How is observer defined in context to quantum physics?does it have

[nouveau_riche]

how is observer defined in context to quantum physics?
does it have to b a conscious observer?

 

[Polyrhythmic]

It doesn't have to be a conscious observer. Contrary to what many popular myths suggest, consciousness is not part of quantum mechanics. What we mean by observation is anything that is mathematically equivalent to applying an operator on a quantum mechanical state. For example, applying the position operator and getting its eigenvalue is equal to measuring a particle's position. An observation might be a collision between two electrons, or a specific measurement conducted in a laboratory.

 

[nouveau_riche]

It doesn't have to be a conscious observer. Contrary to what many popular myths suggest, consciousness is not part of quantum mechanics. What we mean by observation is anything that is mathematically equivalent to applying an operator on a quantum mechanical state. For example, applying the position operator and getting its eigenvalue is equal to measuring a particle's position. .

what makes something(event) as mathematically equivalent to applying an operator?

 

[I like Serena]

The example that I like is the experiment where electrons are shot through a double-slit, impacting on a photographic plate behind it.

In the dark, you'll get an interference pattern.
In the light (with a short wavelength), you'll get a particle impact pattern.

In light with a long wavelength (in which an observer would not be able to distinguish through which slit the electron goes), you'll get an interference pattern again.

This is without anyone actually watching!
It's enough that someone might be able to see through which slit the electron goes. Weird huh! :uhh:

 

[Polyrhythmic]

what makes something(event) as mathematically equivalent to applying an operator?

That's one of the axioms of quantum mechanics, measurements correspond to eigenvalues of hermitian operators. I guess you should read an introductory text to the subject in order to understand what is going on.

 

[nouveau_riche]

The example that I like is the experiment where electrons are shot through a double-slit, impacting on a photographic plate behind it.

In the dark, you'll get an interference pattern.
In the light (with a short wavelength), you'll get a particle impact pattern.

In light with a long wavelength (in which an observer would not be able to distinguish through which slit the electron goes), you'll get an interference pattern again.

This is without anyone actually watching!
It's enough that someone might be able to see through which slit the electron goes. Weird huh! :uhh:

not much...
if an observer is not interested in measuring the position of photon then even the short wavelength will cause an interference...weird?

 

[nouveau_riche]

That's one of the axioms of quantum mechanics, measurements correspond to eigenvalues of hermitian operators. I guess you should read an introductory text to the subject in order to understand what is going on.

i have read some introductory but wasn't sattisfied with the conclusion

 

[I like Serena]

The example that I like is the experiment where electrons are shot through a double-slit, impacting on a photographic plate behind it.

In the dark, you'll get an interference pattern.
In the light (with a short wavelength), you'll get a particle impact pattern.

In light with a long wavelength (in which an observer would not be able to distinguish through which slit the electron goes), you'll get an interference pattern again.

This is without anyone actually watching!
It's enough that someone might be able to see through which slit the electron goes. Weird huh! :uhh:

not much...
if an observer is not interested in measuring the position of photon then even the short wavelength will cause an interference...weird?

There!




Dr. Quantum thinks it's weird!

(Just found this on facebook! )

 

[nouveau_riche]

There!




Dr. Quantum thinks it's weird!

(Just found this on facebook! )

suppose the intereference experiment is set up again,the light used is of very high frequency(so that it's wavelength could match with that of slits) ,now the distance at which interefence pattern is seen is at max distant from slit(max in sense that pattern can b seen nicely).,now one of the observer set up apparatus to see the intereference pattern while the other one set up an apparutus to measure position of photon(to see from which slit it passes through)
question is-will the interference pattern b lost or both will see their portion of reality ?

 

[I like Serena]

suppose the intereference experiment is set up again,the light used is of very high frequency(so that it's wavelength could match with that of slits) ,now the distance at which interefence pattern is seen is at max distant from slit(max in sense that pattern can b seen nicely).,now one of the observer set up apparatus to see the intereference pattern while the other one set up an apparutus to measure position of photon(to see from which slit it passes through)
question is-will the interference pattern b lost or both will see their portion of reality ?

Hmm, afaik the measurement making it observable which slit the electron passes through, collapses the electron's wave function, so regardless of the other observer, the interference pattern would be lost.

Do you have information that this is otherwise?

 

[nouveau_riche]

Hmm, afaik the measurement making it observable which slit the electron passes through, collapses the electron's wave function, so regardless of the other observer, the interference pattern would be lost.

