, to being a wave, and simultaneously exist in all the other undetectable locations that i could assume, according to Fermat's Principle. And as long as i stay undetected, my location is only a probability function.

Edited by AbstractDreamer
Link to comment
Share on other sites

Does the electron collapse its own wavefunction?


mass of an electron is about 1 x [math] 10^{-30} kg [/math]

mass of Earth 6 x [math] 10^{24} kg [/math]

mass of a large human (not me) 1 x [math] 10^2 kg [/math]

 

So if i float in a vacuum, enveloped within a sphere of mass 6 x [math] 10^{56} kg [/math]. I would effectively be gravitationally insignificant, undetectable and most importantly unobservable and unmeasurable, and assume my true waveform!!

Edited by AbstractDreamer
Link to comment
Share on other sites

What if we detect the momentum of the electron very close to emission source (therefore we can figure out which slit it will go through, but have lost super position). Then leave it to bounce off a series of perfect mirrors for a relatively long time and therefore long distance. This distance needs to be as significant as the electrons mass and charge is insignificant compared gravitationally and magnetically to the earth. Due to this distance, the electron should be able to regain super position. We can then experimentally shorten that distance until we reach a point precisely where super position is lost to find out the exact "scale factor of observability for distance" which i will call [math] F_d(E) [/math] and will be a function of its energy. Just like [math] F_g(M) [/math] could be the scale factor of observability for gravition which could be a function of mass, and [math] F_{em}(Q) [/math] could be the scale factor of observability for electromagnetism a function of its charge

 

Alternatively, if we detect the position of the electron close to emission source but far from either slit, does uncertainly principle allow the electron to maintain super position wrt momentum? If we can control the observation of the electron in such a way as maintain super position even as it goes through either slit, then we can adjust that "angle of observation" to see if there is a relationship between angle of observation and observability.

 

What if the beam intensity is increased so its a steady flow of electrons, close enough to interact and lose each other's super position? Is the interference still apparent?

Edited by AbstractDreamer
Link to comment
Share on other sites

Is super position a continuous measure? Or is it black and white; "in super position" or "not in superposition"?

 

If the location of the electron on the detector is a probability function, and if all measurements are inherently uncertain:

 

Surely observation is degree of observation, super position lost is degree of super position lost?

 

In other words, the more certain we are of the electron going through one slit, the less interference would be detected?

 

If its continuous, is this continuity in discreet quanta?

Link to comment
Share on other sites

Might help if you understand the term Superposition. Sean Carroll has a decent coverage (make sure you click the underlined hyperlinks on the key terminologies)

 

http://www.physicsoftheuniverse.com/topics_quantum_superposition.html

 

Put simply superposition is your probabilities once you make a measurement you know the state so its no longer probalistic but determined.

Edited by Mordred
Link to comment
Share on other sites

This makes me wonder about Schrodinger's Cat. I mean, isn't the cat in the box an observer? What difference would it make if we took the cat out of the box and simply wondered if the if the atomic particle decayed or not?

 

Actually, putting the cat in the box, puts two observers in the box, so if we eliminate the cat then we'd still have a detector detecting if there were radiation in the box.

 

Doesn't Schrodinger's thought-experiment make more sense if we eliminated the detector and the cat, and we only had the radioactive particle in the box and wondered whether it decayed or not? If there are already detectors and observers (the cat) inside the box, then the state of whether the cat is dead or alive is determined before any human looks in the box. Therefore the cat is actually alive or dead, not both, and it isn't our observation that determines the state of the cat.

Link to comment
Share on other sites

But you can limit the probabilities, and reduce the positions right?

 

if there are no slits, there are no probabilities for the electron to hit the detector.

if there is 1 slit, the probabilities are particle like.

If there are two slits, the probabilities show wave like interference.

 

So like with observations, as the electron moves through the Earth's gravitic and magnetic fields, the Earth's observations are so tiny they hardly affect the probabilities and the positions of the electron.

 

But when the experimenter sticks his measuring device at the slit, that observation is significant enough to limit the probability and positions to particle-like behaviour.

 

By that reasoning, observation is degree of measurement and on a continuous scale (perhaps quantisable) and it similarly affects super position?

 

So if the experimenter used a really bad measuring device, a very insensitive device, such that he is only 50% sure that the electron has passed through one slit and not the other, would that have any affect on the interference pattern?

