Definition of observing

[Debil]

Everything is at the same time both a wave and a particle, and it only becomes fully one of those when we observe it. Please correct be if I'm wrong here, but if I am right, what does that "observed" mean?

[Sensei]

Observation means measurement.

 

f.e. passing light through polarization filter.

 

f.e. absorbing photons (and converting them to electric impulses that can be counted)

[MigL]

An 'observation' is any and all interactions.

We measure and observe by interacting.

[Enthalpy]

Everything is at the same time both a wave and a particle, and it only becomes fully one of those when we observe it.

 

Where did you get this bizarre idea?

 

Forget any philosophical or rhetoric approach of QM. Go to the maths.

[jaydnul]

Everything is at the same time both a wave and a particle, and it only becomes fully one of those when we observe it. Please correct be if I'm wrong here, but if I am right, what does that "observed" mean?

A "particle" is neither a wave or a particle, it is a quantum object. In some instances, it exhibits properties that are able to be modeled by the math of a standard macroscopic wave, but by no means is it actually a wave.

[IM Egdall]

To help me picture quantum mechanics, I like the simplified view that a particle travels like a wave and is detected (interacts) like a particle. For example, in the double-slit experiment:

 

A single electron is emitted from an electron gun. It speads out and travels like a wave, passing through both slits. This probability wave reaches the bank of detectors.

 

Then only a single detector detects (interacts with) the electron, like a particle. This is the so-called collapse of the wave function.

 

Two more points:

(1) Due to the Uncertainty Principle, we cannot simultaneously know both the exact location and velocity of this detected electron. This uncertainty means the detected electron is also a wave, but a localized one.

(2) When a number of electrons are shot from the gun, an interference pattern forms at the bank of detectors. This implies that each electron wave passing through the two slits produces two waves which interfere with each other, resulting in the interference pattern we see.

[jaydnul]

(1) Due to the Uncertainty Principle, we cannot simultaneously know both the exact location and velocity of this detected electron. This uncertainty means the detected electron is also a wave, but a localized one.

 

I was also trying to make this point in a different thread. It is a common misinterpretation of the uncertainty principle to assume we can never know what the momentum and position of a particle was. If the electron has been detected, then we can easily deduce its position and the momentum it needed to get there. The H.U.P. is if we ran the experiment again, there would be an uncertainty in our ability to predict the momentum or position, which has to satisfy the uncertainty principle.

[Sensei]

It is a common misinterpretation of the uncertainty principle to assume we can never know what the momentum and position of a particle was.

The same can be said about any moving macroscopic object like f.e. airplane.

If we know position of airplane, it must be on land, not flying anymore.

If it's flying in air, its position is constantly changing, and while we're seeing it, it's not there anymore, we just see photons reflected from it after some time. And our recorded position is more or less not actual.

 

If the electron has been detected, then we can easily deduce its position and the momentum it needed to get there.

You're talking about electron in electronic circuit.

 

Double slit experiment for electrons is done in vacuum with free electrons, emitted by electron gun.

 

Detection of electron means it has hit detector. And perhaps created photon, which has been seen, and counted.

Electron slowed down, gave detector it's kinetic energy, or part of it. And it no longer poses that kinetic energy.

Act of "observation", measurement, caused change in object we're measuring.

[jaydnul]

The same can be said about any moving macroscopic object like f.e. airplane.

If we know position of airplane, it must be on land, not flying anymore.

If it's flying in air, its position is constantly changing, and while we're seeing it, it's not there anymore, we just see photons reflected from it after some time. And our recorded position is more or less not actual.

 

 

You're talking about electron in electronic circuit.

 

Double slit experiment for electrons is done in vacuum with free electrons, emitted by electron gun.

 

Detection of electron means it has hit detector. And perhaps created photon, which has been seen, and counted.

Electron slowed down, gave detector it's kinetic energy, or part of it. And it no longer poses that kinetic energy.

Act of "observation", measurement, caused change in object we're measuring.