Do you have information that this is otherwise?

out of many answers that i have received for the above experiment many say that it's impossible to set up an apparatus like that ,one observer have to eliminate the other
like refer

https://www.physicsforums.com/showthread.php?t=434366&page=2

 

[nouveau_riche]

Hmm, afaik the measurement making it observable which slit the electron passes through, collapses the electron's wave function, so regardless of the other observer, the interference pattern would be lost.

the important point to b highlighted here is that both observer aren't aware of the act of measurement made by the other one,so if the pattern get lost ,the other observer get's to know the history of interaction of that particle

 

[danR]

As a layperson, I have trouble enough following mainstream theorizing, without the potential added confusion of 'Dr.' Quantum's entanglement with new age 'physics'.

Experts may be able to parse out the material in those videos that's valid, but then they might rather direct me to videos by established theorists and expositors of QM.

 

[I like Serena]

out of many answers that i have received for the above experiment many say that it's impossible to set up an apparatus like that ,one observer have to eliminate the other
like refer

https://www.physicsforums.com/showthread.php?t=434366&page=2

I've read the thread you mentioned and I have to say I have trouble understanding what is said exactly. As yet I see no reason to change my previous statement.
Here's a section of the wikipedia article that supports my statement:

http://en.wikipedia.org/wiki/Double-slit_experiment#Relational_interpretation

 

[nouveau_riche]

I've read the thread you mentioned and I have to say I have trouble understanding what is said exactly. As yet I see no reason to change my previous statement.
Here's a section of the wikipedia article that supports my statement:

http://en.wikipedia.org/wiki/Double-slit_experiment#Relational_interpretation

i would prefer the paragraph in that wiki link that could justify ur point.

 

[alxm]

the important point to b highlighted here is that both observer aren't aware of the act of measurement made by the other one,so if the pattern get lost ,the other observer get's to know the history of interaction of that particle

"Awareness" has nothing to do with it. Either you perform a measurement or you don't. And by 'measurement' I mean it in the QM sense, which means any interaction with the environment-at-large which could, even if only in principle, be used to determine which path the particle took.

It does not matter if anything or anybody is "aware" of the results of the measurement.

Please use the search function. This question about what constitutes an observer or measurement in QM has been asked and answered here literally hundreds of times already.

 

[DrChinese]

Hmm, afaik the measurement making it observable which slit the electron passes through, collapses the electron's wave function, so regardless of the other observer, the interference pattern would be lost.

Do you have information that this is otherwise?

Just to be clear: It is NOT the effect of the apparatus in and of itself that causes the collapse. It is the overall context of the setup, as to whether knowledge is gained or not.

Take a double slit setup using light. Place one polarizer over the left slit at angle L, another over the right slit at angle R. When L-R=0, there IS interference. When L-R=90 degrees, there is NO interference. Obviously, the presence or absence of a polarizer does not change anything unless which slit information is gained. That becomes progressively more feasible as L-R goes from 0 to 90 degrees.

 

[I like Serena]

i would prefer the paragraph in that wiki link that could justify ur point.

Ah well, my link pointed to the specific paragraph.
But I have to admit that the wikipedia article is not very clear. :(

I have just reread my copy of Feynman's Lectures volume III chapter 1, which I can recommend.
I can hardly quote everything that he said, but I'll outline my interpretation of his chapter.

If you have a bright light source with a sufficient short wavelength that "could" show you through which slit the electron went, then regardless if you're really observing or not, the interaction will collapse the wave function and you will see a particle impact pattern.

If you dim the light, you'll get to see a mix of a particle impact and an interference pattern (since there is only a chance that a photon will actually reveal the location of the electron).

If you use a long wave length, you'll see an interference pattern, since you won't be able to distinguish through which slit the electron went.

Feynman also quotes Heisenberg who said:
"It is impossible to design an apparatus to determine which hole the electron passes through, that will not at the same time disturb the electrons enough to destroy the interference pattern."

 

[DrChinese]

Feynman also quotes Heisenberg who said:
"It is impossible to design an apparatus to determine which hole the electron passes through, that will not at the same time disturb the electrons enough to destroy the interference pattern."

That *may* be true for electrons, but definitely is not for photons as described above. That is definitely NOT why the interference is destroyed, it is the obtaining of which path information that causes that to occur.

 

[I like Serena]

That *may* be true for electrons, but definitely is not for photons as described above. That is definitely NOT why the interference is destroyed, it is the obtaining of which path information that causes that to occur.

Please elaborate? I don't understand.

 

[DrChinese]

Please elaborate? I don't understand.