 

Or if the experimenter used a banana to measure the electron at the slits, that would surely not affect the interference the pattern at all?

Edited by AbstractDreamer
Link to comment
Share on other sites

yes the number of slits affects the probabilities. However wave particle duality and the Heisenburg uncertainty principle has nothing to do with how clever we are in measurement. They are both fundamental aspects of particles in our universe.

 

Here is as close to basic as I can find on bullets (particles) vs waves.

 

http://physics.mq.edu.au/~jcresser/Phys201/LectureNotes/TwoSlitExpt.pdf

 

It doesnt detail constructive vs destructive interferance but we can get into that later.

Edited by Mordred
Link to comment
Share on other sites

How is it that the Earth's gravity field or the Earth's magnetic field do not collapse the wave function the moment the electron is emitted from the source?

 

That they are not being measured by humans is supposedly irrelevant?

That the electron has an infinitesimally small affect on the Earth's fields and is immeasurable with our technology is also supposedly irrelevant?

 

How does the electron maintain super position when the Earth is in perpetual observation?

Link to comment
Share on other sites

well for one thing gravity on Earth simply has negligible effect on an electron. Gravity is simply too weak to affect individual particles.

We can't even measure interferance due to gravity for an electron or any individual particle. Gravity is literally swamped out by electrostatic charge.

 

In case of detail gravity has very little influence at the quantum scale.

 

Try it take the mass of an electron at 9.1×10^-31 and multiply it by 9.8 m/s^2 using f=ma.

 

You should get [latex] 8.9*10^{-30}[/latex] Newtons of force roughly

Edited by Mordred
Link to comment
Share on other sites

... wave particle duality and the Heisenburg uncertainty principle has nothing to do with how clever we are in measurement. They are both fundamental aspects of particles in our universe.

 

Doesn't that mean "very little" and "negligible" are irrelevant?

 

... gravity on Earth simply has negligible effect on an electron. Gravity is simply too weak to affect individual particles.
We can't even measure interferance due to gravity for an electron or any individual particle.

.. gravity has very little influence at the quantum scale.

Try it take the mass of an electron at 9.1×10^-31 and multiply it by 9.8 m/s^2 using f=ma.

You should get [latex] 8.9*10^{-30}[/latex] Newtons of force roughly

 

I thought anything above zero is enough to act as an observer? How large does an influence have to be in order to count as an observer?

Link to comment
Share on other sites

Well for that you would have to understand how constructive and destructive interferance works with intensity. You most likely would have extreme difficulty finding interferance values on an electron for gravity. This is because the Coulomb constant is roughly 10^20 times greater than the gravitational constant. So its literally swamped out by electrostatic charge.

 

There are some very complex experiments for superposition of heavier mass items in freefall to simulate zero g but I don't know the results.

 

An observer has to generate a measurable interferance pattern is about the only answer I can provide on that.

Link to comment
Share on other sites

Appreciate the help. I'm not trying to be obtuse. Just this point of observability is bugging me. Seems like it is relative to the object being measured. If its small enough relative to the object being measured it can be ignored. So that kind of contradicts the uncertainty in measuring principle.

 

That's why i was thinking along the lines of degrees of observation, and degrees of superposition, and some continuous (maybe quantisable) scale.

 

And what about the Earths magnetic field? Does that not influence the (moving) electron in such a way as to act as an observer, determine its position or momentum, collapse the wave function, and make it behave like a particle?

 

The two questions are:

 

How large does an influence have to be in order to count as an observer?

How small does an influence need to be to unable to affect an object enough to make it lose superposition (to any degree).

Edited by AbstractDreamer
Link to comment
Share on other sites

Observer amounts to interferance sufficient to cause decoherence. Decoherence causing a loss of superposition. I would recommend first learning Schrodingers equation. Then look at decoherence equations.

 

The values will depend on your system state but the formulas to run the calcs can be found here.

 

https://en.m.wikipedia.org/wiki/Quantum_decoherence

 

Yes various environment factors can cause decoherence even temperature/gravity/fields etc. Yet the influence must cause sufficient interferance.

 

The questions your asking are not obtuse they are very good questions. Unfortunately also difficult to answer lol

 

(particularly since QM is one of my weaker subjects)

Edited by Mordred
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.