No

 

The best example is a simple single slit experiment. The slit width is the uncertainty in position. If we perform the experiment and see a hit on the detector plate, we know the particle went through the slit, and thus its position. Then we can calculate the momentum it must have had using the time it took to hit the detector and the distance between the slit and the detector. Now we know what the particle's momentum and position were.

 

The HUP still hasn't come in to play yet. When we perform the experiment again and try to predict the momentum with the same slit width, that is when the HUP inequality must be satisfied.

 

The rest of your post seems a little irrelevant to me, but that is probably my own fault in understanding .

[imatfaal]

I was also trying to make this point in a different thread. It is a common misinterpretation of the uncertainty principle to assume we can never know what the momentum and position of a particle was. If the electron has been detected, then we can easily deduce its position and the momentum it needed to get there. The H.U.P. is if we ran the experiment again, there would be an uncertainty in our ability to predict the momentum or position, which has to satisfy the uncertainty principle.

 

Surely not.

 

The uncertainty is not an experimental problem - there is just a maximum level of certainty, there exist no quantum states which are a basis eigenstates for both the position and the momentum. If you measure the momentum of a particle you have an eigenstate for the momentum - mathematically the position is now not in an eigenstate but in a combination of position basis eigenstates and is thus imprecise. You can come to exactly the same conclusion using wave packet size/position.

 

It is not merely that we cannot predict both the momentum and the position precisely, nor even that we cannot measure the momentum or the position precisely, but that the momentum and position do not exist together in precise states.

[jaydnul]

What is the evidence that the momentum and position don't exist together in precise states (genuine question, I'm really quite curious)? Sure the operators don't commute but that is math, which is only used to predict.

 

Are you saying that once we detect a particle on the detector plate, we can calculate the momentum it must have had to get there, but we can't assume that is the correct value?

[Carrock]

I was also trying to make this point in a different thread. It is a common misinterpretation of the uncertainty principle to assume we can never know what the momentum and position of a particle was. If the electron has been detected, then we can easily deduce its position and the momentum it needed to get there. The H.U.P. is if we ran the experiment again, there would be an uncertainty in our ability to predict the momentum or position, which has to satisfy the uncertainty principle.

You can't find the position and the momentum the electron needed to get there.

( eg you would break time symmetry )

 

Surely not.

 

The uncertainty is not an experimental problem - there is just a maximum level of certainty, there exist no quantum states which are a basis eigenstates for both the position and the momentum. If you measure the momentum of a particle you have an eigenstate for the momentum - mathematically the position is now not in an eigenstate but in a combination of position basis eigenstates and is thus imprecise. You can come to exactly the same conclusion using wave packet size/position.

 

It is not merely that we cannot predict both the momentum and the position precisely, nor even that we cannot measure the momentum or the position precisely, but that the momentum and position do not exist together in precise states.

An alternative rather simplistic view, without doing the maths....

 

If you want to find the position of the electron very accurately, detect it with a very short wavelength photon ( gamma ray ). The accuracy, and the energy of the photon, increases as you shorten the wavelength. As you've hit the electron very hard, you don't have much idea of the electron's momentum before the detection event.

 

If you want to know the electron's momentum accurately, detect it with a very long wavelength low energy photon. You can't measure its position to much better than a wavelength.

 

An example of a minimal observation:- in double slit diffraction set up minimally interacting detectors to see which slit each electron goes through without stopping them. Whether or not you look at the detectors' output, you will find the interference pattern has disappeared.

 

There are lots of apparent getouts, many of them suggested by Einstein in thought experiments. They all failed.

 

His nearest success was the EPR paradox; if QM is valid then eg a photon leaving the sun 'knows' what angle your polaroid sunglasses will be at when it hits them eight minutes later.

Photons always get it right.

[jaydnul]

I thought I was going crazy so I dusted off my Feynman Lectures. This is from volume 3, pg. 2-3 (talking about the single slit experiment):

 

"Sometimes people say quantum mechanics is all wrong. When the particle arrived from the left, its vertical momentum was zero. And now that it has gone through the slit, its position is known. Both position and momentum seem to be known with arbitrary accuracy. It is quite true that we can receive a particle, and on reception determine what its position is and what its momentum would have had to have been to have gotten there. That is true, but that is not what the uncertainty relation refers to. [HUP] refers to the predictability of a situation, not remarks about the past. It does no good to say "I knew what the momentum was before it went through the slit, and now i know the position," because now the momentum is lost. We are talking about a predictive theory, not just measurements after the fact. So we must talk about what we can predict."