Take a double slit setup using light. Place one polarizer over the left slit at angle L, another over the right slit at angle R. When L-R=0 (parallel), there IS interference. When L-R=90 degrees (crossed), there is NO interference. Obviously, the presence or absence of a polarizer does not change anything unless which slit information is gained. That becomes progressively more feasible as L-R goes from 0 to 90 degrees. In other words: you *could* know which path (100% certain) by checking the polarization of the detected photon when L-R=90 (crossed), even if you didn't.

In other words, the particle goes through the same apparatus whether the polarizers are parallel or crossed. So clearly the apparatus itself is not altering the particles as they go through more in one case than in another. And yet when they are parallel, no which path information is gained and interference is produced. And vice versa when crossed.

Conceptually, you could do the same thing with electrons (using S-G method) although I doubt it is practical. I am not trying to dispute Heisenberg (or Feynman) but the analogy is not correct regardless. Early quantum theorists used it to help describe how things work, but with the advent of entanglement experiments it is clear that the apparatus definitely is not the source of this effect. It is the overall context (configuration) of the setup and the Heisenberg Uncertainty Principle at work. Which is to say I have no idea what is happening.

 

[IllyaKuryakin]

Take a double slit setup using light. Place one polarizer over the left slit at angle L, another over the right slit at angle R. When L-R=0 (parallel), there IS interference. When L-R=90 degrees (crossed), there is NO interference. Obviously, the presence or absence of a polarizer does not change anything unless which slit information is gained. That becomes progressively more feasible as L-R goes from 0 to 90 degrees. In other words: you *could* know which path (100% certain) by checking the polarization of the detected photon when L-R=90 (crossed), even if you didn't.

In other words, the particle goes through the same apparatus whether the polarizers are parallel or crossed. So clearly the apparatus itself is not altering the particles as they go through more in one case than in another. And yet when they are parallel, no which path information is gained and interference is produced. And vice versa when crossed.

Conceptually, you could do the same thing with electrons (using S-G method) although I doubt it is practical. I am not trying to dispute Heisenberg (or Feynman) but the analogy is not correct regardless. Early quantum theorists used it to help describe how things work, but with the advent of entanglement experiments it is clear that the apparatus definitely is not the source of this effect. It is the overall context (configuration) of the setup and the Heisenberg Uncertainty Principle at work. Which is to say I have no idea what is happening.

Yes, great example. As I understand it, it doesn't matter if someone or several is watching or not. It doesn't matter if some outside force or particle disturbed the stream of photons or not. All that matters is "which path?" detection is made possible, or not. Any method that will definitively say which path the photon took, will destroy the interference pattern.

Maybe it has to do with the transfer of information, or change in entropy, between the particle and it's environment. Something along the lines of de Broglie's formula, Action/h=-entropy/k, or, maybe QM is just so wierd, even if we knew the answer we wouldn't believe it. Most would probably go with the latter. Great fun, eh?

 

[nouveau_riche]

Maybe it has to do with the transfer of information, or change in entropy, between the particle and it's environment.

this is where things changes in quantum world...i think the relationships that bind us to that world is the reason to those uncertainities and decoherence

 

[I like Serena]

Take a double slit setup using light. Place one polarizer over the left slit at angle L, another over the right slit at angle R. When L-R=0 (parallel), there IS interference. When L-R=90 degrees (crossed), there is NO interference. Obviously, the presence or absence of a polarizer does not change anything unless which slit information is gained. That becomes progressively more feasible as L-R goes from 0 to 90 degrees. In other words: you *could* know which path (100% certain) by checking the polarization of the detected photon when L-R=90 (crossed), even if you didn't.

In other words, the particle goes through the same apparatus whether the polarizers are parallel or crossed. So clearly the apparatus itself is not altering the particles as they go through more in one case than in another. And yet when they are parallel, no which path information is gained and interference is produced. And vice versa when crossed.

Conceptually, you could do the same thing with electrons (using S-G method) although I doubt it is practical. I am not trying to dispute Heisenberg (or Feynman) but the analogy is not correct regardless. Early quantum theorists used it to help describe how things work, but with the advent of entanglement experiments it is clear that the apparatus definitely is not the source of this effect. It is the overall context (configuration) of the setup and the Heisenberg Uncertainty Principle at work. Which is to say I have no idea what is happening.

Actually, I think you're saying the same thing as I did.

In my case I gave the example that with a long wavelength the path-distinction can not be made, resulting in an interference pattern.
A low intensity short wavelength will show a partial interference pattern (since there is only a chance that the light will reveal the path).

In your case, the polarizer inhibits the path-distinction as well, either destroying the interference pattern, or showing a partial interference pattern.

Cheers!

 

[nouveau_riche]

suppose the intereference experiment is set up again,the light used is of very high frequency(so that it's wavelength could match with that of slits) ,now the distance at which interefence pattern is seen is at max distant from slit(max in sense that pattern can b seen nicely).,now one of the observer set up apparatus to see the intereference pattern while the other one set up an apparutus to measure position of photon(to see from which slit it passes through)
question is-will the interference pattern b lost or both will see their portion of reality ?

i still don't have a satisfying answer for this experiment...the only thing i am asking is the reason if this experiment is not possible?

 

[I like Serena]

i still don't have a satisfying answer for this experiment...the only thing i am asking is the reason if this experiment is not possible?

Oh sorry, I guess I was not clear enough before. :shy:

The experiment is possible.
The answer is that both observers would always see the same reality.

I take it both observers have their own light source that they use to see the electron?
The actual observers would not play a role - only their light source would.

The first light source would try to illuminate the electron in such a way that it becomes observable through which slit the electron goes.

The second light source would illuminate the electron near the screen, but would not show through which slit the electron went.

It is the first light source that determines everything!
If it's on and bright enough it will destroy the interference pattern.
The second light will show that result.

If the first light is off, there will be an interference pattern.
The second light will show that result.


Does this satisfy you?

 

[nouveau_riche]

Oh sorry, I guess I was not clear enough before. :shy:

The experiment is possible.
The answer is that both observers would always see the same reality.

I take it both observers have their own light source that they use to see the electron?
The actual observers would not play a role - only their light source would.

The first light source would try to illuminate the electron in such a way that it becomes observable through which slit the electron goes.

The second light source would illuminate the electron near the screen, but would not show through which slit the electron went.

It is the first light source that determines everything!
If it's on and bright enough it will destroy the interference pattern.
The second light will show that result.

If the first light is off, there will be an interference pattern.
The second light will show that result.


Does this satisfy you?

i am quite not happy with ur assumption,use a single light source

 

[I like Serena]

i am quite not happy with ur assumption,use a single light source

Well, I'm quite not happy that you are quite not happy.

With a single light source the results will be the same.


Cheers!

 

[nouveau_riche]

Well, I'm quite not happy that you are quite not happy.

With a single light source the results will be the same.


Cheers!

i think both will going to see the same thing:no interference pattern

 

[I like Serena]

i think both will going to see the same thing:no interference pattern

Yep! :)

Any light source that makes it observable through which slit an electron goes, destroys the interference pattern (for all observers).

 

[nouveau_riche]

Yep! :)

Any light source that makes it observable through which slit an electron goes, destroys the interference pattern (for all observers).

but there's another experiment i got which i think could give real weirdness...\

"consider the same interference experiment,the interference pattern produced is used to illuminate the next source in series,which then produces a new interference pattern,this gets cascaded say 10 times and the final interference pattern is obtained on the wall.
now ,the observer, observing the final interference pattern is not known of the fact that there's another observer measuring the particle position in the first stage of interference,and viceversa

"will ur answer b the same(no interference pattern) or something else?
and if u think that the experiment is not possible,state the reason along

 

[I like Serena]

but there's another experiment i got which i think could give real weirdness...\

"consider the same interference experiment,the interference pattern produced is used to illuminate the next source in series,which then produces a new interference pattern,this gets cascaded say 10 times and the final interference pattern is obtained on the wall.
now ,the observer, observing the final interference pattern is not known of the fact that there's another observer measuring the particle position in the first stage of interference,and viceversa

"will ur answer b the same(no interference pattern) or something else?
and if u think that the experiment is not possible,state the reason along

This experiment can be done just fine, although you seem to be mixing electrons with photons.
But that's not really a problem, it doesn't really matter which of them you use.

Cascading through a number of interference patterns will reduce the chance that a photon/electron will reveal which slit an electron/photon goes through.
I believe that the result will be a mixture of an interference pattern and a destroyed interference pattern (for all observers).

 

[nouveau_riche]

This experiment can be done just fine, although you seem to be mixing electrons with photons.
But that's not really a problem, it doesn't really matter which of them you use.

Cascading through a number of interference patterns will reduce the chance that a photon/electron will reveal which slit an electron/photon goes through.
I believe that the result will be a mixture of an interference pattern and a destroyed interference pattern (for all observers).

i think the result will still b the same as it was with previous experiment