 

There's a few possibilities here.

1) There has been sufficient evidence after Feynman's death that refutes him.

2) I am misinterpreting the text and need clarification.

3) The great Richard Feynman was simply wrong. (Which I'm sure is possible, but I suspect someone would have caught the error by now given that the lectures were written in the 60's)

Also read this: http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html

[swansont]

It is quite true that we can receive a particle, and on reception determine what its position is and what its momentum would have had to have been to have gotten there.

 

One must note that these are not simultaneous measurements

[jaydnul]

 

One must note that these are not simultaneous measurements

Ahh I see. The previous momentum has nothing to do with the current position. I think I understand now. Thanks swansont

[swansont]

The previous momentum has nothing to do with the current position.

 

That's not the point. The issue at hand is what the HUP restricts. Reconstructing a trajectory is not a simultaneous measurement.

 

But even this example is problematic, if you think they can be considered simultaneous, because you are determining the momentum at the point of detection, because how does that happen? How do you know the initial momentum precisely, which requires knowing the initial position?

[Carrock]

In jaydnul's post of 4 September 2014 - 10:29 PM he attempted to refute a double slit experiment referred to by IM Egdall on 04 Sept 2014 - 9:32 PM by referring to ( I presume ) Feynman's comments on a single slit experiment.

 

As there was no reference in jaydnul's post to either single slit or double slit or Feynman, I assumed he was referring only to the double slit experiment.

 

In this experiment, it's only possible to know the momentum and position if you know which slit the electron passed through etc and according to Feynman et al you can do that or have the two slit interference pattern.

 

 

As he only referred to 'the experiment' in this post, I think it is unreasonable for jaydnul to criticise me for not having guessed 'the experiment' referred to two different experiments.

 

It would be helpful in this topic if posters made clear which experiment(s) they're referring to.

[jaydnul]

 

That's not the point. The issue at hand is what the HUP restricts. Reconstructing a trajectory is not a simultaneous measurement.

 

But even this example is problematic, if you think they can be considered simultaneous, because you are determining the momentum at the point of detection, because how does that happen? How do you know the initial momentum precisely, which requires knowing the initial position?

Yes, I see now how I was misenterpreting it. So in this post talking about the single slit, paragraphs 6 and 7 (labeled #1,#2), when the particle hits the detector, is the information about the instantaneous momentum lost? Couldn't you measure the momentum given to the detector plate?

 

In jaydnul's post of 4 September 2014 - 10:29 PM he attempted to refute a double slit experiment referred to by IM Egdall on 04 Sept 2014 - 9:32 PM by referring to ( I presume ) Feynman's comments on a single slit experiment.

.

.

.

As he only referred to 'the experiment' in this post, I think it is unreasonable for jaydnul to criticise me for not having guessed 'the experiment' referred to two different experiments.

 

It would be helpful in this topic if posters made clear which experiment(s) they're referring to.

Sorry, should have been more clear but I'm not sure when I critisized you

[swansont]

Yes, I see now how I was misenterpreting it. So in this post talking about the single slit, paragraphs 6 and 7 (labeled #1,#2), when the particle hits the detector, is the information about the instantaneous momentum lost? Couldn't you measure the momentum given to the detector plate?

 

 

Can you? That's just another measurement.

 

I have an issue with #2, in that it makes a classical assumption, that the particle has a definite momentum during its trip. But as that has not been measured, how do we know this? How can we be sure? Zz calls that a reasonable assumption, but I'm not sure it is. It sounds like assuming classical physics holds in a quantum system, which is not a winning bet.

 

The other issue is one I brought up in the other thread where this was discussed: one needs to explain how one can determine the uncertainty/error with a single measurement. You've gotten a very precise number, but what does that actually mean?